Results 1 to 7 of 7

Math Help - Linear First Order PDE- Looking for an example

  1. #1
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461

    Linear First Order PDE- Looking for an example

    Hi All,

    I have an examinable question that I am stuck on and obviously I cannot submit it here.
    Instead I am asking if some one could provide a simple problem that demonstrates the principle. Here goes:

    We have linear first order PDE u(x,y). Asked to find u given intial conditions which I can.
    Next it asks to show that the solution is not defined when y > f(x). (I know what the f(x) of x is but cannot show it)

    Could anyone provide a simple example to demonstrate this or at least what do i do?

    Thanks
    bugatti79
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    How about something like

    2u u_y = -1, \;\;u(x,0) = x^2?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    How about something like

    2u u_y = -1, \;\;u(x,0) = x^2?
    Are we missing an x partial derivative?

    As given above i get u(y)=\sqrt{-y+c}. Cant go any further?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    It's a PDE so integrating gives (taking the positive solution)

    u(x,y) = \sqrt{f(x)-y}.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    It's a PDE so integrating gives (taking the positive solution)

    u(x,y) = \sqrt{f(x)-y}.
    Ok, the constant must be a function of x ie

    2\int{u} du=-\int{y} dy therefore

    u^2=-y+c but c=f(x)

    u(x,y)=+\sqrt{-y+f(x)} using boundary conditions give particular as

    u(x,y)=\sqrt{-y+x^4}

    How do we use this to show the solution is not defined when y> some function of x?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    56
    What if y > x^4?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member bugatti79's Avatar
    Joined
    Jul 2010
    Posts
    461
    Quote Originally Posted by Danny View Post
    What if y > x^4?
    Ok, after some reading and with your help my interpretation is the following:

    there are 2 restrictions: We never divide by 0 and we assume real values functions. Therefore based on this and looking at the above example, the quantity -y+x^4 \not < 0. Therefore x^4 \geq y . Otherwise we get a complex number.

    What is the domain of the function?

    The domain of this function is the set of all real values of x and y whose quantity -y+x^4 \not < 0.
    Is there a better way of stating this mathematically?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear program with higher order non-linear constraints.
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: September 12th 2010, 03:36 AM
  2. 2nd Order Linear O.D.E. Help
    Posted in the Differential Equations Forum
    Replies: 15
    Last Post: October 20th 2008, 01:22 AM
  3. Replies: 4
    Last Post: August 12th 2008, 05:46 AM
  4. Help with third order, non-linear DE.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 9th 2008, 04:45 PM
  5. Replies: 1
    Last Post: May 11th 2007, 04:01 AM

Search Tags


/mathhelpforum @mathhelpforum