How Can I Simplify the Integral in u(t,x) to Solve u_{tt}=u_{xx}?

[FrogPad]

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 (-x,x) [/itex]. I definitely need some help :)<br /> <br /> Thanks in advance.

 

[nrqed]

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&gt;0, -1 for x&gt;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

No. x must be treated as a constant. So dy=dz

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 &gt; z+x \\ -1 \,\, ,0&lt;z+x \end{array}

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.

 

[FrogPad]

No. x must be treated as a constant. So dy=dz

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?

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.

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

you have 0>z+x so z<-x

Thanks man. I appreciate it.

 

[nrqed]

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?

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

 

[FrogPad]

Ahh beautiful. Thank you!