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.
$$F(x) = \begin{cases} x, \ 0 Solution: For an insulated rod the solution $X(x,t)=
a_0/2+\sum B_n cos(n\pi x/L)e^{(n^2π^2α^2)t/L}$ I found $a_0= 1$
and $Bn=
(−(2/nπ)sin(nπ/2)+(2/nπ)2cos(nπ/2)−(2/nπ)^2)$ then just plug in the coefficients into the sum.
I am just not sure if these are the correct values for the $a_0$ and $B_0$ coefficients.
Differential equations, HEAT equation with insulated ends.
2
$\begingroup$
pde
fourier-series
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0Not pertaining to the question, but oh my... Look at the titles of the related pages. They're all the same! Terrible! – 2012-04-10
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0lol because they did not let me use new tags. – 2012-04-10
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1I put it into LaTeX to be more readable. I had to fix a typo too, so I hope it is all what you wanted. – 2012-09-05
1 Answers
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(Note: This looks similar to this question)
(As another user pointed out), your formula for the solution should be written:
$$ 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} $$
Recall $a_0 = (2/L)\int_0^L f(x) dx, \ $ $B_n = (2/L)\int_0^L \cos(n\pi x / L) f(x) dx$. So, with $f(x)$ as above, $L=2$, we have
$$a_0 = (2/2)\int_0^1 x dx = \frac{1}{2}$$
$$b_n = (2/2)\int_0^1 \cos(n\pi x / 2) x dx = [x (2/n\pi) \sin(n \pi x / 2) + (2/n\pi )^2 \cos(n \pi x / 2)]_{x=0}^1$$ $$ = (2/n \pi) \sin(n \pi / 2) + (2/n \pi)^2(\cos(n\pi / 2) - 1) $$ so it looks like you are on the right track, but made a small sign error somewhere.