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Homework Help: Help with PDE please

  1. Jan 10, 2006 #1
    Hi,
    I'm working through 'Partial Differential Equations, an introduction' by Colton and am not finding it as clear as I hoped to.
    I'm working through an example on how to solve a linear 1st order PDE.
    I'll post Colton's example and Italic my questions:
    Find the GS of

    [tex]xu_x-yu_y+u=x[/tex]

    The characteristic equation is

    [tex]\frac{dy}{dx}=-\frac{y}{x}[/tex]

    Integrating gives:

    [tex]logy=-logx+c[/tex]

    or

    [tex]xy=\gamma[/tex]

    Hence setting:

    [tex]\zeta=x , \\

    \eta=xy[/tex]

    in our first order PDE yields:

    [tex]\frac{\partial w}{\partial \zeta}+\frac{1}{\zeta}w=1[/tex]

    Ok, my first question, why set [tex]\zeta=x , \eta=xy[/tex]

    I don't really see how this relates to the original equation, and also eta doesn't seem to 'do' anything, why not set it to y, or xy^2 etx...i don't follow the logic


    ctd...whose solution is:

    [tex]w(\zeta,\eta)=\frac{\zeta}{2}+\frac{1}{\zeta}d(\eta)[/tex]

    next question, this only works for [tex]d(\eta)=0[/tex] so what is the purpose of introducing it?

    If anyone can help I'd be very grateful.
    Rich
     
    Last edited: Jan 10, 2006
  2. jcsd
  3. Jan 10, 2006 #2

    saltydog

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    Your solution contains an arbitrary function of xy. That's the d(xy) part or to avoid confusion with the d standing for derivatives, just call it K(xy) so that the solution in terms of x and y is:

    [tex]u(x,y)=\frac{x}{2}+\frac{K(xy)}{x}[/tex]

    Try back-substituting this into the PDE and use for example the cases:

    [tex]K(xy)=xy[/tex]

    [tex]K(xy)=Sin[xy][/tex]

    [tex]K(xy)=xy+(xy)^2+e^{xy}[/tex]

    Should work.
     
  4. Jan 10, 2006 #3
    Thanks saltydog.

    How about the first part to my question - the (in my 'non-lucid' mind) arbitrary assignment of eta and zeta?
     
  5. Jan 11, 2006 #4

    saltydog

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    Hey Robousy, I'm not clear on the tangent vector explanation of this so I'll explain it this way:

    When we have the first-order PDE:

    [tex]a(x,y)u_x+b(x,y)u_y+u=f[/tex]

    we try to make a change of variables:

    [tex]w=h(x,y)\quad z=g(x,y)[/tex]

    to eliminate one of the partials so that we end up with an equation:

    [tex]v_z+v=f(w,z)\quad\text{with}\; v(w,z)\equiv u(x,y)[/tex]

    or some variant thereof. This then can be treated as an ordinary ODE considering the variable w as a constant which we can then "partially integrate" with respect to z remembering when such partial integrations are done, the arbitrary constants become arbitrary functions of the variables kept constant. So what's:

    [tex]\int f(x,y,z)\partial x[/tex]

    but I digress.

    In order to find suitable change of variables, we substitute v(w,z) into the PDE above and obtain (via the chain rule):

    [tex]a(v_w w_x+v_z z_x)+b(v_w w_y+v_z z_y)+\text{the rest of it}[/tex]

    or:

    [tex]v_w(aw_x+bw_y)+(az_x+bz_y)v_z[/tex]

    Thus, in order to convert the PDE into one in which only one derivative is present, we'd like:

    [tex]aw_x+bw_y=0[/tex]

    right?

    Well, it turns out that this can be done if we let w equal to the implicit solution of the characteristic equation:

    [tex]\frac{dy}{dx}=\frac{b(x,y)}{a(x,y)}[/tex]

    That is, suppose the solution is in the form:

    [tex]h(x,y)=c[/tex]

    Then we let:

    [tex]w=h(x,y)[/tex]

    z can be any function of x and y as this will not affect the part we wish to collapse but we of course choose the simplest one.
     
    Last edited: Jan 11, 2006
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