How Do You Solve the Method of Characteristics for Nonlinear PDEs?

[gtfitzpatrick]

Homework Statement

x \frac{ \partial u}{ \partial x} + y \frac{ \partial u}{ \partial y}= -x^2u^2

Homework Equations

The Attempt at a Solution

characteristics are given by
\frac{ dy}{ dx} = \frac{ y}{ x} (a)

and

\frac{ du}{ dx} = -\frac{x^2u^2 }{ x} (b)

So i integrate both equations

but for (a) do i bring the y across which ends up giving ln(y) = ln(x) + k
or leave it where it is and i get y = -yln(x) + k

??

 

[hunt_mat]

Hello, still doing characteristics? Why not write the characteristic equations as:
<br /> \dot{x}=x,\quad\dot{y}=y,\quad\dot{u}=-x^{2}u^{2}<br />
You won't be able to write down a complete solution as you have no initial condition to work from. For your equation for characteristics, your first answer was correct, the characteristics are given by y=kx for k constant.

 

[nickalh]

To expand on hunt_mat's answer:
Don't we need to separate the variables before integrating?

1/y dy = 1/x dxHmmm, then again you're using partial deriv. notation, which I've only studied Calc. for, not yet Differential Equations.

 

[gtfitzpatrick]

thanks for the replies lads. I guess i never total got to grips with methods of chars.!
I also have initial conditions of u(x,1) = x for -infinity<x<infinity.

y=kx becomes k=y/x

then from (b) we have \frac{ du}{ dx} = -\frac{x^2u^2 }{ x} which when differentiated gives u = \frac{ 2}{ x^2} + F(k)

and transfer in our earlier value of k gives u = \frac{ 2}{ x^2} + F(\frac{ y}{ x})
im getting confused now i think in different methods?

 

[hunt_mat]

There are essentially two ways for the method of characteristics (I recognise your name from a number of MOC posts, did I post my notes on the subject?)
From one of your calculations you have:
<br /> \frac{du}{dx}=-xu^{2}<br />
Integrating this equation shows that:
<br /> \frac{1}{u}=\frac{x^{2}}{2}+F(\xi )<br />
We now paramatrise the initial data, so take (\xi ,1) as the point which the characteristic passes through, this will give the initial values as u(\xi ,1)=\xi, evaluating the characteristic at this point yields 1=k\xi, giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate u at the point (\xi ,1) to obtain
<br /> \frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )<br />
From this you can compute F(\xi ) and then from there you can substitute for \xi by using the equation of the characteristic.
Now to find u you have the solution