Using the Dirac Delta function in PDE's

Question

Solve the diffusion equation on the positive half-line

$\frac{∂u}{∂t}−a^2\frac{∂^2u}{∂x^2}=0,0≤x<∞$ subject to the initial and the boundary condition

$u(x,0)=Qδ(x−x_0),u_x(0,t)=0.$

Where $Q≠0$ and $x_0>0$ are given constants, and $δ(⋅)$ is the Dirac delta-function.

I think that The equation is invariant with respect to $x → −x$ and $v → −v$, and the initial condition is odd in $x$, hence the solution is even in $x$. Would I be correct in saying this? Also I am stuck at this point and unsure where to go from here.

Answer 1

For simplicity I will assume $a=1$. Extend the initial value to make it even as $u_0(x)=Q(\delta(x-x_0)+\delta(x+x_0))$. The solution is then $$ u(x,t)=\frac{1}{\sqrt{4\,\pi\,t}}\int_{-\infty}^\infty e^{-\tfrac{(x-y)^2}{4t}}u_0(y)\,dy=\frac{Q}{\sqrt{4\,\pi\,t}}\Bigl(e^{-\tfrac{(x-x_0)^2}{4t}}+e^{-\tfrac{(x+x_0)^2}{4t}}\Bigr). $$

Answer 2

EDIT: As pointed out by Julián Aguirre my solution does not satisfy the Neumann boundary. I will copy his way and extend the initial value via $u_0(x)=Q(\delta(x-x_0)+\delta(x+x_0))$ to save this. I will adjust the computations so you can see how to derive the solution to your PDE with the ansatz of Fourier transformations.

\begin{align} \partial_t \hat u(k,t) &=- a^2 k^2 \hat u(k,t) \\ \Longrightarrow \hat u(k,t)&=\hat u(k,0)e^{-a^2k^2t}=Q \left(\widehat{\delta(\bullet-x_0)}(k)+\widehat{\delta(\bullet+x_0)}(k) \right)e^{-a^2k^2t}\\ &=\frac{Q}{\sqrt{2\pi}}\left(e^{ikx_0}+e^{-ikx_0}\right)e^{-a^2k^2t} \\ \Longrightarrow u(x,t)&=\frac{Q}{\sqrt{2\pi}} \bigg( \big(\mathscr{F}^{-1}(e^{ix_0\bullet})*\mathscr{F}^{-1}(e^{-a^2t\ \bullet^2})\big)(x) +\big(\mathscr{F}^{-1}(e^{-ix_0\bullet})*\mathscr{F}^{-1}(e^{-a^2t\ \bullet^2})\big)(x) \bigg)\end{align}

Above we used $\sqrt{2\pi}\widehat{\delta(\bullet\mp x_0)}(k)=e^{\pm ikx_0}$ and therefore $\mathscr{F}^{-1}(e^{\pm ix_0\bullet})(x)=\sqrt{2\pi}\delta(x\mp x_0)$. Further the Fourier transform of the Gaussian is well-known i.e. \begin{align}\widehat{e^{-bx^2/2}}(k)&=\sqrt{\frac{2\pi}{b}}e^{-\frac{k^2}{2b}} \\ \Longrightarrow\mathscr{F}^{-1}\left(e^{-\frac{k^2}{2b}}\right)(x)&=\sqrt{\frac{b}{2\pi}}e^{-bx^2/2}\end{align}

Now, applying this to our case (i.e. $b=\frac{1}{2a^2t}$) and noticing that $\delta(\bullet\pm x_0)*f(x)=f(x\pm x_0)$ we finally get

\begin{align}u(x,t)&=\frac{Q}{\sqrt{2\pi}}\left(\sqrt{2\pi}\big(\delta(\bullet-x_0)\big)*\sqrt{\frac{1}{4\pi a^2t}}\exp\left(-\frac{\bullet^2}{4a^2t}\right)\right)(x)\\&~~~+\frac{Q}{\sqrt{2\pi}}\left(\sqrt{2\pi}\big(\delta(\bullet+x_0)\big)*\sqrt{\frac{1}{4\pi a^2t}}\exp\left(-\frac{\bullet^2}{4a^2t}\right)\right)(x) \\ &=\frac{Q}{\sqrt{4\pi ta^2} } \bigg(\exp\left(-\frac{(x-x_0)^2}{4a^2t}\right)+\exp\left(-\frac{(x+x_0)^2}{4a^2t}\right) \bigg)\end{align}