Analytical solution for the diffusion equation Is there a way to compute the solution $u: \mathbb{R}^n \times [0,\infty) \rightarrow \mathbb{R}$  of the heat/diffusion equation
$ u_t - \Delta u = 0, u(x,0) = \exp(h \cdot x) $ with $h \in \mathbb{R}^n$?
I don't want to use the fundamental solution or advanced methods here, if there is another simple possibility. 
Integrating this with respect to the variable $t$ leads to an integral over $\Delta u$ and there i'm stuck with my ideas.
I know the method of Fourier-series, fundamental solutions and Separation of Variables. But there has to be an easy way to get the solution. Does someone know such a simple method?
 A: Let $\hat h=h/|h|$ be the normalized $h$. Every point $x\in \mathbb R^n$ can be written as $x= r \hat h+ y$ where $x\cdot y=0$. Since $u(x,0)=u(x+y,0)$ for any such $y$, the uniqueness of solution implies that $u(x,t)=u(x+y,t)$ holds for all $t>0$. Therefore, $u(x,t)=u(r\hat h,t)$. This is what Anthony Carapetis wrote in a comment already. 
Now that $u$ depends only on one space coordinate (projection onto the direction of $h$), the solution is found from the one-dimensional problem $U_t=U_{rr}$, $U(x,0)=e^{|h|r}$. The convolution with Gaussian kernel (fundamental solution) can be evaluated explicitly by completing the square in the exponent. 
A: $\newcommand{\+}{^{\dagger}}%
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Use Laplace Transform $\pars{~\mbox{this is a}\ {\large\tt\mbox{hint}}\ \mbox{for}\ n = 1\ \mbox{but it can
be straightforward generalized}~}$:
$$
0
=
\int_{0}^{\infty}
\bracks{{\rm u}_{t}\pars{x,t} - {\rm u}_{xx}\pars{x,t}}\expo{-st}\,\dd t
=
-{\rm u}\pars{x,0} + s\,\tilde{\rm u}\pars{x,s} - \tilde{\rm u}_{xx}\pars{x,s} 
$$
where $\tilde{\rm u}\pars{x,s}$ is the Laplace transform of ${\rm u}\pars{x,t}$. Now, you have a 'Helmholtz like' equation with a source:
$$
\tilde{\rm u}_{xx}\pars{x,s} - s\,\tilde{\rm u}\pars{x,s} = -{\rm u}\pars{x,0}
$$
which is easier to solve than the original one. We hope you can take fom here.
