# Using characteristics to transform a transport PDE

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.

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- $$\partial_t[ u (t,\chi(t,y)) ] \overset{\text{above & \chi ODE}}{=} [\partial_t u + a\cdot \nabla u](t,\chi(t,y)) \overset{u \text{ PDE}}= b(t,\chi(t,y))$$
• 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? Jan 15, 2020 at 18:22
• @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 $2x$. Jan 16, 2020 at 2:17