# 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?

• Please, consider edit your question, in order to add additional information, containing your thoughts in the problem, for example, you can add what attempts you have tried to solve the problem, what is your background in this area, etc... – Tomás Nov 4 '13 at 12:05
• I forgot to explain that i'm searching for a simple way, i.e. without fundamental solutions. There is no other information given for this problem. I have no idea how to start. Maybe integrating? – PeterJ Nov 4 '13 at 12:14
• I just need an idea, not a full computation. – PeterJ Nov 4 '13 at 12:24
• Symmetry considerations should convince you that it stays a function of $h \cdot x$. This reduces the problem to the heat evolution of $e^{x}$ on $\mathbb R$, which has an easily found solution. – Anthony Carapetis Nov 4 '13 at 13:37
• Thank you. You mean symmetry of the initial values? – PeterJ Nov 4 '13 at 18:02

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