# Weird Integral (to me)

1. Apr 15, 2006

We are now studying the one space dimension heat equation $u_t = u_{xx} [/tex] The fundemental solution is given as: $$u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy$$ I don't understand where the $y$ comes from. The example in this section is: If $u_0(x)=1$, the temperature stays at $u =1$ for all $t$. I wish I could see the solution, instead of just the answer. But that's the style of the book. I just don't see how to go from the fundemental solution, to the answer. 2. Apr 15, 2006 ### 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. 3. Apr 15, 2006 ### 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. 4. Apr 15, 2006 ### 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). 5. Apr 15, 2006 ### HallsofIvy Staff Emeritus If you did integrals in calculus, you used dummy variables: $$\int_0^1 y dy= \frac{1}{2}y^2 \|_0^1= \frac{1}{2}$ Do you see that there is no "y" in the result? Do you see that it doesn't matter what variable we use? If the solution to you differential equation had been given as $$u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-p)^2/4t}u_0(p)dp$$ or $$u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-z)^2/4t}u_0(z)dz$$ it would still be exactly the same. 6. Apr 16, 2006 ### FrogPad Yeah, I see how the 'y' disappears. And understand that it could be anything disappearing. I guess it's just how you "see" the problem sometimes. In the heat equation, the 'y' just seemed to come out of nowhere. I thought the problem only dealt with one spatial dimension, and then one dimension with time to track the temperatures at various locations. Then 'y' was introduced... and I didn't understand why that was even there. I still have a question though regarding the example given in the book. So we have: $$u(t,x)=\int_{-\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy$$ With BC $u_0(x)=1$$ for all [itex] x$.

So, when evaluting the integral, $u_0(y)$ from $-\infty$ to $\infty$ will evaluate to 1.

Thus the fundental solution drops drops down to:
$$\frac{1}{2\sqrt{\pi t}} \int_{-\infty}^{\infty} e^{-(x-y)^2/4t}dy$$

and this is supposed to evalutate to 1 for all t?

I'm not sure if I'm doing this right. Am I?

Thanks everyone.

Last edited: Apr 16, 2006
7. Apr 16, 2006

### arildno

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