1-D Wave equation with mixed boundary conditions

[KEØM]

Homework Statement

Solve, $$u_{t} = u_{xx}c^{2}$$

given the following boundary and initial conditions

$$u_{x}(0,t) = 0, u(L,t) = 0$$

$$u(x,0) = f(x) , u_{t}(x,0) = g(x)$$

Homework Equations

$$u(x,t) = F(x)G(t)$$

The Attempt at a Solution

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

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

$$a_{n} = \frac{2}{L}\int_0^L f(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx, b_{n} = \frac{4}{(2n-1)c\pi}\int_0^L g(x)cos((n-\frac{1}{2})\frac{\pi}{L}x)dx$$

Can someone please verify this for me?

Thanks in advance,
KEØM

 

[HallsofIvy]

Why does your formula for b_{n[/b] involve cosine rather than sine?}

 

[KEØM]

Thanks for replying HallsofIvy.

Well to get $$b_{n}$$ I applied the initial condition

$$u_{t}(x,0) = g(x).$$

So I first find $$u_{t}$$ which comes to

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

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

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

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

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

where $$\lambda_{n} = (n-\frac{1}{2})\frac{\pi}{L}c$$

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

Please let me know if I did something wrong.

Thanks again,
KEØM