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

Partial Differential Equations Lecture Notes for Math 404 Rouben Rostamian Department of Mathematics and Statistics UMBC Fall 2020

The wave equation as a prototype of hyperbolic equations

The wave equation Hyperbolic equations in applications Instances of use The wave equation The method of characteristics d’Alembert’s solution to the second order wave equation The wave equation Waves in semi-infinite ∂ 2 u ∂ t 2 = c 2 ∂ 2 u domains and reflections from the ∂ x 2 , boundary along with its many variants, is the prototype of a very large class of hyperbolic equations that arise in many applications such as • vibration of solid structures (strings, beams, membranes, plates) • propagation of seismic waves • geological exploration, oil well detection • aerodynamics and supersonic flight • propagation of electromagnetic waves (radiant heat, light, radio waves, microwaves, fiber optics, antennas)

The wave equation The wave equation Instances of use The wave equation The method of characteristics We wish to derive the equation of motion of a stretched string with ends fixed. d’Alembert’s solution to the (Think of a guitar string or cello string). Depending on the manner of excitation, second order wave equation the string may flex in many different ways. See the figure to the right. We write T Waves in semi-infinite domains and for the tensile force within the string, ρ for the mass of string per unit length, reflections from the boundary u ( x , t ) for the lateral displacement of the string, and θ ( x , t for the angle between the string and the equilibrium state at the location x at time t , We assume that the deflection away from equilibrium is small so that we may approximate sin( θ ) ≈ θ and tan( θ ) ≈ θ . u ( x , t ) x

The wave equation The wave equation (continued) Instances of use The wave equation The method of Let us focus on a small segment of the string between locations x and x + ∆ x . characteristics d’Alembert’s The mass of that segment is ρ ∆ x , and its vertical acceleration is ∂ 2 u ∂ t 2 . Therefore, solution to the second order wave equation according to Newton, ρ ∆ x ∂ 2 u ∂ t 2 equals the resultant of vertical forces acting on the Waves in semi-infinite domains and string. But in the diagram below we see that the vertical component of the acting reflections from the boundary forces is T sin θ ( x + ∆ x , t ) − T sin θ ( x , t ). We conclude that ρ ∆ x ∂ 2 u ∂ t 2 = T sin θ ( x + ∆ x , t ) − T sin θ ( x , t ) . T θ ( x + ∆ x , t ) T θ ( x , t ) x x + ∆ x

The wave equation The wave equation (continued) Instances of use The wave equation The method of We divide through by ∆ x characteristics d’Alembert’s solution to the ρ∂ 2 u second order wave ∂ t 2 = T sin θ ( x + ∆ x , t ) − sin θ ( x , t ) equation Waves in ∆ x semi-infinite domains and reflections from the and pass to the limit as ∆ x → 0: boundary ρ∂ 2 u ∂ t 2 = T ∂ � sin θ � . ∂ x However, by our smallness assumption of θ we have ∂ u slope = tan θ ≈ sin θ and ∂ x therefore ρ∂ 2 u = T ∂ 2 u ∂ t 2 = T ∂ � ∂ u � ∂ x 2 . ∂ x ∂ x We let T /ρ = c 2 and cast the equation above into the standard form of the wave equation . It expresses Newton’s law of motion applied to a stretched string: ∂ 2 u ∂ t 2 = c 2 ∂ 2 u ∂ x 2 . (1)

The wave equation The vibrating string Instances of use The wave equation The method of characteristics d’Alembert’s Consider the stretched string depicted in Slide 4. We have seen that its motion solution to the second order wave equation u ( x , t ) is a solution of the wave equation. We supply that equation with initial and Waves in semi-infinite boundary conditions to obtain a well-posed initial boundary value problem : domains and reflections from the boundary u tt = c 2 u xx 0 < x < L , t > 0 , (2a) u (0 , t ) = 0 t > 0 , (2b) u ( L , t ) = 0 t > 0 , (2c) u ( x , 0) = f ( x ) 0 < x < L , (2d) u t ( x , 0) = g ( x ) 0 < x < L . (2e) Note the specification of the initial condition. The condition (2d) specifies the string’s deflection at t = 0. The condition (2e) specifies the string’s velocity at t = 0. In the slides that follow, we will calculate the solution of this initial boundary value problem through. . . what else? Separation of variables!

The wave equation The wave equation — separation of variables Instances of use The wave equation The method of characteristics d’Alembert’s We look for solutions to (2) in the form u ( x , t ) = X ( x ) T ( t ). Plugging this solution to the second order wave equation into (2a) we see that X ( x ) T ′′ ( t ) = c 2 X ′′ ( x ) T ( t ), whence Waves in semi-infinite domains and reflections from the c 2 T ( t ) = X ′′ ( x ) T ′′ ( t ) boundary X ( x ) = − λ 2 , where we have, based on our previous experiences with such matters, picked − λ 2 (a negative number) for the separation constant. Thus, we obtain T ′′ ( t ) + c 2 λ 2 T ( t ) = 0 , X ′′ ( x ) + λ 2 X ( x ) = 0 , X (0) = 0 , X ( L ) = 0 . (3) The last two equations are the consequences of (2b) and (2c). The general solution of the X equation is X ( x ) = A cos λ x + B sin λ x . Applying the condition X (0) = 0 implies that A = 0, and thus we are left with X ( x ) = B sin λ x . Applying the condition X ( L ) = 0 implies that sin λ L = 0, whence λ = n π/ L for all positive integers n . We write these as λ n = n π/ L .

The wave equation The wave equation — separation of variables 2 Instances of use The wave equation The method of characteristics Having determined the values of the separation constant, we write X n ( x ) = sin λ n x d’Alembert’s solution to the for the corresponding solutions. Moreover, in view of the T equation in (3), we see second order wave equation that T n ( t ) = A cos λ n ct + B sin λ n ct . We conclude that Waves in semi-infinite domains and X n ( x ) T n ( t ) = ( A n cos λ ct + B n sin λ n ct ) sin λ n x is a solution of the equations (2a), reflections from the boundary (2b), and (2c) for any positive integer n , and therefore the following infinite linear combination is also a solution: ∞ � u ( x , t ) = ( A n cos λ n ct + B n sin λ n ct ) sin λ n x . (4) n =1 In remains to pick the A ’s and B s in order to satisfy the initial conditions (2d) and (2e). Let’s observe that the velocity of the string at ( x , t ) is obtained by differentiating the displacement u ( x , t ) with respect to t : ∞ � u t ( x , t ) = ( − A n λ n c sin λ n ct + B n λ n c cos λ n ct ) sin λ n x . n =1 We set u ( x , 0) = f ( x ), u t ( x , 0) = g ( x ) and continue into the next slide.

The wave equation The wave equation — separation of variables 3 Instances of use The wave equation The method of characteristics d’Alembert’s solution to the second order wave equation Waves in We see that semi-infinite domains and reflections from the boundary ∞ ∞ � � A n sin λ n x = f ( x ) , B n λ n c sin λ n x = g ( x ) . n =1 n =1 Then A n and B n may be calculated from our old formulas for the Fourier sine series: � L � L A n = 2 2 f ( x ) sin λ n x dx , B n = g ( x ) sin λ n x dx . (5) λ n cL L 0 0 This completes our analysis and solution of the vibrating string problem. The string’s motion is given in (4), where the coefficients A n and B n are calculated according to (5).

The wave equation Vibrating string – An example Instances of use The wave equation The method of Suppose that we deflect the string into the shape of a parabola f ( x ) = x (1 − x L ) characteristics d’Alembert’s and release it without imparting any initial velocity, i.e., g ( x )=0. The motion is solution to the second order wave equation given in (4), with the A s and B s as in (5). Since g ( x ) = 0, we have all B s equal Waves in semi-infinite zero, and the solution is domains and reflections from the boundary ∞ � u ( x , t ) = A n cos λ n ct sin λ n x , n =1 where � L � L π 3 · 1 − ( − 1) n A n = 2 f ( x ) sin λ n x dx = 2 L ) sin λ n x dx = 4 L x (1 − x . n 3 L L 0 0 An animation with L = 1, c = 1 and infinity set to 20.

The wave equation Vibrating string – Another example Instances of use The wave equation The method of characteristics d’Alembert’s � x if x < L / 3 , solution to the second order wave We simulate the plucking of the string by setting f ( x ) = equation 1 2 ( L − x ) if x > L / 3 . Waves in semi-infinite domains and and g ( x )=0. Then u ( x , t ) = � ∞ n =1 A n cos λ n ct sin λ n x , where reflections from the boundary � L π 2 · sin n π A n = 2 f ( x ) sin λ n x dx = 3 L 3 . n 2 L 0 An animation with L = 1, c = 1 and infinity set to 20.