Results 1 to 2 of 2

Math Help - How to solve this pde?

  1. #1
    Newbie
    Joined
    May 2007
    Posts
    3

    How to solve this pde?

    du/dt-t(du/dx)=-u^2

    d is meant to be partial derivatives.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by theoretician View Post
    du/dt-t(du/dx)=-u^2

    d is meant to be partial derivatives.
    I am going to try. I cannot gaurentee my solution is correct.

    \frac{\partial u}{\partial t} - t\cdot \frac{\partial u}{\partial x} = - u^2

    Say we want to solve this subject to some curve x=x(s) \mbox{ and }t=t(s) with x(0)=x_0 \mbox{ and }t(0)=0.

    Then u(x,t) = u(x(s),t(s))
    Use the chain rule,
    \frac{du}{ds} = \frac{\partial u}{\partial x} \cdot \frac{dx}{ds} + \frac{\partial u}{\partial t} \cdot \frac{dt}{ds}
    By the method of charachteristics we set,
    1) \frac{dt}{ds} = 1 \mbox{ with }t(0)=0
    2) \frac{dx}{ds} = -t \mbox{ with }x(0) = x_0
    3) \frac{du}{ds} = -u^2

    The solution to the IVP problems are:
    1) t = s (**)
    2) x = x_0 - ts (*)

    The solution to #3 is simple:
    \frac{du}{dx} = -u^2 \Rightarrow -\frac{1}{u^2}u' = 1
    This is a Seperable equation,
    \frac{1}{u} = s + F(x_0) \Rightarrow u =\frac{1}{s+F(x_0)}
    Where F(x_0) is an arbitrary starting point on the solution curve.
    By (*) we see that x_0 = x+ts.
    Thus by (**),
    u(x,t) = \frac{1}{t+F(x+ts)}
    Where F is any arbitrary differencial function (and not equal to -s).
    ----
    Which seems to work !!! Because if F = 0 everywhere then u(x,t) = \frac{1}{t+0} = \frac{1}{t}. If you substitute that into the PDE you get:
    \frac{\partial u}{\partial t} = - \frac{1}{t^2}  = - u^2
    Which works.
    Last edited by ThePerfectHacker; June 11th 2007 at 12:14 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. need to solve summation equation to solve sum(x2)
    Posted in the Statistics Forum
    Replies: 2
    Last Post: July 16th 2010, 10:29 PM
  2. how do i solve this IVP: y'=y^2 -4
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: February 24th 2010, 11:14 AM
  3. Replies: 1
    Last Post: June 9th 2009, 10:37 PM
  4. how do i solve this?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: August 2nd 2008, 02:58 PM
  5. how to solve ..
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: August 2nd 2008, 08:17 AM

Search Tags


/mathhelpforum @mathhelpforum