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Let $u(x,t)$ be the solution of the following initial value problem:

$u_{tt} - \Delta u = 1\;\forall x \in \mathbb R^3 ,\;t \gt 0\\u(x,0)=0\\u_t(x,0)=0$

Is it true or false that $u(x,t) \neq 0\;\forall x \in \mathbb R^3 and\;\forall t \gt 0$ ?

Well, for a general non-homogeneous wave equation I.V.P. on $\mathbb R^3$,such as: $u_{tt} - \Delta u = f(x,t)\;\forall x \in \mathbb R^3 ,\;t \gt 0\\u(x,0)=φ(x)\\u_t(x,0)=ψ(x)\\$

we know that the solution $u$ is given by this :

$\frac{1}{4\pi t^2} \int_{\partial Β(x,t)} φ(y) + \nabla φ(y) (y-x) +t ψ(y) \;\;dS(y)+\frac{1}{4\pi} \int_{0}^t \int_{\partial Β(x,t-s)} \frac{f(y,s)}{t-s}\; dS(y) \;ds$

So, I substituted $φ=0\;,ψ=0\;and\; f=1\;$ in the previous formula and I concluded to this:

$u(x,t)=\frac {-\vert \partial Β(x,t-s) \vert }{4\pi} [\ln \vert t-s \vert ]_{0}^t \;$ where $\vert \partial Β(x,t-s) \vert =\int_{\partial Β(x,t-s)} dS(y) $ .

I have trouble handling the above...If $t=s$ the $\ln\;$ is not defined..Could somebody help me see what I'm missing.. ? Hints or other solutions than this are also welcome.

I would appreciate any help! Thanks in advance..

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2 Answers 2

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Suppose $u$ is such a solution, and let $v(x,t)=u(x,t)-t$. Then $v$ is a solution of \begin{align} v_{t}-\Delta v &= u_t-\Delta u-1=0 \\ v(x,0)&=u(x,0)-0=0 \\ v_{t}(x,0)& = u_{t}(x,0)-1 = -1. \end{align} It seems to work out better with $f=0$, $\varphi=0$ and $\psi=-1$.

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Ι 've made a mistake in the above calculation. After substitution we get $u(x,t)=\frac{1}{4\pi} \int_{0}^t \int_{\partial Β(x,t-s)} \frac{1}{t-s} dS(y)ds\;=\;\frac{1}{4\pi} \int_{0}^t \frac{1}{t-s} \int_{\partial Β(x,t-s)} \;\;dS(y)ds\;=\;\frac{1}{4\pi} \int_{0}^t \frac{4\pi (t-s)^2}{t-s} \;=\;\int_{0}^t t-s\;\;ds\;=\;[st-\frac{s^2}{2}]_{0}^t\;=\;\frac{t^2}{2}$

Now it is obvious that $u(x,t)\neq 0\;\forall x \in \mathbb R^3\;and\;\forall t \gt 0$

The mistake was that I hadn't seen $\vert \partial Β(x,t-s) \vert\;$ depends on the outer integral.

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