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Solve the IBVP \begin{gather} u_{tt}=u_{xx},\quad 0<x<1,\quad t>0\\ u(x,0)=x, \quad u_{t}(x,0)=0, \quad 0<x<1\\ u(0,t)=t^{2},\quad u(1,t)=\cos{t},\quad t>0 \end{gather}

I'm not sure how to approach this problem. I've done other IBVP's where the initial conditions are homogeneous, and solved them using the separation of variables method. I approached this problem the same way, but I get stuck, so I'm not sure separation of variables is the appropriate method for this problem. Any guidance, tutorials, or sample problems would be greatly appreciated.

So using the hint below this is what I've come up with but I feel like I'm doing something wrong. For $\varphi(x,t)$ I got: $$\varphi(x,t)=t^{2}+x(\cos{t}-t^2)$$ With $\varphi(0,t)=t^2$ and $\varphi(1,t)=\cos{t}.$

My problem comes with working with $v(x,t)$. I have to solve for $v(x,t)$ $$\begin{align} v_{tt}&=v_{xx}\\ v(0,t)&=0\\ v(1,t)&=0\\ v(x,0)&=x-\varphi(x,0)=0\\ v_t(x,0)&=-\varphi_t(x,0)=0 \end{align}$$

Using the separation of variables method I end up with $$\sum\limits_{n=1}^\infty b_{n}\sin{n\pi x}\sin{n\pi t}$$ Looking at the initial conditions $v_t(x,0)=0$ I get: $$\begin{align} v_t(x,t)&=\pi\sum\limits_{n=1}^\infty n b_{n}\sin{n\pi x} \cos{n\pi t}\\ v_t(x,0)&=\pi\sum\limits_{n=1}^\infty n b_{n}\sin{n\pi x} \cos{n\pi (0)}\\ 0&=\pi\sum\limits_{n=1}^\infty n b_{n}\sin{n\pi x} \end{align}$$

Does this mean that $b_n=0$ and if so wouldn't that make $v(x,t)=0$?

user84529
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  • A brief comment on formatting: instead of using a whole bunch of \,s in a row, use larger spaces: \, \: \; \quad \qquad. Also, if you don't actually want to align on those equal signs, replace the align with a gather and drop the &s. – dfeuer Jun 30 '13 at 23:57
  • thanks for the tip. I'm new to latex. – user84529 Jul 01 '13 at 01:00

1 Answers1

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Extended hints: You have non-homogeneous boundary conditions at $u(0,t)$ and $u(1,t)$. Now suppose that $u(x,t)=v(x,t)+\varphi(x,t)$, where $\varphi(x,t)$ satisfies the boundary conditions and $v(x,t)=0$ on the boundaries. Substituting this into the PDE gives $$\begin{align} v_{tt}+\varphi_{tt}&=v_{xx}+\varphi_{xx}\\ v(x,0)+\varphi(x,0)&=x\\ v'(x,0)+\varphi'(x,0)&=0\\ v(0,t)+\varphi(0,t)=\varphi(0,t)&=t\\ v(1,t)+\varphi(1,t)=\varphi(1,t)&=\cos t \end{align}$$ The last two equations have been simplified as we have assumed that $v=0$ on the boundaries. This makes two problems which need to be solved.

  1. Solve for $\varphi(x,t)$ $$\begin{align} \varphi_{tt}&=\varphi_{xx}\\ \varphi(0,t)&=t\\ \varphi(1,t)&=\cos t \end{align}$$

  2. Solve for $v(x,t)$ $$\begin{align} v_{tt}&=v_{xx}\\ v(0,t)&=0\\ v(1,t)&=0\\ v(x,0)&=x-\varphi(x,0)\\ v'(x,0)&=-\varphi'(x,0) \end{align}$$

Daryl
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