Math 220a - Fall 2002 Homework 6 Solutions
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1 Math a - Fall Homework 6 Solutions. Use the method of reflection to solve the initial-boundary value problem on the interval < x < l, u tt c u xx = < x < l u(x, = < x < l u t (x, = x < x < l u(, t = = u(l, t. In particular, calculate the explicit solution of u in regions R, R, R 3 shown below. xt, (x, t R u(x, t = c l l(x + ct + xt, (x, t R c xt l(x + ct + c c l, (x, t R 3. Do the same thing as in problem, except now for the Neumann boundary conditions. That is, use the method of refelection to solve the inital-boundary value problem on the interval < x < l with Neumann boundary conditions, u tt c u xx =, u(x, = u t (x, = x u x (, t = = u x (l, t. < x < l < x < l < x < l Write the explicit solution in the same three regions as shown in problem. a. For x R ( u(x, t = = ( = + ( ct x = (ct + x xdx
2 b. For x R ( u(x, t = = ( = + ( ct x = l tx + (ct + x + l xdx c. For x R 3 ( u(x, t = = ( = + ( ct x + = + (x ct + (x + ct 4c l + xdx 3. Use Duhamel s principle to find the solution of the inhomogeneous wave equation on the half-line with Neumann boundary conditions u tt c u xx = f(x, t, < x < u(x, = φ(x < x < u t (x, = ψ(x < x < u x (, t =. In particular, introducing a new function v = u t, rewrite the equation as the system U t + AU = F < x < U(x, = Φ(x U x (, t = [ ux (, t v x (, t ] = [ ] < x < where [ ] u U = v [ ] F = f [ ] A = c x [ ] φ Φ =. ψ
3 (a Find the solution operator S(t associated with the homogeneous system U t + AU = < x < U(x, = Φ(x [ ] < x < U x (, t =. Extending the initial data φ ψ to be even, we know that the solution of the wave equation on the half-line with Neumann boundary conditions is given as follows: u(x, t = [φ even(x + ct + φ even (x ct] + In particular, for x > ct, we have While for x < ct, we have u(x, t = [φ(x + ct + φ(x ct] + u(x, t = [φ(x + ct + φ(ct x] + ψ(y dy + ψ even (y dy. ψ(y dy. ct x ψ(y dy. Therefore, the solution operator associated with the system above is given by S(tΦ = [ t ( [φ even(x + ct + φ even (x ct] + [φ even(x + ct + φ even (x ct] + ψ even(y dy ψ even(y dy (b Use S(t to construct a solution of the inhomogeneous system. By Duhamel s principle the solution of the inhomogeneous system will be given by S(tΦ + t S(t sf (s ds. Therefore, the solution of the inhomogeneous system is given by [ [φ even(x + ct + φ even (x ct] + x+ct ψ ] ( even(y dy [φ t even(x + ct + φ even (x ct] + x+ct ψ even(y dy t x+c(t s + f x c(t s even(y, s dy ( x+c(t s f ds. t x c(t s even(y, s dy ]. 3
4 (c Use the solution of the inhomogeneous system to solve the inhomogeneous wave equation on the half-line with Neumann boundary conditions. Therefore, the solution of the inhomogeneous wave equation on the half-line with Neumann boundary conditions is given by the first component of the vector-valued function found in part (b, u(x, t = [φ even(x + ct + φ even (x ct] + In particular, for x > ct, + t x+c(t s u(x, t = [φ(x + ct + φ(x ct] + + x c(t s t x+c(t s x c(t s while for x < ct, defining t such that x c(t t =, u(x, t = [φ(x + ct + φ(ct x] + + t x+c(t s 4. Use separation of variables to solve f(y, s dy ds + + ψ even (y dy f even (y, sdyds ψ(y dy f(y, sdyds. ψ(y dy + t c(t s x t x+c(t s t x c(t s ct x ψ(y dy f(y, s dy ds f(y, s dy ds. u tt c u xx = < x < l, t > u(x, = x(x l < x < l u t (x, = < x < l u(, t = u x (l, t = letting u(x, t = X(xT (x we have, Therefore X(x is of the form, T c T = X X = λ 4
5 X(x = C cos(βx + D sin(βx boundary conditions imply X( = X (l = which yields that C = β = (n + π Therefore ( (n + π X(x = sin x Now T (t satisfies Therefore for each λ n, T takes the form, T = c λt ( ( (n + πc (n + πc T (t = A n cos t + B n sin t Using the boundary condition T ( = we have B n = So we have for u(x, t u(x, t = n= [ ( ( ] (n + πc (n + π A n cos t sin x using the boundary conditions for u(x, orthogonality arguments, we have for A n A n = l ( (n + π sin x x(x l dx 5. Consider the eigenvalue problem, X = λx < x < X ( + ax( = X( = When we are looking for positive eigenvalues, we can take λ = β. We are then looking for solutions to X + β X = 5
6 Solutions to the above are of the form X = C cos(βx + D sin(βx Now using the boundary conditions X ( + ax( =, Dβ + ac = C cos(β + D sin(β = Simplyfing yields, tan(β = β a Graphically, we can show that the above has infinite number of solutions by plotting both, tan(β β for all values of β. One would see that the graphs of both a the functions intersect infinite number of times, thereby indicating infinite number of solutions. Next, when we are looking for negative eigne values, we use λ = β follow similar arguments as above arrive at the final equation for β as, tanh(β = β a One can see by plotting the graphs of tanh(β β, that when a there is no a intersection when a > there is exactly intersection. 6. Use seperation of variables to solve u tt c u xx + γ u = < x < l, t > u(x, = φ(x u t (x, = ψ(x u(, t = u(l, t = where γ > letting u(x, t = X(xT (x we have, T c T = X X c = λ Therefore X(x solves X = (λ + γ c X is of the form, 6
7 X(x = C cos(βx + D sin(βx boundary conditions imply X( = X(l = which yields that C = with Now T (t satisfies β = nπ l λ n = n π l c Therefore for each λ n, T takes the form, T (t = A n cos ( n π l T = c λt ( c ct n π + B n sin l c ct So we have for u(x, t [ ( n π u(x, t = A n cos n= l ( ] c ct n π + B n sin ( nπ l c ct sin l x using the boundary conditions for u(x, orthogonality arguments, we have for A n B n A n = ( nπ sin l l x φ(xdx B n = lc ( n π l / ( nπ sin c l x ψ(xdx 7
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