I am given an inhomogeneous transport equation

$$ \partial_t u(t,x) + a(t,x)\cdot\nabla u(t,x)=b(t,x), $$

with $x\in\mathbb{R}^3$,

$\nabla u(t,x)=\sum_{i=1}^3 \partial_{x_1}u(t,x)$,

$a : [0,\infty)\times \mathbb{R}^3\to\mathbb{R}^3$,

$b : [0 \infty)\times\mathbb{R}^3\to\mathbb{R}$ and

$u : [0, \infty)\times\mathbb{R}^3\to\mathbb{R}$, $u(0,x)=u_0(x)$.

and also given a solution $\chi: [0,\infty)\times\mathbb{R}^3\to\mathbb{R}^3$ to the characteristic equation

$$ \partial_t\chi(t,y)=a(t,\chi(t,y)), $$

with $\chi(0,y)=y$.

I then have to show that

$$ \partial_t u(t,\chi(t,y)) = b(t,\chi(t,y)),\quad u(t,\chi(t,y))=u_0(y), $$

whenever $u$ is a solution to the transport equation.

Does this mean that I need to prove that $a(t,x)\cdot \nabla u(t,x) = 0$ for $x=\chi(t, y)$, and thus all $y$?

My work:

We can expand the transport equation to

$$ u_t + a_1 u_{x_1} + a_2 u_{x_2} + a_3 u_{x_3} = b. $$

I think this is a linear first order PDE, and by this (Solution transport equation) answer we get

$$ \frac{dt}{1} = \frac{dx_1}{a_1(t,x)} = \frac{dx_2}{a_2(t,x)} = \frac{dx_3}{a_3(t,x)} = \frac{dw}{b(t,x)}, $$

though my case is different, since my $a$ depends on $x$ and $t$.

From here I do not understand how this ties into the characteristic equation given.

I wanted to follow the Wikipedia example on Method of characteristics, but I immediately don't understand how to link my problem to it. There they're using that the variables $x$ and $t$ on which $u$ depends, only depend on one variable $s$. But we have that $\chi(t,y)$ depends on two variables, so I cannot do something like $\frac{d}{ds} u(t,\chi(t, y))$, because I don't see how $t$ and $\chi(t,y)$ only depend on one other variable.


1 Answer 1


I assume that by $\partial u(t,\chi(t,y))$ you mean $\partial_{\color{red}t} \color{red}(u(t,\chi(t,y))\color{red})$ . Then its nothing but the chain rule, $$ \partial_{\color{red}t} \color{red}(u(t,\chi(t,y))\color{red}) = (\partial_t u)(t,\chi(t,y)) + \partial_t\chi(t,y)\cdot\nabla u(t,\chi(t,y))$$ plug what $\partial_t\chi(t,y)$ is equal to from the characteristics ODE, $$\partial_t\chi(t,y)=a(t,\chi(t,y)),$$ and use the equation for $u(t,\bullet)$. Done- $$\partial_t[ u (t,\chi(t,y)) ] \overset{\text{above & $\chi$ ODE}}{=} [\partial_t u + a\cdot \nabla u](t,\chi(t,y)) \overset{u \text{ PDE}}= b(t,\chi(t,y)) $$

  • $\begingroup$ Thanks, that seems to do it. How do you see the difference between when you take $\partial_t$ with respect to the function only, as in the equation at the top of my question, and $\partial_t$ using the chain rule, as in your answer? $\endgroup$ Commented Jan 15, 2020 at 18:22
  • 1
    $\begingroup$ @TheCodingWombat I think that's largely a notation problem, its something like $$ \frac{d}{dx}[ f(2x)] = 2\frac{df}{dx}(2x)$$ here $d/dx$ is an operator acting on a function $x\mapsto f(2x)$, and $df/dx$ is a function, which is evaluated at $2x$. $\endgroup$ Commented Jan 16, 2020 at 2:17

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