How do i find domain of influnce?

[gtfitzpatrick]

Homework Statement

determine the char. curve and gen. sol. for x$$^{3}$$ $\frac{\partial{u}}{\partial{x}}$ - $\frac{\partial{u}}{\partial{y}}$ = 0

Homework Equations

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

show sol is not defined when y> $$\frac{1}{2x^2}$$

The Attempt at a Solution

so $$\frac{\partial{y}}{\partial{x}} = \frac{-1}{x^3}$$

and $$\frac{\partial{u}}{\partial{x}} = 0$$

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



then sol. at u(x,0) = $$\frac{1}{1+x^2}$$

$$\frac{1}{1+x^2}$$ = $$F( \frac{1}{2x^2} - y)$$
= $$F( \frac{1}{2x^2})$$

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

 

[hunt_mat]

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

 

[gtfitzpatrick]

i don't understand what you are doing?

 

[gtfitzpatrick]

$$<br /> \dot{x}=x^{3},\quad\dot{y}=-1\quad\dot{u}=0<br />$$

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

but then...

 

[hunt_mat]

From my equations you have:
$$-\frac{1}{2x^{2}}=s+r,y=-s,u=\frac{1}{1+r^{2}}$$
so substitute to get u=u(x,y)

 

[gtfitzpatrick]

im getting


$$<br /> u(x,y)=\frac{1}{1+(y - \frac{1}{2x^2})^{2}}<br />$$

but I am still not sure about this, but this is the particular solution right?

 

[hunt_mat]

This is the correct solution, well done.

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

 

[gtfitzpatrick]

Thanks hunt_mat for your help, and patiance!

Im still not sure of a few things,
so the characteristics are $$<br /> \frac{dx}{dt} = \dot{x}=x^{3}, \frac{dy}{dt}=\dot{y}=-1, \frac{du}{dt}=\dot{u}=0<br />$$

the general solution is
$$u(x,y)=y - \frac{1}{2x^2}$$

and the particular solution is

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

but I am not sure to about the domain of influence or how to
show sol is not defined when y> $$\frac{1}{2x^2}$$

 

[gtfitzpatrick]

also in the book it gives the solution as u = $$\frac{1-2x^2y}{1+x^2-2x^2y}$$ which try as i might i cannot seem to make mine fit

 

[hunt_mat]

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?

 

[gtfitzpatrick]

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...

 

[gtfitzpatrick]

is the dom. of influ. just?

$$<br /> u(x+t,y+t)=\frac{1}{1+((y+t) - \frac{1}{2(x+t)^2})^{2}}<br />$$