We know for
$\left\{\begin{array}{cccc}\dfrac{1}{c^2}u_{tt}&=&u_{xx}\hspace{0.25cm}x\in [0,L]&\\ u(0,t)&=&u(L,t)=0\\ u(x,0)&=&f(x)\\ u_t(x,0)&=&g(x) \end{array}\right.$
$$u(x,t)=\sum_{k=1}^{\infty}\sin\left(\frac{k\pi x}{L}\right)\left[a_k\cos\left(\frac{ck\pi t}{L}\right)+b_k\sin\left(\frac{ck\pi t}{L}\right)\right]$$ and $$a_k=\frac{2}{L}\int_0^Lf(x)\sin\left(\frac{k\pi x}{L}\right)dx\hspace{1cm}b_k=\frac{2} {ck\pi}\int_0^Lg(x)\sin\left(\frac{k\pi x}{L}\right)dx$$
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My question is for the case when $x\in[-L,L]$, what is $u(x,t)$? And how could I find $a_k$ and $b_k$
What I did
$\left\{\begin{array}{cccc}\dfrac{1}{c^2}u_{tt}&=&u_{xx}\hspace{0.25cm}x\in [-L,L]&\\ u(-L,t)&=&u(L,t)=0\\ u(x,0)&=&f(x)\\ u_t(x,0)&=&g(x) \end{array}\right.$
$u(x,t)=\Psi(x)\theta(t)\\ u_{tt}=\Psi(x)\ddot{\theta}(t)\\ u_{xx}=\theta(t)\ddot{\Psi}(x)$
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$\Psi(x)=e^{rx}\\ \Psi'(x)=re^{rx}\\ \Psi''(x)=r^2e^{rx}$
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$r^2e^{rx}=-\lambda \Psi(x)\Rightarrow r=\sqrt{\lambda}i\Rightarrow \Psi(x)=e^{\sqrt{\lambda}i}\Rightarrow \Psi(x)=c_1\cos(\sqrt{\lambda}x)+c_2\sin(\sqrt{\lambda}x)\\ \Psi(L)=c_1\cos(\sqrt{\lambda}L)+c_2\sin(\sqrt{\lambda}L)\\ \Psi(-L)=c_1\cos(\sqrt{\lambda}(-L))+c_2\sin(\sqrt{\lambda}(-L))$ \ Thanks for your help