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Homework Help: Basic partial differential problem

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the solution of each of the following partial differential equation

    [itex]\frac{\partial^{2}u}{\partial x^{2}} = 0[/itex]

    2. Relevant equations
    assume the product form?
    [itex] u(x,y) = f(x)g(y)[/itex] ? (not 100% sure)

    3. The attempt at a solution
    Hello, I'm only new to PDEs and I was wondering what kind of algorithm is used to solve a function such as this (with no boundary conditions).

    I have the solution which states that
    [itex]\frac{\partial u}{\partial x} = f(y)[/itex]

    [itex]u(x,y) = xf(y)+g(y)[/itex]

    but I cannot see the intuition behind this, does anyone know?
  2. jcsd
  3. May 14, 2014 #2


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    Homework Helper

    Where do constants of integration come from? From the fact that the derivative of a constant is zero.

    So here, except that "derivative of a constant is zero" is replaced by "partial derivative with respect to [itex]x[/itex] of a function which doesn't vary with [itex]x[/itex] is zero".
  4. May 14, 2014 #3


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    Science Advisor

    You understand, I presume, that if the problem were an ordinary differential equation, [itex]\frac{d^2}{dx^2}= 0[/itex], you would solve it by integrating twice: since the second derivative is 0, the first derivative must be a constant: [itex]\frac{du}{dx}= C[/itex]. And now, integrating again, [itex]u(x)= Cx+ D[/itex] where D is the constant of integration.

    It's the same basic idea with partial derivatives except that the constant of integration may be a function of the other variables.

    [itex]\frac{\partial^{2}u}{\partial x^{2}} = 0[/itex] can be read as
    [tex]\frac{\partial }{\partial x}\left(\frac{\partial u}{\partial x}\right)= 0[/tex]

    That is, that [itex]\partial u/\partial x[/itex] does not depend upon x. It does NOT, itself, say what variables the partial derivative does depend on. If we are given that the only variable are x and y, then [itex]\partial u/\partial x[/itex] must be a function of y only- and it could be any function of y: [itex]\partial u/\partial u= f(y)[/itex].

    The point is that taking the partial derivative with respect to one variable, we treat the other variables as constants. So going the other way, taking the anti-derivative with respect to that variable, the ''constant of integration" may actually depend upon those other variables. If g(y) is a function of y only, [itex]\frac{\partial g}{\partial x}= 0[/itex].

    Once we have
    [tex]\frac{\partial u}{\partial x}= f(y)[/tex]
    integrating again (with respect to x, holding y constant) gives [tex]u(x, y)= f(y)x+ g(y)[/tex] where again, the "constant of integration" can be an arbitrary function of y.

    If the independent variables were x, y, and z, the general solution to [itex]\frac{\partial^2 u}{\partial x^2}= 0[/itex] would be u(x, y, z)= f(y,z)x+ g(y, z) where, now, f and g can be any functions of y and z. You should be able to see by differentiating that these do satisfy the equation.

    If you had [itex]\frac{\partial^2 u}{\partial x\partial y}= 0[/itex], again assuming that the only variables are x and y, integrating first, with respect to y would give [itex]\partial u/\partial y= f(y)[/itex] where f(y), the "constant of integration" can be any function of y. Integrating again, this time with respect to y, would give u= F(y)+ G(x) where F(y) is an anti-derivative of f(y) (since f(y) was arbitrary so is F(y)) and G(x) is the "constant of integration", an arbitrary function of x.
  5. May 14, 2014 #4
    Ahhhh, I see now. If I was doing this with in two dimensions, the integral of 0 is just constant c, however because I'm doing this in 3d (w.r.t 2 variables) my constant is now replaced with an arbitrary function. Who has it's function in terms the other variable. Right?
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