1-D Wave equation with mixed boundary conditions

  • Thread starter KEØM
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Homework Statement


Solve, [tex]u_{t} = u_{xx}c^{2}[/tex]

given the following boundary and initial conditions

[tex]u_{x}(0,t) = 0, u(L,t) = 0[/tex]

[tex]u(x,0) = f(x) , u_{t}(x,0) = g(x)[/tex]


Homework Equations



[tex]u(x,t) = F(x)G(t)[/tex]

The Attempt at a Solution


I solved it, I am just not sure if it is right.

[tex]u(x,t) = \sum_{n=1}^\infty(a_{n}cos(\lambda_{n}t) + b_{n}sin(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x)

, \lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c [/tex]

[tex]a_{n} = \frac{2}{L}\int_0^L f(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx,

b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx
[/tex]

Can someone please verify this for me?

Thanks in advance,
KEØM
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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Why does your formula for bn[/b] involve cosine rather than sine?
 
  • #3
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Thanks for replying HallsofIvy.

Well to get [tex]b_{n}[/tex] I applied the initial condition

[tex]u_{t}(x,0) = g(x).[/tex]

So I first find [tex]u_{t}[/tex] which comes to

[tex]u_{t} = \sum_{n=1}^\infty(-a_{n}\lambda_{n}sin(\lambda_{n}t) + b_{n}\lambda_{n}cos(\lambda_{n}t))cos((n-\frac{1}{2})\frac{\pi}{L}x) [/tex]

now evaluating [tex]u_{t}[/tex] at t = 0 and setting it equal to g(x) we get,

[tex]u_{t}(x,0) = \sum_{n=1}^\infty b_{n}\lambda_{n}cos((n-\frac{1}{2})\frac{\pi}{L}x) = g(x) [/tex]

But that is just the Fourier cosine series of g(x) just with the extra [tex]\lambda_{n}[/tex] and with the zero term equal to zero

so [tex]b_{n} = \frac{2}{\lambda_{n}}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

where [tex]\lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c
[/tex]

[tex] \Rightarrow b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx[/tex]

Please let me know if I did something wrong.

Thanks again,
KEØM
 

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