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the problem states to solve the forced string problem with

$$\frac{\partial^2u}{\partial t^2}=\frac{\partial^2u}{\partial x^2} - \cos( \pi t)\sin( \pi x)$$

the boundary conditions for the string are $u(0,t)=u(1,t)=0$ and the initial conditions are $u(x,0)=0, \frac{du}{dt}(x,0)=0$

I have no clue how to solve this, any help or even pointers in the right directions to sources, etc will be extremely useful.

I thank you in advance

Dmoreno
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Filip
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  • How far did you get with separation of variables? – muaddib Jun 19 '15 at 23:23
  • I usually use U=X(x)T(t), then rearrange for a proportion X''/X=T''/T=-$\lambda$ – Filip Jun 19 '15 at 23:27
  • Can you see that $u_h = A \sin{n \pi x}$ (where $n$ is a nonzero integer) is a nontrivial solution of the homogenous problem when no forcing is considered? Are you familiar with Fredholm's alternative? – Dmoreno Jun 20 '15 at 03:08

1 Answers1

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Consider the change of variables $\xi=x+t$, $\eta=x-t$. Then

\begin{align} \frac{\partial^2 }{\partial t^2}=\frac{\partial^2}{\partial\xi^2}-2\frac{\partial^2}{\partial\xi\partial\eta}+\frac{\partial^2}{\partial \eta^2} \end{align}

and

\begin{align} \frac{\partial^2 }{\partial x^2}=\frac{\partial^2}{\partial\xi^2}+2\frac{\partial^2}{\partial\xi\partial\eta}+\frac{\partial^2}{\partial \eta^2}. \end{align}

Then the equation now becomes

\begin{align} 4\frac{\partial^2 u}{\partial\xi\partial\eta}=\cos\left[\frac{\pi}{2}(\xi-\eta)\right]\sin\left[\frac{\pi}{2}(\xi+\eta)\right] \end{align}

since $t=\frac{1}{2}(\xi-\eta)$ and $x=\frac{1}{2}(\xi+\eta)$. The general solution to this PDE is given by

\begin{align} u(\xi,\eta) &= \frac{1}{4}\left\{c_2(\xi)+\xi c_1(\eta)+\int_0^\xi\int_0^\eta\cos\left[\frac{\pi}{2}(s-r)\right]\sin\left[\frac{\pi}{2}(s+r)\right]\,dr\,ds\right\} \\ &= \frac{1}{4}\left\{c_2(\xi)+\xi c_1(\eta)+\frac{1}{2\pi}[\eta+\xi-\xi\cos(\pi\eta)-\eta\cos(\pi\xi)] \right\}\\ \end{align}

(just integrate with respect to both variables). Substituting our original variables, we have

\begin{align} u(x,t)=\frac{1}{4}\left\{c_2(x+t)+(x+t)c_1(x-t)+\frac{1}{2\pi}[2x-(x+t)\cos(\pi x-\pi t)-(x-t)\cos(\pi x+\pi t)]\right\}. \end{align}

Use the initial conditions to determine expressions for $c_1$ and $c_2$.

user31415926535
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