Diffusion equation on the positive half-line with Dirac delta-function boundary condition.

Question

The diffusion equation $u_{t} = k^{2}u_{xx}$ has initial boundary conditions $u(x,0) = A \delta(x - x_{0})$, $u(0,t) = 0$ where $A\neq0$ and $x_{0} > 0$ are given constants, and $\delta(\cdot)$ is the Dirac delta-function. I am not sure how to solve this without having another boundary condition. I believe I am supposed to use an (odd) extension to solve this but I'm not sure how to go about doing that.

Answer 1

Consider the same equation on $(-\infty,\infty)\times[0,\infty)$ with initial value $u(x,0)=-A\,\delta(x+x_0)+A\,\delta(x-x_0)$. Since the initial value is odd, the solution will be odd and satisfy the boundary condition $u(0,t)=0$.

Answer 2

The diffusion equation $u_{t} = k^{2}u_{xx}$ has initial boundary conditions $u(x,0) = A \delta(x - x_{0})$, $u(0,t) = 0$ where $A\neq0$ and $x_{0} > 0$ are given constants, and $\delta(\cdot)$ is the Dirac delta-function. I am not sure how to solve this without having another boundary condition. I believe I am supposed to use an (odd) extension to solve this but I'm not sure how to go about doing that.

The initial conditions are Derichlet conditions, and you don't need additional boundary conditions (well technically you also need the fact that the solution is square integrable to infinity in $x$, to exclude possible solutions that blow up).

The solution will be obtained by the following steps:

If the initial conditions were $v(y,0) = f(y)$ then we could, using our fundamental solution $u(x,t)$ form the gerneric solution $$ v(x,t) = \int_{y=-\infty}^{+\infty} u(x-y,t)f(y)\,dy$$

Let $\tilde{u}(s,t)$ be the Fourier transform of $u(x,t)$, then we have $$ \frac{d\tilde{u}}{dt} = -\frac{1}{k^2} \tilde{u} $$ and the initial conditions transform to $$ \tilde{u}(s-y,0)= \frac{1}{2\pi}e^{isy} $$ From there we use the formulas for the inverse Fourier transform $$ F^{-1} \left\{ \frac{1}{2\pi} e^{-iky} \right\} =\delta(x-y)\mbox{ and} \\ F^{-1} \left\{ \frac{1}{2\pi} e^{-\frac{s^2}{k^2}t} \right\} =\frac{k}{\sqrt{4\pi t}}e^{-k^2(x-y)^2/(4t)} $$ You end up with the fundamenal solution $$ u(x,t) = \frac{1}{\sqrt{4\pi k^2 t}} e^{-\frac{x^2}{4k^2 t}}$$

I have been a bit careless about my powers of $k$ here so you should check the work, but this is the basic way the Green's function for the heat equation is obtained. You can get a reasonable exposition in BUtkov's *Mathematical Physics$, section 12.5.