Convolution solution to the heat equation

Solution to heat equation by convolution

The heat equation describes the evolution in time of the temperature of an object. So the 1-dimensional heat equation could be used to model the temperature of a rod. The 1-dimensional heat equation is

u_t = \alpha u_{xx}

In class, we took a Fourier transform of the heat equation with respect to x. We then solved this transformed equation. Taking an inverse transform gave us the following exact solution of the heat equation.

u(x,t)= u(x,0) \ast \frac{e^{-x^2/4\alpha t}}{\sqrt{4\pi \alpha t}}

The function \frac{e^{-x^2/4\alpha t}}{\sqrt{4\pi \alpha t}} is called the heat kernel. So the solution to the heat equation on an infinite interval is the convolution of the initial temperature data and the heat kernel. Notice that all of the time dependence comes from the heat kernel. You can think of the heat kernel as the solution to the heat equation when the initial temperature distribution is a Dirac functional at x=0; that is, one unit of energy is concentrated at a single point.

Now suppose the initial temperature distribution is

u(x,0)= \begin{cases} 1 & \text{ if }0 \leq x \leq 1 \\ 0 & \text{ else. }\end{cases}

Here's an amination of our solution. The first is of the time interval from 0.001 to 0.501 with frames each 0.005. Notice that our convolution formula for the solution is only valid for time strictly greater than zero.

Now let's look at a slower animation to see how the temperature distribution changes shape near t=0. The animation shows the time interval from 0.0001 to 0.0050 with frames each 0.0001.