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I want to solve the following integral equation

$$\int_0^\tau \ddot{\Psi}(t-t')g(t')dt'= g(\tau^-)\dot{\Psi}(\tau-t)+g(0^+)\dot{\Psi}(t),$$

where $\Psi$ is a even function, $g(\tau^-)\neq g(0^+)$ and $0<t<\tau$. This problem arrives from the optimization of a functional.

I know integral equations are hard to solve, but I tried to give me a chance and use Laplace transform $\mathcal{L}$ in the variable $t$ to solve that problem. In this manner,

$$\mathcal{L}\left[\int_0^\tau \ddot{\Psi}(t-t')g(t')dt'\right]= \mathcal{L}\left[g(\tau^-)\dot{\Psi}(\tau-t)+g(0^+)\dot{\Psi}(t)\right]$$

leads to

$$\int_0^\tau e^{-st'}g(t')dt' = \frac{(g(0^+)-g(\tau^-)e^{-s\tau})(s\tilde{\Psi}(s)-\Psi(0^+))}{s^2\tilde{\Psi}(s)-s\Psi(0^+)-\dot{\Psi}(0^+)}.$$

Using the inverse Laplace Transform $\mathcal{L}^{-1}$, we have

$$g(t) = \int_0^\tau \delta(t'-t)g(t')dt' = \mathcal{L}^{-1}\left[\frac{(g(0^+)-g(\tau^-)e^{-s\tau})(s\tilde{\Psi}(s)-\Psi(0^+))}{s^2\tilde{\Psi}(s)-s\Psi(0^+)-\dot{\Psi}(0^+)}\right].$$

However, when I tried to use that formula in a simple example I got a wrong answer. Below my code from Mathematica for $\Psi(t)=\cos{(t)}:$

RelaxF[t_] = Cos[t];
DRelaxF[t_] = D[RelaxF[t], t];

LaplaceRelaxF[s_] = LaplaceTransform[RelaxF[t], t, s];
P[s_] = Simplify[(A - B Exp[-s \[Tau]]) (s LaplaceRelaxF[s] - 
         Limit[RelaxF[x], x -> 0, 
          Direction -> "FromAbove"])/(s^2 LaplaceRelaxF[s] - 
      s Limit[RelaxF[x], x -> 0, Direction -> "FromAbove"] - 
      Limit[DRelaxF[x], x -> 0, Direction -> "FromAbove"])];

(* A = g[0]   ;   B = g[\[Tau]] *)

g[A_, B_, t_] = FullSimplify[InverseLaplaceTransform[P[s], s, t]];

whose result is

$$g(t) = g(0^+)-g(\tau^-)\text{Heaviside}(t-\tau),$$

which can not satisfy the first equation. Can anyone show me where I did wrong?

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