I have the following boundary and initial value problem on $0\leq x\leq 1$: $u_{xx}=u_{tt}$,

$ f(x)=u(x,0) = \begin{cases} 0& \textrm{ if $0\leq x\leq 1/4$} \\ x-1/4& \textrm{ if $1/4\leq x\leq 1/2$} \\ 3/4-x &\textrm{if $1/2\leq x\leq 3/4$}\\ 0 &\textrm{if $3/4\leq x\leq 1$} \end{cases} $

Furthermore $u_{t}(x,0)=0$ and $u_x(0,t)=u_x(1,t)=0$. If I look at the D'Alembertian we have a solution on the whole real line without the neumann boundary condition given by $u(t,x)=\frac{f(x-t)+f(x+t)}{2}$, meaning that the hat-function splits in to two halves where one goes to $-\infty$ and the other one goes to $+\infty$ as $t\rightarrow \infty$. In the book of Peter Olver (Introduction to partial differential equations) on page 148 he uses this to solve the problem with the neumann boundary condition. He says that we should extend the initial condition periodically and symmetrically on the whole real line, i.e. $f(x+1)=f(x)$ and $f(-x)=f(x)$.

Here is then my problem: If this half hat funtion travels to the left to hit $x=0$, the we should see another hat-function appearing from the left going to the right, i.e. the wave reflects as it hits the line $x=0$. But how is this solving the neumann boundary problem since then the derivative with respect to $x$ at $x=0$ will not be 0, but both 1 and -1.

I have been struggling with this for some time, but I still dont know how to solve this. Thanks.


1 Answer 1



Let $u(x,t)=\sum\limits_{n=0}^\infty C(n,t)\cos4n\pi x$ so that it automatically satisfies $u_x(0,t)=u_x(1,t)=0$ and the size of the intervals are $\dfrac{1}{4}$ and start from $0$ ,

Then $-\sum\limits_{n=0}^\infty16n^2\pi^2C(n,t)\cos4n\pi x=\sum\limits_{n=0}^\infty C_{tt}(n,t)\cos4n\pi x$

$\sum\limits_{n=0}^\infty(C_{tt}(n,t)+16n^2\pi^2C(n,t))\cos4n\pi x=0$

$\therefore C_{tt}(n,t)+16n^2\pi^2C(n,t)=0$

$C(n,t)=A(n)\sin4n\pi t+B(n)\cos4n\pi t$

$\therefore u(x,t)=\sum\limits_{n=0}^\infty A(n)\cos4n\pi x\sin4n\pi t+\sum\limits_{n=0}^\infty B(n)\cos4n\pi x\cos4n\pi t$

$u_t(x,t)=\sum\limits_{n=0}^\infty4n\pi A(n)\cos4n\pi x\cos4n\pi t-\sum\limits_{n=0}^\infty4n\pi B(n)\cos4n\pi x\sin4n\pi t$

$u_t(x,0)=0$ :

$\sum\limits_{n=0}^\infty4n\pi A(n)\cos4n\pi x=0$


$\therefore u(x,t)=\sum\limits_{n=0}^\infty B(n)\cos4n\pi x\cos4n\pi t$


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