Results 1 to 9 of 9

Math Help - First Order Semi-Linear PDE

  1. #1
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132

    First Order Semi-Linear PDE

    2yu_x - xu_y = -4xyu^{\frac{1}{2}}

    Where u_x and u_y denotes partial differentiation with respect to x and y respectively of u(x,y)

    For the above PDE how do I find the characteristic curves in the xy-plane?

    How do I show that along these characteristics u(x,y) = (y^2 +C)^2 ? (C is a constant)

    I think I found that \frac{dx}{dt} = 2y, and \frac{dy}{dt} = -x and \frac{du}{dt} = -4xyu^{\frac{1}{2}}

    and that on the characterisitcs y^2 + \frac{x^2}{2} = C_1 but I don't know where to go from here

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Aug 2008
    Posts
    903
    This is how I'd approach it:

    I'd first write it in standard form for ``quasi-linear PDEs'':

    2yu_x-xu_y-(-4xyu^{1/2})=0

    We now first solve the equation:

    2yw_x-xw_y-4xyz^{1/2}w_z=0;\quad w(x,y,z)

    The solution of the first is then found to be:

    w(x,y,u)=0

    Solving for the characteristic curves:

    \frac{dy}{dx}=-\frac{x}{2y};\quad\quad \frac{dz}{dx}=-\frac{4xyz^{1/2}}{2y}=-2xz^{1/2}

    I get:

    y^2=-\frac{x^2}{2}+a\quad\quad z=\left(b/2-x^2/2\right)^2

    Solving for a and b, I get the characteristic curves are then the intersections of the surfaces:

    a=1/2(x^2+2y^2),\quad  b=3x^2/4

    Been a while for me though. Need to double check this.
    Last edited by shawsend; November 14th 2008 at 07:21 AM. Reason: corrected expressions for a and b
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132
    We now first solve the equation:

    2yw_x-xw_y-4xyz^{1/2}w_z=0;\quad w(x,y,z)
    I don't understand how are getting this equation. Please explain.


    The solution of the first is then found to be:

    w(x,y,u)=0
    What do you mean here?


    I get:

    y^2=-\frac{x^2}{2}+a\quad\quad z=\left(b/2-x^2/2\right)^2

    Solving for a and b, I get the characteristic curves are then the intersections of the surfaces:

    a=1/2(x^2+2y^2),\quad  b=3x^2/4
    How did you solve for b here? I don't see how you arrived at that answer for b.

    So how do I get back to u(x,y) ?

    Thanks for your help
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132
    I think I understand how to do this now.

    \frac{dx}{dt} = 2y, and \frac{dy}{dt} = -x and \frac{du}{dt} = -4xyu^{\frac{1}{2}}

    so \frac{du}{dx} = -2xu^{\frac{1}{2}}

    thus u = (\frac{-x^2}{2} + c_1)^2

    and y^2 + \frac{x^2}{2} = c_2

    from \frac{dy}{dx}

    so u(x,y) = (y^2 + C)^2

    with the initial condition u(x,0) = x^4

    u(x,y) = (y^2 + x^2)^2
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Hi. The solution of:

    a(x,y,u)u_x+b(x,y,u)u_y-c(x,y,u)=0;\quad u=u(x,y)

    can be expressed implicitly as w(x,y,u)=0 where w(x,y,z) is a solution of the following:

    a(x,y,z)w_x+b(x,y,z)w_y+c(x,y,z)w_z=0

    However I'm having problems implementing it. Also, I used Mathematica to solve for a and b: Personally I prefer to get the answer any sleazy way I can get it, then go back and figure out how to do it for the purist in the class.

    See "Basic Partial Differential Equations" by Bleecker and Csordas. This is covered and is what I'm using for this problem.
    Last edited by shawsend; November 14th 2008 at 08:26 AM. Reason: implicitly
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Alright, alright, I made a mistake: I'm relying too much on Mathematica. We need to solve the equations:

    y^2=-x^2/2+a

    z^{1/2}=-x^2/2+b/2

    for a and b. This by inspection is obviously:

    a=1/2(x^2+2y^2)

    b=2(z^{1/2}+x^2/2)

    I don'w know what Mathematica is doing. Anyway, when I use those substitutions, the requisite terms cancel as expected, leaving the desired w_z term. I think we got this one . . .
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member
    Joined
    Aug 2008
    Posts
    903
    I'd like to finish this as best as I can (it's not rigorous, some holes remain but it's a start):

    We now make the transformation:

    r=y^2+\frac{x^2}{2}

    s=2z^{1/2}+x^2

    t=z

    and let v(r,s,t)\equiv w(x,y,z)

    and express the PDE in terms of v(r,s,t):

    <br />
\begin{aligned}<br />
2yw_x-xw_y-4xyz^{1/2}w_z&=2y\left(v_r r_x+v_s s_x+v_t t_x\right)\\<br />
&-x\left(v_r r_y+v_s s_y+v_t t_y\right)\\<br />
&-4xyz^{1/2}\left(v_r r_z+v_s s_z+v_t t_z\right)=0\end{aligned}

    =2xyv_r+4xyv_s-2xyv_r-4xyv_s-4xyz^{1/2}v_t=0

    -4xyz^{1/2}v_t=0

    Note I left x, y, z untransformed into the r,s, and t variables since this is not necessary at this point.

    I'm left then with the PDE: v_t=0 which is easily integrated to a constant function in the variables r and s:

    v(r,s,t)=C(r,s)

    A solution for the PDE in terms of w is then:

    w(x,y,x)=C(y^2+x^2/2,2z^{1/2}+x^2/2) where C is any C^1 function. For example, the function:

    w(x,y,z)=y^2+x^2/2 is a solution. Another one is:

    w(x,y,z)=C(\alpha,\beta)=\alpha+\beta=y^2+x^2/2+2z^{1/2}+x^2/2

    The solution of the original PDE is then w(x,y,u)=0 In the last case, that would be:

    y^2+3/2 x^2+2u^{1/2}=0 is an implicit solution to the PDE. We can check this easily:

    u^{1/2}=-\frac{y^2+3/2x^2}{2}
    (slight problem here with the negative sign but let's continue anyway). Let a solution to the original PDE be:

    u=\frac{(y^2+3/2x^2)^2}{4}

    back-substituting I get:

    \frac{2y}{4}\left[2(y^2+3/2x^2)3x\right]-\frac{x}{4}\left[2(y^2+3/2x^2)2y\right]-\frac{4xy}{2}(y^2+3/2x^2)=0
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member Jason Bourne's Avatar
    Joined
    Nov 2007
    Posts
    132
    Quote Originally Posted by Jason Bourne View Post
    I think I understand how to do this now.

    \frac{dx}{dt} = 2y, and \frac{dy}{dt} = -x and \frac{du}{dt} = -4xyu^{\frac{1}{2}}

    so \frac{du}{dx} = -2xu^{\frac{1}{2}}

    thus u = (\frac{-x^2}{2} + c_1)^2

    and y^2 + \frac{x^2}{2} = c_2

    from \frac{dy}{dx}

    so u(x,y) = (y^2 + C)^2

    with the initial condition u(x,0) = x^4

    u(x,y) = (y^2 + x^2)^2
    Thanks for your help.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member
    Joined
    Aug 2008
    Posts
    903
    Quote Originally Posted by Jason Bourne View Post
    I think I understand how to do this now.

    \frac{dx}{dt} = 2y, and \frac{dy}{dt} = -x and \frac{du}{dt} = -4xyu^{\frac{1}{2}}

    so \frac{du}{dx} = -2xu^{\frac{1}{2}}

    thus u = (\frac{-x^2}{2} + c_1)^2

    and y^2 + \frac{x^2}{2} = c_2

    from \frac{dy}{dx}

    so u(x,y) = (y^2 + C)^2

    with the initial condition u(x,0) = x^4

    u(x,y) = (y^2 + x^2)^2
    Ok, I see what you mean. Sorry about that.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] 1st Order Semi Linear PDE xu_x+yu_y=u+3
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: May 6th 2011, 12:45 AM
  2. [SOLVED] Form of Semi Linear Linear PDE's
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: April 22nd 2011, 07:51 AM
  3. [SOLVED] Semi Linear PDE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: February 15th 2011, 03:55 AM
  4. Replies: 11
    Last Post: August 8th 2010, 03:31 AM
  5. Third Order Semi Implicit Runge Kutta Method
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: May 2nd 2009, 08:33 AM

Search Tags


/mathhelpforum @mathhelpforum