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$$ \begin{cases} v''(t)-v(t)=f(t) & 0<t<h\\ v(0)=0=v(h) \end{cases}, $$ where $f\in L^{2}$.

Is there a explicit solution of integral form?


For the case of $$ v''(t)-v(t)=0, $$ the solution is $v(t)=C_{0}e^{t}+C_{1}e^{-t}$ for constants $C_{0},$ $C_{1}$. But I can't go further with the general nonhomogenous term $f(t)$.

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  • $\begingroup$ Do you know about Green's functions? $\endgroup$ – Chappers Oct 15 '17 at 14:11
  • $\begingroup$ @Chappers No. Is this related to that? $\endgroup$ – kayak Oct 15 '17 at 14:15
  • $\begingroup$ Yes, Green's functions are used to solve a nonhomogeneous differential equation with specified boundary conditions and a general function on the right hand side. $\endgroup$ – Chappers Oct 15 '17 at 14:20
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    $\begingroup$ The solution will be the same as the variation of parameters one (although possibly in a different form), but the technique is much more general than just the second-order ODEs variation of parameters is normally used for. $\endgroup$ – Chappers Oct 15 '17 at 14:22
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Consider the solutions of the associated homogeneous problem $$y_1(t)=\sinh(t),y_2(t)=\sinh(t-h),$$ satisfying $y_1(0)=y_2(h)=0$. Variation of parameters gives some particular solutions: $$v(t)=\sinh(t)\int_{t_1}^t \frac{-\sinh(\tau-h)f(\tau)}{\sinh(h)} \mathrm{d} \tau+ \sinh(t-h)\int_{t_2}^t \frac{\sinh(\tau) f(\tau)}{\sinh(h)} \mathrm{d} \tau .$$ In order to satisfy the boundary condition $v(0)=0$ we take $t_2=0$, and in order to satisfy $v(h)=0$ we take $t_1=h$.

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