# Verifying the 2-dimensional fundamental solution of the wave equation

I'm trying to verify that $$u(t,x)=H(t-|x|)(t^2-|x|^2)^{-1/2}$$ is the fundamental solution of the 2-dimensional wave equation; that is, $\Box u = u_{tt}-\Delta u = \delta_{0}$. I know there are tricks that make this calculation easier (involving realizing distributions as limits) but I'd really like to figure out how to do this with straightforward integration techniques.

I need to show that $$\langle u(t,x),\Box\varphi(t,x)\rangle = \langle\delta_0,\varphi(t,x)\rangle = \varphi(0,0,0),$$ where $\varphi\in C_{0}^{\infty}(\mathbb{R}^{1+2})$. So far I have:

$$\langle u(t,x),\Box\varphi(t,x)\rangle =\int_{-\infty}^{\infty}\int_{\mathbb{R}^2}H(t-|x|)(t^2-|x|^2)^{-1/2}\Box\varphi(t,x)dAdt$$

$$=\int_{-\infty}^{\infty}\int_{0}^{2\pi}\int_{0}^{\infty}H(t-r)(t^2-r^2)^{-1/2}r\Box_{\text{pol}}\varphi(t,r,\theta)drd\theta dt$$ $$=\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{t}\partial_{r}\left((t^2-r^2)^{\frac{1}{2}}\right)\left( \varphi_{tt} - \Delta_{\text{pol}}\varphi\right)drd\theta dt$$ $$=\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{t}\partial_{r}\left((t^2-r^2)^{\frac{1}{2}}\right)\left( \varphi_{tt}-\frac{1}{r}\partial_{r}\left(r\varphi_r\right)-\frac{1}{r^2}\varphi_{\theta\theta} \right)drd\theta dt.$$

Can anyone point me in the right direction? I'm sure integration by parts will be important here, but I'm not sure how to proceed. Any help?

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Fixed a small mistake...

I think the usual way is to express the fourier transfom of $u(x, t)$ as the (weak) limit of some "normal" function. This is how the $i \epsilon$ appears in the physics books. So what you should show is that

$$u(x, t) = \mathcal{S}'-\lim_{\epsilon \rightarrow 0} \int \frac{dk^2 d\omega}{(2 \pi)^3} \frac{e^{-i(wt - (k, x))}}{|k|^2 - \omega^2 \pm i \epsilon}\, .$$

Note that the sign of the $i \epsilon$ term is important. Different ways of moving the poles away from the real axis give different fundamental solutions (moving forward and backward in time). Follands book about fourier analysis shows how actual calculations are done.

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Thanks for that approach. I planned on reading more about distribution theory soon. But does anyone have any idea if the "straightforward" integration method is even doable? –  Patch Mar 22 '12 at 1:57
I think the "straightforward" approach is "in principle" do-able, meaning that you can probably find some version of it written somewhere... that may or may not be an actual proof, etc. Further, I find such a "direct" approach not very convincing, because it's hard to see that one is verifying what one says one is. Gearing up to understand tempered distributions is the saner approach, by far. Well worth the trouble. (Hilbert-Courant may do "classical" computations of this sort...) –  paul garrett Jul 26 '12 at 16:52

I think the trick is to integrate by parts.

We rewrite $H(t)(t^{2}-|x|^{2})^{-1/2}$ to be $$\lim_{\epsilon\rightarrow 0+}Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}$$

Then we need to verify that

$$F(0,0)=\lim_{\epsilon\rightarrow 0+}\int_{\mathbb{R}^{2}}\int^{\infty}_{0}Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}(\partial_{t}^{2}-\Delta)F(t,x)dtdx$$

If we mark $Im(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=u$, $(\partial_{t}^{2}-\Delta)F(t,x)=dv$. Then the $uv$ boundary part vanish because when $t=0$, $u=0$, and other boundary parts disappears. So we need to evaluate

$$\lim_{\epsilon\rightarrow 0+}\int_{\mathbb{R}^{2}}\int^{\infty}_{0}Im(\partial_{t}+\partial_{x})(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}(\partial_{t}-\partial_{x})F(t,x)dtdx$$

We integrate by parts again. If we let $$Im(\partial_{t}+\partial_{x})(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=u', (\partial_{t}-\partial_{x})F(t,x)=dv$$ Then the $du'v$ term involves $$(\partial_{t}^{2}-\Delta)(|x|^{2}-(t+i\epsilon)^{2})^{-1/2}=0$$ So the only term really matters is the boundary term $u'v$. The $\partial_{x}$ part vanishes on the boundary, and what we are really integrating is

$$\int_{\mathbb{R}^{2}}\frac{\partial}{\partial t}(Im (|x|^{2}-(t+i\epsilon)^{2})^{-1/2})|_{t=0}F(0,x)dx$$ But this is the Possion integral on the upper half plane:

$$\int_{\mathbb{R}^{2}}\frac{\epsilon}{(|x|^{2}+\epsilon^{2})^{3/2}}F(0,x)dx$$ which tends to $F(0,0)$ times a constant when $\epsilon\rightarrow 0$.

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