I would like to determine the solution to the 1D heat equation where the initial condition is a Delta function at the boundary
- $$ \frac{\partial u}{\partial t} = D \frac{\partial^2 u}{\partial x^2}$$
- $$x \in [0, L], t \in [0, \infty]$$
- $$u(0,x) = \beta \delta(x)$$
- $$ u_{x} (0,t) = u_{x}(L,t) =0$$
Physically, a finite amount of mass $\beta$ is concentrated at one end of an insulated bar of length $L$ at time $t=0$ and I want to know the evolution of this mass distribution throughout time. The total amount of mass in the bar should always remain $\beta,$ and the steady solution should be $u(x,\infty) = \frac{\beta}{L}.$
Is this initial condition equivalent to just solving find the Green's function? If so, how do I proceed?
The fundamental solution (Green's function) in an unbounded domain is $$ u_0(x,t) = \frac{1}{2\sqrt{\pi Dt}} \exp\left(-\frac{x^2}{4Dt}\right). $$
Note that this satisfies the BC at $x=0$ but not at $x=L$. To build that in use method of images: $$ u(x,t) = \beta\sum_{n=-\infty}^\infty u_0(x+2nL, t) = \frac{\beta}{2\sqrt{\pi D t}}\sum_{n=-\infty}^\infty \exp\left(-\frac{(x+2nL)^2}{4Dt}\right). $$ That is, imagine an initial condition with a delta function at $x=0, x=\pm 2L, x=\pm 4L, \ldots$. By symmetry the BC at $x=L$ will now also be satisfied.
Note that the average mass from $0$ to $L$ will not be $\beta/L$, but rather $\beta/(2L)$. Roughly speaking only the right 'half' of the delta function is in the domain (if you like imagine the initial condition being a very tall Gaussian peak centred at zero. Only half of that peak is in the domain).