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I hope to get some insights into expressing Partial Differential Equation (PDE) in terms of Separable Variables.

Given the wave equation under Dirichlet Boundary Conditions, \begin{align} u_{tt}&=c^{2}u_{xx}, 0<x<l, t>0 \\ u(0,t)&=u(l,t)=0, \\ u(x,0)&=\phi(x), u_{t}(x,0)=\psi(x) \end{align} Assuming $u(x,t)=X(x)T(t)$. When we solve the ODE separately, we get $X(x)=B\sin(\sqrt{\lambda}x)$. But the solution given, after solving $T(t)$ is \begin{equation} u(x,t)=\sum_{n=1}^{\infty} (C_{n}\cos(c\sqrt{x}t)+D_{n}\sin(c\sqrt{x}t))\sin(\sqrt{\lambda}x) \end{equation} I like to ask, why is $B_{n}$ not necessary to be considered into the solution?

Gary
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user3497
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1 Answers1

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$B_n$ is redundant, since it can be absorbed into $C_n$ and $D_n$.

That is, if you include $B_n$, you would get \begin{equation} \begin{aligned} u(x,t) &= \sum_{n=1}^{\infty} \bigl( C_{n} \cos(c\sqrt{x}t) + D_{n} \sin(c\sqrt{x}t) \bigr) B_n \sin(\sqrt{\lambda}x) \\ &= \sum_{n=1}^{\infty} \bigl( B_n C_{n} \cos(c\sqrt{x}t) + B_n D_{n} \sin(c\sqrt{x}t) \bigr) \sin(\sqrt{\lambda}x) , \end{aligned} \end{equation} but here you can let $\tilde C_n = B_n C_n$ and $\tilde D_n = B_n D_n$ to get rid of $B_n$.

Hans Lundmark
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