Showing solution satisfies PDE (chain rule help)

Question

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 $$\partial_t u + a\cdot Du = - a \cdot \nabla g(x-at) +f(x,t) + \int_0^t \nabla f(x+(s-t)a, s) \cdot (-a) ds \\ + a \cdot \nabla g(x-at) + a\cdot \int_0^t \nabla f(x+(s-t)a, s) ds$$ $$= f(x, t)$$ 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.

Answer 1

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...

Answer 2

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)\Big\vert_{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.

Answer 3

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

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

$\frac{\partial g}{\partial t}=\frac{\partial g}{\partial X}\frac{\partial X}{\partial t}=-a\frac{\partial g}{\partial X}$

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

$\frac{\partial f}{\partial t}=\frac{\partial f}{\partial \Psi}\frac{\partial \Psi}{\partial t}=-a\frac{\partial f}{\partial \Psi}$

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

$\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)-a\int_0^t \frac{\partial f}{\partial \Psi}ds $

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

$\Psi=x+a(s-t)=x+a(t-t)=x\quad$ and $\quad\Phi=s=t$ .

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

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

$$\frac{\partial u }{\partial t}+a\frac{\partial u }{\partial x}= \left(-a\frac{\partial g}{\partial X} +f(x,t)-a\int_0^t \frac{\partial f}{\partial \Psi}ds \right) + a\left( \frac{\partial g}{\partial X}+\int_0^t \frac{\partial f}{\partial \Psi}ds\right)$$ After simplification : $\quad\frac{\partial u }{\partial t}+a\frac{\partial u }{\partial x}=f(x,t)$

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