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General solution to PDEs-arbitrary constants

  1. Sep 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Claim:
    For the partial differential equation ∂u/∂x + ∂u/∂y = 0, the general solution is u(x,y) = f(x-y)

    2. Relevant equations
    N/A

    3. The attempt at a solution
    I remember that the number of arbitrary constants in the general solution should be the same as the order of the differential equation. For example, for 1st order ODEs, there should be one arbitrary constant in the general solution; for 2nd order ODEs, there should be two arbitrary constants in the general solution.

    In this case, the order of the differential equation is one, so it should have one arbitrary constant in the general solution, right? However, the above claims that u(x,y) = f(x-y) is the general solution. I am puzzled...Where is the arbitrary constant?

    Thanks for any help!:smile:
     
  2. jcsd
  3. Sep 9, 2009 #2

    nicksauce

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    In PDEs you don't have arbitrary constants... you can actually have arbitrary functions, as is the case in this example.
     
  4. Sep 9, 2009 #3
    Then for PDEs, is there any general difference between the general solutions to, say, 1st and 2nd order PDEs?

    Also, they have f(x-y), which is a function of x-y, is this allowed for general solutions to 1st order PDEs? I've seen arbitrary functions like k(x), c(y) for general solutions to 1st order PDEs, but not something like f(x-y).
     
  5. Sep 10, 2009 #4

    HallsofIvy

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    Yes, there are enormous differences between first and second order partial differential equations just as there are for ordinary differential equations!

    If u(x,y)= f(x- y) where f is any differentiable function of one variable, then, by the chain rule, taking z= x- y,
    [tex]\frac{\partial f}{\partial x}= \frac{df}{dz}\frac{\partial z}{\partial x}= f' (1)= f'[/tex]
    [tex]\frac{\partial f}{\partial y}= \frac{df}{dz}\frac{\partial z}{\partial y}= f' (-1)= -f'[/tex]
    so that
    [tex]\frac{\partial f}{\partial x}+ \frac{\partial f}{\partial y}= 0[/tex]
    no matter what f is.

    The general solution for the second order differential equation
    [tex]\frac{\partial^2 u}{\partial x^2}= \frac{\partial^2 u}{\partial y^2}[/tex]
    is F(x-y)+ G(x+y) where F and G can be any twice differentiable functions.
     
  6. Sep 11, 2009 #5
    Is it correct to write the following? (I am asking this because you somehow didn't write the (x-y) part after the f ', you just wrote f ' and I am not sure why)
    ∂u/∂x = f '(x-y) (1)
    ∂u/∂y = f '(x-y) (-1)

    In other words, if we let z=x-y, is it correct to write df/dz=f '(x-y) ?


    So first order PDEs have ONE arbitrary function (instead of one arbitrary constant) in the general solution, and second order PDEs have TWO arbitrary functions (instead of two arbitrary constants) in the general solution, is this the general rule?

    Thanks for your help!!!
     
  7. Sep 13, 2009 #6
    Is there a general pattern that typifies the general solutions to first order PDEs?
     
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