A Conceptual Solution of a First Order PDE

1. May 13, 2017

Gallo

Hello I would like to check my reasoning about solutions of first order PDE. I've spell out (almost) all details.

I'll consider the following problem: find $u=u(t,x)$ s.t. :
$$\partial_t u(t,x) + a(x) \cdot \nabla u(x) =0 \qquad \qquad u(0,x) = u_0(x)$$
say with smooth coefficient and initial condition (I'm not too much worried about local existence etc..).

Next I consider the following ordinary differential equation:
$$\dot{X}(s) = a(X(s)) \qquad X(0) = y$$
whose solution $X(s,p)$ satisfies:
$$X(t+s,y) = X(t,X(s,y))$$
and $X(0,y)=y$.

Then If I define $w(s) = u(s,X(s))$, it must be:
$$\frac{d w(s)}{d s} = \partial_t u(s,X(s)) + a(X(s)) \cdot \nabla u(X(s)) = 0$$
that is on the curves $X(s)$:
$$u(t,X(t,y)) = w(t) = w(0) = u(0,X(0,y)) = u_0(y) .$$

Now given $x$ I need to find $y$ such that $x = X(t,y)$. If I consider $X(-t,x)$, then:
$$X(-t,x) = X(-t,X(t,y)) = X(t-t,y) = y$$
This means that $u$ is given by:
$$u(t,x) = u_0(X(-t,x))$$

Moreover if I define $\tilde{X}(s,y) = X(-s,y)$, then it solves:
$$\dot{\tilde{X}}(s) = - a(\tilde{X}(s)) \qquad \tilde{X}(0) = y$$
so that the final form of the solution can also be given by
$$u(t,x) = u_0(\tilde{X}(t,x))$$

Is this correct?
Thank you

2. May 13, 2017

zwierz

It looks correct. Indeed if $g^t(x)$ is a flow generated by the vector field $a(x)$ then a function $u(t,x)=u_0(g^{-t}(x))$ is a solution to the following Cauchy problem
$$u_t+a^i\frac{\partial u}{\partial x^i}=0,\quad u\mid_{t=0}=u_0(x).$$
To obtain this fact it is sufficient to differentiate the equality
$$u(t,g^t(y))=u_0(y)$$ in time and then put $y=g^{-t}(x)$

$u(t,x)$ is a first integral to the system $\dot x=a(x)$

Last edited: May 13, 2017
3. May 13, 2017

Gallo

Also it can also be said that $g^{-t}(x)$ is the flow generated by $-a(x)$, correct?

4. May 13, 2017

yes