Consider heat equation
$$ u_{xx} = u_t $$
on infinite domain $-\infty< x < \infty$ and $t>0$ with initial condition $u(x,0) = g(x)$. We know can solve this and obtain
$$ u(x,t) = \frac{1}{\sqrt{4 \pi t} } \int\limits_{- \infty}^{\infty} g( \xi ) e^{ \frac{ (x- \xi)^2 }{4t }} d \xi $$
Im trying to solve the following:
For (a), it is easy to plot this on software, Im just looking for a simple way to graph without using a graphing device.
For part b), We have
$$ u(x,t) = \frac{1}{4 \pi \sqrt{t} } \int\limits_{- \infty}^{\infty} \exp \left( – \frac{ (\xi +2 )^2 }{4} + \frac{ (x – \xi)^2 }{4t}\right) – \exp \left( – \frac{ (\xi -2 )^2 }{4} + \frac{ (x – \xi)^2 }{4t}\right)$$
This is really cumbersome to compute. In what easier way can we compute this integral by hand?
Best Answer
For a, you should recognize this as the sum of two Gaussian distributions, a positive one with mean $2$ and standard deviation $\sqrt 2$ and a negative one with mean $-2$ and standard deviation $\sqrt 2$. That should be enough to generate a reasonable sketch.
For b, I would leave it as the integral. You might be expected to "evaluate it" by using the error function, but the rest of the question makes me feel we are just looking for the general idea here.
For c, you can use the linearity of the equation. If you had just one Gaussian, it would spread out as the time increases, staying centered at the same point. Here you add the two. In between them, it will go to zero reasonably quickly. You will have a positive pulse running off to $+\infty$ followed by a negative pulse of the same shape $4$ units behind it. Going toward $-\infty$ is just the negative of that. The pulses widen as time goes on.
I don't have anything in the same vein for d except to note that the max/min of the starting condition is $\pm \frac 1{\sqrt {4\pi}}$