Math 220 Quiz 11 Spring 2015 Name: Put all work on this sheet in the space provided below and show details for your solution and attach additional pages if necessary. You must hand in this rst page. You may use your own notes but not somebody else's notes. Do your own quiz and not somebody else's. Date due: Friday May 1, 2015 1. Solve the vibrating string problem governed by the nonhomogeneous initial-boundary value problem @ 2u @ 2u (x, t) = (x, t) + x cos t, 0 < x < , t > 0. @t2 @x2 u(0, t) = 0, u(, t) = 0, t > 0. u(x, 0) = 0, 0 x . @u (x, 0) = 3 sin 4x 5 sin 6x, @t (1) (2) (3) 0 x . (4) We shall classify the given data in our problem: Note that the PDE, the boundary conditions, and the initial condition, are nonhomogeneous. We can use the principle of superposition to decompose this problem into two sub problems The sub problems for v and w are. The solution of the original problem is u(x; t) = v(x; t) + w(x; t): We solve separately, the v and the w problems. First, find the steady-state position s(x) (independent of t) so that s(0) =0, s() = 0and The function s is easily found s(x) = 0 Second, write a BVP for the function y(x; t) = v(x; t) s(x) (assuming that v solves the v-problem). We have yt = vt 0 = vt And yxx = vxx s(x) = vxx : Hence y(x; t) satisfies the wave equation yt = yxx. The boundary conditions for y are y(0; t) = v(0; t) s(0) = 0 y(L; t) = v(L; t) s(L) = 0 The initial condition for y is: y(x; 0) = v(x; 0) s(x) = f(x) 0 The BVP problem for the function y(x; t) is therefore the familiar problem yt = yxx y(0; t) = 0 y(; t) = 0 y(x; 0) = f(x) Third, this is a problem that we can solve by using separation of variables. We find An is the n-th Fourier sine coefficient of the f(x) s(x): ) Conclude that the solution of the v-problem is: V(x; t) = y(x; t) + s(x): V(x; t) = 0+ 0 The wproblem: Indicate how to use Eigen function expansion to construct a formal series solution. The Eigen functions to be used are those of the associated SL-problem (In this case X + _X = 0, X (0) = X () = 0): Sin (_nx); n Z+ First, expand the nonhomogeneous term F(x; t) into a Fourier sine series in x (The variable t is considered as a parameter). We have; Second, write the solution w(x; t), has the Fourier sine series in x given by where the Fourier coefficients wn(t) are function of t to be determined. Find ODEs for the coefficients wn(t). For this rewrite the PDE wt(x; t) wxx(x; t) = F(x; t) By using the Fourier expansions for w and F. We obtain: ; u(x; t) = v(x; t) + w(x; t): u(x; t) = 0 + We shall classify the given data in our problem: 2 Wave equation: u 2 2 2 (x, t) u 2 2 (x, t) x cos t 00 t x Boundary condition: u(0, t) 0, u(, t) 0, t 0 Initial Condition: u(x, 0) 0, 0 x u x, 0 3sin4x-5sin6x, 0 x t Note that the PDE, the boundary conditions, and the initial condition, are nonhomogeneous. We can use the principle of superposition to decompose this problem into two sub problems The sub problems for v and w are. vt vxx wt wxx F(x, t) v(0, t) 0 w(0, t) 0 and v(, t) 0 w(, t) 0 v(x, 0) f (x) w(x, 0) 0 vt vxx wt wxx x cos t v(0, t) 0 w(0, t) 0 and v(, t) 0 w(, t) 0 v(x, 0) 3sin4x-5sin6x w(x, 0) 0 The solution of the original problem is u(x; t) = v(x; t) + w(x; t): We solve separately, the v and the w problems. First, find the steady-state position s(x) (independent of t) so that s(0) =0, s() = 0and The function s is easily found s(x) = 0 Second, write a BVP for the function y(x; t) = v(x; t) s(x) (assuming that v solves the v-problem). We have yt = vt 0 = vt And yxx = vxx s(x) = vxx : Hence y(x; t) satisfies the wave equation yt = yxx. The boundary conditions for y are y(0; t) = v(0; t) s(0) = 0 y(L; t) = v(L; t) s(L) = 0 The initial condition for y is: y(x; 0) = v(x; 0) s(x) = f(x) 0 The BVP problem for the function y(x; t) is therefore the familiar problem yt = yxx y(0; t) = 0 y(; t) = 0 y(x; 0) = f(x) Third, this is a problem that we can solve by using separation of variables. We find y(x, t) Ane v 2 t n sin(vnx) n 1 n n An is the n-th Fourier sine coefficient of the f(x) s(x): 2 nx An (f (x) s(x))sin dx ) L0 n 2 An f (x)sin nxdx L0 An 2 (3sin x 5sin 6x) sin nxdx 0 This computes out to be equal to zero after inputting the limits. 0 y(x, t) 0 e v 2 t n sin(vnx) n 1 0 n n n Conclude that the solution of the v-problem is: V(x; t) = y(x; t) + s(x): V(x; t) = 0+ 0 The w-problem: Indicate how to use Eigen function expansion to construct a formal series solution. The Eigen functions to be used are those of the associated SL-problem (In this case X + _X = 0, X (0) = X () = 0): Sin (_nx); n Z+ First, expand the nonhomogeneous term F(x; t) into a Fourier sine series in x (The variable t is considered as a parameter). We have; F(x, t) Fn (t)sin(vnx), x (0,L),t>0, n 1 Fn (t) 2 nx F(x, t)sin dx L0 F(x, t) Fn (t) sin(vnx), x (0,L),t>0, n 1 2 Fn (t) x cos t sin nxdx 0 n 2 (1) ( )x n n 2 (1) .x n Second, write the solution w(x; t), has the Fourier sine series in x given by w(x, t) wn (t)sin n 1 nx where the Fourier coefficients wn(t) are function of t to be determined. Find ODEs for the coefficients wn(t). For this rewrite the PDE wt(x; t) wxx(x; t) = F(x; t) By using the Fourier expansions for w and F. We obtain: wn(t)sin(vnx) wn(t)(v n 1 n 1 n 1 2 n n 1 t wn (t) v 2 w n n )sin(vnx) Fn(t)sin(vnx) n 1 sin(vnx) Fn(t)sin(vnx) n 1 wn (t)sin(vnx) Fn(t)sin(vnx) ; n 1 wn(t) Fn (t) wn (t) Fn (t)dt wn (t) Fn (t)dt n 2 (1) t.x n u(x; t) = v(x; t) + w(x; t): u(x; t) = 0 + n 2 (1) t.x n