Solving PDE Heat Equation for Temperature Distribution

[josftx]

Homework Statement

Find the distribution of temperatures in the rod of length L with the follow BC and NC

Homework Equations

u_{t}=\alpha u_{xx}\,\,\,x\in]\frac{-L}{2},\frac{L}{2}
u(\frac{-L}{2},t)=u(\frac{L}{2},t)=700
u(x,0)=300\,\,\,x\in]\frac{-L}{2},\frac{L}{2}

The Attempt at a Solution

With a change of variable v(x,t)=u(L(x-\frac{1}{2}))+700 with the bounday conditions now the new problem its.
v_{t}=\frac{\alpha}{L^{2}}u_{xx}\,\,\,x\in]0,1
v(0,t)=v(1,t)=0
v(x,0)=-400

now, the final solution will be u(x,t)=v(\frax{x}{L}+\frac{1}{2})+700 now my teacher says that i have must find the odd extension of f(x) but i can't resolve. Anyone can't help me for find the Final Solucion with all the changes of variables?

 

[LCKurtz]

What is f(x)? I'm guessing it is v(x,0).

When you use separation of variables on your v(x,t) equation and get to the point where you are evaluating v(x,0), you should have v(x,0) = a Fourier sine series in x. Presumably you will use a half-range expansion for the coefficients, which amounts to using the odd extension of v(x,0).

So proceed with separation of variables.

 

[josftx]

Yes, f(x) means v(x,0). with MSV i find it the general expression for the series.
v(\hat{x},t)={\displaystyle \sum_{n=1}^{\infty}a_{n}sin(n\pi x)e^{-n^{2}\pi^{2}ct}} with c=\frac{\alpha}{L^{2}} , but the next step i have doubts.

look..

f(x)_{odd}=\begin{cases}<br /> -400 &amp; -1&lt;x\leq0\\<br /> 400\,\ &amp; 0\leqx&lt;1\end{cases}
so..

a_{n}={\displaystyle 2\int_{-1}^{0}-400sin(n\pi x)dx+2\int_{0}^{1}400sin(n\pi x)dx}=\frac{1600}{n\pi}\,\,\,..n=1,3,5...

thats right?

 

[LCKurtz]

You have

v(x,0) = \sum_{n=1}^\infty a_n \sin(n\pi x) = f(x),\ 0 &lt; x &lt; 1

Use the half-range sine expansion formula for the constants:

a_n = 2\int_0^1 f(x) \sin(n\pi x) \,dx = 2\int_0^1 (-400) \sin(n\pi x) \, dx

 

[josftx]

You have

v(x,0) = \sum_{n=1}^\infty a_n \sin(n\pi x) = f(x),\ 0 &lt; x &lt; 1

Use the half-range sine expansion formula for the constants:

a_n = 2\int_0^1 f(x) \sin(n\pi x) \,dx = 2\int_0^1 (-400) \sin(n\pi x) \, dx

so the a_{n}=\frac{-1600}{n\pi}

and the solution its. u(x,t)=-400+\sum_{n=1}^\infty \frac{-1600}{n\pi} \sin(n\pi \hat{x})e^{n^{2}\pi^{2}ct} + 700

but now i have \hat{x}[\tex] ,,, with \hat{x}=L(x-\frac{1}{2}) <br /> <br /> and the solution of the problem its u(x,t)=v(\frac{x}{L}+\frac{1}{2})+700<br /> <br /> so u(x,t)=300-\sum_{n=1}^\infty \frac{1600}{n\pi} \sin(n\pi (\frac{x}{L}+\frac{1}{2})x)e^{n^{2}\pi^{2}\frac{\alpha}{L^{2}}t}

 

[LCKurtz]

I didn't check your change of variable stuff; I just assumed you have done that correctly. But you might want to check your calculations because a positive exponential in t doesn't make any physical sense.

 

[josftx]

I didn't check your change of variable stuff; I just assumed you have done that correctly. But you might want to check your calculations because a positive exponential in t doesn't make any physical sense.

Sorry its a negative exponential. i might wrong in the transcription to the forum. but i don't know if the solutions its fine, because in any book or something i find a similar exercise with these domain ]-L/2,L/2[ or with the change of variable.

 

[LCKurtz]

Sorry its a negative exponential. i might wrong in the transcription to the forum. but i don't know if the solutions its fine, because in any book or something i find a similar exercise with these domain ]-L/2,L/2[ or with the change of variable.

Since you didn't show the work for the substitution, I couldn't say about that. What I can tell you is that I would have begun with the substitution w(x,t) = u(x + L/2, t) to convert it to a problem on [0,L]. The fact that you did something different doesn't make yours wrong, just different, and as long as you did it correctly there should be no problem.

[Edit] x + L/2, not x + 1/2 as it was.