Consider the wave equation: $$\frac{\partial^2 u}{\partial t^2}=c^2\frac{\partial^2 u}{\partial x^2}.$$

The general solution is given by $$u(x,t)=f(x-ct)+g(x+ct).$$

However, separation of variables $u(x,t)=X(x)T(t)$ leads to $$ u(x,t)=\int_{-\infty}^{+\infty} \left( A_k \cos(kx)+B_k\sin(kx) \right) \left( C_k \cos(ckt)+ D_k \sin(ckt) \right) dk $$

where $A_k, B_k, C_k, D_k$ are independent of $x$ and $t$.

How are these two forms related? Is there more general context to this? How can they be transformed into another?

  • 1
    $\begingroup$ Hint: express $\cos(kx) \cos(ckt)$ (and the same with one or both $\cos$ replaced by $\sin$) in terms of functions of $x+ct$ and $x-ct$. $\endgroup$ Jun 9, 2020 at 17:52
  • 1
    $\begingroup$ I agree with the below answer. In short, one is the Fourier Series of the other. $\endgroup$
    – K.defaoite
    Jun 9, 2020 at 17:56

1 Answer 1


Using simple trigonometry you can write $$u(x,t)=\int_{-\infty}^\infty[\alpha_k\cos(kx-kct)+\beta_k\sin(kx-kct)+\gamma_k\cos(kx+kct)+\delta_k\sin(kc+kct)]dk\\=\int_{-\infty}^\infty[\alpha_k\cos(kx-kct)+\beta_k\sin(kx-kct)]dk+\int_{-\infty}^\infty[\gamma_k\cos(kx+kct)+\delta_k\sin(kc+kct)]dk$$ The first term is the Fourier representation of $f(x-ct)$ and the second term is the Fourier representation for $g(x+ct)$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .