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Two solutions of the equation are the Mathieu sine and cosine and they form an orthogonal basis. The equation has arisen from the wave equation for an infinite string with a sinusoidal cross-section. The initial values are the displacement of the string at $t=0$ and the velocity of the sting at $t=0$, which are both functions of $x$.

Wave equation:

$\frac{\partial^2u(x,t)}{\partial x^2}-(1-\kappa\cos{2x})\frac{\partial^2u(x,t)}{\partial t^2}=0$, $\kappa$ is a parameter.

Initial condition:

$u(x,t=0)=f(x),\quad u'(x,t=0)=g(x)$.

Separation of variables:

$u(x,t)=A(x)T(t)$

New equation:

$\frac{1}{A}\frac{\partial^2A}{\partial x^2}\frac{1}{-1+\kappa\cos{2x}}=-\frac{1}{T}\frac{\partial^2T}{\partial t^2}=constant=\lambda$

$\lambda$ must be greater than zero so that T is harmonic and not exponential. The equation for A is the Mathieu DE and its solutions are

$A(x)=C_1 C(\lambda,\frac{\kappa \lambda}{2},x)+C_2 S(\lambda,\frac{\kappa \lambda}{2},x)$, $C$ and $S$ are the Mathieu cosine and sine respectively.

So my question is how do I satisfy the initial conditions $f(x)$ and $g(x)$. IIRC there is a sum with respect to $\lambda$ if where are also boundary conditions but in this case there is an integral? Any help would be appreciated.

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If $ F(x,s)$ is the Laplace transform of $u(x,t)$ in the $t$ variable (assuming $u(x,t)$ is exponentially bounded in $t$), you get an inhomogeneous Mathieu ODE for $F(x,s)$ (at each fixed $s$): $$ s^2 \left( k \cos \left( 2\,x \right) -1 \right) F \left( x,s \right) +{\frac {{\partial}^{2}}{{\partial}{x}^{2}}}F \left( x,s \right) = \left( g \left( x \right) +sf \left( x \right) \right) \left( k\cos \left( 2\,x \right) -1 \right) $$ Solve this inhomogeneous differential equation and take the inverse Laplace transform. For typical $f$ and $g$, I doubt that it can be done in closed form.

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