Q:

Partial DE Question

I have a quick question. 

Question 1). Hence deduce that
$$
u(x,t)=F(x+at)+G(x-at)
$$
satisfies the original PDE for any twice-differentiable functions $F$ and $G$.  
c) Hence find the solution of the PDE which satisfies the initial conditions
$$
u(x,0)=e^{-x}\quad\text{and}\quad\dfrac{\partial u}{\partial t}(x,0)=0\quad,
$$
d) If the initial conditions are $~u(x,0)=r(x);\,\,\dfrac{\partial u}{\partial t}(x,0)=s(x)~$ for some functions $r(x)$ and $s(x)$,
show that
$$
F(x)=\tfrac12r(x)+\tfrac1{2a}\Big(\int\limits_0^xs(w)\,\mathrm dw\Big)+k
\quad\text{and}\quad
G(x)=\tfrac12r(x)-\tfrac1{2a}\Big(\int\limits_0^xs(w)\,\mathrm dw\Big)-k
$$
for some arbitrary constant $k$

On part d, there's a bit I'm not too sure of. My workings are: 
$$u(x,t) = F(x+at) + G(x-at)$$
$$u(x,0) = F(x) + G(x) = r(x)$$
$$u_t(x,0) = a(F'(x) - G'(x)) = s(x) \Rightarrow F(x) - G(x) =\frac {1}{a} \int s(x) dx + k$$
From there I can use simulataneous equation and solve etc. But in the solution they have a change of variable to change the integral from $s(x)dx$ to $ \int_0^x s(w) dw$. I'm not sure why you can do this. Thanks!

A:

Your formula $F(x) - G(x) =\frac {1}{a} \int s(x) dx + k$ isn't quite right. On the left, $x$ is an actual variable that is plugged into the function. On the right it's the dummy variable of integration. Mixing the two, as in 
$$F'(x)=C+\int_0^x F'(x)\,dx$$
is to be avoided. "Integrate with respect to $x$ from $x=0$ to $x=x$?"
The correct notational approach is to proceed from
$$ F'(x) - G'(x) = \frac{1}{a} s(x)$$
to $$F(x)-G(x)=k+\int_0^x(F'(w)-G'(w))\,dw = k+\frac{1}{a} \int_0^xs(w)\,dw$$
and then solve for $F$ and $G$ algebraically. 

