Understanding the Heat Equation and its Fundamental Solution

[FrogPad]

We are now studying the one space dimension heat equation u_t = u_{xx} [/tex]<br /> <br /> The fundamental solution is given as:<br /> u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy<br /> <br /> I don&#039;t understand where the y comes from. <br /> <br /> The example in this section is:<br /> If u_0(x)=1, the temperature stays at u =1 for all t.<br /> <br /> I wish I could see the solution, instead of just the answer. But that&#039;s the style of the book. I just don&#039;t see how to go from the fundamental solution, to the answer.

 

[TD]

I'm not familiar with the heat equation but in this integral, y is just a dummy variable. You integrate with respect to y (although you can give it any other name, just not t or x to avoid confusion), but the y will vanish in the answer of course.

 

[krab]

The y is a dummy variable; it is integrated out. If you want insight into the solution, try some other easily integrable cases. For example, what if u_0 is a delta function? Or try plotting the integrand for a sequence of different y's, then imagine them all added together.

 

[FrogPad]

Weird so it's kind of like a loop for the expression contained in the integral?

I'll have to play around with this. It's new to me. (or at least I can't remember doing this before).

 

[HallsofIvy]

If you did integrals in calculus, you used dummy variables:
\int_0^1 y dy= \frac{1}{2}y^2 \|_0^1= \frac{1}{2}[/itex]<br /> Do you see that there is no &quot;y&quot; in the result? Do you see that it doesn&#039;t matter what variable we use?<br /> <br /> If the solution to you differential equation had been given as<br /> u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-p)^2/4t}u_0(p)dp<br /> or <br /> u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-z)^2/4t}u_0(z)dz<br /> it would still be exactly the same.

 

[FrogPad]

Yeah, I see how the 'y' disappears. And understand that it could be anything disappearing. I guess it's just how you "see" the problem sometimes. In the heat equation, the 'y' just seemed to come out of nowhere. I thought the problem only dealt with one spatial dimension, and then one dimension with time to track the temperatures at various locations. Then 'y' was introduced... and I didn't understand why that was even there.

I still have a question though regarding the example given in the book.
So we have:
u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy

With BC u_0(x)=1 [/tex] for all x.<br /> <br /> So, when evaluting the integral, u_0(y) from -\infty to \infty will evaluate to 1.<br /> <br /> Thus the fundental solution drops drops down to:<br /> \frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} e^{-(x-y)^2/4t}dy<br /> <br /> and this is supposed to evalutate to 1 for all t?<br /> <br /> I&#039;m not sure if I&#039;m doing this right. Am I?<br /> <br /> Thanks everyone.

 

[arildno]

Correct.
Change your dummy variable y to u by
u=\frac{y-x}{2\sqrt{t}}
We then have:
\frac{dy}{du}=2\sqrt{t}
whereby your integral becomes:
\frac{1}{\sqrt{\pi}}\int_{-\infty}^{\infty}e^{-u^{2}}du

 

[FrogPad]

Cool, thankyou!
Also, I was going to ask. The term "dummy variable"... that's not the "technical" name is it?