# Integral (Simplifying it)

1. Apr 18, 2006

I don't understand this simplification given in this problem:

Q: At $t=0$ a hot gas is on one side and a cold gas is on the other: $u_0 = 1$ for $x>0$, $-1$ for $x>0$. Write down the solution:
$$u(t,x)=\int_{\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy$$

to $u_{tt}=u_{xx}$ and with $z = y-x$ simplify to:
$$u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz$$

So my attempt thus far (with very little progress):

$$u(t,x)=\int_{\infty}^{\infty} \frac{1}{2\sqrt{\pi t}}e^{-(x-y)^2/4t}u_0(y)dy$$
$$u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz$$

$$z = y-x \Rightarrow -z = -y+x \Rightarrow y = z+x$$, $$dy=dz+dx$$

$$u(t,x)=\frac{1}{2\sqrt{\pi t}} \int_{\infty}^{\infty} e^{-(-z)^2/4t}u_0(z+x)(dz+dx)$$

$$u(t,x)=\frac{1}{2\sqrt{\pi t}} \int_{\infty}^{\infty} e^{-z^2/4t}u_0(z+x)(dz+dx)$$

Things get sketchy here (assuming things are even right above):
$$u_0(z+x) = \left\{ \begin{array}{c} 1 \,\,\,\, ,0 > z+x \\ -1 \,\, ,0<z+x \end{array}$$

$$u(t,x) = \frac{1}{\sqrt{\pi t}}\left( \int_{-\infty}^{0}e^{-z^2/4t}(-1)(dz+dx) + \int_{0}^{\infty}e^{-z^2/4t}(dz+dx) \right)$$

and that's it folks...

I don't understand how the bounds of the integration made the jump from $(\infty,-\infty)$ to [tex] (-x,x) [/itex]. I definitely need some help :)

Last edited: Apr 18, 2006
2. Apr 18, 2006

### nrqed

No. x must be treated as a constant. So dy=dz
So in the first case, you have 0>z+x so z<-x. So you must integrate from -infinity to -x. In the second case, z>-x so you integrate from -x to plus infinity. a quick glance at the integrals gives me the impression that after rewriting the integrals a little and using the parity of the integrands, you will get a difference of two integrals that will cancel outside of the interval [-x,x]. But I did not work it out explicitly.

3. Apr 18, 2006

Why is this so? Sorry, I'm not seeing it. I understand why differentiating a constant is 0. But why is it held as a constant?

I'll work through it. I'm slapping myself in the face for this part:

Thanks man. I appreciate it.

Last edited: Apr 18, 2006
4. Apr 18, 2006

### nrqed

Because you are evaluating u(t,x). So you must think of the integral as providing the value of u(t,x) for a certain fixed value of x and t.

Patrick

5. Apr 18, 2006