Results 1 to 11 of 11

Math Help - Another multivariable differntials problem

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    51

    Another multivariable differntials problem

    Here is what I did.



    I think the method is very lame, but the idea should be this way. Could any one correct it for me please ?

    Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Two comments/corrections:

    1. Because of the regularity conditions on f (that is, it is a continuously differentiable function), you can use Clairaut's Theorem to equate the mixed partial derivatives.
    2. Your g(y) and h(x) are not unique. Almost any ol' functions will do there! But this does not hurt your final result, since you're not asked to find unique functions.

    Other than these comments, it looks fine to me.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    51
    Sorry for multiple posts.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Apr 2009
    Posts
    51
    Quote Originally Posted by Ackbeet View Post
    Two comments/corrections:

    1. Because of the regularity conditions on f (that is, it is a continuously differentiable function), you can use Clairaut's Theorem to equate the mixed partial derivatives.
    2. Your g(y) and h(x) are not unique. Almost any ol' functions will do there! But this does not hurt your final result, since you're not asked to find unique functions.

    Other than these comments, it looks fine to me.
    There are 3 things.

    1. A previous problem in my book said:

    Let f: R^n --> R , differentiable on the open ball centering at a = (a1, ...,an) and radius r and f'_n = 0 Then there is a UNIQUE g: R^n-1 centering at (a1,...,an-1) and radius r such that f(x1,...,xn) = g(x1,...,xn-1).

    2. I am mostly concerned with the bottom line beginning with

    Df(x,y) ....

    The integrals? Is it correct that \int Df(x,y) = f(x,y) ?

    Is it correct to just put \int in front of h(x)dx ??
    I think integral must come with \int .... dx

    3. Clairaut's Theorem ??? Never heard it before. Is it the same as Young Theorem ?

    Thanks a lot
    Follow Math Help Forum on Facebook and Google+

  5. #5
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    1. I get you. Jolly good, then.

    2. Yeah, you might have a point here. Thinking...

    3. Clairaut's Theorem.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Apr 2009
    Posts
    51
    Clairaut's Theorem... I know it as Young's theorem.

    Nevertheless, I am waiting for your corrections.

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Why not just do this:

    \displaystyle\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)=0\quad\Rightarrow

    \displaystyle\frac{\partial f}{\partial y}=\int\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial y}\right)\,\partial x=g(y)\quad\Rightarrow

    \displaystyle\int\frac{\partial f}{\partial y}\,\partial y=\int g(y)\,dy\quad\Rightarrow

    \displaystyle f=\int g(y)\,dy+h(x).

    Let f_{1}(x)=h(x), and let

    \displaystyle f_{2}(y)=\int g(y)\,dy,

    and you're done, right?

    Note that I'm using the \partial x and \partial y to denote "partial integration", as opposed to a total integration where y=y(x).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Apr 2009
    Posts
    51
    How could you explain this step? I cannot see the reason why h(x) came from.


    \displaystyle f=\int g(y)\,dy+h(x).


    and, is there a way without using partial integration ? The book I am studying does not mention it. (I try doing problems without resorting to other definitions outside the book.)

    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Not sure there's a way to do this problem without partial integration. In any case, when you partially integrate w.r.t. one independent variable, the "constant" of integration must be allowed to be a function of the other independent variables. You can verify this by taking the partial derivative w.r.t. the variable of partial integration. The "constant" of integration gets killed by the partial differentiation. Remember finding harmonic functions? Or solving exact ODE's? It's the same idea here.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Apr 2009
    Posts
    51
    I know nothing about harmonic functions or ODE. I am not exactly mathematic majors, but I will study what you mentioned later.
    Thank you very much.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're welcome!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 8th 2010, 06:56 PM
  2. Multivariable limit problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 8th 2010, 03:09 PM
  3. Multivariable Limit Problem
    Posted in the Calculus Forum
    Replies: 1
    Last Post: September 23rd 2010, 07:08 PM
  4. Replies: 1
    Last Post: November 17th 2008, 01:06 PM
  5. Differntials??
    Posted in the Calculus Forum
    Replies: 4
    Last Post: November 12th 2008, 07:40 PM

Search Tags


/mathhelpforum @mathhelpforum