Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is the question, I have solved it but I need someone to double check my solution. Question: Find the temperature $u(x, t)$ in a rod of length $L$ if the initial temperature is $f(x)$ throughout and if the ends $x=0$ and $x=L$ are insulated. $F(x)= x, 0$

Solution:

For an insulated rod the solution $$ X(x,t)= \frac{a_0}{2}+\sum B_0 \frac{\cos(n\pi)x}{Le}−\frac{n^2\pi^2\alpha^2}{L}t $$ I found $a_0= 1$ and $$ B_n= −\frac{2}{n\pi}\sin\left(\frac{n\pi}{2}\right)+\frac{2}{n\pi}2\cos\left(\frac{n\pi}{2}\right)−\left(\frac{2}{n\pi}\right)^2 $$ then just plug in the coefficients into the sum. I am just not sure if these are the correct values for the $a_0$ and $B_0$ coefficients.

share|improve this question
4  
Your solution can't be right because it has no dependence on the initial condition, f(x). Also, you should edit your question. There are a couple of typos, e.g. in your solution the e shouldn't be in the denominator of the first fraction, it should be outside the fraction with the second fraction being its argument. –  in_wolframAlpha_we_trust Apr 13 '12 at 6:59
add comment

1 Answer 1

(As another user pointed out), your formula for the solution should be written:

$$ X(x,t)= \frac{a_0}{2}+\sum B_{\mathbf{n}} \frac{\cos(n\pi x)}{L}\mathbf{e}^{−\frac{n^2\pi^2\alpha^2}{L}t} $$

Also, you don't have the correct formulas for the coefficients.

Recall $a_0 = (2/L)\int_0^L f(x) dx, \ $ $B_n = (2/L)\int_0^L \cos(n\pi x / L) f(x) dx$. So, with $f(x) = x$, we have

$$a_0 = (2/L)\int_0^L x dx = L$$

$$b_n = (2/L)\int_0^L \cos(n\pi x / L) x dx = (2/L)[x (L/n\pi) \sin(n \pi x / L) + (L/n\pi )^2 \cos(n \pi x / L)]_{x=0}^L$$ $$ = 2L/(n\pi)^2(\cos(n\pi) - 1) $$ ( = 0 if n even, $-4L/(n\pi)^2$ if n odd)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.