I have a problem with the following exercise (I know how to proof for $u_{tt}$ but for $u_t$ I met a problem, for $u_{tt}$ proof works as we can assume that $v(x,t,t)=0$ and we add it artificially to the equation, when here this trick doesn't work...)
Prove Duhamels principle:
for $s> 0$ let $v(x; t; s)$ be the solution of the following initial-boundary problem (which depends on the parameter s): $$ v_t - v_{xx} = 0; 0 < x < L; t > s;$$ $$ v(0; t; s) = v(L; t; s) = 0; t > s;$$ $$ v(x; s; s) = F(x; s); 0<x<L$$
Prove that the function $$ u(x; t) = \int_0^t v(x; t; s)ds$$ is a solution of the nonhomogeneous problem: $$u_t - u_{xx} = F(x; t); 0 < x < L; t > 0;$$ $$u(0; t) = u(L; t) = 0 t > 0;$$ $$u(x; 0) = 0; 0<x<L$$
Differentiating $u$ twice with respect to x we get
$$u_{xx}(x,t)= \int_{0}^{t}v_{xx}(x,t,s)ds.$$
Differentiating $u$ once with respect to t we get
$$u_{t}(x,t)= \int_{0}^{t}v_{t}(x,t,s)ds+v(x,t,t),$$
where the second term is a consequence of Leibniz's rule for differentiating an integral with variable limits.
Subtracting, we get
$$u_{t}(x,t)-u_{xx}(x,t)= \int_{0}^{t}[v_{t}(x,t,s)-v_{xx}(x,t,s)]ds+v(x,t,t).$$
and
$$u_{t}(x,t)-u_{xx}(x,t)= F(x,t),$$
because the integrand vanishes and $v$ has initial condition $v(x,t,t)=F(x,t)$.