Differentiation of a 2-variable function Let $$u(x,t)=\frac{1}{2c}\int_0^t \int_{x-ct+cs}^{x+ct-cs}f(y,s)\,dy\,ds$$
then, this satisfies the PDE : $$u_{tt}=c^2u_{xx}+f \quad \text{and} \quad u(x,0)=u_t(x,0)=0$$
Can you help me get $u_{tt},u_{xx}$ plz?
 A: You have to use Leibniz rule for differentiation under the sign of the integral consecutively. Is a little messy but is straightforward. 
$$u_t=\frac{1}{2c}\{\int_{x}^{x}f(y,s)dy+0+\int_{0}^{t}\frac{d}{dt}\int_{x-ct+cs}^{x+ct-cs}f(y,s)\,dy\,ds\}=$$
$$=0+0+\frac{1}{2c}\int_{0}^{t}[c\cdot f(x+ct-cs,s)+c\cdot f(x-ct+cs,s)]\,ds\}=$$
$$=\frac{1}{2}\int_{0}^{t}[f(x+ct-cs,s)+f(x-ct+cs,s)]\,ds\}$$
Now
$$u_{tt}=\frac{1}{2}\{f(x)+f(x)+\int_0^t[c\cdot f_x(x+ct-cs,s)-c\cdot f_x(x-ct+cs,s)]\,ds\}$$
And
$$u_x=\frac{1}{2c}\{0+0+\int_{0}^{t}\frac{d}{dx}\int_{x-ct+cs}^{x+ct-cs}f(y,s)\,dy\,ds\}=$$
$$=\frac{1}{2c}\int_{0}^{t}[f(x+ct-cs,s)-f(x-ct+cs,s)]ds+0\}$$
Finally
$$u_{xx}=\frac{1}{2c}\int_{0}^{t}[f_x(x+ct-cs,s)-f_x(x-ct+cs,s)]ds\}$$
Now just multiply $u_{tt}$ times $c^2$, substract this expression to $u_{tt}$ and the equality is proven. In case you don't understand some step just look at the mentioned formula and you will see that is the only thing i use:
$$\frac{d}{dx}\left[\int_{a\left(x\right)}^{b\left(x\right)}f\left(x,t\right)dt=f\left(x,b\left(x\right)\right)b'\left(x\right)-f\left(x,a\left(x\right)\right)a'\left(x\right)+\int_{a\left(x\right)}^{b\left(x\right)}\frac{\partial}{\partial x}f\left(x,t\right)\, dt\right]$$
