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For the domain $(0,1)$ find a solution for $$u_{tt}=9u_{xx}, u_x(t,0)=0,u(t,1)=0,u(0,x)=2\cos(3\pi x/2)$$

This problem is really throwing me off because I understand how I would go about solving the wave equation for homogeneous Neumann or Dirichlet boundaries, but how do I handle both $u_x(t,0)$ and $u(t,1)$ boundary conditions in the same problem? I believe it is Robin boundary conditions. Additionally $u(0,x)=f(x)$ is throwing me off because I always worked with integer coefficients of $\pi$ inside the trig functions. Is there some identity I can use so that there's only integer coefficients?

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The condition given is not a Robin type of condition. Robin conditions involve the value of the function and the first derivative at the same endpoint. You have a Dirichlet and a Neumann condition, as well as an initial condition.

This problem can be solved by separation of variables. $$ \frac{T''(t)}{9T(t)} = \lambda, \;\; \lambda = \frac{X''(x)}{X(x)} $$ with spatial conditions $$ X'(0)=0,\;\; X(1)=0. $$ The eigenfunctions of this problem have the form $$ X_n(x) = \cos(\lambda_n x),\;\;\;\lambda_n = \frac{\pi}{2}+n\pi,\;\; n=0,1,2,\cdots. $$ The corresponding solutions $T_n$ are $$ T_n(t) = A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t). $$ So the general solution is $$ u(t,x)=\sum_{n=1}^{\infty}(A_n\cos(3\lambda_n t)+B_n\sin(3\lambda_n t))\cos(\lambda_n x). $$ There are not enough conditions to uniquely solve the problem because the one $t=0$ condition does not involve the constants $B_n$. However, one solution is obtained by setting all $B_n=0$, which is equivalent to imposing the additional condition $u_t(0,x)=0$. The remaining condition is $$ 2\cos(3\pi x/2) = \sum_{n=1}^{\infty} A_n\cos(\lambda_n x) = \sum_{n=1}^{\infty}A_n\cos((n+1/2)\pi x). $$ The eigenfunctions $\{ \cos((n+1/2)\pi x) \}_{n=1}^{\infty}$ are mutually orthogonal on $[0,1]$, which allows you to isolate the $A_n$ using the above condition.

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Write the equation as $$u_{tt}-9u_{xx}=(\partial_t-3\partial_x)(\partial_t+3\partial_x)u=0$$ so u has to vanish if acted on by either of those operators, and $$u=f(x+3t)+g(x-3t)$$ By writing the boundary conditions in this form, you should be able to find $f$ and $g$

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  • $\begingroup$ Are we allowed to factor derivatives that way? $\endgroup$ – Peetrius Oct 23 '17 at 3:14

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