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First order Linear PDE, Method of Characteristics

  1. Oct 4, 2009 #1

    PAR

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    1. The problem statement, all variables and given/known data
    [tex]x*u_{x} + y*u_{y}= 1 + y^2[/tex]
    u(x,1) = 1+ x; -infinity < x < +infinity

    Solve this parametrically and in terms of x and y


    2. Relevant equations

    We are supposed to solve this using the method of characteristics



    3. The attempt at a solution

    My problem is that solving the equation parametrically and with x and y give two different solutions.

    I will first try to solve this parametrically, using variables s and t.

    let x = s along the initial line y = 1 Is this assumption correct, or should the initial line be y = 0? I have only solved problems where the initial line as been explicitly stated. I chose y = 1 because it made the equations work out, and because the initial condition was u(x,1), so I am not sure if that is correct.

    So,

    dx/dt = x => x = s * e^t

    dy/dt = y => y = Ce^t, since the initial line is y = 1, C= 1; y = e^t

    du/dt = 1 + y^2 => u = t + [tex]\frac{e^{2t}}{2}[/tex] + D

    Now apply the initial condition:

    u(x,1) = 1 + x = 0 + 1/2 + D => D = x + 1/2

    So,

    u = t + [tex]\frac{e^2t}{2}[/tex] + x + 1/2

    since y = e^t that means t = ln|y| (|| is absolute value... I would also like to know if y is always positive so I can remove the absolute value.)

    So the solution for u is:

    u = ln|y| + (y^2)/2 + x + 1/2

    That is solving the equation parametrically, now I am a supposed so solve the equation treating x as a parameter variable and s as the other parameter variable:

    Rewrite the equation to:

    [tex]u_{x}+ (y/x)*u_{y} = (1 + y^2)/x [/tex] (divided through by x)

    so,

    dy/dt = y/x => y = k(s)x (k is a function of s)

    du/dt = (1+y^2)/x => u = (1+y^2)ln(x) + C(s) = (1+y^2)ln(x) + G(y/x)

    Applying the initial condition:

    u(x,1) = 1+x = 2ln(x) + G(1/x)

    G(1/x) = 1+x - 2ln(x)

    This is where I get confused:

    First, the solutions I derived don't look at all like each other, and second for the later solution, I don't see how G(1/x) will tell me what G(y/x) is. I would appreciate any and all help, Thank You.
     
  2. jcsd
  3. Oct 5, 2009 #2

    PAR

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    I've done the entire problem a second time through and I get the same answer. My main concerns are:

    What is the initial curve and why? y = 1?

    When solving the problem in terms of x,y (the later solution), what is G(y/x)? I've tried applying the initial condition but I just get an equation for G(1/x). Again, thanks in advance.
     
  4. Oct 19, 2009 #3
    The correct answer is:

    u=x/y + ln y + y^2/2 + 1/2

    You can check that this satisfies both the PDE and the initial condition:

    u(x,1)=1+x

    If you're still interested in this I'll try to write more. I haven't done this stuff for 30 years so I'm a little rusty.
     
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