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Solving the simples Partial Differential Equations

  1. Sep 26, 2010 #1

    I joined a class on Partial Differential Equations 3 or 4 lectures late, so I have missed the classes outlining how to solve the simplest forms of them. I am trying to understand the textbook, but so far it is not going so well for me, so I just need to ask some questions about the whole process.

    First, let's start with something of the form a*ux + b*uy = 0 (where ux and uy are the partial derivatives of u)
    We have a geometric method (find equations of all lines parallel to the vector (a, b) and assume the function u is constant on those lines (given that the directional derivative in the direction of (a, b) is 0
    so, solving for the constant (in the expression for all lines parallel to the vector), we find u = f (bx - ay)

    This method I understand somewhat alright - I managed to apply it to some other cases, such as x*ux + y*uy = 0 , by finding the equations of all curves with the vector (x,y) tangent to them, and assuming the function u is constant on those curves.

    However, in the same chapter, we have some examples where the right side is not 0
    such as ux + uy = 1
    Here, I can no longer assume that the function u is a constant on the curves with V = (1,1) as a tangent (since the derivative is not 0). How do I extend this geometric method to cases when the right hand side is not zero?

    Also, we have a 2nd method outlined- Coordinate Method.
    For example, in the a*ux + b*uy = 0 case, we change coordinates such that the x axis is alligned with the direction of the vector of the directional derivative (a, b)

    so x' = ax + by and y' = bx - ay

    (Aside.... to change to prime coordinates that align the x-axis to any vector (c, d), do we just have to carry out A*x = x' where

    A = | c d |
    | d -c |
    Can someone confirm if this is how we rotate coordinate axes to align with a vector?
    End of Aside)

    Then, using these primed coordinates, we rewrite the original equation, solve for ux and uy in terms of ux' and uy', plug them in, and solve.
    this gives us ux' = 0 ------> u = f(y') = f (bx-ay)

    Now, this method is a bit harder for me to carry out. I assume this method only works when the coefficient before ux and uy are constants (contain no x or y terms), so that the vector to which we need to rotate is constant.

    I tried using this method to solve the above mentioned question ux + uy = 1, which i couldnt' solve using the geometric method (or at least I don't know how to)

    giving me x' = x + y , y' = x - y
    then ux = du/dx' * dx'/dx + du/dy'*dy'/dx
    and likewise for uy
    solving for these i get ux = ux' + uy' and uy = ux' - uy'
    now ux + uy = 2ux' = 1

    or ux' = 1/2

    or u = (1/2) x' + f(y')
    or u = (1/2) (x+y) + f(x-y) + C

    with f an arbitrary function

    then ux = 1/2 + f'(x-y) and uy = 1/2 -f' (x-y)

    ux + uy = 1, which is what i needed, but I still don't know if the method I arrived at it via is right ... I thought my solution was supposed to be some kind of arbitrary function f, but in this case, my solution has both a specific function in it (1/2x + 1/2 y) and an arbitrary function f(x-y) .... is this normal for PDEs?

    sorry for the long post....I hope someone can help clear up all the small confusions regarding this topic that I have mentioned. Thank you in advance :)
    Last edited: Sep 26, 2010
  2. jcsd
  3. Sep 27, 2010 #2
    One can find a particular solution and then use the general solution to ux + uy = 0.
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