I don't understand this simplification given in this problem:(adsbygoogle = window.adsbygoogle || []).push({});

Q: At [itex] t=0 [/itex] a hot gas is on one side and a cold gas is on the other: [itex] u_0 = 1 [/itex] for [itex] x>0 [/itex], [itex] -1 [/itex] for [itex] x>0 [/itex]. Write down the solution:

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

to [itex] u_{tt}=u_{xx} [/itex] and with [itex] z = y-x [/itex] simplify to:

[tex] u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz [/tex]

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

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

[tex] u(t,x) = \frac{1}{2\sqrt{\pi t}}\int_[-x}^{x} e^{-z^2/4t}dz [/tex]

[tex] z = y-x \Rightarrow -z = -y+x \Rightarrow y = z+x[/tex], [tex] dy=dz+dx [/tex]

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

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

Things get sketchy here (assuming things are even right above):

[tex]u_0(z+x) = \left\{ \begin{array}{c} 1 \,\,\,\, ,0 > z+x \\ -1 \,\, ,0<z+x \end{array}[/tex]

[tex]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)[/tex]

and that's it folks...

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

Thanks in advance.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Integral (Simplifying it)

**Physics Forums | Science Articles, Homework Help, Discussion**