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?

Thanks

Alan

2. Originally Posted by yungman
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?

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?

3. Originally Posted by Jhevon
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?

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?

4. Originally Posted by yungman

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?

5. Originally Posted by yungman

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?

6. 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.

7. Originally Posted by yungman
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...