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Region of plane where the solution is uniquely determined

  • Thread starter roldy
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  • #1
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1. (a) Solve the equation yux+xuy=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)ux+b(x,y)uy=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=w1/2

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

Therefore...u(x,y)=e-(y2-x2)=ex2-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?

 

Answers and Replies

  • #2
fzero
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The fact that [tex]u(x,y)[/tex] is even in [tex]x[/tex] and [tex]y[/tex] is important.
 
  • #3
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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.
 
  • #4
fzero
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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.
 
  • #5
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Ok, I see the plot
[PLAIN]http://www4c.wolframalpha.com/Calculate/MSP/MSP38419e3dddc3c76gi4d000010dg3gf927260efa?MSPStoreType=image/gif&s=63&w=200&h=197 [Broken]

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.
 
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  • #6
fzero
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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.
 

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