TITLE: Fourier series and differential equations (vibrating string)
QUESTION [0 upvotes]: We consider a vibrating string. Its movement $u(x,t)$ satisfying the wave equations : $$
 \dfrac{d^2u}{dt^2} = \dfrac{d^2 u}{dx^2},\qquad u(-\pi,t)=u(\pi,t), \qquad u(x,0)= u_0(x),\qquad \dfrac{du}{dt}(x,0)=0.
$$
where the initial condition $u_0(x)$ is the restriction at $[-\pi,\pi]$ of a periodic $C^{\infty}$ function.
I have to "cut" $u(.,t)$ in Fourier serie to give a solution of the wave equation.
I wrote $u(.,t) = \dfrac{a_0}{2} + \sum_{k \geq 1} a_k \cos(kt) + b_k \sin(kt)$ and its derivative $u'(.,t) = \sum_{k \geq 1} kb_k \cos(kt) - k a_k \sin(kt)$ and finally, $u''(.,t)= \sum_{k \geq 1} -k^2 a_k \cos(kt) - k^2 b_k \sin(kt)$.
Someone could help me ? I usually work with one variable and not with two variables, so I don't know how to approach the problem. Could someone help me ? I hope my English is understandable.

REPLY [1 votes]: I assume what you actually need to do is to solve this partial differential equation (PDE) by the Fourier method. The method is based on assuming that some particular solution can be represented as:
$$u_p(x,t)=f(x)g(t)$$
Then we have from the PDE:
$$f g''=f'' g$$
Dividing by $f g$ we have:
$$\frac{g''}{g}=\frac{f''}{f}$$
Because the left hand side depends only on $t$, while the right hand side depends only on $x$, they both should be equal to some constant, which we can choose freely (I name it $- \omega^2$):
$$\frac{g''}{g}=\frac{f''}{f}=- \omega^2$$
The solutions can be written as (I use complex exponentials instead of trigonometric functions, it doesn't really matter):
$$g(t)=A(\omega) e^{i \omega t}+B(\omega) e^{-i \omega t}$$
$$f(x)=C(\omega) e^{i \omega x}+D(\omega) e^{-i \omega x}$$
Because we are restricted to the functions with period $2 \pi$, we can only have integer frequencies:
$$\omega = n=0, \pm 1, \pm 2, \pm 3, \ldots$$
So:
$$g_n(t)=A_n e^{i n t}+B_n e^{-i n t}$$
$$f_n(x)=C_n e^{i n x}+D_n e^{-i n x}$$
So the general solution should look like a kind of Fourier series:

$$u(x,t)=\sum_{n = - \infty}^\infty \left(A_n e^{i n t}+B_n e^{-i n t} \right) \left(C_n e^{i n x}+D_n e^{-i n x} \right)$$

We can set one of the amplitudes to $1$ without the loss of generality.
Now let's look at the last initial condition:
$$\frac{\partial u}{\partial t}=\sum_{n = - \infty}^\infty i n \left(A_n e^{i n t}-B_n e^{-i n t} \right) \left(C_n e^{i n x}+D_n e^{-i n x} \right)=$$
For $t=0$ we have:
$$=\sum_{n = - \infty}^\infty i n \left(A_n -B_n \right) \left(C_n e^{i n x}+D_n e^{-i n x} \right)=0$$
Which can only work if for every $n$ we have:
$$A_n=B_n$$
Which means $g_n(t)$ are cosines and we can rewrite the function as:

$$u(x,t)=\sum_{n = - \infty}^\infty \cos nt  \left(C_n e^{i n x}+D_n e^{-i n x} \right)$$

We set $A_n=\frac{1}{2}$ without the loss of generality and used the Euler formula $e^{ia}=\cos a+i \sin a$.
Now the final initial condition is:
$$u(x,0)=\sum_{n = - \infty}^\infty \left(C_n e^{i n x}+D_n e^{-i n x} \right)=u_0(x)$$
Now we just need to expand $u_0(x)$ in a Fourier series and match the coefficients to find $C_n$ and $D_n$.
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