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Thread: Please help with partial derivative.

  1. #1
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    Please help with partial derivative.

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

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

    From the book:

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

    $\displaystyle u_x=2xv_x + 2yv_y$

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

    $\displaystyle u_y=2yv_x + 2xv_y$

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

    $\displaystyle \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:

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

    $\displaystyle \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} $

    $\displaystyle \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} $

    $\displaystyle \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) } ] $

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




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

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


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

    Please help explain to me.

    Thanks

    Alan
    Last edited by yungman; Oct 10th 2010 at 04:12 PM.
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  2. #2
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    Quote Originally Posted by yungman View Post
    Show if v is harmonic ie. $\displaystyle \; \nabla^2v=0 \;$ , then $\displaystyle \nabla^2u=0 \hbox { where } u(x,y)=v(x^2-y^2,2xy) $

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

    From the book:

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

    $\displaystyle u_x=2xv_x + 2yv_y$

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

    $\displaystyle u_y=2yv_x + 2xv_y$

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

    $\displaystyle \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:

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

    $\displaystyle \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} $

    $\displaystyle \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} $

    $\displaystyle \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) } ] $

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




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

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


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

    Please help explain to me.

    Thanks

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

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

    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 $\displaystyle \displaystyle v$, they had to show it happens for $\displaystyle \displaystyle u$.

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

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

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

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

    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 $\displaystyle \displaystyle v$, they had to show it happens for $\displaystyle \displaystyle u$.

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

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

    capice?

    Thanks for the reply.

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

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

    My understanding is:

    $\displaystyle 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
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    Quote Originally Posted by yungman View Post
    Thanks for the reply.

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

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

    My understanding is:

    $\displaystyle 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 $\displaystyle \displaystyle t = x^2 - y^2$ and $\displaystyle \displaystyle s = 2xy$, so that $\displaystyle \displaystyle u (x,y) = v(t,s)$.

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

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

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

    so that

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

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

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

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

    My understanding is:

    $\displaystyle 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 $\displaystyle \displaystyle t = x^2 - y^2$ and $\displaystyle \displaystyle s = 2xy$, so that $\displaystyle \displaystyle u (x,y) = v(t,s)$.

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

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

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

    so that

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

    Now it seems they assumed $\displaystyle \displaystyle v_{tt} + v_{ss} = 0$. 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 $\displaystyle 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; Oct 11th 2010 at 04:28 PM.
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  7. #7
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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 $\displaystyle 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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