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Homework Help: PDE question

  1. Aug 27, 2012 #1

    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


    dy/dt = 1 => y=t+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

    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


    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
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