1
$\begingroup$

So this PDE is a transport equation given as:

$u_x + xu_y = 0$ with initial condition that $u(x,0) = \exp(x)$.

My approach: Like usual I tried it to interpret this as a direction derivation, and found the curve $C_1$ (which guides me the direction to take for differenciation). Now since RHS is $0$, the value of $u$ of this curve should remain constant, but my calculation shows me the this curve intersect $x$-axis twice at different $x$ and so, the constant value should be $\exp(x_1)$ and $\exp(x_2)$ for $x_1 \neq x_2$. This appears to me as clear contradiction and hence I suspect the initial data. Could someone confirm my suspicion?

P.S If my explanation appears unclear, then try solving the above PDE by using characteristics method.

$\endgroup$
  • 1
    $\begingroup$ I didn't yet try to solve the PDE. But the problem, you describe can well happen. The solutions to a first order PDE are only defined locally (such that the characteristic lines are invertible). $\endgroup$ – Fabian Jan 30 '17 at 18:17
1
$\begingroup$

That is correct. Any solution must be constant on parabolas $x^2 - 2 y = $ constant, and thus the values at $(x,0)$ and $(-x,0)$ must be equal.

$\endgroup$
  • $\begingroup$ So that means that the PDE with the initial condition as given does not have a solution (I guess). $\endgroup$ – Fabian Jan 30 '17 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.