Analytical solution for the diffusion equation

Question

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?

Answer 1

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.

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ 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.