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Function in terms of its partial derivatives

  1. Jan 16, 2012 #1
    Hi,

    I remember having read in basic calculus that the following is true, but I dont know what this property is called and am having a hard time finding a reference to this.
    [tex]d u(x,y) = \frac{\partial u}{\partial x} dx + \frac{\partial u}{\partial y} dy[/tex]
    Ques: Is this true ? Is this true for all functions? Or is there a condition that the function u should be separable ?

    I also think that if this is true, then I should be able to reconstruct the function u by taking the anti-derivative of the above equation:
    [tex]\int d u(x,y) = \int \frac{\partial u}{\partial x} dx + \int \frac{\partial u}{\partial y} dy[/tex]
    However, this fails when
    [tex]u(x,y) = x^2y^2[/tex]
    Am I missing something here ? If the above holds only for separable functions, is there a way I can reconstruct a function from its partial derivatives ?

    Any guidance is appreciated. Thanks.
     
  2. jcsd
  3. Jan 16, 2012 #2

    chiro

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    Hello bhatiaharsh and welcome to the forums.

    The first expression you listed is known as the total differential. Take a look at this page:

    http://en.wikipedia.org/wiki/Total_derivative

    In terms of your integral expression, you need some sort of seperation of variables: i.e. the term in the integral has to be in an explicit form and not implicit.

    This is not just a property for these kinds of problems, it is a general requirement for calculating integral expressions. If there is an implicit connection to the different variables then you need to transform it into an explicit expression in terms of the variables you are dealing with in terms of the actual integration.

    One way of doing this is to find a correct parameterization of the system if that is possible. This can be very very ugly, but it can be a good way to attack this problem if the implicit form of the representation (function whether it describes a line/curve, surface, volume etc) is not easy to separate.
     
  4. Jan 17, 2012 #3
    Thanks for the pointer chiro. I followed on the link you gave and reached exact differential equations. Hopefully learning about that should give me a more clear understanding. But I can already see how separability can be an issue.
     
  5. Jan 19, 2012 #4
    So, if I know [itex]\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial u}{\partial z}[/itex] can someone help me how can I find [itex]u[/itex] ? Any pointers are appreciated.
     
  6. Jan 19, 2012 #5

    Char. Limit

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    If you have the partial derivatives of u, finding u is easy, assuming that all the partials lead to the same u. Take [itex]\frac{\partial u}{\partial x}[/itex] for example. Partial integration shows:

    [tex]\int \frac{\partial u}{\partial x} \partial x = f(x,y,z) + C(y,z)[/tex]

    where C(y,z) is an arbitrary function that's constant with respect to x. Repeating this with [itex]\frac{\partial u}{\partial y}[/itex] and [itex]\frac{\partial u}{\partial z}[/itex] will give you three functions as shown:

    [tex]f(x,y,z) + C_1(y,z)[/tex]

    [tex]g(x,y,z) + C_2(x,z)[/tex]

    [tex]h(x,y,z) + C_3(x,y)[/tex]

    And setting all three of those equal to each other will allow you to figure out what the three Constant functions are. All three of those are equal to u(x,y,z). A simple example with u_x = 2 x y^2 z^2, u_y = 2 x^2 y z^2, u_z = 2 x^2 y^2 z:

    [tex]\int 2 x y^2 z^2 dx = x^2 y^2 z^2 + C_1(y,z)[/tex]

    [tex]\int 2 x^2 y z^2 dy = x^2 y^2 z^2 + C_2(x,z)[/tex]

    [tex]\int 2 x^2 y^2 z dz = x^2 y^2 z^2 + C_3(x,y)[/tex]

    Since all three of those are equal to each other, we get [itex]C_1 = C_2 = C_3[/itex] and it's relatively easy to prove that the only function that could fit that profile (it can't be a function of x, because of [itex]C_1[/itex], and symmetry arguments show it can't be a function of y and z either) is the constant function C. So we get that [itex]u = x^2 y^2 z^2 + C[/itex]. With an initial value of u, we could get C as well.
     
    Last edited: Jan 19, 2012
  7. Jan 20, 2012 #6
    Well, isnt the no of unknowns greater than the no of equations ?
    [tex]u = f_1 + c_1[/tex]
    [tex]u = g_1 + c_2[/tex]
    [tex]u = h_1 + c_3[/tex]
    There are 4 unknowns and 3 equations.
     
    Last edited: Jan 20, 2012
  8. Jan 21, 2012 #7

    HallsofIvy

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    No. f1, f2, and f3 are all related to u. They are not independent.

    I would have done the problem Char. Limit posed a little differently.

    From [itex]u_x= 2xy^2z^2[/itex] we get [itex]u= x^2y^2z^2+ f(y,z)[/itex]
    (since the partial derivative with respect to x treats y and z a constants, the "constant" of integration may be a function of y and z)

    Now differentiate that with respect to y: [itex]u_y= 2x^2yz^2+ f_y[/itex].
    Comparing that to the given [itex]u_y= 2x^2yz^2[/itex] we see that we must have f_y= 0. That means that f is not a function of y but a function of z only. But then differentiating [itex]u= x^2y^2z^2+ f(z)[/itex] with respect to z we get [itex]u_z= 2x^2y^2z+ f_z[/itex] and comparing that to the given equation, we have [itex]f_x= 0[/itex] so that f is, in fact, a constant.

    The problem would be a little more interesting if it were [itex]u_x= 2xy^2z^2[/itex], [itex]u_y= 2x^2yz^2+ 4y[/itex], [itex]u_z= 2x^2y^2z+ e^z[/itex]. Try solving those for u.
     
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