If we have $k(x,t)= \frac {1}{(4t)^{\frac{n}{2}}} \exp\left(\frac{-|x|^2}{4t}\right)$ is the fundamental solution of heat equation. If we consider $n \ge 3 $, I would like to show that $\int_0^\infty k(x,t) dt$ is the fundamental solution of lLaplace equation. I would like some hints. I thought of integrating but don't know how to approach. Thank you ie , i need to arrive to a form like $\frac{1}{B} \frac{1}{|x|^{n-1}}$ $B $ is a constant depending on the measure of the space.
Some kind of relation between classical heat equation and Laplace .
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0@ArturoMagidin : thank you for suggestion . – 2012-07-03
2 Answers
We use the substitution $s=\frac t{|x|^2}$ (then $dt=|x|^2ds$) to get \begin{align} \int_0^{+\infty}k(x,t)dt&=\int_0^{+\infty}\frac 1{(4t)^{n/2}}\exp\left(-\frac{|x|^2}{4t}\right)dt\\ &=\frac 1{4^{n/2}}\int_0^{+\infty}\frac 1{(s|x|^2)^{n/2}}\exp\left(-\frac 1{4s}\right)|x|^2ds\\ &=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}\frac 1{s^{n/2}}\exp\left(-\frac 1{4s}\right)ds\\ &=\frac 1{4^{n/2}}|x|^{2-n}\int_0^{+\infty}y^{n/2-2}\exp(-y/4)dy\\ &=c_n|x|^{2-n}. \end{align} We have to show that $f\colon x\mapsto |x|^{2-n}=\left(\sum_{k=1}^nx_k^2\right)^{1-n/2}$ is harmonic. Let $j\in\{1,\dots,n\}$. We have $\partial_jf(x)=\left(\sum_{k=1}^nx_k^2\right)^{-n/2}2x_j\left(1-\frac n2\right)$ and $\partial_{jj}f(x)=2\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2}-\left(1-\frac n2\right)nx_j\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}(2x_j).$ Summing that, we get \begin{align} \Delta f(x)&=\sum_{j=1}^n\partial_{jj}f(x)\\ &=\left(1-\frac n2\right)\left(\sum_{k=1}^nx_k^2\right)^{-n/2-1}\left(2n|x|^2-n\cdot 2\cdot |x|^2\right)\\ &=0. \end{align}
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0Right, a factor $2$ was missing. This makes a link between Laplace and heat equation. – 2012-07-03
Let $e(x)=\int_0^\infty k(x,t)\,\mathrm{d}t$, and let us compute the action of the distribution $\Delta e$ on the test function $\phi$ (which is a compactly supported smooth function in $\mathbb{R}^n$) as $ \begin{split} \langle\Delta e,\phi\rangle &= \int_{\mathbb{R}^n} e(x) \Delta\phi(x)\,\mathrm{d}x = \int_{\mathbb{R}^n} \int_0^\infty k(x,t) \Delta\phi(x)\,\mathrm{d}t\,\mathrm{d}x = \int_0^\infty \int_{\mathbb{R}^n} k(x,t) \Delta\phi(x)\,\mathrm{d}x\,\mathrm{d}t \\ &= \int_0^\infty \int_{\mathbb{R}^n} \Delta k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t = \int_0^\infty \int_{\mathbb{R}^n} \partial_t k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t \\ &= \int_0^\infty \int_{\mathbb{R}^n} \partial_t k(x,t) \phi(x)\,\mathrm{d}x\,\mathrm{d}t = -\lim_{t\to0}\int_{\mathbb{R}^n} k(x,t) \phi(x)\,\mathrm{d}x = - \phi(0), \end{split} $ where we have used the fact that $\Delta k =\partial_tk$ and that $\int_{\mathbb{R}^n} k(x,t) \phi(x)\,\mathrm{d}x\to\phi(0)$ as $t\to0$. Thus we have $\Delta e = - \delta$. Some would say that $e$ is not exactly a fundamental solution, but that $-e$ is one.