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Partial Differential Equations Lecture Notes for Math 404 Rouben - PowerPoint PPT Presentation

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Partial Differential Equations Lecture Notes for Math 404 Rouben Rostamian Department of Mathematics and Statistics UMBC Fall 2020 Introduction The heat Introduction to PDEs equation Instances of use You are already familiar with Ordinary

  1. Introduction The heat Remarks equation Instances of use The formulation of the heat conduction as a system of first order PDEs Heat conduction across a refrigerator wall The derivation of ∂ u ∂ t = − 1 ∂ q q = − k ∂ u the heat equation ∂ x , (2) Initial/boundary c ρ ∂ x value problems for the heat equation seems to be equivalent to the single second order PDE Separation of variables ∂ u ∂ t = 1 ∂ k ∂ u � � Homogeneous equations (3) Insulated boundary c ρ ∂ x ∂ x Equations with heat source Prescribed but there are subtle and significant differences. temperature at the boundary A compact notation for partial In (3) the diffusion coefficient k is under a differentiation sign while in (2) it is not. derivatives Inhomogeneous If k is a constant or a smoothly varying function, that’s not a big deal, but what if boundary conditions Newton’s Law of k is discontinuous? cooling The Fourier sine series in 2D Recall the example of heat conduction through a refrigerator wall. The wall Heat conduction in two dimensions consists of a metal layer on the outside, a plastic layer on the inside, and styrofoam From Cartesian to polar The Fourier series filling in between. The conductivities of these materials are drastically different, Steady-state heat conduction in a disk therefore k varies discontinuously as we move through the wall.

  2. Introduction The heat Remarks (continued) equation Instances of use Heat conduction There are various ways of handling discontinuous k at theoretical and computational levels. across a refrigerator wall The derivation of • [Theoretical] Generalize the classical definitions of functions and their derivative to the heat equation non-smooth functions. This leads to the theory of generalized functions and Initial/boundary value problems distributions. Dirac’s delta function falls in that category. for the heat equation • [Theoretical] Formulate differentiation as an operator in a function space. This leads Separation of to Sobolev spaces and weak formulations of PDEs . variables Homogeneous equations • [Computational] In the weak formulation of a PDE, replace the infinite-dimensional Insulated boundary Sobolev space with an appropriate finite-dimensional approximation. This leads to Equations with heat source Prescribed Galerkin’s formulation and the method of finite elements . temperature at the boundary • [Computational] Approximate the derivatives in (2) through difference quotients. This A compact notation for partial derivatives leads to a finite difference formulation of the problem. Inhomogeneous boundary conditions • [Computational] Approximate the derivatives in (3) through difference quotients. A Newton’s Law of cooling The Fourier sine naive implementation will produce junk since it will attempt to differentiate k . series in 2D Heat conduction in Special-purpose finite difference schemes are available for producing correct results. two dimensions From Cartesian to polar • [Computational] Apply (3) separately within each layer where k is differentiable. The Fourier series Connect the layers through equations that enforce the conservation of energy. Steady-state heat conduction in a disk

  3. Introduction The heat The domain of u ( x , t ) equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation t Initial/boundary value problems for the heat equation u (0 , t ) = α ( t ) Separation of u ( L , t ) = β ( t ) variables Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous x 0 L boundary conditions u ( x , 0) = φ ( x ) Newton’s Law of cooling The Fourier sine series in 2D Domain of solution: Heat conduction in The graph of temperature u ( x , t ) within two dimensions From Cartesian to 0 < x < L , T > 0 the refrigerator’s wall, as a function of x polar The Fourier series and t . Steady-state heat conduction in a disk

  4. Introduction The heat Initial/boundary value problems for the heat equation Instances of use equation Heat conduction across a refrigerator wall Prescribed boundary temperature: The derivation of the heat equation Initial/boundary ∂ t = κ∂ 2 u  ∂ u value problems ∂ x 2 + f ( x , t ) 0 < x < L , t > 0   for the heat   equation    u (0 , t ) = α ( t ) t > 0 Separation of variables  u ( L , t ) = β ( t ) t > 0  Homogeneous  equations    Insulated boundary u ( x , 0) = φ ( x ) 0 < x < L  Equations with heat source Prescribed Prescribed boundary flux at one end: temperature at the boundary A compact notation for partial  ∂ t = κ∂ 2 u ∂ u derivatives ∂ x 2 + f ( x , t ) 0 < x < L , t > 0  Inhomogeneous  boundary conditions    Newton’s Law of   cooling u (0 , t ) = α ( t ) t > 0   The Fourier sine  series in 2D − k ∂ u � Heat conduction in two dimensions �  = γ ( t ) t > 0  � From Cartesian to  ∂ x  polar �  x = L  The Fourier series    u ( x , 0) = φ ( x ) 0 < x < L Steady-state heat  conduction in a disk

  5. Separation of variables for homogeneous equations

  6. Introduction The heat The separation of variables trick equation Instances of use Heat conduction across a refrigerator The simplest initial/boundary value problem: wall The derivation of the heat equation ∂ t = κ∂ 2 u ∂ u Initial/boundary 0 < x < L , t > 0 (4a) value problems ∂ x 2 for the heat equation u (0 , t ) = 0 t > 0 (4b) Separation of variables u ( L , t ) = 0 t > 0 (4c) Homogeneous equations u ( x , 0) = φ ( x ) 0 < x < L (4d) Insulated boundary Equations with heat source Prescribed Try for a solution of the form u ( x , t ) = X ( x ) T ( t ): temperature at the boundary A compact notation for partial ⇒ T ′ ( t ) κ T ( t ) = X ′′ ( x ) derivatives X ( x ) T ′ ( t ) = κ X ′′ ( x ) T ( t ) (5a) Inhomogeneous X ( x ) boundary conditions Newton’s Law of cooling X (0) T ( t ) = 0 ⇒ X (0) = 0 (5b) The Fourier sine series in 2D Heat conduction in X ( L ) T ( t ) = 0 ⇒ X ( L ) = 0 (5c) two dimensions From Cartesian to X ( x ) T (0) = φ ( x ) ⇒ ? (will worry about this one later) (5d) polar The Fourier series Steady-state heat conduction in a disk

  7. Introduction The heat The separation of variables trick – part 2 equation Instances of use Heat conduction Equation (5a) implies that across a refrigerator wall The derivation of the heat equation κ T ( t ) = X ′′ ( x ) T ′ ( t ) X ( x ) = some constant, say η (6) Initial/boundary value problems for the heat equation The constant η may be positive, zero, or negative Separation of variables Spoiler! Turns out that only η < 0 leads to anything interesting. Homogeneous equations Insulated boundary Case η = λ 2 > 0: From (6) we get: Equations with heat source Prescribed temperature at the T ′ ( t ) = κλ 2 T ( t ) , X ′′ ( x ) = λ 2 X ( x ) boundary A compact notation for partial derivatives From the second equation above we get X ( x ) = A sinh λ x + B cosh λ x , and Inhomogeneous boundary conditions Newton’s Law of therefore X (0) = B . Then from (5b) we get B = 0. Thus, we are left with cooling The Fourier sine X ( x ) = A sinh λ x , and therefore X ( L ) = A sinh λ L . Then from (5c) we get series in 2D Heat conduction in A sinh λ L = 0. Since λ L � = 0, we must have A = 0, and therefore the solution is two dimensions From Cartesian to polar X ( x ) = 0 for all x . Not interesting. The Fourier series Steady-state heat conduction in a disk Case η = 0: You do it. (conclusion: Not interesting)

  8. Introduction The heat The separation of variables trick – part 3 equation Case η = − λ 2 < 0: From (6) we get: Instances of use Heat conduction across a refrigerator wall The derivation of T ′ ( t ) + κλ 2 T ( t ) = 0 , X ′′ ( x ) + λ 2 X ( x ) = 0 (7) the heat equation Initial/boundary value problems From the second equation above we get X ( x ) = A sin λ x + B cos λ x , and therefore for the heat equation X (0) = B . Then from (5b) we get B = 0. Thus, we are left with X ( x ) = A sin λ x , Separation of and therefore X ( L ) = A sin λ L . Then from (5c) we get A sin λ L = 0. We don’t variables Homogeneous want A to be zero (not interesting) so we get sin λ L = 0 and therefore λ L = n π , equations Insulated boundary for any integer n , will do. We let Equations with heat source Prescribed λ n = n π temperature at the L , n = 1 , 2 , . . . (8) boundary A compact notation for partial derivatives Inhomogeneous and thus, X ( x ) = A sin λ n x . boundary conditions Newton’s Law of cooling Furthermore, from the first equation in (7) we get T ( t ) = Ce − κλ 2 n t , and therefore The Fourier sine series in 2D n t sin λ n x as a solution that satisfies the we arrive at u ( x , t ) = ACe − κλ 2 Heat conduction in two dimensions equations (5a), (5b), and (5c). and consequently, equations (4a), (4b), and (4c). From Cartesian to polar The Fourier series We have not yet accounted for equation (5d) (or equivalently, equation (4d)). We Steady-state heat conduction in a disk turn to that issue now.

  9. Introduction The heat The separation of variables trick – part 4 equation Instances of use Heat conduction Equations (4a)–(4c) are linear and homogeneous , which is the technical way of across a refrigerator wall saying that if u 1 ( x , t ) and u 2 ( x , t ) satisfy those equations, then any linear The derivation of the heat equation combination c 1 u 1 ( x , t ) + c 2 u 2 ( x , t ) with constant coefficients c 1 and c 2 , also Initial/boundary value problems satisfy those equations. (Verify this for yourself; it’s not hard!) for the heat equation n t sin λ n x satisfies the In the previous slide (slide 18) we saw that u ( x , t ) = e − κλ 2 Separation of variables equations (4a)–(4c) for any integer n . Therefore, so does the (infinite) linear Homogeneous equations combination Insulated boundary ∞ n t sin λ n x Equations with heat a n e − κλ 2 � u ( x , t ) = (9) source Prescribed temperature at the n =1 boundary A compact notation where the choice of the (constant) coefficients a n is at our disposal. We are going for partial derivatives to choose those coefficients so that u ( x , t ), expressed as (9), satisfies the one last Inhomogeneous boundary conditions remaining requirement, that is, the equation (4d). Newton’s Law of cooling The Fourier sine series in 2D From (9) we have u ( x , 0) = � ∞ n =1 a n sin λ n x , and therefore from (4d) we get Heat conduction in two dimensions From Cartesian to ∞ polar � a n sin λ n x = φ ( x ) . (10) The Fourier series Steady-state heat conduction in a disk n =1

  10. Introduction The heat The separation of variables trick – part 5 equation Instances of use Question: Can any function φ be expressed as the infinite sum in (10) ? Heat conduction across a refrigerator wall The answer is yes! provided that φ satisfies certain regularity conditions such as The derivation of the heat equation sufficient continuity and integrability. (We won’t get into those conditions in this Initial/boundary value problems course, but for practical purposes it is safe to assume that those are satisfied.) If for the heat equation so, we multiply (10) by sin λ m x and integrate over the interval (0 , L ): Separation of � L � L ∞ variables � a n sin λ m x sin λ n x dx = φ ( x ) sin λ m x dx (11) Homogeneous equations 0 0 Insulated boundary n =1 Equations with heat source It is left to you as an exercise to show that for λ s defined as in (8), and any two Prescribed temperature at the integers m and n : boundary A compact notation for partial � L derivatives � 0 if m � = n Inhomogeneous sin λ m x sin λ n x dx = boundary conditions L / 2 if m = n Newton’s Law of 0 cooling The Fourier sine series in 2D and therefore in the infinite sum in (11) only one term survives and we arrive at Heat conduction in two dimensions � L From Cartesian to L polar 2 a m = φ ( x ) sin λ m x dx . The Fourier series 0 Steady-state heat conduction in a disk This tells us the value of a m for all m , since the initial condition φ is known.

  11. Introduction The heat Summary of the two preceding slides equation Instances of use Heat conduction A function φ defined in the interval (0 , L ) may be expressed as the infinite sum across a refrigerator wall The derivation of the heat equation Initial/boundary ∞ value problems � φ ( x ) = a n sin λ n x , for the heat (12) equation n =1 Separation of variables where Homogeneous equations � L Insulated boundary a n = 2 Equations with heat φ ( x ) sin λ n x dx . source (13) L Prescribed 0 temperature at the boundary A compact notation and where for partial λ n = n π derivatives L , n = 1 , 2 , . . . (14) Inhomogeneous boundary conditions Newton’s Law of cooling The expression on the right-hand side of (12) is called the Fourier sine series The Fourier sine series in 2D representation of the function φ . The coefficients a n are called the corresponding Heat conduction in two dimensions Fourier coefficients (named after the French mathematician Joseph Fourier, From Cartesian to polar 1767–1830). The Fourier series Steady-state heat conduction in a disk

  12. Introduction The heat How good is the Fourier series? equation Instances of use Heat conduction across a refrigerator In these demos, the original function φ is plotted in blue, while the approximations wall The derivation of the heat equation by the first N terms of the Fourier series are plotted in red. Initial/boundary value problems for the heat equation Separation of variables Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine φ ( x ) = x ( x − 1 / 3)(1 − x ) φ ( x ) = 1 / 2 − | x − 1 / 2 | series in 2D � sin n π x Heat conduction in � 5( − 1) n + 4 sin n π two dimensions ∞ ∞ 2 sin n π x 4 = 4 � From Cartesian to � = polar 3 π 3 n 3 π 2 n 2 The Fourier series n =1 n =1 Steady-state heat conduction in a disk

  13. Introduction The heat The separation of variables trick – part 6 and equation Instances of use conclusion Heat conduction across a refrigerator wall The derivation of the heat equation Summary: Initial/boundary value problems In the previous slides we have developed the bits and pieces needed for calculating for the heat the solution u ( x , t ) of the initial/boundary value problem (4). In (9) we saw that equation Separation of variables ∞ n t sin λ n x a n e − κλ 2 Homogeneous � u ( x , t ) = (15a) equations Insulated boundary n =1 Equations with heat source Prescribed and we learned that the coefficients a n are obtained from (13) temperature at the boundary A compact notation for partial � L derivatives a n = 2 Inhomogeneous φ ( x ) sin λ n x dx , (15b) boundary conditions L 0 Newton’s Law of cooling The Fourier sine where series in 2D λ n = n π Heat conduction in two dimensions L . (15c) From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  14. Introduction The heat A fully worked-out example equation Instances of use Equations (15) on the previous slide present the solution u ( x , t ) of the Heat conduction across a refrigerator wall initial/boundary value problem (4) (on slide 17) for an arbitrary initial condition The derivation of the heat equation u ( x , 0) = φ ( x ). Initial/boundary value problems Calculating the solution for a specific φ is a matter of carrying out the integration for the heat equation in (15b). Here is a sketch of the calculations. Separation of variables Homogeneous equations Insulated boundary Equations with heat � x if x < L / 2 φ ( x ) = L � x − L source � � 2 − � = Prescribed � � temperature at the 2 L − x if x > L / 2 boundary A compact notation for partial derivatives Inhomogeneous The graph of φ ( x ) with L = 1 boundary conditions Newton’s Law of cooling The Fourier sine � L series in 2D a n = 2 2 2 sin λ n L � � Heat conduction in φ ( x ) sin λ n x dx = − sin λ n L (from Quiz #1) two dimensions λ 2 L n L 2 0 From Cartesian to polar 2 L 2 sin n π n 2 π 2 sin n π 4 L The Fourier series � � = 2 − sin n π = 2 . (from (15c)) Steady-state heat conduction in a disk n 2 π 2

  15. Introduction The heat The solution equation Instances of use Heat conduction across a refrigerator n t sin n π wall ∞ u ( x , t ) = 4 L 2 sin λ n x e − κλ 2 The derivation of � the heat equation π 2 n 2 Initial/boundary n =1 value problems = 4 L e − κ ( π/ L ) 2 t sin π x L − 1 3 2 e − κ (3 π/ L ) 2 t sin 3 π x + 1 5 2 e − κ (5 π/ L ) 2 t sin 5 π x for the heat � � − · · · equation π 2 L L Separation of variables Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat The solution u ( x , t ) evaluated with L = 1, κ = 1 and truncated as � 19 n =1 (ten conduction in a disk terms)

  16. Insulated boundary � − k ∂ u � � x = L = 0 � ∂ x �

  17. Introduction The heat Insulated boundary at x = L equation Instances of use Heat conduction across a refrigerator − k ∂ u u = 0 ∂ x = 0 wall The derivation of the heat equation Initial/boundary value problems for the heat L equation ∂ t = κ∂ 2 u Separation of ∂ u variables (16a) ∂ x 2 Homogeneous equations u (0 , t ) = 0 (16b) Insulated boundary Equations with heat source − k ∂ u � Prescribed x = L = 0 (16c) temperature at the � boundary ∂ x � A compact notation for partial u ( x , 0) = φ ( x ) (16d) derivatives Inhomogeneous boundary conditions Newton’s Law of Separate the variables: u ( x , t ) = X ( x ) T ( t ). Then X ( x ) T ′ ( t ) = κ X ′′ ( x ) T ( t ) and cooling The Fourier sine therefore series in 2D Heat conduction in two dimensions 1 T ′ ( t ) T ( t ) = X ′′ ( x ) From Cartesian to X ( x ) = − λ 2 , polar X ′ ( L ) = 0 X (0) = 0 , The Fourier series κ Steady-state heat conduction in a disk

  18. Introduction The heat Separation of variables equation Instances of use Heat conduction across a refrigerator wall T ′ ( t ) = − κλ 2 T ( t ) , The derivation of the heat equation X ′′ ( x ) + λ 2 X ( x ) = 0 , Initial/boundary X ′ ( L ) = 0 X (0) = 0 , value problems for the heat equation The general solution of the X equation is X ( x ) = A sin λ x + B cos λ x . Applying Separation of the boundary condition X (0) = 0, we get B = 0. Therefore X ( x ) = A sin λ x . variables Homogeneous Then X ′ ( x ) = λ A cos λ x . Therefore applying the boundary condition X ′ ( L ) = 0 we equations Insulated boundary get cos λ L = 0. We conclude that λ L is an odd multiple of π/ 2, that is Equations with heat source λ n L = (2 n − 1) π 2 , and therefore Prescribed temperature at the boundary A compact notation λ n = (2 n − 1) π T n ( t ) = e − κλ 2 t for partial , X n ( x ) = sin λ n x , n = 1 , 2 , . . . (17) derivatives 2 L Inhomogeneous boundary conditions and Newton’s Law of cooling The Fourier sine ∞ ∞ series in 2D n t sin λ n x a n e − κλ 2 � � u ( x , t ) = a n X n ( x ) T n ( t ) = Heat conduction in two dimensions From Cartesian to n =1 n =1 polar ∞ � 2 The Fourier series t sin (2 n − 1) π � a n e − κ (2 n − 1) π/ (2 L ) � Steady-state heat = x . conduction in a disk 2 L n =1

  19. Introduction The heat Separation of variables continued equation Instances of use Heat conduction across a refrigerator The coefficients a n are determined by applying the initial condition u ( x , 0) = φ ( x ): wall The derivation of the heat equation ∞ Initial/boundary � u ( x , 0) = a n X n ( x ) = φ ( x ) value problems for the heat n =1 equation Separation of Exercise: Show that for any integer m and n , and λ n defined as in (17), we have: variables Homogeneous equations � L � L Insulated boundary � 0 if m � = n Equations with heat X m ( x ) X n ( x ) dx = sin λ m x sin λ n x dx = source L / 2 if m = n Prescribed 0 0 temperature at the boundary A compact notation for partial Therefore derivatives Inhomogeneous boundary conditions � L � L a n = 2 φ ( x ) X n ( x ) dx = 2 Newton’s Law of cooling φ ( x ) sin λ n x dx The Fourier sine L L 0 0 series in 2D � L Heat conduction in = 2 φ ( x ) sin (2 n − 1) π x two dimensions dx From Cartesian to polar L 2 L 0 The Fourier series Steady-state heat conduction in a disk

  20. Introduction The heat The modal shapes and an animation equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation Initial/boundary value problems for the heat equation Separation of variables Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The solution u ( x , t ) with the initial condition φ ( x ) = x / L , evaluated with L = 1, The Fourier series Steady-state heat κ = 1 and truncated as � 10 conduction in a disk n =1 (ten terms)

  21. Equations with heat source . . . but zero boundary conditions ∂ t = κ∂ 2 u ∂ u ∂ x 2 + f ( x , t ) u (0 , t ) = 0 u ( L , t ) = 0 u ( x , 0) = φ ( x )

  22. Introduction The heat Eigenfunction expansion equation Instances of use We are going to solve the initial/boundary value problem Heat conduction across a refrigerator wall The derivation of ∂ t = κ∂ 2 u ∂ u the heat equation ∂ x 2 + f ( x , t ) 0 < x < L , t > 0 (18a) Initial/boundary value problems for the heat u (0 , t ) = 0 t > 0 (18b) equation u ( L , t ) = 0 t > 0 (18c) Separation of variables u ( x , 0) = φ ( x ) 0 < x < L (18d) Homogeneous equations Insulated boundary Equations with heat On slide 21 we saw that any function of x defined in the interval 0 < x < L may be source Prescribed expanded into a Fourier sine series. We let temperature at the boundary A compact notation for partial ∞ ∞ ∞ derivatives ¯ ¯ � � � u ( x , t ) = a n ( t ) sin λ n x , f ( x , t ) = f n ( t ) sin λ n x , φ ( x ) = φ n sin λ n x , Inhomogeneous boundary conditions Newton’s Law of n =1 n =1 n =1 cooling The Fourier sine series in 2D where the coefficients a n ( t ) are unknown, but ¯ f n ( t ) and ¯ φ n may be calculated from: Heat conduction in two dimensions From Cartesian to � L � L polar f n ( t ) = 2 φ n = 2 ¯ ¯ The Fourier series f ( x , t ) sin λ n x dx , φ ( x ) sin λ n x dx . Steady-state heat L L conduction in a disk 0 0

  23. Introduction The heat Reducing the PDE into a set of infinitely many equation Instances of use ODEs Heat conduction across a refrigerator Substitute the expansions into equations (18a) and (18d): wall The derivation of the heat equation ∞ ∞ ∞ Initial/boundary ¯ � � ( − λ 2 � a ′ n ( t ) sin λ n x = κ n ) a n ( t ) sin λ n x + f n ( t ) sin λ n x , value problems for the heat n =1 n =1 n =1 equation ∞ ∞ Separation of ¯ � � a n (0) sin λ n x = φ n sin λ n x , variables Homogeneous n =1 n =1 equations Insulated boundary and groups the summands Equations with heat source Prescribed temperature at the ∞ boundary � � n a n ( t ) − ¯ n ( t ) + κλ 2 � a ′ f n ( t ) sin λ n x = 0 , A compact notation for partial derivatives n =1 Inhomogeneous boundary conditions ∞ � � Newton’s Law of a n (0) − ¯ � φ n sin λ n x = 0 . cooling The Fourier sine series in 2D n =1 Heat conduction in two dimensions � ∞ � sin λ n x Since n =1 is a basis, it follows that From Cartesian to polar The Fourier series n a n ( t ) = ¯ a n (0) = ¯ Steady-state heat n ( t ) + κλ 2 a ′ f n ( t ) , φ n , n = 1 , 2 , . . . (19) conduction in a disk

  24. Introduction The heat Calculating the coefficients a n ( t ) equation Instances of use Heat conduction Equations (19) express a set of infinitely many initial value problems for ODEs in across a refrigerator wall the unknowns a n ( t ). which may be solved with the integrating factor method The derivation of the heat equation learned in a course in ODEs. Initial/boundary value problems n t and combine terms: So we multiply through by the integrating factor e κλ 2 for the heat equation Separation of � ′ � e κλ 2 = e κλ 2 n t ¯ variables n t a n ( t ) f n ( t ) , Homogeneous equations Insulated boundary and integrate: Equations with heat source � t s = t Prescribed �� � e κλ 2 e κλ 2 temperature at the n s ¯ n s a n ( s ) � = f n ( s ) ds . boundary � A compact notation � 0 s =0 for partial derivatives but Inhomogeneous boundary conditions s = t �� Newton’s Law of � e κλ 2 = e κλ 2 n t a n ( t ) − a n (0) = e κλ 2 n t a n ( t ) − ¯ n s a n ( s ) � cooling φ n , � The Fourier sine � series in 2D s =0 Heat conduction in two dimensions therefore From Cartesian to � t polar e κλ 2 e κλ 2 n t a n ( t ) − ¯ n s ¯ φ n = f n ( s ) ds . The Fourier series Steady-state heat 0 conduction in a disk

  25. Introduction The heat Calculation of a n ( t ): Conclusion equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation From the previous slide: Initial/boundary value problems for the heat � t equation e κλ 2 e κλ 2 n t a n ( t ) − ¯ n s ¯ φ n = f n ( s ) ds . Separation of 0 variables Homogeneous therefore equations � t Insulated boundary n t ¯ a n ( t ) = e − κλ 2 e − κλ 2 n ( t − s ) ¯ φ n + f n ( s ) ds . Equations with heat source 0 Prescribed temperature at the boundary We conclude that the solution u ( x , t ) of the initial/boundary value problem (18) is A compact notation for partial derivatives � t Inhomogeneous ∞ � � boundary conditions n t ¯ e − κλ 2 e − κλ 2 n ( t − s ) ¯ � u ( x , t ) = φ n + f n ( s ) ds sin λ n x . Newton’s Law of cooling 0 n =1 The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  26. Introduction The heat A worked-out example equation Instances of use Heat conduction across a refrigerator wall Let’s solve the initial/boundary value problem The derivation of the heat equation Initial/boundary ∂ t = κ∂ 2 u  ∂ u value problems ∂ x 2 + σ sin ω t , 0 < x < L , t > 0 for the heat    equation     Separation of u (0 , t ) = 0 t > 0 (20) variables Homogeneous  u ( L , t ) = 0 t > 0  equations   Insulated boundary   u ( x , 0) = 0 0 < x < L  Equations with heat source Prescribed temperature at the This corresponds to f ( x , t ) = σ sin ω t , and therefore boundary A compact notation for partial derivatives � L � L f n ( t ) = 2 σ sin ω t sin λ n x dx = 2 σ sin ω t Inhomogeneous ¯ boundary conditions sin λ n x dx Newton’s Law of L L 0 0 cooling The Fourier sine � 1 − ( − 1) n � 1 − ( − 1) n = 2 σ sin ω t · L = 2 σ � � series in 2D sin ω t . Heat conduction in two dimensions L π n π n From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  27. Introduction The heat A worked-out example (continued) equation Instances of use Then equations (19) on slide 33 take the form Heat conduction across a refrigerator wall � 1 − ( − 1) n n a n ( t ) = 2 σ � The derivation of n ( t ) + κλ 2 a ′ sin ω t , a n (0) = 0 , n = 1 , 2 , . . . the heat equation π n Initial/boundary value problems which may be solved with an integrating factor as before, but in this case it is for the heat equation quicker to express the solution as the sum of homogeneous and particular solutions, Separation of as is done in a course in ODEs. variables Homogeneous n a n ( t ) = 0, whence a n ( t ) = Ce − κλ 2 equations n ( t ) + κλ 2 n t . The homogeneous equation is a ′ Insulated boundary Equations with heat source Look for a particular solution of the form a n ( t ) = A cos ω t + B sin ω t . Prescribed temperature at the � 1 − ( − 1) n boundary = 2 σ � � � � � + κλ 2 A compact notation − A ω sin ω t + B ω cos ω t A cos ω t + B sin ω t sin ω t , for partial n π n derivatives Inhomogeneous boundary conditions � 1 − ( − 1) n cos ω t = 2 σ � � � � � Newton’s Law of − A ω + B κλ 2 B ω + A κλ 2 sin ω t + sin ω t , cooling n n π n The Fourier sine series in 2D Heat conduction in two dimensions Q n ω  From Cartesian to � 1 − ( − 1) n  n = 2 σ � A = − polar  − A ω + B κλ 2 ω 2 + κ 2 λ 4  ≡ Q n The Fourier series     n π n Steady-state heat ⇒ conduction in a disk Q n κλ 2 A κλ 2 n   B = n + B ω = 0   ω 2 + κ 2 λ 4   n

  28. Introduction The heat A worked-out example (continued) equation Instances of use Particular solution: Heat conduction across a refrigerator wall Q n κλ 2 � 1 − ( − 1) n Q n ω where Q n = 2 σ The derivation of � n the heat equation a n ( t ) = − cos ω t + sin ω t , ω 2 + κ 2 λ 4 ω 2 + κ 2 λ 4 π n Initial/boundary n n value problems for the heat General solution: equation Separation of Q n κλ 2 Q n ω variables n t − a n ( t ) = Ce − κλ 2 n cos ω t + sin ω t . Homogeneous ω 2 + κ 2 λ 4 ω 2 + κ 2 λ 4 equations n n Insulated boundary Equations with heat source Initial condition: Prescribed temperature at the boundary Q n ω Q n ω A compact notation a n (0) = 0 ⇒ 0 = C − ⇒ C = for partial ω 2 + κ 2 λ 4 ω 2 + κ 2 λ 4 derivatives n n Inhomogeneous boundary conditions Newton’s Law of cooling Q n κλ 2 Q n ω Q n ω The Fourier sine n t − e − κλ 2 n series in 2D a n ( t ) = cos ω t + sin ω t ω 2 + κ 2 λ 4 ω 2 + κ 2 λ 4 ω 2 + κ 2 λ 4 Heat conduction in two dimensions n n n From Cartesian to polar Q n � � n t − ω cos ω t + κλ 2 ω e − κλ 2 The Fourier series = n sin ω t ω 2 + κ 2 λ 4 Steady-state heat conduction in a disk n

  29. Introduction The heat A worked-out example (conclusion) equation Instances of use Heat conduction across a refrigerator wall The derivation of ∞ the heat equation � u ( x , t ) = a n ( t ) sin λ n x Initial/boundary value problems n =1 for the heat ∞ equation Q n � � n t − ω cos ω t + κλ 2 ω e − κλ 2 � = n sin ω t sin λ n x ω 2 + κ 2 λ 4 Separation of variables n n =1 Homogeneous ∞ 1 − ( − 1) n equations = 2 σ � � n t − ω cos ω t + κλ 2 ω e − κλ 2 � Insulated boundary n sin ω t sin λ n x n ( ω 2 + κ 2 λ 4 Equations with heat π n ) source n =1 Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous An animation of u ( x , t ) boundary conditions Newton’s Law of evaluated as � 10 n =1 (five terms) cooling The Fourier sine series in 2D Note the transient behavior. Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  30. Prescribed temperature at the boundary ∂ t = κ∂ 2 u ∂ u ∂ x 2 + f ( x , t ) u (0 , t ) = α ( t ) u ( L , t ) = β ( t ) u ( x , 0) = φ ( x )

  31. Introduction The heat Prescribed temperature at the boundary equation Instances of use Up to now all of our boundary conditions have been of the form u = 0 (zero Heat conduction across a refrigerator temperature) of ∂ u wall ∂ x = 0 (zero flux). We sought solutions in the form The derivation of the heat equation u ( x , t ) = � ∞ n =1 a n ( t ) X n ( t ), where X n ( x ) were selected expressly to satisfy those Initial/boundary zero boundary conditions. As a result, the sum satisfies the those zero boundary value problems for the heat conditions and we are done. equation Separation of But what if the boundary conditions are other than zero? There is no use in variables Homogeneous changing the X n s to satisfy those boundary conditions because even if each X n equations Insulated boundary satisfies a nonzero boundary condition, it does not follow that the sum Equations with heat source � ∞ n =1 a n ( t ) X n ( t ) also satisfies that boundary condition. (This clearly shows that a Prescribed temperature at the zero boundary condition is something very special!) boundary A compact notation for partial derivatives Here is a bright idea: Split u ( x , t ) into a sum u ( x , t ) = v ( x , t ) + ξ ( x , t ). For Inhomogeneous boundary conditions ξ ( x , t ) pick a function, any function, that satisfies the problem’s boundary Newton’s Law of cooling conditions. Since u ( x , t ) also satisfies those boundary conditions, it follows that The Fourier sine series in 2D v ( x , t ) satisfies the corresponding zero boundary conditions! Heat conduction in two dimensions From Cartesian to In the PDE, replace u ( x , t ) by v ( x , t ) + ξ ( x , t ). This will yield a PDE involving v . polar The Fourier series But v satisfies zero boundary conditions, and therefore we may calculate it through Steady-state heat conduction in a disk our previous techniques. Once we have v , we add ξ to it to obtain u .

  32. Introduction The heat Temperature prescribed at the boundaries equation Instances of use Heat conduction Heat condition in a rod with prescribed temperatures at the ends: across a refrigerator wall The derivation of the heat equation ∂ t = κ∂ 2 u  ∂ u Initial/boundary ∂ x 2 + f ( x , t ) 0 < x < L , t > 0  value problems    for the heat   equation  u (0 , t ) = α ( t ) t > 0 (21) Separation of  u ( L , t ) = β ( t ) t > 0 variables    Homogeneous  equations  u ( x , 0) = φ ( x ) 0 < x < L  Insulated boundary Equations with heat For the function ξ ( x , t ) we pick source Prescribed temperature at the boundary 1 − x α ( t ) + x � � A compact notation ξ ( x , t ) = L β ( t ) . (22) for partial L derivatives Inhomogeneous boundary conditions and note that ξ (0 , t ) = α ( t ), ξ ( L , t ) = β ( t ). Newton’s Law of cooling The Fourier sine series in 2D Then substitute 1 − x α ( t ) + x Heat conduction in � � two dimensions u ( x , t ) = v ( x , t ) + L β ( t ) From Cartesian to L polar The Fourier series into (21). Steady-state heat conduction in a disk

  33. Introduction The heat Equation with homogeneous boundary conditions equation Instances of use Heat conduction The v equation:  across a refrigerator wall L β ′ ( t ) = κ∂ 2 v The derivation of ∂ v 1 − x α ′ ( t ) + x � � the heat equation ∂ t + ∂ x 2 + f ( x , t )   Initial/boundary L   value problems    v (0 , t ) = 0 for the heat  equation v ( L , t ) = 0 Separation of    variables  1 + x α (0) + x  � �  Homogeneous v ( x , 0) + L β (0) = φ ( x )  equations  L Insulated boundary Equations with heat source Rearrange: ∂ t = κ∂ 2 v Prescribed ∂ v 1 − x α ′ ( t ) − x  � � temperature at the L β ′ ( t ) ∂ x 2 + f ( x , t ) −  boundary  L  A compact notation   for partial  derivatives  v (0 , t ) = 0  Inhomogeneous (23) boundary conditions v ( L , t ) = 0 Newton’s Law of  cooling    1 + x α (0) − x The Fourier sine  � �  series in 2D v ( x , 0) = φ ( x ) − L β (0)   Heat conduction in L two dimensions From Cartesian to polar So going from u equations in (21) to the v equations in (23) amounts to modifying The Fourier series the heat source function f and the initial condition φ . Steady-state heat conduction in a disk

  34. Introduction The heat The heat equation with oscillating temperature at equation Instances of use the boundary Heat conduction across a refrigerator wall Oscillatory temperature imposed at the right-hand end: The derivation of the heat equation ∂ t = κ∂ 2 u  ∂ u Initial/boundary 0 < x < L , t > 0  value problems  ∂ x 2  for the heat    equation  u (0 , t ) = 0 t > 0 (24) Separation of variables u ( L , t ) = σ sin ω t t > 0    Homogeneous   equations  u ( x , 0) = 0 0 < x < L  Insulated boundary Equations with heat source Prescribed This is a special case of the problem (21) on slide 42. The ξ function in (22) is temperature at the boundary ξ ( x , t ) = x L σ sin ω t , and therefore u ( x , t ) = v ( x , t ) + x L σ sin ω t and then A compact notation for partial derivatives problem (23) takes the form Inhomogeneous boundary conditions ∂ t = κ∂ 2 u  ∂ v ∂ x 2 − x Newton’s Law of cooling L σω cos ω t 0 < x < L , t > 0   The Fourier sine   series in 2D   Heat conduction in  v (0 , t ) = 0 t > 0 two dimensions (25) From Cartesian to polar  v ( L , t ) = 0 t > 0  The Fourier series    Steady-state heat  v ( x , 0) = 0 0 < x < L conduction in a disk 

  35. Introduction The heat Solution continued equation Instances of use Heat conduction across a refrigerator wall The derivation of The initial/boundary value problem (25) is quite similar to the system (20) on the heat equation Initial/boundary slide 36. Solving it is left to you as homework. When you work out the details, you value problems for the heat will find that: equation Separation of ∞ ( − 1) n v ( x , t ) = 2 σω variables n t + κλ 2 � n e − κλ 2 � − κλ 2 � n cos ω t + ω sin ω t sin λ n x . Homogeneous n ( ω 2 + κ 2 λ 4 equations π n ) Insulated boundary n =1 Equations with heat source and therefore Prescribed temperature at the boundary A compact notation u ( x , t ) = x for partial L σ sin ω t derivatives Inhomogeneous boundary conditions ∞ ( − 1) n Newton’s Law of + 2 σω n t + κλ 2 � n e − κλ 2 � cooling − κλ 2 � n cos ω t + ω sin ω t sin λ n x . The Fourier sine n ( ω 2 + κ 2 λ 4 π n ) series in 2D n =1 Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  36. Introduction The heat Animation of the solution equation Instances of use We animate the solution with the parameter values Heat conduction across a refrigerator wall The derivation of L = 1 , ω = 1 , σ = 1 , κ = 0 . 02 , the heat equation Initial/boundary value problems and truncate the series at the tenth term. for the heat equation Separation of variables Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  37. A compact notation for partial derivatives u xx = ∂ 2 u u t = ∂ u u x = ∂ u ∂ x 2 ∂ t ∂ x u x ( L , t ) = ∂ u � � � ∂ x � x = L

  38. Introduction The heat A compact notation for partial derivatives equation Instances of use Heat conduction Initial/boundary value problem in the expanded notation: across a refrigerator wall The derivation of the heat equation ∂ t = κ∂ 2 u  ∂ u Initial/boundary  ∂ x 2 + f ( x , t ) 0 < x < L , t > 0  value problems    for the heat   equation  u (0 , t ) = α ( t ) t > 0   Separation of − k ∂ u � variables � = γ ( t ) t > 0   Homogeneous �  ∂ x equations  �  x = L  Insulated boundary   Equations with heat  u ( x , 0) = φ ( x ) 0 < x < L  source Prescribed temperature at the boundary A compact notation The same problem in compact notation: for partial derivatives Inhomogeneous boundary conditions  u t = κ u xx + f ( x , t ) 0 < x < L , t > 0 Newton’s Law of  cooling   The Fourier sine u (0 , t ) = α ( t ) t > 0   series in 2D Heat conduction in − ku x ( L , t ) = γ ( t ) t > 0 two dimensions  From Cartesian to   polar  u ( x , 0) = φ ( x ) 0 < x < L  The Fourier series Steady-state heat conduction in a disk

  39. Handling inhomogeneous boundary conditions u t = κ u xx + f ( x , t ) α 1 ( t ) u (0 , t ) + α 2 ( t ) u x (0 , t ) = α ( t ) β 1 ( t ) u ( L , t ) + β 2 ( t ) u x ( L , t ) = β ( t ) u ( x , 0) = φ ( x )

  40. Introduction The heat Handling inhomogeneous boundary conditions equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation Initial/boundary value problem with inhomogeneous boundary conditions: Initial/boundary value problems for the heat u t = κ u xx + f ( x , t ) (26a) equation α 1 ( t ) u (0 , t ) + α 2 ( t ) u x (0 , t ) = α ( t ) (26b) Separation of variables β 1 ( t ) u ( L , t ) + β 2 ( t ) u x ( L , t ) = β ( t ) (26c) Homogeneous equations Insulated boundary u ( x , 0) = φ ( x ) (26d) Equations with heat source Prescribed temperature at the Introduce a new unknown v ( x , t ) through boundary A compact notation for partial derivatives u ( x , t ) = v ( x , t ) + c 1 ( t ) + c 2 ( t ) x (27) Inhomogeneous boundary conditions Newton’s Law of cooling and eliminate u in favor of v in the problem. Then, pick c 1 ( t ) and c 2 ( t ) so as to The Fourier sine series in 2D Heat conduction in eliminate the inhomogeneous terms α ( t ) and β ( t ) in (26b) and (26c). two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  41. Introduction The heat Eliminating the inhomogeneous terms equation Instances of use Heat conduction across a refrigerator wall Substituting u ( x , t ) from (27) into (26b) and (26c) we get The derivation of the heat equation � + α 2 � = α, Initial/boundary � v (0 , t ) + c 1 � v x (0 , t ) + c 2 α 1 value problems � + β 2 for the heat � v ( L , t ) + c 1 + c 2 L � v x ( L , t ) + c 2 � β 2 = β. β 1 equation Separation of variables whence Homogeneous equations Insulated boundary α 1 v (0 , t ) + α 2 v x (0 , t ) = α − α 1 c 1 − α 2 c 2 (28a) Equations with heat source Prescribed β 1 v ( L , t ) + β 2 v x ( L , t ) = β − β 1 c 1 − ( β 1 L + β 2 ) c 2 (28b) temperature at the boundary A compact notation for partial derivatives To get homogeneous boundary conditions on v , set the right-hand sides to zero: Inhomogeneous boundary conditions Newton’s Law of cooling α 1 c 1 + α 2 c 2 = α, (29a) The Fourier sine series in 2D β 1 c 1 + ( β 1 L + β 2 ) c 2 = β (29b) Heat conduction in two dimensions From Cartesian to polar and solve the system for the unknowns c 1 and c 2 . The Fourier series Steady-state heat conduction in a disk

  42. Introduction The heat Eliminating the inhomogeneous terms – continued equation Instances of use Heat conduction across a refrigerator wall c 1 = ( β 1 L + β 2 ) α − α 2 β α 1 β − β 1 α The derivation of , c 2 = . (30) the heat equation α 1 ( β 1 L + β 2 ) − α 2 β 1 α 1 ( β 1 L + β 2 ) − α 2 β 1 Initial/boundary value problems for the heat Observation: Since α , α 1 , α 2 , β , β 1 , β 2 are generally functions of time, c 1 and c 2 equation calculated above are also functions of time. Occasionally we will write c 1 ( t ) and Separation of variables c 2 ( t ) to stress that. Homogeneous In view of (29), the boundary conditions (28) on v reduce to equations Insulated boundary Equations with heat source α 1 v (0 , t ) + α 2 v x (0 , t ) = 0 , (31a) Prescribed temperature at the boundary β 1 v ( L , t ) + β 2 v x ( L , t ) = 0 (31b) A compact notation for partial derivatives Inhomogeneous which are homogeneous by design. boundary conditions Newton’s Law of cooling To obtain a PDE on v , substitute u ( x , t ) from (27) into (26a) and we get The Fourier sine series in 2D v t + c ′ 1 ( t ) + c ′ 2 ( t ) x = κ v xx + f ( x , t ) , that is, Heat conduction in two dimensions From Cartesian to polar v t = κ v xx + f ( x , t ) − c ′ 1 ( t ) − c ′ 2 ( t ) x (32) The Fourier series Steady-state heat conduction in a disk

  43. Introduction The heat Eliminating the inhomogeneous terms – continued equation Instances of use Heat conduction To obtain the initial condition on v , substitute u ( x , t ) from (27) into (26d). We across a refrigerator wall get v ( x , 0) + c 1 (0) + c 2 (0) x = φ ( x ) , that is The derivation of the heat equation Initial/boundary value problems v ( x , 0) = φ ( x ) − c 1 (0) − c 2 (0) x . (33) for the heat equation In summary, the change of variables (27) with c 1 and c 2 selected as in (30), Separation of variables converts the inhomogeneous boundary conditions in (26) into homogeneous Homogeneous equations boundary conditions in the modified equation: Insulated boundary Equations with heat source Prescribed v t = κ v xx + f ( x , t ) − c ′ 1 ( t ) − c ′ 2 ( t ) x (34a) temperature at the boundary A compact notation α 1 ( t ) v (0 , t ) + α 2 ( t ) v x (0 , t ) = 0 (34b) for partial derivatives Inhomogeneous β 1 ( t ) v ( L , t ) + β 2 ( t ) v x ( L , t ) = 0 (34c) boundary conditions Newton’s Law of cooling v ( x , 0) = φ ( x ) − c 1 (0) − c 2 (0) x . (34d) The Fourier sine series in 2D Heat conduction in two dimensions Observation: Going from (26) to (34) amounts to (a) zeroing the inhomogeneous From Cartesian to polar parts of the boundary conditions; (b) replacing f ( x , t ) by f ( x , t ) − c ′ 1 ( t ) − c ′ 2 ( t ) x ; The Fourier series Steady-state heat and (c) replacing φ ( x ) by φ ( x ) − c 1 (0) − c 2 (0) x . conduction in a disk

  44. Introduction The heat Special case: Dirichlet boundary conditions equation Instances of use The initial/boundary value problem Heat conduction across a refrigerator wall u t = κ u xx + f ( x , t ) (35a) The derivation of the heat equation u (0 , t ) = α ( t ) (35b) Initial/boundary value problems for the heat u ( L , t ) = β ( t ) (35c) equation u ( x , 0) = φ ( x ) (35d) Separation of variables Homogeneous is a special case of (26) with α 1 ( t ) = 1 , α 2 ( t ) = 0 , β 1 ( t ) = 1 , β 2 ( t ) = 0. From (30) equations Insulated boundary � β ( t ) − α ( t ) � / L and then (27) and (34) reduce to we get c 1 = α ( t ) , c 2 = Equations with heat source Prescribed u ( x , t ) = v ( x , t ) + α ( t ) + β ( t ) − α ( t ) temperature at the x (36) boundary L A compact notation for partial derivatives and Inhomogeneous v t = κ v xx + f ( x , t ) − α ′ ( t ) − β ′ ( t ) − α ′ ( t ) boundary conditions x (37a) Newton’s Law of cooling L The Fourier sine series in 2D v (0 , t ) = 0 (37b) Heat conduction in two dimensions From Cartesian to v ( L , t ) = 0 (37c) polar The Fourier series v ( x , 0) = φ ( x ) − α (0) − β (0) − α (0) Steady-state heat x (37d) conduction in a disk L

  45. Introduction The heat Special case: Dirichlet and Neumann boundary equation Instances of use conditions Heat conduction across a refrigerator The initial/boundary value problem wall The derivation of the heat equation u t = κ u xx + f ( x , t ) Initial/boundary value problems u (0 , t ) = α ( t ) for the heat equation u x ( L , t ) = β ( t ) Separation of variables u ( x , 0) = φ ( x ) Homogeneous equations Insulated boundary is a special case of (26) with α 1 ( t ) = 1 , α 2 ( t ) = 0 , β 1 ( t ) = 0 , β 2 ( t ) = 1. Equations with heat source From (30) we get c 1 = α ( t ) , c 2 = β ( t ) and then (27) and (34) reduce to Prescribed temperature at the boundary A compact notation u ( x , t ) = v ( x , t ) + α ( t ) + β ( t ) x for partial derivatives Inhomogeneous boundary conditions and Newton’s Law of cooling v t = κ v xx + f ( x , t ) − α ′ ( t ) − β ′ ( t ) x The Fourier sine series in 2D Heat conduction in v (0 , t ) = 0 two dimensions From Cartesian to polar v ( L , t ) = 0 The Fourier series Steady-state heat v ( x , 0) = φ ( x ) − α (0) − β (0) x conduction in a disk

  46. Introduction The heat Special case: Neumann and Robin boundary equation Instances of use conditions Heat conduction across a refrigerator The initial/boundary value problem wall The derivation of the heat equation u t = κ u xx + f ( x , t ) Initial/boundary value problems u x (0 , t ) = α ( t ) for the heat equation β 1 ( t ) u ( L , t ) + β 2 ( t ) u x ( L , t ) = β ( t ) Separation of variables u ( x , 0) = φ ( x ) Homogeneous equations Insulated boundary is a special case of (26) with α 1 ( t ) = 0 , α 2 ( t ) = 1. From (30) we get Equations with heat source � � β ( t ) − β 1 ( t ) L + β 2 ( t ) α ( t ) Prescribed c 1 ( t ) = , c 2 ( t ) = α ( t ) and then (27) and (34) reduce to temperature at the β 1 ( t ) boundary A compact notation for partial u ( x , t ) = v ( x , t ) + c 1 ( t ) + c 2 ( t ) x derivatives Inhomogeneous boundary conditions and Newton’s Law of cooling v t = κ v xx + f ( x , t ) − c ′ 1 ( t ) − c ′ 2 ( t ) x The Fourier sine series in 2D Heat conduction in v (0 , t ) = 0 two dimensions From Cartesian to polar v ( L , t ) = 0 The Fourier series Steady-state heat v ( x , 0) = φ ( x ) − c 1 (0) − c 2 (0) x conduction in a disk

  47. Introduction The heat Exceptional cases equation Instances of use The change in (27) from u ( x , t ) to the v ( x , t ) works for reducing inhomogeneous Heat conduction across a refrigerator wall boundary conditions to homogeneous ones in most cases, but not always. That’s The derivation of the heat equation because the equations in (30) fail to provide values for c 1 and c 2 when their Initial/boundary denominators vanish. Once such instance occurs when Neumann boundary value problems for the heat conditions are specified at both ends: equation Separation of u x (0 , t ) = α ( t ) , u x ( L , t ) = β ( t ) . (38) variables Homogeneous equations Insulated boundary That’s a special case of (26b) and (26c) with Equations with heat source Prescribed α 1 ( t ) = 0 , α 2 ( t ) = 1 , β 1 ( t ) = 0 , β 2 ( t ) = 1 . temperature at the boundary A compact notation for partial Calculating c 1 and c 2 in this case fails since the denominators in (30) vanish. derivatives Inhomogeneous boundary conditions A little experimentation shows that we can make things work by replacing the Newton’s Law of cooling change of variables (27) by The Fourier sine series in 2D Heat conduction in two dimensions u ( x , t ) = v ( x , t ) + c 1 ( t ) x + c 2 ( t ) x 2 . (39) From Cartesian to polar The Fourier series Determining the proper choices for these c 1 ( t ) and c 2 ( t ) is left as a homework Steady-state heat conduction in a disk problem.

  48. Newton’s Law of cooling � � − ku x ( L , x ) = γ u ( L , t ) − u ∞ or equivalently γ u ( L , t ) + ku x ( L , x ) = γ u ∞

  49. Introduction The heat Newton’s Law of cooling – Example 1 equation Instances of use Rod with prescribed temperature at the left, Newton’s cooling on the right. Heat conduction across a refrigerator wall The derivation of ⇒ � u ( L , t ) − u ∞ � u (0 , t ) = α ( t ) − ku x ( L , t ) = γ the heat equation Initial/boundary value problems for the heat  u t = κ u xx + f ( x , t ) 0 < x < L , t > 0 equation    Separation of u (0 , t ) = α ( t ) t > 0   variables (40) Homogeneous γ u ( L , t ) + ku x ( L , x ) = γ u ∞ t > 0 equations   Insulated boundary   u ( x , 0) = φ ( x ) 0 < x < L  Equations with heat source Prescribed temperature at the boundary The initial/boundary value problem (40) matches (26) on slide 50 with A compact notation for partial α 1 = 1 , α 2 = 0 , β 1 = γ, β 2 = k , β = γ u ∞ . Thus, from (30) we obtain derivatives Inhomogeneous boundary conditions � u ∞ − α ( t ) � c 2 = γ Newton’s Law of c 1 = α ( t ) , cooling γ L + k The Fourier sine series in 2D Heat conduction in and therefore (27) takes the form two dimensions From Cartesian to polar � u ∞ − α ( t ) � u ( x , t ) = v ( x , t ) + α ( t ) + γ The Fourier series x . (41) Steady-state heat conduction in a disk γ L + k

  50. Introduction The heat Newton’s Law of cooling – Example 1 (continued) equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation Plugging (41) into (40), and having the Observation on slide 53 in mind, we arrive Initial/boundary value problems at for the heat equation v t = κ v xx + f ( x , t ) − γ ( L − x ) + k  Separation of α ′ ( t ) 0 < x < L , t > 0  variables  γ L + k   Homogeneous   equations   v (0 , t ) = 0 t > 0 Insulated boundary   Equations with heat (42) source γ v ( L , t ) + kv x ( L , t ) = 0 t > 0 Prescribed  temperature at the   boundary  � u ∞ − α (0) � α (0) + γ  � � A compact notation   v ( x , 0) = φ ( x ) − x 0 < x < L for partial   derivatives γ L + k  Inhomogeneous boundary conditions Newton’s Law of Now that we have homogeneous boundary conditions, we may solve for v through cooling The Fourier sine series in 2D eigenfunction expansion as usual, and then obtain u from (41). Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  51. Introduction The heat Newton’s Law of cooling – Example 2 equation Instances of use Heat conduction in a rod with forced flux at the left, Newton’s cooling on the right. Heat conduction across a refrigerator wall The derivation of ⇒ � u ( L , t ) − u ∞ � − ku x (0 , t ) = α ( t ) − ku x ( L , t ) = γ the heat equation Initial/boundary value problems for the heat  u t = κ u xx + f ( x , t ) 0 < x < L , t > 0 equation    Separation of − ku x (0 , t ) = α ( t ) t > 0   variables (43) � u ( L , t ) − u ∞ � Homogeneous − ku x ( L , t ) = γ t > 0 equations   Insulated boundary   u ( x , 0) = φ ( x ) 0 < x < L  Equations with heat source Prescribed temperature at the Rearrange the terms in the right boundary condition as boundary A compact notation for partial γ u ( L , t ) + ku x ( L , t ) = γ u ∞ . Then (43) matches (26) on slide 50 with derivatives Inhomogeneous α 1 = 0 , α 2 = − k , β 1 = γ, β 2 = k , β = γ u ∞ . Thus, from (30) we obtain boundary conditions Newton’s Law of cooling c 1 = u ∞ + α ( t ) + L α ( t ) c 2 = − α ( t ) The Fourier sine , series in 2D γ k k Heat conduction in two dimensions From Cartesian to and therefore (27) takes the form polar The Fourier series u ( x , t ) = v ( x , t ) + α ( t ) ( L − x ) + α ( t ) Steady-state heat conduction in a disk + u ∞ . (44) k γ

  52. Introduction The heat Newton’s Law of cooling – Example 2 (continued) equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation Plugging (44) into (43), and having the Observation on slide 53 in mind, we arrive Initial/boundary value problems at for the heat equation  � L − x + 1 � Separation of α ′ ( t ) v t = κ v xx + f ( x , t ) − 0 < x < L , t > 0  variables   k γ  Homogeneous   equations   Insulated boundary v x (0 , t ) = 0 t > 0   Equations with heat (45) source γ v ( L , t ) + kv x ( L , t ) = 0 t > 0 Prescribed  temperature at the   boundary  � L − x + 1  � A compact notation   v ( x , 0) = φ ( x ) − α (0) − u ∞ 0 < x < L for partial   derivatives k γ  Inhomogeneous boundary conditions Newton’s Law of Now that we have homogeneous boundary conditions, we may solve for v through cooling The Fourier sine series in 2D eigenfunction expansion as usual, and then obtain u from (44). Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  53. Introduction The heat The eigenfunctions of problem (42) equation Instances of use Here we give the details of solving problem (42). The solution of problem (45) is Heat conduction across a refrigerator wall along similar lines and is left as a homework problem. The derivation of the heat equation We begin by examining the homogeneous PDE corresponding to (42), and the Initial/boundary value problems associated boundary conditions: for the heat equation  v t = κ v xx 0 < x < L , t > 0 Separation of  variables   v (0 , t ) = 0 t > 0 (46) Homogeneous equations  Insulated boundary  γ v ( L , t ) + kv x ( L , t ) = 0 t > 0  Equations with heat source Prescribed temperature at the boundary We look for a separable solution of the form v ( x , t ) = X ( x ) T ( t ). We get: A compact notation for partial derivatives Inhomogeneous T ′ ( t ) X ( x ) = κ T ( t ) X ′′ ( x ) , γ X ( L ) T ( t ) + kX ′ ( L ) T ( t ) = 0 X (0) T ( t ) = 0 , boundary conditions Newton’s Law of cooling which simplifies to The Fourier sine series in 2D Heat conduction in two dimensions κ T ( t ) = X ′′ ( x ) T ′ ( t ) From Cartesian to hX ( L ) + X ′ ( L ) = 0 X ( x ) , X (0) = 0 , (47) polar The Fourier series Steady-state heat conduction in a disk where h = γ/ k .

  54. Introduction The heat The eigenfunctions of problem (42) – slide 2 equation Instances of use The first of equations (47) implies that Heat conduction across a refrigerator wall κ T ( t ) = X ′′ ( x ) T ′ ( t ) The derivation of X ( x ) = − λ 2 the heat equation Initial/boundary value problems for the heat for some constant λ . Therefore T ′ ( t ) + κλ 2 T ( t = 0 and equation Separation of X ′′ ( x ) + λ 2 X ( x ) = 0 , hX ( L ) + X ′ ( L ) = 0 , X (0) = 0 , (48) variables Homogeneous equations whence Insulated boundary T ( t ) = Ce − κλ 2 t , Equations with heat X ( x ) = A sin λ x + B cos λ x . source Prescribed temperature at the The boundary condition X (0) = 0 implies that B = 0. Therefore X ( x ) = A sin λ x . boundary A compact notation The boundary condition at x = L says that hA sin λ L + λ A cos λ L = 0 , that is, for partial derivatives tan λ L = − 1 h λ . We rewrite this as tan λ L = − 1 hL λ L and then let µ = λ L and arrive Inhomogeneous boundary conditions at tan µ = − 1 Newton’s Law of hL µ . cooling The Fourier sine series in 2D Conclusion: Need to solve the transcendental equation Heat conduction in two dimensions tan µ = − 1 From Cartesian to polar hL µ (49) The Fourier series Steady-state heat conduction in a disk numerically to determine µ . Then λ = µ/ L .

  55. Introduction The heat The eigenfunctions of problem (42) – slide 3 equation Instances of use Heat conduction across a refrigerator wall The derivation of The graphs of tan µ and − 1 hL µ plotted together. We the heat equation have taken L = 1, h = 1 for the purposes of this Initial/boundary value problems illustration. The intersection of the graphs mark the for the heat equation solutions of (49). The first five positive roots are Separation of µ = 2 . 0288 , 4 . 9132 , 7 . 9787 , 11 . 0855 , 14 . 2074. variables Homogeneous equations Insulated boundary Equations with heat source We write µ n , n = 1 , 2 , . . . for the roots of the equation (49). The corresponding Prescribed temperature at the boundary values of λ are λ n = µ n / L , and the solution of (48) are X n ( x ) = sin λ n x . A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  56. Introduction The heat The eigenfunctions of problem (42) – slide 4 equation Instances of use The Sturm–Liouville Theory. The problem (48) that we just solved, is a special case of Heat conduction across a refrigerator what is know as the Sturm–Liouville problem : wall The derivation of the heat equation � ′  � p ( x ) X ′ ( x ) + q ( x ) X ( x ) + λ w ( x ) X ( x ) = 0 , Initial/boundary    value problems (50) for the heat α 1 X ( a ) + α 2 X ′ ( a ) = 0 , equation   β 1 X ( b ) + β 2 X ′ ( b ) = 0 .  Separation of variables The Sturm–Liouville Theory, dating back to 1837, states that under certain conditions (see Homogeneous equations Wikipedia for the precise requirements) the boundary value problem (50) has infinitely Insulated boundary Equations with heat many eigenvalues λ n which may be ordered as source Prescribed temperature at the λ 1 < λ 2 < · · · < λ n < · · · → ∞ , boundary A compact notation for partial and corresponding to each λ n there is a unique (up to a multiplicative constant) nonzero derivatives Inhomogeneous eigenfunctions X n ( x ). The eigenfunctions, after appropriate scaling, satisfy the boundary conditions Newton’s Law of orthogonality condition cooling The Fourier sine � b � series in 2D 0 if m � = n Heat conduction in w ( x ) X m ( x ) X n ( x ) dx = two dimensions 1 if m = n From Cartesian to a polar The Fourier series Any function φ ( x ) on the interval ( a , b ) may be expressed as the infinite sum Steady-state heat conduction in a disk � b φ ( x ) = � ∞ n =1 c n X n ( x ), where c n = a w ( x ) φ ( x ) X n ( x ) dx .

  57. The Fourier sine series in 2D

  58. Introduction The heat The Fourier sine series in 2D equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation In Slide 21 we learned how to expand a function φ ( x ) into the Fourier sine series. Initial/boundary value problems Here we generalize the idea to functions of two variables. Specifically, let us for the heat equation consider a function φ ( x , y ) on the square ( O , L ) × (0 , L ). For any fixed value of y , Separation of this is a function of the single variable x , and therefore we may apply the variables Homogeneous formulas (12), (13), and (14) on Slide 21 to obtain: equations Insulated boundary Equations with heat ∞ source � Prescribed φ ( x , y ) = b n ( y ) sin λ n x , (51) temperature at the boundary n =1 A compact notation for partial derivatives where Inhomogeneous boundary conditions � L b n ( y ) = 2 λ n = n π Newton’s Law of φ ( x , y ) sin λ n x dx , and L . (52) cooling The Fourier sine L 0 series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  59. Introduction The heat The Fourier sine series in 2D – continued equation Instances of use The function b n ( y ) itself may be expanded into a Fourier sine series, as in Heat conduction across a refrigerator wall The derivation of ∞ the heat equation � b n ( y ) = a mn sin λ m y (53) Initial/boundary value problems m =1 for the heat equation where Separation of � L a mn = 2 variables b n ( y ) sin λ m y dy Homogeneous L equations 0 Insulated boundary Equations with heat Substituting for b n ( y ) from (52), this becomes source Prescribed temperature at the � L � L boundary a mn = 4 A compact notation φ ( x , y ) sin λ n x sin λ m y dx dy . for partial L 2 derivatives 0 0 Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine Furthermore, substituting b n ( y ) from (53) into (51) we see that series in 2D Heat conduction in two dimensions ∞ ∞ From Cartesian to polar � � φ ( x , y ) = a mn sin λ n x sin λ m y . The Fourier series Steady-state heat n =1 m =1 conduction in a disk

  60. Introduction The heat The Fourier sine series in 2D – summary equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation To summarize the calculations of the previous two slides: A function φ ( x , y ) on the Initial/boundary value problems square (0 , L ) × (0 , L ) may be expanded into two-dimensional Fourier sine series as for the heat equation Separation of variables ∞ ∞ � � Homogeneous φ ( x , y ) = a mn sin λ n x sin λ m y . equations (54) Insulated boundary n =1 m =1 Equations with heat source Prescribed temperature at the where boundary A compact notation � L � L a mn = 4 for partial derivatives φ ( x , y ) sin λ n x sin λ m y dx dy . (55) Inhomogeneous L 2 boundary conditions 0 0 Newton’s Law of cooling The Fourier sine series in 2D These are the two-dimensional versions of the formulas on Slide 21. Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  61. Heat conduction in two dimensions ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 + f ( x , y ) = 0

  62. Introduction The heat Heat conduction in two dimensions equation Instances of use The equation of heat conduction ∂ u /∂ t = κ∂ 2 u /∂ x 2 + f ( x , t ) generalizes to two Heat conduction across a refrigerator wall spatial dimensions as The derivation of the heat equation Initial/boundary � ∂ 2 u ∂ x 2 + ∂ 2 u ∂ u � value problems ∂ t = κ + f ( x , y , t ) , for the heat ∂ y 2 equation Separation of where the temperature u is a function of three variables, u = u ( x , y , t ). variables Homogeneous When the heat generation term f ( x , y , t ) and the boundary conditions are equations Insulated boundary independent of time t , the temperature stabilizes to the steady state distribution, Equations with heat source u ( x , y ), and therefore ∂ u /∂ t drops out and we are left with Prescribed temperature at the boundary A compact notation � ∂ 2 u ∂ x 2 + ∂ 2 u for partial � derivatives κ + f ( x , y ) = 0 . Inhomogeneous ∂ y 2 boundary conditions Newton’s Law of cooling The Fourier sine Dividing through κ and renaming 1 κ f ( x , y ) as f ( x , y ), we arrive at: series in 2D Heat conduction in two dimensions From Cartesian to ∂ 2 u ∂ x 2 + ∂ 2 u polar ∂ y 2 + f ( x , y ) = 0 . (Poisson’s equation) The Fourier series Steady-state heat conduction in a disk

  63. Introduction The heat Solving the heat equation in 2D equation Instances of use Let us look at the heat conduction problem in the square S = (0 , L ) × (0 , L ) with Heat conduction across a refrigerator wall zero boundary conditions along the edges: The derivation of the heat equation ∂ x 2 + ∂ 2 u ∂ 2 u Initial/boundary value problems ∂ y 2 + f ( x , y ) = 0 in S , (56a) for the heat equation u ( x , 0) = u ( x , L ) = u (0 , y ) = u ( L , y ) = 0 for all 0 < x < L , 0 < y < L . (56b) Separation of variables Homogeneous To solve that boundary value problem, we expand the known function f ( x , y ) and equations Insulated boundary the unknown function u ( x , y ) into Fourier sine series according to (54) Equations with heat source Prescribed temperature at the ∞ ∞ ∞ ∞ boundary � � � � u ( x , y ) = a mn sin λ n x sin λ m y , f ( x , y ) = c mn sin λ n x sin λ m y , A compact notation for partial derivatives n =1 m =1 n =1 m =1 Inhomogeneous boundary conditions Newton’s Law of cooling The coefficients c mn are calculated according to (55): The Fourier sine series in 2D Heat conduction in � L � L two dimensions c mn = 4 From Cartesian to f ( x , y ) sin λ n x sin λ m y dx dy , (57) polar L 2 The Fourier series 0 0 Steady-state heat conduction in a disk but the coefficients a mn are unknown and are to be determined.

  64. Introduction The heat Solving the heat equation in 2D – continued equation Instances of use To determine the coefficients a mn in the expansion of u ( x , y ), we calculate the Heat conduction across a refrigerator wall partial derivatives of that expansion, as in The derivation of the heat equation ∂ 2 u ∞ ∞ Initial/boundary � � − λ 2 ∂ x 2 = n a mn sin λ n x sin λ m y , value problems for the heat n =1 m =1 equation ∂ 2 u ∞ ∞ Separation of − λ 2 � � ∂ y 2 = m a mn sin λ n x sin λ m y , variables Homogeneous n =1 m =1 equations Insulated boundary and substitute these, along with the series expansion of f ( x , y ), into the Equations with heat source PDE (56a). We get Prescribed temperature at the boundary A compact notation ∞ ∞ ∞ ∞ for partial − λ 2 − λ 2 � � � � derivatives n a mn sin λ n x sin λ m y + m a mn sin λ n x sin λ m y Inhomogeneous boundary conditions n =1 m =1 n =1 m =1 Newton’s Law of cooling ∞ ∞ The Fourier sine � � + c mn sin λ n x sin λ m y = 0 . series in 2D Heat conduction in two dimensions n =1 m =1 From Cartesian to polar ∞ ∞ � � The Fourier series − ( λ 2 n + λ 2 � � m ) a mn + c mn sin λ n x sin λ m y = 0 . Steady-state heat conduction in a disk n =1 m =1

  65. Introduction The heat Solving the heat equation in 2D – continued equation Instances of use Heat conduction across a refrigerator wall The derivation of the heat equation Initial/boundary value problems It follows that − ( λ 2 n + λ 2 m ) a mn + c mn = 0, and therefore for the heat equation c mn Separation of a mn = . variables λ 2 n + λ 2 Homogeneous m equations Insulated boundary Equations with heat source Consequently Prescribed temperature at the ∞ ∞ c mn boundary � � u ( x , y ) = sin λ n x sin λ m y , (58) A compact notation λ 2 n + λ 2 for partial derivatives m n =1 m =1 Inhomogeneous boundary conditions where the coefficients c mn are given in (57). Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  66. Introduction The heat A worked out example equation Instances of use Heat conduction across a refrigerator wall Let us calculate the the temperature u ( x , y ) in problem (56) on Slide 73 where the The derivation of the heat equation heat generation is uniform f ( x , y ) = 1 throughout the square. Initial/boundary value problems We begin with calculating the coefficients c mn of the Fourier expansion of f ( x , y ) for the heat equation through the formula (55): Separation of variables � L � L Homogeneous c mn = 4 equations 1 × sin λ n x sin λ m y dx dy Insulated boundary L 2 0 0 Equations with heat source �� L ��� L = 4 � Prescribed temperature at the sin λ n x dx sin λ m y dy boundary L 2 0 0 A compact notation for partial derivatives = 4 − 1 − 1 � L �� L � � � Inhomogeneous cos λ n x cos λ n y � � boundary conditions L 2 λ n λ m � � 0 0 Newton’s Law of cooling 4 The Fourier sine � �� � series in 2D = − cos λ n L + 1 − cos λ m L + 1 . Heat conduction in λ m λ n L 2 two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  67. Introduction The heat A worked out example – continued equation Instances of use Then according to (58) we get Heat conduction across a refrigerator wall ∞ ∞ u ( x , y ) = 4 (1 − cos λ m L )(1 − cos λ n L ) The derivation of � � the heat equation sin λ n x sin λ m y . L 2 λ m λ n ( λ 2 n + λ 2 m ) Initial/boundary n =1 m =1 value problems for the heat Considering that λ n = n π/ L and that cos λ n L = cos n π = ( − 1) n , this takes the equation Separation of form variables Homogeneous � 1 − ( − 1) m �� 1 − ( − 1) n � equations u ( x , y ) = 4 L 2 ∞ ∞ sin n π x sin m π y Insulated boundary � � . Equations with heat mn ( m 2 + n 2 ) π 4 L L source n =1 m =1 Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Newton’s Law of Movie made with L = 1 and ∞ set to 8 cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  68. Introduction The heat Exercise equation Instances of use Calculate the solution u ( x , y ) of the steady-state heat conduction problem (56) on Heat conduction across a refrigerator wall Slide 73, assuming that heat is generated only in the lower-left quarter of the The derivation of the heat equation domain, that is, Initial/boundary value problems � 1 if 0 < x < L / 2 and 0 < y < L / 2 , for the heat f ( x , y ) = equation 0 otherwise , Separation of variables Here is what the solution looks like: Homogeneous equations Insulated boundary Equations with heat source Prescribed temperature at the boundary A compact notation for partial derivatives Inhomogeneous boundary conditions Movie made with L = 1 and ∞ set to 8 Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  69. From Cartesian to polar

  70. Introduction The heat Change of coordinates from Cartesian to polar equation Instances of use Heat conduction The point P at ( x , y ) in Cartesian across a refrigerator wall y coordinates is represented as ( r , θ ) in The derivation of the heat equation polar coordinates where r , called the P ( x , y ) Initial/boundary value problems radial coordinate or the radius , is the r for the heat equation distance of the point from the origin O , Separation of and θ , called the angular coordinate or variables θ Homogeneous the polar coordinate , is the rotation equations x Insulated boundary angle, measured counterclockwise, of the Equations with heat source x = r cos θ, y = r sin θ ray OP away from the positive x axis Prescribed temperature at the boundary 1 = ∂ r ∂ x cos θ − r sin θ ∂θ ∂/∂ x A compact notation x = r cos θ = ⇒ for partial derivatives ∂ x Inhomogeneous boundary conditions 0 = ∂ r ∂ x sin θ + r cos θ ∂θ ∂/∂ x Newton’s Law of y = r sin θ = ⇒ cooling ∂ x The Fourier sine series in 2D Solve for ∂ r ∂ x and ∂θ∂ x : Heat conduction in two dimensions From Cartesian to ∂ r ∂ x = − 1 ∂θ ∂ r ∂θ ∂ y = 1 polar ∂ x = cos θ, r sin θ, and similarly ∂ y = sin θ, r cos θ. The Fourier series Steady-state heat conduction in a disk

  71. Introduction The heat First derivatives in polar coordinates equation Instances of use Heat conduction across a refrigerator wall A function u ( x , y ) expressed in the Cartesian coordinates may be evaluated at the The derivation of the heat equation corresponding polar coordinates as u ( r cos θ, r sin θ ). The result of the evaluation is Initial/boundary a function U ( r , θ ), where value problems for the heat equation u ( x , y ) = u ( r cos θ, r sin θ ) = U ( r , θ ) . Separation of variables Homogeneous equations Then by the chain rule Insulated boundary Equations with heat source ∂ u ∂ x = ∂ U ∂ x + ∂ U ∂ r ∂ x = ∂ U ∂θ ∂ r cos θ − 1 ∂ U Prescribed ∂θ sin θ temperature at the ∂ r ∂θ r boundary A compact notation for partial derivatives ∂ u ∂ y = ∂ U ∂ y + ∂ U ∂ r ∂ y = ∂ U ∂θ ∂ r sin θ + 1 ∂ U Inhomogeneous ∂θ cos θ boundary conditions ∂ r ∂θ r Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions From Cartesian to Coming up next. . . the calculation of ∂ 2 u ∂ x 2 and ∂ 2 u ∂ 2 u polar ∂ y 2 and (homework) ∂ x ∂ y . The Fourier series Steady-state heat conduction in a disk

  72. Introduction The heat The derivative u xx in polar coordinates equation Instances of use Heat conduction across a refrigerator wall The derivation of ∂ 2 u ∂ x 2 = ∂ � ∂ u � the heat equation Initial/boundary ∂ x ∂ x value problems for the heat = ∂ � ∂ U ∂ r cos θ − 1 ∂ U � equation ∂θ sin θ ∂ x r Separation of variables = ∂ � ∂ U ∂ r cos θ − 1 ∂ U cos θ − 1 ∂ � ∂ U ∂ r cos θ − 1 ∂ U � � Homogeneous equations ∂θ sin θ ∂θ sin θ sin θ Insulated boundary ∂ r r r ∂θ r Equations with heat source � ∂ 2 U ∂ 2 U ∂ r 2 cos θ + 1 ∂ U ∂θ sin θ − 1 � Prescribed = ∂ r ∂θ sin θ cos θ temperature at the boundary r 2 r A compact notation � ∂ 2 U for partial ∂ 2 U derivatives − 1 ∂ r ∂θ cos θ − ∂ U ∂ r sin θ − 1 ∂θ 2 sin θ − 1 ∂ U � Inhomogeneous ∂θ cos θ sin θ boundary conditions r r r Newton’s Law of cooling = ∂ 2 U ∂ 2 U ∂ 2 U The Fourier sine ∂ r 2 cos 2 θ − 2 ∂ r ∂θ sin θ cos θ + 1 ∂θ 2 sin 2 θ series in 2D Heat conduction in r 2 r two dimensions From Cartesian to + 1 ∂ U ∂ r sin 2 θ + 2 ∂ U polar ∂θ sin θ cos θ The Fourier series r 2 r Steady-state heat conduction in a disk

  73. Introduction The heat The derivative u yy in polar coordinates equation Instances of use Heat conduction across a refrigerator wall The derivation of ∂ 2 u ∂ y 2 = ∂ � ∂ u � the heat equation Initial/boundary ∂ y ∂ y value problems for the heat = ∂ � ∂ U ∂ r sin θ + 1 ∂ U � equation ∂θ cos θ ∂ y r Separation of variables = ∂ � ∂ U ∂ r sin θ + 1 ∂ U sin θ + 1 ∂ � ∂ U ∂ r sin θ + 1 ∂ U � � Homogeneous equations ∂θ cos θ ∂θ cos θ cos θ Insulated boundary ∂ r r r ∂θ r Equations with heat source � ∂ 2 U ∂ 2 U ∂ r 2 sin θ − 1 ∂ U ∂θ cos θ + 1 � Prescribed = ∂ r ∂θ cos θ sin θ temperature at the boundary r 2 r A compact notation � ∂ 2 U for partial ∂ 2 U derivatives + 1 ∂ r ∂θ sin θ + ∂ U ∂ r cos θ + 1 ∂θ 2 cos θ − 1 ∂ U � Inhomogeneous ∂θ sin θ cos θ boundary conditions r r r Newton’s Law of cooling = ∂ 2 U ∂ 2 U ∂ 2 U The Fourier sine ∂ r 2 sin 2 θ + 2 ∂ r ∂θ sin θ cos θ + 1 ∂θ 2 cos 2 θ series in 2D Heat conduction in r 2 r two dimensions From Cartesian to + 1 ∂ U ∂ r cos 2 θ − 2 ∂ U polar ∂θ sin θ cos θ The Fourier series r 2 r Steady-state heat conduction in a disk

  74. Introduction The heat The Laplacian in polar coordinates equation Instances of use Heat conduction across a refrigerator The expression ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 and its three-dimensional version ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 + ∂ 2 u wall ∂ z 2 are The derivation of the heat equation ever-present in mathematical models stemming from physics, and encompass heat Initial/boundary value problems conduction, fluid and solid mechanics, electromagnetism, relativity, and cosmology. for the heat That expression is called the Laplacian of a function u and is written ∆ u (notation equation popular among mathematicians) and ∇ 2 u (notation popular among engineers and Separation of variables physicists). We have seen how the Laplacian plays a fundamental role in describing Homogeneous equations heat conduction. So far we have dealt with the Laplacian expressed in Cartesian Insulated boundary Equations with heat source coordinates. Equipped with the calculations of the preceding two slides, we may Prescribed express the Laplacian in polar coordinates by summing the expressions for ∂ 2 u /∂ x 2 temperature at the boundary and ∂ 2 u /∂ y 2 calculated there. There is great deal of cancellation/simplification A compact notation for partial derivatives and we arrive at Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine ∆ u = ∇ 2 u = ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 = ∂ 2 U ∂ 2 U series in 2D ∂ r 2 + 1 ∂ U ∂ r + 1 Heat conduction in two dimensions (59) r 2 ∂θ 2 r From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  75. The Fourier series ∞ � � f ( x ) = A 0 + A n cos nx + B n sin nx � n =1

  76. Introduction The heat The Fourier series equation Instances of use Heat conduction across a refrigerator Up to this point we have focused on the Fourier sine series which was formally wall The derivation of the heat equation defined on Slide 21. The Fourier sine series works best with functions f ( x ) defined Initial/boundary on an interval (0 , L ) that satisfy zero boundary conditions, that is f (0) = f ( L ) = 0. value problems for the heat equation In this section we introduce the general Fourier series which works for all functions, Separation of regardless of any boundary conditions. To simplify the algebra, we limit the variables Homogeneous presentation to functions on the interval ( − π, π ). Extending the conclusions to equations Insulated boundary arbitrary intervals ( a , b ) is pretty straightforward. Equations with heat source Prescribed Here is the general Fourier series for functions defined on the interval ( − π, π ): temperature at the boundary A compact notation for partial ∞ derivatives � � � f ( x ) = A 0 + A n cos nx + B n sin nx . (60) Inhomogeneous boundary conditions Newton’s Law of n =1 cooling The Fourier sine series in 2D We skip the technical details here, but suffice to say that such a representation is Heat conduction in two dimensions possible for just about any function f ( x ) that you would normally run across. In From Cartesian to polar the next few slides we focus on how to determine the A s and B s for a given f . The Fourier series Steady-state heat conduction in a disk

  77. Introduction The heat Calculating the Fourier series’ coefficients equation Instances of use Heat conduction across a refrigerator wall The value of A 0 is easy to determine: just integrate (60) over ( − π, π ) and note The derivation of the heat equation that for any positive integer n we have Initial/boundary value problems � π for the heat π cos nx dx = 1 = 1 � equation � � � n sin nx sin n π − sin( − n π ) = 0 , (61a) � n Separation of − π � − π variables � π π sin nx dx = − 1 = − 1 Homogeneous � � � equations � n cos nx cos n π − cos( − n π ) = 0 . (61b) Insulated boundary � n − π � Equations with heat − π source Prescribed temperature at the � π boundary Consequently, − π f ( x ) dx = 2 π A 0 , and therefore A compact notation for partial derivatives � π Inhomogeneous A 0 = 1 boundary conditions f ( x ) dx . Newton’s Law of cooling 2 π − π The Fourier sine series in 2D Heat conduction in Remark: It is worth noting that A 0 calculated above is precisely the average value two dimensions From Cartesian to of f ( x ) over the interval ( − π, π ). polar The Fourier series Steady-state heat conduction in a disk

  78. Introduction The heat Calculating the Fourier series’ coefficients equation Instances of use (continued) Heat conduction across a refrigerator wall To calculate the remaining A s and B s, we observe that for all positive integers m The derivation of the heat equation and n we have Initial/boundary value problems � π � 0 if m � = n for the heat cos mx cos nx dx = (62a) equation π if m = n − π Separation of variables � π � 0 if m � = n Homogeneous equations sin mx sin nx dx = (62b) Insulated boundary π if m = n − π Equations with heat source � π Prescribed temperature at the sin mx cos nx dx = 0 (62c) boundary − π A compact notation for partial derivatives Inhomogeneous Going back to (60), multiply both sides by cos mx , where m is a positive integer, boundary conditions Newton’s Law of and integrate. We get cooling The Fourier sine series in 2D � π � π Heat conduction in two dimensions f ( x ) cos mx dx = A 0 cos mx dx From Cartesian to − π − π polar � π � π The Fourier series ∞ � � Steady-state heat � + A n cos mx cos nx dx + B n cos mx sin nx dx . (63) conduction in a disk − π − π n =1

  79. Introduction The heat Calculating the Fourier series’ coefficients equation Instances of use (continued) Heat conduction across a refrigerator wall The derivation of the heat equation Initial/boundary value problems The coefficients of A 0 and B n in (63) are zero due to (61a) and (62c). The for the heat coefficients of A n are all zero by (62a) except when n = m in which case the equation � π Separation of coefficient is π . Thus, (63) collapses to − π f ( x ) cos mx dx = A m π, whence variables Homogeneous � π equations A n = 1 Insulated boundary f ( x ) cos nx dx , n = 1 , 2 , . . . . Equations with heat π source − π Prescribed temperature at the boundary To determine the coefficients B n , we multiply (60) by sin mx and integrate. A compact notation for partial Repeating the reasoning above, we arrive at derivatives Inhomogeneous boundary conditions � π Newton’s Law of B n = 1 cooling f ( x ) sin nx dx , n = 1 , 2 , . . . . The Fourier sine π series in 2D − π Heat conduction in two dimensions From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

  80. Introduction The heat The Fourier series – Summary equation Instances of use Here we summarize the findings of this section. Heat conduction across a refrigerator wall For all practical purposes, any function f ( x ) defined in the interval ( − π, π ) may be The derivation of the heat equation expressed as Initial/boundary value problems for the heat equation ∞ � � � Separation of f ( x ) = A 0 + A n cos nx + B n sin nx , variables (64a) n =1 Homogeneous equations Insulated boundary Equations with heat where source � π Prescribed A 0 = 1 temperature at the boundary f ( x ) dx , (64b) A compact notation 2 π − π for partial derivatives Inhomogeneous boundary conditions � π A n = 1 Newton’s Law of cooling f ( x ) cos nx dx , n = 1 , 2 , . . . , (64c) The Fourier sine π series in 2D − π Heat conduction in two dimensions From Cartesian to � π B n = 1 polar The Fourier series f ( x ) sin nx dx , n = 1 , 2 , . . . . (64d) Steady-state heat π − π conduction in a disk

  81. Steady-state heat conduction in a disk

  82. Introduction The heat Steady-state heat conduction in a disk equation Instances of use Consider a thin circular disk of radius a , insulated Heat conduction across a refrigerator wall on its flat faces, and exposed all around its y The derivation of the heat equation peripheral edge. u = h ( θ ) Initial/boundary value problems We install a polar coordinate system ( r , θ ) in the for the heat equation r plane of the disk, with the origin at the disk’s Separation of θ variables center, and we impose a prescribed temperature x Homogeneous f ( θ ), − π < θ < π around the edge and wait until equations Insulated boundary the temperature stabilizes to a steady-state Equations with heat source u ( r , θ ). Mathematically, this is described as a Prescribed temperature at the boundary boundary value problem: A compact notation for partial derivatives ∂ 2 u ∂ 2 u  ∂ r 2 + 1 ∂ u ∂ r + 1 Inhomogeneous boundary conditions ∂θ 2 = 0 , 0 < r < a , − π < θ < π   r 2 Newton’s Law of r   cooling   The Fourier sine  u ( a , θ ) = h ( θ ) . − π < θ < π (65) series in 2D Heat conduction in two dimensions  u ( r , θ ) is 2 π -periodic in θ  From Cartesian to   polar   u (0 , θ ) is finite  The Fourier series Steady-state heat conduction in a disk

  83. Introduction The heat Separation of variables equation Instances of use We look for a solution u ( r , θ ) = R ( t )Ψ( θ ). Plugging this into the PDE we obtain Heat conduction across a refrigerator wall The derivation of R ′′ ( t )Ψ( θ ) + 1 r R ′ ( t )Ψ( θ ) + 1 the heat equation r 2 R ( t )Ψ ′′ ( θ ) = 0 , Initial/boundary value problems for the heat equation and then we separate the variables: Separation of variables r 2 R ′′ ( r ) + rR ′ ( r ) R ( r ) = − Ψ ′′ ( θ ) Homogeneous (66) equations R ( r ) Ψ( θ ) Insulated boundary Equations with heat source Prescribed The left-hand side involves r only, and the right-hand side involves θ only. temperature at the boundary Therefore each side is a constant. The constant may be negative, zero, or positive. A compact notation for partial derivatives A negative constant, say − λ 2 , is not interesting since the Ψ equation becomes Inhomogeneous boundary conditions − Ψ ′′ ( θ ) Newton’s Law of Ψ( θ ) = − λ 2 , that is, Ψ ′′ ( θ ) − λ 2 Ψ( θ ) = 0 whose general solution is cooling The Fourier sine Ψ( θ ) = A cosh λθ + B sinh λθ . But such a function is not periodic in θ , and series in 2D Heat conduction in two dimensions therefore the periodicity condition in (65) cannot be met. From Cartesian to polar The Fourier series On the other hand, the zero or positive choices for the separation constant are both Steady-state heat conduction in a disk viable and lead to interesting results.

  84. Introduction The heat The case of a zero separation constant equation Instances of use Let’s consider the case where the separation constant, that is, the common value of Heat conduction across a refrigerator wall the two sides of (66), is zero. Then we would have The derivation of the heat equation Initial/boundary r 2 R ′′ ( r ) + rR ′ ( r ) = 0 . Ψ ′′ ( θ ) = 0 , value problems for the heat equation The solution of the Ψ equation is Ψ( θ ) = A θ + B . The periodicity requirement on Separation of Ψ forces A to be zero, therefore we are left with Ψ( θ ) = B . In other words, Ψ( θ ) is variables Homogeneous any constant function. That certainly satisfies the periodicity condition. equations Insulated boundary Equations with heat To solve the R equation, we rewrite it as r 2 R ′′ ( r ) + rR ′ ( r ) = 0, and thus source Prescribed R ′′ ( r ) R ′ ( r ) = − 1 temperature at the r , and integrate and get ln R ′ ( r ) = ln c 1 − ln r which simplifies to boundary � = ln c 1 , that is rR ′ ( r ) = c 1 . Therefore R ′ ( r ) = c 1 / r and consequently A compact notation for partial � rR ′ ( r ) ln derivatives Inhomogeneous boundary conditions Newton’s Law of R ( r ) = c 1 ln r + c 2 . (67) cooling The Fourier sine series in 2D Heat conduction in We are forced to take c 1 = 0, otherwise the function would blow up as r two dimensions From Cartesian to approaches zero, violating the finiteness requirement stated in (65). polar The Fourier series Conclusion: When the separation constant is zero, the only acceptable solution is Steady-state heat conduction in a disk Ψ( θ ) = constant, R ( r ) = constant, and therefore u ( r , θ ) = constant.

  85. Introduction The heat The case of a positive separation constant equation Instances of use Let’s consider the case where the separation constant, that is, the common value of Heat conduction across a refrigerator wall the two sides of (66) is positive, say λ 2 . Then we would have The derivation of the heat equation Initial/boundary Ψ ′′ ( θ ) + λ 2 Ψ( θ ) = 0 , r 2 R ′′ ( r ) + rR ′ ( r ) − λ 2 R ( r ) = 0 . (68) value problems for the heat equation The general solution of the Ψ equation is Ψ( θ ) = A cos λθ + B sin λθ , whence Separation of variables Ψ ′ ( θ ) = − A λ sin λθ + B λ cos λθ . Homogeneous equations We are interested in the range − π < θ < π . The solution u ( r , θ ) will be continuous Insulated boundary Equations with heat and smooth across the negative x axis if Ψ( − π ) = Ψ( π ) and Ψ ′ ( − π ) = Ψ ′ ( π ), source Prescribed that is temperature at the boundary A compact notation for partial derivatives A cos( − λπ ) + B sin( − λπ ) = A cos( λπ ) + B sin( λπ ) , Inhomogeneous boundary conditions − A λ sin( − λπ ) + B λ cos( − λπ ) = − A λ sin( λπ ) + B λ cos( λπ ) . Newton’s Law of cooling The Fourier sine series in 2D These two equations simplify to B sin λπ = 0 and A sin λπ = 0, respectively. If Heat conduction in two dimensions From Cartesian to sin λπ is nonzero, then both A and B are zero, and that results in the trivial polar The Fourier series solution Ψ( θ ) = 0. We conclude that sin λπ = 0, and therefore λπ = n π for all Steady-state heat conduction in a disk positive integers n .

  86. Introduction The heat The case of a positive separation constant equation Instances of use (continued) Heat conduction across a refrigerator wall The derivation of We conclude that the Ψ functions of interest are Ψ n ( θ ) = A n cos n θ + B n sin n θ , the heat equation Initial/boundary n = 1 , 2 , . . . . value problems for the heat equation We return to (68) now and evaluate the R equation with λ = n . We get r 2 R ′′ ( r ) + rR ′ ( r ) − n 2 R ( r ) = 0. This ODE is called Euler’s equation and there is a Separation of variables well-know trick for solving it. Specifically, We try a solution of the form R ( r ) = r α Homogeneous equations Insulated boundary for a yet unspecified exponent α . Plugging this into the ODE we see that Equations with heat α ( α − 1) r α + α r α − n 2 r α = 0, whence α ( α − 1) + α − n 2 = 0, which simplifies to source Prescribed α 2 = n 2 . We conclude that α = ± n , and therefore the general solution of Euler’s temperature at the boundary A compact notation equation is for partial derivatives R ( r ) = c 1 r − n + c 2 r n . Inhomogeneous (69) boundary conditions Newton’s Law of cooling We are forced to take c 1 = 0, otherwise the function would blow up as r The Fourier sine series in 2D approaches zero, violating the finiteness requirement stated in (65). Heat conduction in two dimensions From Cartesian to Conclusion: When the separation constant is positive, it has to be an integer, and polar The Fourier series the function u ( r , θ ) = r n ( A n cos n θ + B n sin n θ ) satisfies the PDE in (65). Steady-state heat conduction in a disk

  87. Introduction The heat A Fourier series representation of the solution equation Instances of use Putting together the results of the preceding slides, we arrive at the following Heat conduction across a refrigerator wall candidate for the solution of the boundary value problem (65): The derivation of the heat equation ∞ Initial/boundary r n � � � u ( r , θ ) = A 0 + A n cos n θ + B n sin n θ . (70) value problems for the heat n =1 equation Separation of This solution candidate satisfies the PDE, the periodicity, and finiteness variables Homogeneous requirements. It remains to pick the A s and B s in order for meet the prescribed equations Insulated boundary boundary condition u ( a , θ ) = h ( θ ), that is, Equations with heat source Prescribed ∞ temperature at the a n � � � boundary h ( θ ) = A 0 + A n cos n θ + B n sin n θ . A compact notation for partial n =1 derivatives Inhomogeneous boundary conditions The form of this expression happens to match precisely that of the general Fourier Newton’s Law of cooling series formalism summarized on Slide 90. Applying equations (64) to the case at The Fourier sine series in 2D hand, we see that; Heat conduction in two dimensions � π � π � π From Cartesian to A 0 = 1 1 1 polar h ( θ ) d θ, A n = h ( θ ) cos n θ d θ, B n = h ( θ ) sin n θ d θ. The Fourier series π a n π a n 2 π Steady-state heat − π − π − π conduction in a disk (71)

  88. Introduction The heat A worked out problem equation Instances of use Let’s solve the boundary value problem (70) when Heat conduction across a refrigerator wall � The derivation of 1 if | θ | < π/ 6 , the heat equation h ( θ ) = Initial/boundary 0 otherwise . value problems for the heat We evaluate the A s and B s according to (71) with the given h . We obtain: equation Separation of A 0 = 1 n π a n sin n π 2 variables 6 , A n = 6 , B n = 0 . Homogeneous equations Insulated boundary Then the solution (70) takes the form Equations with heat source � n ∞ Prescribed u ( r , θ ) = 1 6 + 2 � 1 n sin n π � � r temperature at the � cos n θ. boundary π 6 a A compact notation for partial n =1 derivatives Inhomogeneous boundary conditions Newton’s Law of cooling The Fourier sine series in 2D Heat conduction in two dimensions Illustrations made with From Cartesian to polar a = 1, and ∞ set to 50 The Fourier series Steady-state heat conduction in a disk

  89. Introduction The heat Heat conduction on an annulus equation Instances of use Consider a thin annulus of inner and outer radii a Heat conduction across a refrigerator wall and b , respectively, insulated on its flat faces, and The derivation of u = g ( θ ) the heat equation exposed on its inner and outer peripheral edges Initial/boundary value problems where the temperature is fixed at h ( θ ) around the for the heat a b equation inner edge, and g ( θ ) around the outer edge. Here Separation of θ is the angular coordinate in a polar coordinate variables system affixed to the annulus at its center. Homogeneous equations Insulated boundary The resulting steady-state temperature field, Equations with heat u = h ( θ ) source u ( r , θ ), is the solution of the boundary value Prescribed temperature at the boundary problem: A compact notation for partial derivatives ∂ 2 u ∂ 2 u  ∂ r 2 + 1 ∂ u ∂ r + 1 Inhomogeneous boundary conditions ∂θ 2 = 0 , a < r < b , − π < θ < π   r 2 Newton’s Law of r   cooling   The Fourier sine  u ( a , θ ) = h ( θ ) . − π < θ < π (72) series in 2D Heat conduction in two dimensions  u ( b , θ ) = g ( θ ) . − π < θ < π  From Cartesian to   polar   u ( r , θ ) is 2 π -periodic in θ  The Fourier series Steady-state heat conduction in a disk

  90. Introduction The heat Heat conduction on an annulus – continued equation Instances of use Heat conduction across a refrigerator wall We separate the variables in (72) just as we did in the case of heat conduction on a The derivation of the heat equation disk. In fact, most of the calculation there carries over here with only small Initial/boundary value problems changes. for the heat equation Specifically, recall that in Slide 94 we dismissed the logarithmic term in (67) to Separation of avoid blowup at r = 0. But that is of no concern in an annulus since r = 0 varies variables Homogeneous from a to b , and does not hit zero. Therefore we retain the full solution given equations Insulated boundary in (67) in the current calculation. Equations with heat source Similarly, on Slide 96 we dismissed the r − n in (69), but we retain it in the current Prescribed temperature at the boundary calculation since r does not approach zero. Then, the equivalent of the A compact notation for partial derivatives representation (70) in the case of annulus becomes Inhomogeneous boundary conditions Newton’s Law of cooling ∞ �� � A n r n + B n r − n � � C n r n + D n r − n � The Fourier sine � u ( r , θ ) = A 0 + B 0 ln r + cos n θ + sin n θ . (73) series in 2D Heat conduction in two dimensions n =1 From Cartesian to polar The Fourier series Steady-state heat conduction in a disk

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