1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

PDE question

  1. Aug 27, 2012 #1
    Hey,

    I'm trying to solve the following pde,

    u(x,y) u_x + u_y =0 with u(x,0) = p(x) for some known p(x)

    where u_x defines the partial derivative of u(x,y) wrt x

    after finding the characteristic curves and the first integrals i get the general solution is

    F(x^2 - zy^2, z) = 0

    (note z=u(x,y))

    At this point i'm not sure what to do next,

    usually you can rewrite this as

    f(x^2-zy^2) + z =0, however as z is contained in the argument you cant solve this for z without knowing what f is.

    One thing i was thinking was just to make up some F and continue on, but that does not seem correct,

    can anyone push me in the right direction?
     
  2. jcsd
  3. Aug 27, 2012 #2
    So what are the characteristic curves and how did you get them?
     
  4. Aug 28, 2012 #3
    dx/dt = u = z => x=zt + a

    t=(x-a)/c

    dy/dt = 1 => y=t+b

    t=y-b

    dz/dt = 0 => z=c


    then dx/dt - z dy/dt = 0 => 1/2 d/dt(x^2 + zy^2) = 0

    so x^2 + zy^2 = k


    unless i made a dumb error, i thought it was pretty straight forward so left it out
     
  5. Aug 28, 2012 #4
    I get dx/dy = u and du/dy = 0, which together give u = const and x - uy = const, so u(x, y) = F(x - uy).

    I don't understand how you end up with the squares.
     
  6. Aug 28, 2012 #5
    oh ok,

    the way they taught our class was what i posted above

    dx/dt = u = z

    dy/dt = 1

    dz/dt = 0 => z=c


    then dx/dt - z dy/dt = 0 => 1/2 d/dt(x^2 + zy^2) = 0

    since dx/dt - z dy/dt= 1/2 d/dt(x^2 + zy^2)

    so d/dt(x^2 + zy^2) = 0

    integrating both sides gives x^2 + zy^2 = k

    but yea there's lots of possibly first integrals,

    the problem im having is the next step,

    so using what you got

    the general solution would be

    F(x-uy, u) = 0

    how can i solve this for u?

    this is like

    h(x-uy) + g(u) = 0 or something right?

    im having trouble working out how to seperate u from everything else without having to define F

    is it possible?
     
  7. Aug 28, 2012 #6
    I am afraid in this equation it is possible only if u(x, 0) = p(x) is nice.
     
  8. Aug 28, 2012 #7
    Oh,

    Sorry I didn't mention it states in the question that p(x) is a smooth function,

    Does this mean that you apply the initial condition to F(....)=0?

    So for x=x y=0 you get


    F(x,p(x)) = 0

    f(x) + p(x) = 0

    f(x) = -p(x)

    f(x-uy) = -p(x-uy)

    or something like that? i still see myself getting stuck alone the line
     
  9. Aug 28, 2012 #8
    If u(x, 0) = p(x), then what happens with u(x, y) = F(x - uy) when y = 0? You get p(x) = F(x), so u(x, y) = p(x - uy).

    For some particular p(x), you might be able to express u explicitly as a function of (x, y), but I do not see what can be done further in the general case. I may be missing something, though.
     
  10. Aug 29, 2012 #9
    how come you write it as F(x-uy) and not F(x-uy,u) like i did below?

    F(x-uy, u)=0

    f(x-uy) = -u

    at y=0

    f(x)=-p(x)

    i guess you can absorb the negative into p(x)

    f(x-uy) = p(x-uy)

    i guess one important question that im not sure about is,

    what is the significance of saying that p(x) is smooth?
     
  11. Aug 29, 2012 #10
    That's just differences in notation. We have two integrals x - uy = const and u = const, so we can immediately conclude that u = F(x - uy). Put another way, for any quasilinear homogeneous equation we know that u = const is an integral, so we know in advance that u is a function of the other integral, so we can skip the more general form.

    Because u = p(x - uy), p has to have a continuous derivative. This is guaranteed if it is smooth.
     
  12. Aug 29, 2012 #11
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: PDE question
  1. PDE question (Replies: 1)

  2. PDE questions (Replies: 34)

  3. PDE Question (Replies: 24)

  4. PDE Question (Replies: 3)

  5. PDE question. (Replies: 2)

Loading...