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A Conceptual Solution of a First Order PDE

  1. May 13, 2017 #1
    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. jcsd
  3. May 13, 2017 #2
    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
  4. May 13, 2017 #3
    Also it can also be said that ##g^{-t}(x)## is the flow generated by ##-a(x)##, correct?
     
  5. May 13, 2017 #4
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