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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.

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Please make titles informative as to the content of the post. "Interesting problem" is subjective and says nothing about the mathematics contained in the post. Also, "Need help" is all about you, not about the mathematics of the post. Please consider finding a better title. – Arturo Magidin Jul 3 '12 at 16:57
@ArturoMagidin : thank you for suggestion . – Theorem Jul 3 '12 at 16:59
up vote 1 down vote accepted

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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Somehow the sum doesn't seem to get to zero. :P i am missing something here . – Theorem Jul 3 '12 at 18:11
One more question , what may be the significance of the problem ? – Theorem Jul 3 '12 at 18:14
Right, a factor $2$ was missing. This makes a link between Laplace and heat equation. – Davide Giraudo Jul 3 '12 at 18:19

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.

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