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(a) Solve the initial value problem ${{u}_{xx}}+2{{u}_{xt}}-3{{u}_{tt}}=0$ with $u\left( x,0 \right)={{e}^{x}}$ and ${{u}_{t}}\left( x,0 \right)={{x}^{2}}$ .
(b) Assume that $f\left( x \right)=g\left( x \right)=0$ for all $x\in \left[ 0,1 \right]$. Let $u$ satisfy the wave equation given in (a) with $u\left( x,0 \right)=\text{ }f\text{ and }{{u}_{t}}\left( x,0 \right)=g$ . Find the largest posible region in $\left( t,x \right)$ on which $u\left( x,t \right)$ must vanish identically.

My attempt:

Part (a) is easy.

$u\left( x,t \right)=f\left( x-3t \right)+g\left( x+t \right)$ .

After applying the initial condition, we get

$f\left( x \right)=\frac{-1}{12}{{x}^{3}}+\frac{1}{4}{{e}^{x}}-\frac{1}{4}C$ , and $g\left( x \right)=\frac{1}{12}{{x}^{3}}+\frac{3}{4}{{e}^{x}}+\frac{1}{4}C$ .

Then

$u\left( x,t \right)=\frac{-1}{12}{{\left( x-3t \right)}^{3}}+\frac{1}{4}{{e}^{\left( x-3t \right)}}+\frac{1}{12}{{\left( x+t \right)}^{3}}+\frac{3}{4}{{e}^{\left( x+t \right)}}$

I do not know what to do with part (b). Please help.

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  • $\begingroup$ That seems very strangely worded to me. As the 0 function seems to meet the initial conditions, so the largest possible region would be the whole of time and space.....Unless I'm (likely) reading it wrong $\endgroup$ – Alan Nov 20 '15 at 7:54

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