Unique solution for PDE?

Question

How can one tell if a solution is existent or unique? For example:

$yu_y+uu_x=u-y$

$u(x,1)=x$

I've found the solution to be $u=x+1-y$, but have been told there are infinitely many solutions.

Is there a condition that must be satisfied for a unique solution?

For nonlinear equations this depends on the initial condition. Draw the characteristics. As long as they do not cross and leave no region uncovered, you are safe in the class of analytical solutions. As to further theory, you can look up something on the Burgers equation. — uvs, Oct 28 '14 at 13:12
Interesting, ok. Didn't know that was all there is to it. Thanks! — bfletch, Oct 30 '14 at 13:31

doraemonpaul · Accepted Answer · 2014-11-01T00:01:03.620

$\dfrac{dy}{dt}=y$ , letting $y(0)=1$ , we have $y=e^t$

$\dfrac{du}{dt}=u-y=u-e^t$ , letting $u(0)=u_0$ , we have $u=(u_0-t)e^t=(u_0-\ln y)y$

$\dfrac{dx}{dt}=u=(u_0-t)e^t$ , we have $x=(u_0+1-t)e^t+f(u_0)=u+y+f\left(\dfrac{u}{y}+\ln y\right)$ , i.e. $\dfrac{u}{y}+\ln y=F(x-y-u)$

Consider the general solution $x=u+y+f\left(\dfrac{u}{y}+\ln y\right)$ :

$u(x,1)=x$ :

$f(x)=-1$

$\therefore x=u+y-1$ , i.e. $u=x-y+1$

But consider the general solution $\dfrac{u}{y}+\ln y=F(x-y-u)$ :

$u(x,1)=x$ :

$F(-1)=x$ , which is impossible.

$\therefore$ There is no solution.

Hence the PDE has the unique solution $u=x-y+1$ .

1 Answers1