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Solving equations

  1. Feb 25, 2006 #1
    Hi, can someone check/fix my errors in the following question?

    Q. Consider the equations

    u = x^3 + 2xy + y^2, v = x^2 + y

    (i) Show that these equations can be solved locally for x and y as C^1 functions of u and v near the point where x = 1, y = 1, u = 4, v = 2.

    (ii) Find [tex]\frac{{\partial x}}{{\partial u}}[/tex] and [tex]\frac{{\partial x}}{{\partial v}}[/tex] at this point.

    (i) u = x^3 + 2xy + y^2, v = x^2 + y. I rewrite these equations as

    [tex]f\left( {u,v,x,y} \right) = \left( {x^3 + 2xy + y^2 - u,x^2 + y - v} \right) = \left( {0,0} \right)[/tex] where [itex]f:R^4 \to R^2 [/itex] is a C^1 function.

    I verified that f(4,2,1,1) = (0,0).

    [tex]
    \det \left[ {\begin{array}{*{20}c}
    {\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
    {\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
    \end{array}} \right]
    [/tex]

    [tex]
    = \left| {\begin{array}{*{20}c}
    {3x^2 + 2y} & {2x + 2y} \\
    {2x} & 1 \\
    \end{array}} \right|
    [/tex]

    [tex]
    = \left| {\begin{array}{*{20}c}
    5 & 4 \\
    2 & 1 \\
    \end{array}} \right| = - 3 \ne 0
    [/tex]

    at (u,v,x,y) = (4,2,1,1).

    So is that enough (i)?

    (ii) I think I use the general formula for the derivative in problems of this type to obtain the derivative of (x,y) = h(u,v) in matrix form and then take the relevant parts.

    By (i) there is a function of the form [itex]\left( {x,y} \right) = h\left( {u,v} \right)[/itex] near (u,v,x,y) = (4,2,1,1).

    [tex]
    h'\left( {u,v} \right) = \left[ {\begin{array}{*{20}c}
    {\frac{{\partial f_1 }}{{\partial x}}} & {\frac{{\partial f_1 }}{{\partial y}}} \\
    {\frac{{\partial f_2 }}{{\partial x}}} & {\frac{{\partial f_2 }}{{\partial y}}} \\
    \end{array}} \right]^{ - 1} \left[ {\begin{array}{*{20}c}
    {\frac{{\partial f_1 }}{{\partial u}}} & {\frac{{\partial f_1 }}{{\partial v}}} \\
    {\frac{{\partial f_2 }}{{\partial u}}} & {\frac{{\partial f_2 }}{{\partial v}}} \\
    \end{array}} \right]
    [/tex]

    [tex]
    = \left[ {\begin{array}{*{20}c}
    {\frac{{\partial x}}{{\partial u}}} & {\frac{{\partial x}}{{\partial v}}} \\
    {\frac{{\partial y}}{{\partial u}}} & {\frac{{\partial y}}{{\partial v}}} \\
    \end{array}} \right]
    [/tex] where the last equality follows from the definition of (x,y) = h(u,v).

    Do I plug the values of u,v,x,y into the above? I've been having some problems using the implicit function theorem so I don't really know how to do (i) and (ii). Any comments would be good thanks.
     
    Last edited: Feb 25, 2006
  2. jcsd
  3. Feb 25, 2006 #2

    AKG

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    (i) looks fine. For (ii), look at the proof you're given for the implicit function theorem. There must be some sort of construction for h. Use this construction to find the partials. As it stands, I don't know where you're getting the first equality from.
     
  4. Feb 25, 2006 #3

    saltydog

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    I'll do a simple one first:

    [tex]F:\mathbb{R}^2\rightarrow\mathbb{R}^2,\quad F(x,y)=(2x+y,x-y)=(u,v)[/tex]

    and apply the Implicit Function Theorem near the point (x,y)=(1,2).

    The Jacobian of F is:

    [tex]\mathcal{J}\left\{F\right\}=

    \left|\begin{array}{cc}2 & 1 \\
    1 & -1 \end{array}\right|\neq 0[/tex]

    Therefore by the Inverse Function Theorem:

    [tex]F^{-1}:\mathbb{R}^2\rightarrow\mathbb{R}^2,\quad F^{-1}(u,v)=(g_1(u,v),g_2(u,v))=(x,y)[/tex]

    near the point (1,2). Thus:

    [tex]\left(F^{-1}\right)^{'}(4,-1)=\left[F^{'}\right]^{-1}(1,2)[/tex]

    or:

    [tex]\left(F^{-1}\right)^{'}(4,-1)=\left[\begin{array}{cc}1/3 & 1/3 \\
    1/3 & -2/3 \end{array}\right][/tex]

    That is:

    [tex]x_u=1/3[/tex]

    [tex]x_v=1/3[/tex]

    [tex]y_u=1/3[/tex]

    [tex]y_v=-2/3[/tex]
     
  5. Feb 25, 2006 #4
    Thanks for the help guys.

    I appreciate the help and I hope I don't sound arrogant but I don't think my set up is incorrect at this stage. Although I did leave out a negative sign in the first equality (the part with the product of two matrices). I used the same method on Saltydog's example (this time with the negative sign included as it should be according to my notes) and I obtained the same answer as him. I'll go over the theorem again to see if I've missed something.
     
  6. Feb 26, 2006 #5

    saltydog

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    Good for you Benny. Didn't wish to imply you were doing anything wrong. Just wanted to work a simple problem. Now, can you verify the values of the derivatives for you problem directly without relying on the Implicit Function Theorem? What for? Whatever.

    Oh yea, it could happen. Got Mathematica? Here's part of the code:

    Code (Text):

    [tex]\text{array}=\left\{x,y\right\}/.Solve[\left\{u==x^3+2xy+y^2,v==
    x^2+y\right\},\left\{x,y\right\}][/tex]
     
    Note that quintics are involved so necessarilly Mathematica returns 4 sets of (very messy) solutions, 3 of which are extraneous. Find the good one, calculate the partials and verify:

    [tex]x_u(4,2)=-1/3\quad x_v(4,2)=4/3[/tex]

    [tex]y_u(4,2)=2/3\quad y_v(4,2)=-5/3[/tex]
     
    Last edited: Feb 26, 2006
  7. Feb 27, 2006 #6
    Thanks for assisting me further Saltydog. When I get around to finishing off that question I'll see if I obtain the same values as the ones you have posted.
     
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