Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Chapter 12: Boundary-Value Problems in Rectangular Coordinates Department of Electrical Engineering National Taiwan University ihwang@ntu.edu.tw December 21, 2013 1 / 50 DE Lecture 14 王奕翔 王奕翔
Separation of Variables and Classical PDE’s Wave Equation 2 / 50 Laplace Equation Wave Equation Heat Equation Types of Equations : Method : Separation of variables – convert a PDE into two ODE’s DE Lecture 14 useful particular solutions. Instead of solving the general solutions, we are only interested in finding In this lecture, we focus on solving some classical partial differential Summary Laplace’s Equation equations in boundary-value problems . We focus on linear second order PDE: ( A , · · · , G : functions of x , y ) Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G . ∂ 2 u notation: for example, u xy := ∂ x ∂ y . 王奕翔
Separation of Variables and Classical PDE’s Homogeneous 3 / 50 Elliptic Parabolic Hyperbolic Wave Equation Nonhomogeneous DE Lecture 14 Laplace’s Equation 1 Homogeneous vs. Nonhomogeneous Summary Classification of Linear Second Order PDE Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G . ∂ 2 u notation: for example, u xy := ∂ x ∂ y . ⇐ ⇒ G = 0 ⇐ ⇒ G ̸ = 0 . 2 Hyperbolic, Parabolic, and Elliptic: A , B , C , · · · , G : constants, ⇒ B 2 − 4 AC > 0 ⇐ ⇒ B 2 − 4 AC = 0 ⇐ ⇒ B 2 − 4 AC < 0 ⇐ 王奕翔
Separation of Variables and Classical PDE’s k 4 / 50 combination can be an infinite series Note : We shall assume without rigorous argument that the linear is also a solution. Wave Equation DE Lecture 14 PDE, then a linear combination Theorem Superposition Principle Summary Laplace’s Equation If u 1 ( x , y ) , u 2 ( x , y ) , . . . , u k ( x , y ) are solutions of a homogeneous linear ∑ u ( x , y ) := c n u n ( x , y ) n =1 ∞ ∑ u ( x , y ) := c n u n ( x , y ) n =1 王奕翔
Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary 1 Separation of Variables and Classical PDE’s 2 Wave Equation 3 Laplace’s Equation 4 Summary 5 / 50 DE Lecture 14 王奕翔
Separation of Variables and Classical PDE’s product of a x -function and a y -function: 6 / 50 Note : Derivatives are with respect to different independent variables. Wave Equation DE Lecture 14 Summary Separation of Variables Laplace’s Equation To find a particular solution of an PDE, one method is separation of variables , that is, assume that the solution u ( x , y ) takes the form of a u ( x , y ) = X ( x ) Y ( y ) . Then, with the following, sometimes the PDE can be converted into an ODE of X and an ODE of Y : dx Y = X ′ Y , dy = XY ′ u x = dX u y = XdY u xx = d 2 X u yy = Xd 2 Y dx 2 Y = X ′′ Y , dy 2 = XY ′′ , u xy = X ′ Y ′ For example, X ′ := dX dx . 王奕翔
Separation of Variables and Classical PDE’s = 7 / 50 independent of x and y function of y , independent of x . Hence, the above is equal to something Left-hand side is a function of x , independent of y ; Right-hand side is a separation constant Wave Equation = Example Use separation of variables to convert the PDE below into two ODE’s. Convert a PDE into Two ODE’s Summary Laplace’s Equation DE Lecture 14 x 2 u xx + ( x + 1) u y + ( x + xy ) u = 0 With u ( x , y ) = X ( x ) Y ( y ) , the original PDE becomes x 2 X ′′ Y + ( x + 1) XY ′ + ( x + 1) yXY = 0 ⇒ x 2 X ′′ Y = − ( x + 1) X ( Y ′ + yY ) x 2 X ′′ ( x + 1) X = − Y ′ ⇒ Y − y = λ 王奕翔
Separation of Variables and Classical PDE’s = 8 / 50 = separation constant Wave Equation = DE Lecture 14 Summary Use separation of variables to convert the PDE below into two ODE’s. Example Laplace’s Equation Convert a PDE into Two ODE’s x 2 u xx + ( x + 1) u y + ( x + xy ) u = 0 With u ( x , y ) = X ( x ) Y ( y ) , the original PDE becomes x 2 X ′′ Y + ( x + 1) XY ′ + ( x + 1) yXY = 0 ⇒ x 2 X ′′ Y = − ( x + 1) X ( Y ′ + yY ) x 2 X ′′ ( x + 1) X = − Y ′ ⇒ Y − y = λ { x 2 X ′′ ( x ) − λ ( x + 1) X ( x ) = 0 ⇒ Y ′ ( y ) + ( y + λ ) Y ( y ) = 0 王奕翔
Separation of Variables and Classical PDE’s Wave Equation Laplace’s Equation Summary Some Remarks 1 The method of separation of variables can only solve for some linear second order PDE’s, not all of them. 2 For the PDE’s considered in this lecture, the method works. 9 / 50 DE Lecture 14 3 The method may work for both homogeneous ( G = 0 ) and nonhomogeneous ( G ̸ = 0 ) PDE’s Au xx + Bu xy + Cu yy + Du x + Eu y + Fu = G . 王奕翔
Separation of Variables and Classical PDE’s mechanics, and engineering: 10 / 50 3 (Two-Dimensional) Laplace Equation 2 (One-Dimensional) Wave Equation/Telegraph Equation Wave Equation 1 (One-Dimensional) Heat Equation/Diffusion Equation following three classical PDE’s that arise frequently in physics, In this lecture we focus on solving boundary-value problems of the Three Classical PDE’s Summary Laplace’s Equation DE Lecture 14 ku xx = u t , k > 0 a 2 u xx = u tt u xx + u yy = 0 王奕翔
Separation of Variables and Classical PDE’s Heat only flows in x -direction. 11 / 50 = Wave Equation No heat is generated in the rod. No heat escapes from the surface. DE Lecture 14 Assumptions : Heat Transfer within a Thin Rod: Heat Equation Laplace’s Equation Summary Cross section of area A 0 x x + ∆ x L x Rod is homogeneous with density ρ . Let u ( x , t ) denote the temperature of the rod at location x at time t . Consider the quantity of heat with [ x , x + dx ] : ( γ : 比熱 ) dQ = γ ( ρ Adx ) u = ⇒ Q x = γρ Au = ⇒ Q xt = γρ Au t Heat transfer rate through the cross section = − KAu x , and hence the net heat rate inside [ x , x + dx ] is dQ t = − KAu x ( x , t ) − ( − KAu x ( x + dx , t )) dQ t = KAu ( u x ( x + dx , t ) − u x ( x , t )) = KAu xx dx ⇒ Q tx = KAu xx 王奕翔
Separation of Variables and Classical PDE’s No heat escapes from the surface. 12 / 50 = = Wave Equation Hence, No heat is generated in the rod. DE Lecture 14 Heat only flows in x -direction. Heat Transfer within a Thin Rod: Heat Equation Laplace’s Equation Summary Assumptions : Cross section of area A 0 x x + ∆ x L x Rod is homogeneous with density ρ . Let u ( x , t ) denote the temperature of the rod at location x at time t . { Q xt = γρ Au t ( K ) � � ⇒ γρ Au t = K ⇒ u xx = u t Au xx = γρ Q tx = KAu xx ⇒ ku xx = u t , k > 0 王奕翔
Separation of Variables and Classical PDE’s Boundary Conditions : 13 / 50 Heat is lost from the right end. The right end is insulated. u x : (Neumann condition), for example, Temperature at the right end is held at constant. Wave Equation DE Lecture 14 Provides the spatial distribution of the Heat Equation: Initial and Boundary Conditions Initial Condition : Laplace’s Equation Summary Cross section of area A temperature at time t = 0 . u ( x , 0) = f ( x ) , 0 < x < L 0 x x + ∆ x L x At the end points x = 0 and x = L , give the constraints on u : (Dirchlet condition), for example, ( u 0 : constant) u ( L , t ) = u 0 u x ( L , t ) = 0 u x + hu : (Robin condition), for example, ( h > 0 , u m : constants) u x ( L , t ) = − h { u ( L , t ) − u m } 王奕翔
Separation of Variables and Classical PDE’s Heat 14 / 50 condition Initial condition Boundary equation Heat condition Initial condition Wave Equation equation Boundary DE Lecture 14 A problem involving both initial and Examples : Laplace’s Equation Summary Heat Equation: Boundary-Value Problems one can use different kinds of conditions. boundary-value problem boundary conditions is called a Cross section of area A At the two boundaries x − 0 and x = L , 0 x x + ∆ x L x ku xx = u t , 0 < x < L , t > 0 u (0 , t ) = u 0 , u x ( L , t ) = − h { u ( L , t ) − u m } , t > 0 u ( x , 0) = f ( x ) , 0 < x < L ku xx = u t , 0 < x < L , t > 0 u x (0 , t ) = 0 , u ( L , t ) = 0 , t > 0 u ( x , 0) = f ( x ) , 0 < x < L 王奕翔
Separation of Variables and Classical PDE’s Wave Equation 15 / 50 of the string at location x at time t . Slope of the curve is very small at all points. the same at all points. Tension force is large compared to gravity and is No external force. Assumptions : DE Lecture 14 Laplace’s Equation Summary Dynamics of a String Fixed at Two Ends: Wave Equation u ∆ s u ( x, t ) 0 x x x + ∆ x L Vertical displacement ≪ string length. (a) Segment of string String has mass per unit length ρ . u T 2 Let u ( x , t ) denote the vertical position (displacement) θ 2 ∆ s θ 1 Consider the string in [ x , x + dx ] . Net vertical force is T 1 x x x + ∆ x T ( sin θ 2 − sin θ 1 ) ≈ T ( tan θ 2 − tan θ 1 ) = T { u x ( x + dx , t ) − u x ( x , t ) } = Tu xx dx 王奕翔
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