Question

Problem with partial differential equation.

Solution included:

Answer 1

boundary conditions of u(x, t):

u(0, t) = 0
u(pi, t) = 0
u(x, 0) = 0
\[\frac{ \delta u }{ \delta t } |_{t = 0} = sinx\]
delta is partial derivative (because o.s. doesn't have the symbol yet)

this is a wave equation problem:

\[a^2\frac{ \delta^2 u }{ \delta x^2 } = \frac{ \delta^2 u }{ \delta t^2 }\]

solution is here: http://img23.imageshack.us/img23/6848/l3tj.jpg

i understand everything, and I have solved others.

my only problem is the B1 term they introduced when solving for Bn (which turned out to = 0) using the fourier series method.

I don't understand how it was obtained.

Please explain or point me to the right resource :)

Answer 2

@Euler271 "openstudy" just uses MathJax which most definitely supports \(\partial\)

Answer 3

thanks for letting me know ^_^

Answer 4

@Euler271 \(\partial=\text{\partial}\)

Answer 5

Oh that's what it is o_o

Answer 6

If you see something on OpenStudy and don't know how to produce it in LaTex, just select the offending math, right click and do Show Math As>TeX Commands

Answer 7

This solution manual text really sucks... :-p

Answer 8

forget the solution manual ... do it on your own. there are couple of inconsistencies with what you mention. for eg; you have u(x,0) = 0, solution assumes u_t(x,0) = 0

Answer 9

i don't understand what you see wrong with it.
solution in textbook respects u_t(x, 0) = sinx

I just don't understand what B1 is.

Bn is solved using fourier sine series. How do you find B1?

Answer 10

You find it via comparison ... there is \(T'(0) = 0 \) . it's supposed to be \(T(0) = 0 \)

Answer 11

the solution is of the form X*T ... all those values of n consist of your solution. so you using superposition principle. Use boundary condition on X and T to simplify.