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Math Help - Please help with partial derivative.

  1. #1
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    Please help with partial derivative.

    Show if v is harmonic ie. \; \nabla^2v=0 \; , then \nabla^2u=0 \hbox { where } u(x,y)=v(x^2-y^2,2xy)

     \nabla^2u=0 \;\Rightarrow\; u_{xx}+u_{yy} = 0

    From the book:

    For u(x,y)=v(x^2-y^2,2xy)

    u_x=2xv_x + 2yv_y

     u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x

    u_y=2yv_x + 2xv_y

     u_{yy} = 4y^2v_{xx} - 8 xyv_{xy} + 4x^2v_{yy} - 2v_x

     \nabla^2u = u_{xx} + u_{yy} = (4x^2+4y^2)(v_{xx}+v_{yy}) = 0




    My question

    I don't understand the solution the book gave. This is my work:

    \nabla^2u = \nabla \cdot \nabla u = \nabla \cdot \nabla v(x^2-y^2,2xy)

    \nabla v(x^2-y^2,2xy) = [ \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x }]\hat{x} \;+\; [\frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial y } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial y }]\hat{y}

    \nabla v(x^2-y^2,2xy) = [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\hat{x} \;\;+\;\; [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ] \;\hat{y}

     \nabla^2 v = \nabla \cdot \nabla v = \frac{\partial}{\partial x} [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\;+\;\; \frac{\partial}{\partial y} [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ]

    \nabla^2 v = 4(x^2+y^2) [ \frac{\partial v}{\partial (x^2-y^2) } + \frac{\partial v}{\partial (2xy) }]




    I don't even understand how u_x=2xv_x + 2yv_y

    and  u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x


    What is v_x,\; v_{xx},\; v_y  \hbox { and } v_{yy} here stand for?

    Please help explain to me.

    Thanks

    Alan
    Last edited by yungman; October 10th 2010 at 05:12 PM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yungman View Post
    Show if v is harmonic ie. \; \nabla^2v=0 \; , then \nabla^2u=0 \hbox { where } u(x,y)=v(x^2-y^2,2xy)

     \nabla^2u=0 \;\Rightarrow\; u_{xx}+u_{yy} = 0

    From the book:

    For u(x,y)=v(x^2-y^2,2xy)

    u_x=2xv_x + 2yv_y

     u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x

    u_y=2yv_x + 2xv_y

     u_{yy} = 4y^2v_{xx} - 8 xyv_{xy} + 4x^2v_{yy} - 2v_x

     \nabla^2u = u_{xx} + u_{yy} = (4x^2+4y^2)(v_{xx}+v_{yy}) = 0




    My question

    I don't understand the solution the book gave. This is my work:

    \nabla^2u = \nabla \cdot \nabla u = \nabla \cdot \nabla v(x^2-y^2,2xy)

    \nabla v(x^2-y^2,2xy) = [ \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x }]\hat{x} \;+\; [\frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial y } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial y }]\hat{y}

    \nabla v(x^2-y^2,2xy) = [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\hat{x} \;\;+\;\; [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ] \;\hat{y}

     \nabla^2 v = \nabla \cdot \nabla v = \frac{\partial}{\partial x} [2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) } ] \;\;+\;\; \frac{\partial}{\partial y} [-2y\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2x \frac{\partial v}{\partial (2xy) } ]

    \nabla^2 v = 4(x^2+y^2) [ \frac{\partial v}{\partial (x^2-y^2) } + \frac{\partial v}{\partial (2xy) }]




    I don't even understand how u_x=2xv_x + 2yv_y

    and  u_{xx} = 4x^2v_{xx} + 8 xyv_{xy} + 4y^2v_{yy} + 2v_x


    What is v_x,\; v_{xx},\; v_y  \hbox { and } v_{yy} here stand for?

    Please help explain to me.

    Thanks

    Alan
    It's notation: \displaystyle v_x means the first partial derivative of \displaystyle v with respect to \displaystyle x, that is, \displaystyle \frac {\partial v}{\partial x}

    while, \displaystyle v_{xx} = \frac {\partial ^2 v}{\partial x^2}, the second partial derivative with respect to \displaystyle x.

    you can figure out the rest.


    I'm afraid your work is incorrect and completely misses the point. In fact, you almost go backwards....or maybe you're just mislabeling u and v...? at the very least, what you did was completely confusing. what are you differentiating with respect to?? functions of x and y, rather than just x and y themselves??



    Harmonic means, as they say, the second order un-mixed partials sum to zero. so, assuming this happens for \displaystyle v, they had to show it happens for \displaystyle u.

    Hence, the book found \displaystyle u_{xx} by differentiating the \displaystyle u(x,y) equation with respect to \displaystyle x twice, and then \displaystyle u_{yy} in a similar way.

    they then summed these two equations and found that it was zero. which means \displaystyle u(x,y) is harmonic.

    capice?
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    It's notation: \displaystyle v_x means the first partial derivative of \displaystyle v with respect to \displaystyle x, that is, \displaystyle \frac {\partial v}{\partial x}

    while, \displaystyle v_{xx} = \frac {\partial ^2 v}{\partial x^2}, the second partial derivative with respect to \displaystyle x.

    you can figure out the rest.


    I'm afraid your work is incorrect and completely misses the point. In fact, you almost go backwards....or maybe you're just mislabeling u and v...? at the very least, what you did was completely confusing. what are you differentiating with respect to?? functions of x and y, rather than just x and y themselves??



    Harmonic means, as they say, the second order un-mixed partials sum to zero. so, assuming this happens for \displaystyle v, they had to show it happens for \displaystyle u.

    Hence, the book found \displaystyle u_{xx} by differentiating the \displaystyle u(x,y) equation with respect to \displaystyle x twice, and then \displaystyle u_{yy} in a similar way.

    they then summed these two equations and found that it was zero. which means \displaystyle u(x,y) is harmonic.

    capice?

    Thanks for the reply.

    I understand  v_x=\frac{\partial v}{\partial x}. But how about  v_x=\frac{\partial v(x^2-y^2,2xy)  }{\partial x} ?

    It is by definition that \nabla^2 v = \nabla \cdot \nabla v. If you carry out the calculation the result will still equal to v_{xx}+v_{yy}

    My understanding is:

     v_x = \frac{\partial v(x^2-y^2,2xy)}{\partial x}  = \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x } = 2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) }

    Can you tell me what I did wrong?
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yungman View Post
    Thanks for the reply.

    I understand  v_x=\frac{\partial v}{\partial x}. But how about  v_x=\frac{\partial v(x^2-y^2,2xy)  }{\partial x} ?

    It is by definition that \nabla^2 v = \nabla \cdot \nabla v. If you carry out the calculation the result will still equal to v_{xx}+v_{yy}

    My understanding is:

     v_x = \frac{\partial v(x^2-y^2,2xy)}{\partial x}  = \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x } = 2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) }

    Can you tell me what I did wrong?
    Oh! Haha, I didn't see the comma in the definition of u, i thought it was a plus. so then, you are more on the right track than the book i'd say. you did nothing wrong so far.

    to make writing (and not to mention typing) easier. say \displaystyle t = x^2 - y^2 and \displaystyle s = 2xy, so that \displaystyle u (x,y) = v(t,s).

    Now, lets apply the chain rule with a bit less pain. you would get

    \displaystyle u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}, and

    \displaystyle u_{yy} = -2v_t + 4y^2v_{tt} - 8xyv_{ts} + 4x^2v_{ss}

    so that

    \displaystyle u_{xx} + u_{yy} = (4x^2 + 4y^2)(v_{tt} + v_{ss})

    Now it seems they assumed \displaystyle v_{tt} + v_{ss} = 0. why that's acceptable, i'm not sure. what do you know about v?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yungman View Post
    Thanks for the reply.

    I understand  v_x=\frac{\partial v}{\partial x}. But how about  v_x=\frac{\partial v(x^2-y^2,2xy)  }{\partial x} ?

    It is by definition that \nabla^2 v = \nabla \cdot \nabla v. If you carry out the calculation the result will still equal to v_{xx}+v_{yy}

    My understanding is:

     v_x = \frac{\partial v(x^2-y^2,2xy)}{\partial x}  = \frac{\partial v}{\partial (x^2-y^2) } \frac{\partial (x^2-y^2) }{\partial x } + \frac{\partial v}{\partial (2xy) } \frac{\partial (2xy) }{\partial x } = 2x\frac{\partial v}{\partial (x^2-y^2) } \;+\; 2y \frac{\partial v}{\partial (2xy) }

    Can you tell me what I did wrong?
    Oh! Haha, I didn't see the comma in the definition of u, i thought it was a plus. so then, you are more on the right track than the book i'd say. you did nothing wrong so far.

    to make writing (and not to mention typing) easier. say \displaystyle t = x^2 - y^2 and \displaystyle s = 2xy, so that \displaystyle u (x,y) = v(t,s).

    Now, lets apply the chain rule with a bit less pain. you would get

    \displaystyle u_{xx} = 2v_t + 4x^2 v_{tt} + 8xyv_{ts} + 4y^2 v_{ss}, and

    \displaystyle u_{yy} = -2v_t + 4y^2v_{tt} - 8xyv_{ts} + 4x^2v_{ss}

    so that

    \displaystyle u_{xx} + u_{yy} = (4x^2 + 4y^2)(v_{tt} + v_{ss})

    Now it seems they assumed \displaystyle v_{tt} + v_{ss} = 0. why that's acceptable, i'm not sure. what do you know about v?
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  6. #6
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    Thanks very much for the reply. I think the book make it more confuse to transform x\rightarrow (x^2-y^2) \hbox { and } y\rightarrow 2xy . If they have use t and s like you, then it would be a lot clearer.


    Thanks.
    Last edited by yungman; October 11th 2010 at 05:28 PM.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by yungman View Post
    Thanks very much for the reply. I think the book make it more confuse to transform x\rightarrow (x^2-y^2) \hbox { and } y\rightarrow 2xy . If they have use t and s like you, then it would be a lot clearer.


    Thanks.
    Well, it would seem so...but that would be illegal. you can't use a variable that's already in use to make a substitution with...
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