Region of plane where the solution is uniquely determined

[roldy]

1. (a) Solve the equation yu_x+xu_y=0 with u(0,y)=e^-y2

(b) In which region of the xy-plane is the solution uniquely determined?

2. Homework Equations ...none



3. (a) The characteristic curves of a(x,y)u_x+b(x,y)u_y=0 are given by dy/dx=b(x,y)/a(x,y)

So according to this, I have dy/dx=x/y
or ydy=xdx

Integrating this I get...1/2y²+c₁=1/2x²+c₂

Multiplying by 1/2 and absorbing the two constants into one to solve for y yields...

y=+-(x²+c)^1/2

General Solution:

u(x,y)=f(y²-x²)

e^-y2=u(0,y)=f(y²-0²)=f(y²)
e^-y2=f(y²)

Let w=y²

y=w^1/2

So substitute back in
f(w)=e^-(w1/2)2=e^-w

Therefore...u(x,y)=e^-(y2-x2)=e^x2-y2

This is the correct answer

(b) This is where I get confused. How do I sketch the region on the xy-plane that show where the solution is uniquely determined? What does this look like?

 

[fzero]

The fact that $$u(x,y)$$ is even in $$x$$ and $$y$$ is important.

 

[roldy]

Yes, that makes sense. But I still can't picture the graph of this. I need to be able to show on a graph what this looks like.

 

[fzero]

It's easy to figure out what the 3d graph looks like by making 2d plots for fixed x or y. You can also get a 3d plot at wolframalpha.com.

 

[roldy]

Ok, I see the plot
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP38419e3dddc3c76gi4d000010dg3gf927260efa?MSPStoreType=image/gif&s=63&w=200&h=197

By observation I would have to say that the boundary bounded by y=x and y=-x would be the region where the solution is uniquely determined. Is this correct? Or am I missing something? Great site by the way. Thanks for that and your help.

 

[fzero]

I was missing something. The initial condition defines u on the y-axis, so the solution is uniquely determined everywhere the characteristic curves (contours) cross the y-axis. This seems to agree with your result.