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Consider the wave equation $ u_{tt}=a^2 u_{xx} $ and a separated solution

$u(x,t)=T(t) \varphi (x) $, with boundary condition $u(0,t)=c, \ u(1,t)=d$.

Then I want to show that all separated solutions are periodic in both $x$ and $t$. From lectures I know its true for $c=d=0$ so I can assume one is non zero. Thus assume $c \neq 0$.

By plugging in the separated solution to our pde we get

$- \frac{ \varphi''(x)}{ \varphi (x)}=-\frac{T''(t)}{a^2 T(t)}=\lambda $

This gives two associated odes

$\varphi''(x)+\lambda \varphi(x)=0 \quad T''(t)+\lambda a^2 T(t)=0 $

one can easily show that $\lambda >0$ thus set $\lambda = \beta^2$. One can also easily show that our two solutions are

$\varphi(x)=A \cos(\beta x)+B \sin(\beta x), \quad T(t)=C \cos(\beta a t)+D \sin(\beta at),$

My problem is that my boundary conditions don't give me much to work with, they give the following:

$u(0,t)=A(t) \varphi (0)=A \cdot T(t)=c \ $ and $ \ u(1,t)=\varphi (1) T(t)=d$

I cant really use these to much, and I get the hint that I should see that $T(t)$ is a periodic function or $T(t) \rightarrow \pm \infty$ for $t \rightarrow \infty$. I believe my problems lies in understanding the boundaries conditions, or maybe I have made some mistake?

Soren123
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1 Answers1

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For the pde \begin{align} u_{tt} = a^{2} u_{xx} \hspace{10mm} u(0,t) = c, \hspace{5mm} u(1,t) = d \end{align} consider the separation function $u(x,t) = T(t) \phi(x)$ can be used to obtain the differential equations \begin{align} \phi(x) &= A_{0} \cos(\beta x) + B_{0} \sin(\beta x) \\ T(t) &= C_{0} \cos(\beta a t) + D_{0} \sin(\beta at). \end{align} From the boundary conditions of $u(x,t)$ it is given that $\phi(0) = c$ and $\phi(1) = d$. This leads to \begin{align} \phi(0) = c &= A_{0} \\ \phi(1) = d &= A_{0} \cos(\beta) + B_{0} \sin(\beta). \end{align} This process "seems" to lead to much more confusion. So, try a different method.

Consider changing the boundary conditions by a function that has a value of zero for a second order derivative. In this view consider the form \begin{align} u(x,t) = c + (d-c)x + \phi(x) \psi(t). \end{align} Notice that the values of the boundary conditions are built in to the solution. This can be checked as a valid solution. What is changed is that now it can be stated that the conditions become \begin{align} \phi(0) = 0 \hspace{10mm} \phi(1) = 0. \end{align} Now the $x$ differential equation can be seen to follow \begin{align} \phi(0) = 0 &= A_{0} \\ \phi(1) = 0 &= B_{0} \sin(\beta). \end{align} Either $B_{0} = 0$, which is possible, or $\sin(\beta) = 0$ yielding $\beta = n \pi$. With this the solution then becomes \begin{align} u(x,t) = c + (d-c) x + \sum_{n=1}^{\infty} \sin(n \pi x) \left( A_{n} \cos(n \pi t) + B_{n} \sin(n \pi t) \right). \end{align}

In order to further reduce the general solution some condition for $t$ is needed.

Leucippus
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  • Ty for the answer. I can see that the linear transformation gives a better general solution. But does it help me answer the question that $u(x,t)=\varphi(x) T(t)$ shoul be periodic with respect to both variables. Because for the general solution that you have come up with it is not periodic in the variable of $x$ – Soren123 Mar 22 '15 at 21:19
  • @Soren123 $\sin(n \pi (x+1)) = (-1)^{n} \sin(n \pi x)$ and $\sin(n \pi (x+2)) = \sin(n \pi x)$ for $n \in \mathbb{N}_{0}$ which makes it periodic. The same applies for $\cos(n \pi t), \sin(n \pi t)$. This leads to the ability to say that the complete solution is periodic in both variables. One can also define the function $v(x,t) = u(x,t) - c - (d-c) x$ and reformulate the statement – Leucippus Mar 22 '15 at 22:16