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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: enter image description here

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?

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  • $\begingroup$ You are probably going to have to use the error function or its complement. I assume this is in an Intro to PDE class so I dont think any higher level integration would help you out. $\endgroup$ – Hushus46 Mar 29 at 23:50
  • $\begingroup$ In all your integrals, there should be a "minus" sign in front of the exponent $\frac {(x - \xi)^2} {4t}$. Can you compute the integrals with this correction? $\endgroup$ – Alex M. Mar 30 at 18:55
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No wonder that you were not able to compute $u(x,t)$ - there is a "minus" sign that should appear in front of the exponent $\frac {(x - \xi)^2} {4t}$ in all your formulae! The correct formula is

$$u(x,t) = \frac 1 {\sqrt {4 \pi t}} \int \exp \left( \color{red} - \frac {(x - \xi)^2} {4t} \right) g(\xi) \ \mathrm d \xi \ .$$

The function $h(t,x,y) = \frac 1 {\sqrt {4 \pi t}} \exp \left( - \frac {(x - \xi)^2} {4t} \right)$ is called "the heat kernel" and has "the convolution property" (or "the semigroup property") that

$$\int h(s,x,y) \ h(t,y,z) \ \mathrm d y = h(t+s, x, z) \ .$$

Noticing that $(x-\xi)^2 = (\xi-x)^2$, i.e. $h(t,x,\xi) = h(t,\xi,x)$ (i.e. that $h$ is "symmetric"), and that $u(x,0) = h(1,x,-2) - h(1,x,2)$, it follows that

$$u(x,t) = \int h(t,x,\xi) \ h(1, \xi, -2) \ \mathrm d \xi - \int h(t,x,\xi) \ h(1, \xi, 2) \ \mathrm d \xi = h(t+1, x, -2) + h(t+1, x, 2) = \\ = \frac 1 {\sqrt {4 \pi (t+1)}} \exp \left( - \frac {(x + 2)^2} {4(t+1)} \right) - \frac 1 {\sqrt {4 \pi (t+1)}} \exp \left( - \frac {(x - 2)^2} {4(t+1)} \right) \ .$$

This explicit formula for $u(x,t)$ should be enough to help you solve the problem. Unfortunately, the statements are a bit vague, and without your lectures notes available, to put them into context, it is difficult to see what they mean. (For instance, what does it mean to "sketch the trend"? What (numerical?) methods have you been taught for the finding of appproximate extrema? "Approximate" in what sense?)

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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 $\pm \frac 1{\sqrt {4\pi}}$

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  • $\begingroup$ I disagree with your answer regarding (b). Take a good look at $u(x,0)$: what do you recognize? :) $\endgroup$ – Alex M. Mar 30 at 20:55

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