I'm trying solve this wave equation using Fourier method, but I am stuck...
$${ u }_{ tt } ={ c }^{ 2 }{ u }_{ xx } - \alpha{ u } =0, \ 0<x\le L, t >0 $$ $${ u }( 0,t) = { u }( L,t) = 0$$ $${ u }( x,0) = f(x), { u }_{ t }( x,0) = g(x) $$ I know that first I have to use variable separation: $${ u }( x,t) = T(t)X(x) $$ Making the calculations $$\frac{T''+ \alpha T}{c^{2}T} = \frac{X''}{X} = -\lambda $$ I guess I'm right at this point?Ok? Now I have to solve: $$X'' + \lambda X = 0$$ and $$\frac{T'' + \alpha T}{c^{2}T} = -\lambda$$
I don't know how to solve the second equation and how I add the two equations to solve the first problem.
I will be very grateful for the help!!!!
For the equation $$u_{tt} = c^{2} \, u_{xx} -\alpha u, \, 0<x\leq L, t>0$$ $$u(0,t) = u(L,t) = 0$$ $$u(x,0) = f(x), \, u(x,L) = g(x)$$ let $u(x,t) = F(x) G(t)$ for which \begin{align} \frac{1}{c^{2}} \left( \frac{G''}{G} + \alpha \right) = - \lambda^{2} = \frac{F''}{F} \end{align} or \begin{align} F'' + \lambda^{2} \, F &= 0 \\ G'' + (\alpha + \lambda^{2} c^{2}) G &= 0 \end{align} The solutions are \begin{align} F(x) &= A_{1} \, \cos(\lambda x) + B_{1} \, \sin(\lambda x) \\ G(t) &= A_{2} \, \cos(\sqrt{\alpha + \lambda^{2} c^{2}} \, t) + B_{2} \, \sin(\sqrt{\alpha + \lambda^{2} c^{2}} \, t) \end{align} The conditions are given by $F(0) = F(L) = 0$ for which \begin{align} F(x) = B_{1} \sin\left(\frac{n \pi x}{L}\right). \end{align} The general series solution is of the form \begin{align} u(x,t) = \sum_{n=1}^{\infty} \left(A_{n} \, \cos(\sqrt{\alpha + \lambda^{2} c^{2}} \, t) + B_{n} \, \sin(\sqrt{\alpha + \lambda^{2} c^{2}} \, t) \right) \, \sin\left(\frac{n \pi x}{L}\right). \end{align} Since $u(x,0) = f(x)$ then \begin{align}\tag{1} f(x) = \sum_{n=1}^{\infty} A_{n} \, \sin\left(\frac{n\pi x}{L}\right) \end{align} for which the coefficients $A_{n}$ may be obtained. The remaining condition is $u(x,L) = g(x)$ for which \begin{align}\tag{2} g(x) = \sum_{n=1}^{\infty} \left( A_{n} \, \cos(\sqrt{\alpha L^{2} + n^{2} \pi^{2}}) + B_{n} \, \sin(\sqrt{\alpha L^{2} + n^{2} \pi^{2}}) \right) \, \sin\left(\frac{n \pi x}{L}\right). \end{align} Since $A_{n}$ can be obtained from (1) the coefficients $B_{n}$ can be obtained from (2). It may be more usefull to "shift" the coefficients in (2) by making use of $C_{n} = A_{n} \, \cos(\sqrt{\alpha L^{2} + n^{2} \pi^{2}})$ and $D_{n} = B_{n} \, \sin(\sqrt{\alpha L^{2} + n^{2} \pi^{2}})$ for which $P_{n} = C_{n} + D_{n}$ and \begin{align}\tag{3} g(x) = \sum_{n=1}^{\infty} P_{n} \, \sin\left(\frac{n \pi x}{L}\right) \end{align}
Coefficients By use of Fourier Sine series it can be seen that \begin{align} A_{n} &= \frac{2}{L} \, \int_{0}^{L} f(x) \, \sin\left(\frac{n \pi x}{L}\right) \, dx \\ P_{n} &= \frac{2}{L} \, \int_{0}^{L} g(x) \, \sin\left(\frac{n \pi x}{L}\right) \, dx. \end{align} $B_{n}$ is found to be \begin{align} B_{n} &= \frac{2}{L} \, \int_{0}^{L} \left( \cot(\sqrt{\alpha L^{2} + n^{2} \pi^{2}}) \, f(x) + \csc(\sqrt{\alpha L^{2} + n^{2} \pi^{2}}) \, g(x) \right) \, \sin\left(\frac{n \pi x}{L}\right) \end{align}