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Example of the method of characteristics

  1. May 10, 2014 #1
    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[itex]\in[/itex]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, =[itex]\eta[/itex]
    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 [itex]\eta[/itex] and s.

    The problem with respect to new variables:
    u,when s=0, =sin([itex]\eta[/itex])
    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([itex]\eta[/itex]) 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. jcsd
  3. May 10, 2014 #2

    LCKurtz

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    Science Advisor
    Homework Helper
    Gold Member

    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
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