Hello I would like to check my reasoning about solutions of first order PDE. I've spell out (almost) all details.(adsbygoogle = window.adsbygoogle || []).push({});

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

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# A Conceptual Solution of a First Order PDE

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