Dtyjlui

We have $$u(x, t) = g(x-at, 0) + int_0^t f(x+a(s-t), s)ds$$ and want to show it satisfies the PDE $$partial_t u + a cdot Du = f$$

The solution given goes as follows $$partial_t u + acdot Du = - a cdot nabla g(x-at) +f(x,t) + int_0^t nabla f(x+(s-t)a, s) cdot (-a) ds \ + a cdot nabla g(x-at) + acdot int_0^t nabla f(x+(s-t)a, s) ds$$ $$= f(x, t)$$ as required

However, I have no idea how these terms have appeared. In particular, I'm not sure why we have a grad appearing for $g$ and the $f(x, t)$ seems to come from nowhere...

Any insight is appreciated.

The first three terms arise from the chain rule to calculate $u_t$. First, you dot-multiply the gradient of $g$ by the derivative of the function inside $g$ with respect to $t$, which is the vector $-a$, next you have to take the $t$-derivative of an integral that contains $t$ both in the limit of integration and in the integrand. For that you have to combine the fundamental theorem, setting $s=t$ in the integrand, and a "convective" term including the $t$ derivative of the integrand $-a$. The last two terms arise in a similar way, but now to take the $x$ - gradient of the integral with fixed limits you just take it on the integrand.

It is just too long to write down all details...

Hint: Use Leibniz's integral rule to show that

begin{align}
frac{partial}{partial t}int_0^t f(x+a(s-t),s) ds &= f(x+a(s-t),s)Bigvert_{s=t} - int_0^t frac{partial}{partial t}f(x+a(s-t),s) ds \
&= f(x,t) - int_0^t (-a)f'(x+a(s-t),s) ds
end{align}

The gradient here just means the first derivative of a single variable, i.e.

$$ frac{partial}{partial t} g(x-at) = -ag'(x-at) $$

Try it your self with the remaining derivatives.

Consider the function :
$$u(x, t) = g(X, Y) + int_0^t f(Psi, Phi)ds$$
where $$begin{cases}
X=x-at & frac{partial X}{partial x}=1 quad & frac{partial X}{partial t}=-a\
Y=0 \
Psi=x+a(s-t) & frac{partial Psi}{partial x}=1quad & frac{partialPsi}{partial t}=-a\
Phi=s end{cases}$$

$frac{partial g}{partial x}=frac{partial g}{partial X}frac{partial X}{partial x}=frac{partial g}{partial X}$

$frac{partial g}{partial t}=frac{partial g}{partial X}frac{partial X}{partial t}=-afrac{partial g}{partial X}$

$frac{partial f}{partial x}=frac{partial f}{partial Psi}frac{partial Psi}{partial x}=frac{partial f}{partial Psi}$

$frac{partial f}{partial t}=frac{partial f}{partial Psi}frac{partial Psi}{partial t}=-afrac{partial f}{partial Psi}$

$frac{partial }{partial x}int_0^t f(Psi, Phi)ds =int_0^t frac{partial f}{partial x}ds =int_0^t frac{partial f}{partial psi}ds$

$frac{partial }{partial t}int_0^t f(Psi,Phi)ds = left(f(Psi,Phi)right)_{s=t}+int_0^t frac{partial f}{partial t}ds = f(x,t)-aint_0^t frac{partial f}{partial Psi}ds $

Because $f(Psi,Phi)=f(x,t)$ for $s=t$ :

$Psi=x+a(s-t)=x+a(t-t)=xquad$ and $quadPhi=s=t$ .

$$frac{partial u }{partial x}=frac{partial g }{partial x}+frac{partial }{partial x}int_0^t f(Psi, Phi)ds= frac{partial g}{partial X}+int_0^t frac{partial f}{partial Psi}ds$$

$$frac{partial u }{partial t}=frac{partial g }{partial t}+frac{partial }{partial t}int_0^t f(Psi, Phi)ds=-afrac{partial g}{partial X} +f(x,t)-aint_0^t frac{partial f}{partial Psi}ds $$

$$frac{partial u }{partial t}+afrac{partial u }{partial x}=
left(-afrac{partial g}{partial X} +f(x,t)-aint_0^t frac{partial f}{partial Psi}ds right) +
aleft( frac{partial g}{partial X}+int_0^t frac{partial f}{partial Psi}dsright)$$
After simplification :
$quadfrac{partial u }{partial t}+afrac{partial u }{partial x}=f(x,t)$

Thus the function $quad u(x, t) = g(x-at, 0) + int_0^t f(x+a(s-t), s)dsquad$
satisfies the PDE $quad partial_t u + a: partial_x u = f$ .