# Homework Help: Example of the method of characteristics

1. May 10, 2014

### prehisto

1. The problem statement, all variables and given/known data
Hi,guys I have a example, i understand almost everything but i have problems understanding some steps.
Example:
2u't+3u'x=0,x$\in$R,t>0,u(x,0)=sin(x),u=u(x,y)

3. The attempt at a solution
I rewrite the example
(1)2u't+3u'x=u's
(2)u'tt's+u'xx's=u's

From comparing (1) (2) we can obtain
(3)t's=2
(4)x's=3

x=3s+C
t=2s+A
I suppose these equations are obtained by just integrating (3) and (4)

And here is the confusing par for me:
x,when s=0, =$\eta$
t,when s=0, =0
I do not understand how these conditions are made.

Using these conditions we can calculate constants C,A.
Then express new variables $\eta$ and s.

The problem with respect to new variables:
u,when s=0, =sin($\eta$)
u's=0

How I sad, i do not understand how conditions (3) and (4) are made and then the "new problem":
u,when s=0, =sin($\eta$) is confusing ,it seems that s=t.
Could someone ,please,can explain this to me?
It would help a lot.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. May 10, 2014

### LCKurtz

Maybe this will be of some help, maybe not, but...

You have 3 variables $t,x,y$ on this line. And strange notation like $u'_x$. So I am going to rewrite it in a form that makes sense to me:$$2u_x(x,y) + 3u_y(x,y) = 0,~u(x,0) = \sin x$$

I guess, from what you wrote, that you are looking for a curve $\vec R(s) = \langle x(s),y(s)\rangle$ such that $u(s) = u(x(s),y(s))$ is a solution. So you want$$u'(s) = \nabla u \cdot \frac{d\vec R}{ds}=u_x x'(s) + u_y y'(s) = 3u_x + 2u_y = 0$$

From here I will veer away from what you did. With $x'(s)=3, y'(s)=2$ you have the slope of the curve is $\frac {dy}{dx}= \frac{y'(s)}{x'(s)}=\frac 2 3$. This says $u(s)$ is constant on lines with slope $\frac 2 3$. So the value of $u(x,y)$ at any point in the plane is the value of $u$ where the line through $(x,y)$ with slope $\frac 2 3$ intercepts the $x$ axis. See if you can show that point would be $(x-\frac 3 2 y, 0)$. And the value of $u(x,y)$ at points on the $x$ axis is given.

Hopefully this may be of some help to you. Disclaimer: I am not a PDE guy.

Last edited: May 10, 2014