Register to reply

Method of characteristics

by gtfitzpatrick
Tags: characteristics, method
Share this thread:
gtfitzpatrick
#1
Jul21-11, 01:22 PM
P: 376
1. The problem statement, all variables and given/known data

[itex] x \frac{ \partial u}{ \partial x} + y \frac{ \partial u}{ \partial y}= -x^2u^2[/itex]

2. Relevant equations



3. The attempt at a solution

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

and

[itex] \frac{ du}{ dx} = -\frac{x^2u^2 }{ x} [/itex] (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

??
Phys.Org News Partner Science news on Phys.org
Security CTO to detail Android Fake ID flaw at Black Hat
Huge waves measured for first time in Arctic Ocean
Mysterious molecules in space
hunt_mat
#2
Jul21-11, 05:01 PM
HW Helper
P: 1,583
Hello, still doing characteristics? Why not write the characteristic equations as:
[tex]
\dot{x}=x,\quad\dot{y}=y,\quad\dot{u}=-x^{2}u^{2}
[/tex]
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 [itex]y=kx[/itex] for k constant.
nickalh
#3
Jul22-11, 12:51 AM
P: 72
To expand on hunt_mat's answer:
Don't we need to separate the variables before integrating?

1/y dy = 1/x dx


Hmmm, then again you're using partial deriv. notation, which I've only studied Calc. for, not yet Differential Equations.

gtfitzpatrick
#4
Jul29-11, 08:07 AM
P: 376
Method of characteristics

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 [itex] \frac{ du}{ dx} = -\frac{x^2u^2 }{ x}[/itex] which when differentiated gives [itex] u = \frac{ 2}{ x^2} + F(k) [/itex]

and transfer in our earlier value of k gives [itex] u = \frac{ 2}{ x^2} + F(\frac{ y}{ x}) [/itex]
im getting confused now i think in different methods?
hunt_mat
#5
Jul29-11, 01:21 PM
HW Helper
P: 1,583
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:
[tex]
\frac{du}{dx}=-xu^{2}
[/tex]
Integrating this equation shows that:
[tex]
\frac{1}{u}=\frac{x^{2}}{2}+F(\xi )
[/tex]
We now paramatrise the initial data, so take [itex](\xi ,1)[/itex] as the point which the characteristic passes through, this will give the initial values as [itex]u(\xi ,1)=\xi[/itex], evaluating the characteristic at this point yields [itex]1=k\xi[/itex], giving a value for k which can then be inserted back into the equation for the characteristic. We now evaluate [itex]u[/itex] at the point [itex](\xi ,1)[/itex] to obtain
[tex]
\frac{1}{\xi}=\frac{\xi^{2}}{2}+F(\xi )
[/tex]
From this you can compute [itex]F(\xi )[/itex] and then from there you can substitute for [itex]\xi[/itex] by using the equation of the characteristic.
Now to find u you have the solution


Register to reply

Related Discussions
Method of characteristics Calculus & Beyond Homework 1
Method of characteristics Calculus & Beyond Homework 0
Method of Characteristics Calculus & Beyond Homework 0
Method of characteristics Calculus & Beyond Homework 0
Method of characteristics Differential Equations 0