Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Find the general solution of the PDE: $$ xu_x-xyu_y -u=0 $$

I have found it to be: $u(x,y)=-xf(ye^x)$

This PDE has the property that $u(0,y)=0$. Therefore, $u(0,y)$ cannot be arbitrarily prescribed, even though the $y$-axis crosses each characteristic curve only once.

How do I explain this apparent discrepancy by giving the characteristic curves their preferred parametrizations $(x(t),y(t))$ with

$$ x'(t)=x(t)\quad \mbox{ and } \ y'(t)=-x(t)y(t) $$

Any help will be appreciated, thank you.

share|improve this question
    
I don't quite understand what you're trying to explain. Can you clarify? –  Tunococ Jan 25 '13 at 3:40
add comment

1 Answer

The problem is already apparent in the equation: $xu_x-xyu_y -u=0$: when $x=0$, both coefficients of derivatives vanish, the PDE drops in order from $1$st to $0$th (which is not even a PDE at all). Whenever you see a differential equation changing order like that, expect weirdness in solutions.

As a model, consider the ODE $xu'-u=0$ where $u$ is an unknown function of $x$. The solutions are $u(x)=cx$, for any constant $c$. They all vanish at $x=0$. The theorems about existence and uniqueness for initial value problems do not apply at $x=0$, because they are for equations of the form $u'=F(x,u)$. Here $F(x,u)=u/x$ which is not continuous at $0$.

This is what happens when you work along the characteristics of your PDE. They all get to $x=0$, but since the PDE is degenerate there, one can't solve the initial value problem. What to do: prescribe the values of $u$ somewhere else.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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