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PDE problem

  1. Jan 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Find all functions v(x,y) such that both v_x(x,y) = x^2 + y and v_y(x,y) = x - y^3.

    2. Relevant equations



    3. The attempt at a solution

    I have no idea how to start. My instructor says to guess the solution, which I have tried, but failed.
     
  2. jcsd
  3. Jan 26, 2008 #2

    nicksauce

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    Integrate V_x with respect to x, then you have V(x,y) = x^3/3 + xy + f(y), where f is an arbitrary function. Then differentiate that expression with respect to y and compare to V_y.
     
  4. Jan 26, 2008 #3

    Defennder

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    He gave you THAT advice? I really don't know what to make of it, but guessing definitely doesn't work for me.

    As nicksauce said, the best way to do this is to treat to find F(x,y) such that its partial derivatives wrt x,y are given by your question. Have you learnt what are conservative vector fields/functions yet? The method we're using here is the same as that used to find the potential function, given a conservative vector field.
     
  5. Jan 27, 2008 #4

    HallsofIvy

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    "Guessing a solution" (and then checking) is a good way to solve a problem if you have lots of experience!

    Notice, by the way, that if vx= x2+ y then, differentiating again with respect to y, vyx= 1. If vy= x- y3 then, differentiating again with respect to x, vxy= 1. The fact that those are the same (so that the "mixed" derivatives are the same) tells us we can find such a v!

    Follow nicksauce's advice.
     
  6. Jan 27, 2008 #5

    jambaugh

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    It may help to look at the problem in differential form:

    [tex] d v = \frac{\partial v}{\partial x} dx + \frac{\partial v}{\partial y}dy[/tex]

    (This differential formula can be though of as the definition of the partial derivatives!)


    then integration yields:
    [tex] v+C = \int dv = \int\left(\frac{\partial v}{\partial x}\right) dx + \int\left(\frac{\partial v}{\partial y}\right)dy[/tex]

    (Trust the notation!)

    Note that when you integrate with respect to say [itex]x[/itex] your constant of integration is only "constant" with respect to [itex]x[/itex] and so can be any function of [itex]y[/itex] alone. (Since x and y are both independent variables you treat y as a constant when integrating with respect to x and vise versa).

    Say you do the above integration and get:

    [tex] v= F(x,y) + C_1(y) \quad +\quad G(x,y) + C_2(x)[/tex]

    (You can absorb the left hand side constant into the other "constant" functions.)
    You'll know F and G but must figure out [itex] C_1[/itex] and [itex] C_2[/itex]. You can do this by differentiating again once with each of x or y.
     
  7. Jan 30, 2008 #6

    jambaugh

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    Oops!!!! I erred Terribly:
    I should have said that integration yields:

    [tex] v+C = \int dv = \int \left( \frac{\partial v}{\partial x}\right) dx = \int \left( \frac{\partial v}{\partial y}\right)dy[/tex]

    you would then get something like:

    [tex] v = F(x,y) + C_1(y) = G(x,y)+C_2(x)[/tex]
     
    Last edited: Jan 30, 2008
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