$$\frac{\partial^2}{\partial t^2}\Big\{ \int_0^t \int_{x-c(t-\tau))}^{x+c(t-\tau)} f(s, \tau)ds \, d\tau \Big\}$$
I have seen examples as a start point, but i can't find one like it. Any tip will be apreciated
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Upon taking the first partial derivative we obtain
$$\begin{align} I&=\frac{\partial^2}{\partial t^2} \int_0^t \int_{x-c(t-\tau))}^{x+c(t-\tau)} f(s, \tau)ds \, d\tau \\\\ &=\frac{\partial}{\partial t}\int_{0}^{t} c\left(f(x+c(t-\tau),\tau)+f(x-c(t-\tau),\tau)\right)\,d\tau\ \end{align}$$
since $\int_x^x f(s, \tau)\,ds=0$. Then, taking another partial derivative we find
$$\begin{align} I&=2cf(x,t)+\int_{0}^{t} c^2\left(f_1(x+c(t-\tau),\tau)-f_1(x-c(t-\tau),\tau)\right)\,d\tau\ \end{align}$$
where $f_1$ designates the partial derivative with respect to the first variable.
Mark Viola
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Hi, Dr.MV. I've done the first derivative, but i don't understand why you say $$\int_x^x f(s, \tau),ds=0$$ – Juan Camilo Acosta Arango Sep 20 '15 at 20:10
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When applying Leibniz's Rule, we have to consider both the integration limits in the context of the FOC along with bringing the derivative inside the integral. When evaluating $x\pm c(t-\tau)$ at $\tau =t$, we have that the upper and lower limits are $x$. – Mark Viola Sep 20 '15 at 20:27