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How do i find domain of influnce?

  • #1

Homework Statement



determine the char. curve and gen. sol. for x[tex]^{3}[/tex] [itex]
\frac{\partial{u}}{\partial{x}}
[/itex] - [itex]
\frac{\partial{u}}{\partial{y}}
[/itex] = 0

Homework Equations



find sol and domain of influence when u(x,0) = [tex] \frac{1}{1+x^2} [/tex]

show sol is not defined when y> [tex] \frac{1}{2x^2} [/tex]


The Attempt at a Solution



so [tex] \frac{\partial{y}}{\partial{x}} = \frac{-1}{x^3} [/tex]

and [tex] \frac{\partial{u}}{\partial{x}} = 0 [/tex]

which gives u(x,y) = [tex]F( \frac{1}{2x^2} - y) [/tex] is the gen solution right?



then sol. at u(x,0) = [tex] \frac{1}{1+x^2} [/tex]

[tex] \frac{1}{1+x^2} [/tex] = [tex]F( \frac{1}{2x^2} - y) [/tex]
= [tex]F( \frac{1}{2x^2}) [/tex]

which gives x = +/- 1 am i right in this???? and how do i find domain of influnce?
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,739
18


Much better way to to set:
[tex]
\dot{x}=x^{3},\quad\dot{y}=-1\quad\dot{u}=0
[/tex]
with the initial condition:
[tex]
x(0)=r\quad u(0)=\frac{1}{1+r^{2}}
[/tex]
 
  • #3


i dont understand what you are doing?
 
  • #4


[tex]

\dot{x}=x^{3},\quad\dot{y}=-1\quad\dot{u}=0

[/tex]

Is that not what i did, then solved [tex]
\frac{\dot{y}}{\dot{x}} = \frac{-1}{x^3}
[/tex]

but then...
 
  • #5
hunt_mat
Homework Helper
1,739
18


From my equations you have:
[tex]
-\frac{1}{2x^{2}}=s+r,y=-s,u=\frac{1}{1+r^{2}}
[/tex]
so substitute to get u=u(x,y)
 
  • #6


im getting


[tex]

u(x,y)=\frac{1}{1+(y - \frac{1}{2x^2})^{2}}

[/tex]

but im still not sure about this, but this is the particular solution right?
 
  • #7
hunt_mat
Homework Helper
1,739
18


This is the correct solution, well done.

what happens in first order PDE is that your initial condition is propagated along the characteristics.
 
  • #8


Thanks hunt_mat for your help, and patiance!

Im still not sure of a few things,
so the characteristics are [tex]

\frac{dx}{dt} = \dot{x}=x^{3}, \frac{dy}{dt}=\dot{y}=-1, \frac{du}{dt}=\dot{u}=0

[/tex]

the general solution is
[tex]
u(x,y)=y - \frac{1}{2x^2}
[/tex]

and the particular solution is

[tex]
u(x,y)=\frac{1}{1+(y - \frac{1}{2x^2})^{2}}
[/tex]

but im not sure to about the domain of influence or how to
show sol is not defined when y> [tex]
\frac{1}{2x^2}
[/tex]
 
  • #9


also in the book it gives the solution as u = [tex] \frac{1-2x^2y}{1+x^2-2x^2y} [/tex] which try as i might i cannot seem to make mine fit
 
  • #10
hunt_mat
Homework Helper
1,739
18


Hi, I wrote some notes on the method of characteristics which I have included in this post. It is possible that the book you have is giving you trhe wrong answer, this happens sometimes. I just checked the answer with mathematica and you have the right answer.

What d you understand about domain of dependance?
 

Attachments

  • #11


thanks Mat for the notes, there is a lot of stuff in there that our lecturer hasn't covered yet. I didnt see anything in there about the domain of dependance, and i have nothing in my notes on it either and the more i try to find about it on www the more confused i get....
 
  • #12


is the dom. of influ. just?

[tex]

u(x+t,y+t)=\frac{1}{1+((y+t) - \frac{1}{2(x+t)^2})^{2}}

[/tex]
 

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