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Unfortunately I can't find a solution $u:[0,\infty)\times[0,\infty)\rightarrow\mathbb R$ of this PDE: \begin{array}{rl} u_{tt}&=&u_{xx} &\text{in }(0,\infty)\times(0,\infty)\\ u(0,t)&=&0 &\forall t>0\\ u(x,0)&=&\varphi(x) &\forall x\ge0\\ u_t(x,0)&=&\psi(x) &\forall x\ge0\\ 0=\varphi(0)&=&\psi(0) \end{array}

Can someone help me? Thank you very much!

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2 Answers 2

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Note that when without the condition $u(0,t)=0$ this is in fact a just-determining problem and the solution can be expressed by using D’Alembert’s formula $u(x,t)=\dfrac{\varphi(x+t)+\varphi(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}\psi(s)~ds$ , where $\varphi(s)$ and $\psi(s)$ are any functions that satisfly $\varphi(0)=\psi(0)=0$ .

Since $\dfrac{\varphi(t)+\varphi(-t)}{2}+\dfrac{1}{2}\int_{-t}^t\psi(s)~ds$ does not necessarily equal to $0$ , so when with the condition $u(0,t)=0$ this becomes an overdetermining problem. Therefore the solution should be classified into two cases:

Case $1$: $\dfrac{\varphi(t)+\varphi(-t)}{2}+\dfrac{1}{2}\int_{-t}^t\psi(s)~ds=0$

$u(x,t)=\dfrac{\varphi(x+t)+\varphi(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}\psi(s)~ds$ , where $\varphi(s)$ and $\psi(s)$ are any functions that satisfly $\varphi(0)=\psi(0)=0$

Case $2$: $\dfrac{\varphi(t)+\varphi(-t)}{2}+\dfrac{1}{2}\int_{-t}^t\psi(s)~ds\neq0$

Piecewise solution should be unavoiable. The trick in http://eqworld.ipmnet.ru/en/solutions/lpde/lpde201.pdf should be used:

$u(x,t)=\begin{cases}\dfrac{\varphi(x+t)+\varphi(x-t)}{2}+\dfrac{1}{2}\int_{x-t}^{x+t}\psi(s)~ds&\text{when}~x>t\\\dfrac{\varphi(x+t)-\varphi(t-x)}{2}+\dfrac{1}{2}\int_{t-x}^{x+t}\psi(s)~ds&\text{when}~x<t\end{cases}$ , where $\varphi(s)$ and $\psi(s)$ are any functions that satisfly $\varphi(0)=\psi(0)=0$

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If you didn't have the condition $u(0,t)=0$ and the domain for the $x$ variable was $\mathbb{R}$ instead of $(0,\infty)$, then you just would use D'Alembert's formula. The trick in your situation is to think of $u$, $\varphi$ and $\psi$ to be extended from $(0,\infty)$ to $\mathbb{R}$ by odd reflection, that is, $u(-x,t)=-u(x,t)$, and likewise for $\varphi$ and $\psi$. The extended $u$ satisfies the wave equation with extended initial conditions $\varphi$ and $\psi$ (check this!), and thus you can apply D'Alembert's formula in this situation.

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