How to Proceed with Analytical Functions Proof?

[malawi_glenn]

Homework Statement

See pdf - file

Homework Equations

See pdf - file

The Attempt at a Solution

See pdf - file

Is it right so far? How can I proceed?

 

[StatusX]

Can you relate [itex]\partial u(\bar z)/\partial x[/itex] and [itex]\partial u(z)/\partial x[/itex], or [itex]\partial u(\bar z)/\partial y[/itex] and [itex]\partial u(z)/\partial y[/itex]?

 

[malawi_glenn]

yeah, that is what I am trying to figure out HOW :P

[itex]\partial u(\bar z)/\partial x[/itex] = [itex]\partial u(z)/\partial x[/itex]

I think

and

[itex]\partial u(\bar z)/\partial y[/itex] = [itex]- \partial u(z)/\partial y[/itex]



?? =)

 

[StatusX]

Not quite. The notation is a little confusing here, since the same symbol is used to show the variable being differentiated and the place the derivative is evaluated. To be clear, you should specify where the derivative is evaluated separately.

In this case we have:

[tex]\frac{\partial u(\bar z)}{\partial x} =\frac{\partial u(x-iy)}{\partial x}[/tex]

Now we should rewrite this as:

[tex]= \frac{\partial u(x'-iy')}{\partial x'} \left|_{x'=x, y'=y}[/tex]

This might seem stupid, but it allows us to get what you need as follows:

[tex]= \frac{\partial u(x'+iy')}{\partial x'} \left|_{x'=x, y'=-y}[/tex]

[tex]= \frac{\partial u(z')}{\partial x'} \left|_{z'=\bar z}[/tex]

and similarly for the derivative with respect to y, although there's one more step there.

 

[malawi_glenn]

Not quite. The notation is a little confusing here, since the same symbol is used to show the variable being differentiated and the place the derivative is evaluated. To be clear, you should specify where the derivative is evaluated separately.

In this case we have:

[tex]\frac{\partial u(\bar z)}{\partial x} =\frac{\partial u(x-iy)}{\partial x}[/tex]

Now we should rewrite this as:

[tex]= \frac{\partial u(x'-iy')}{\partial x'} \left|_{x'=x, y'=y}[/tex]

This might seem stupid, but it allows us to get what you need as follows:

[tex]= \frac{\partial u(x'+iy')}{\partial x'} \left|_{x'=x, y'=-y}[/tex]

[tex]= \frac{\partial u(z')}{\partial x'} \left|_{z'=\bar z}[/tex]

and similarly for the derivative with respect to y, although there's one more step there.

I am sorry, the notation really confused me :S

 

[Dick]

Try think of it this way. If f(z)=f(x+iy)=u(x,y)+i*v(x,y). So g(z)=u(x,-y)-i*v(x,-y). So using the CR equations for u and v, prove the CR equations for U(x,y)=u(x,-y) and V(x,y)=-v(x,-y).

 

[malawi_glenn]

I still don't get it, what is worng? :S

[tex]U(x,y)=u(x,-y) and V(x,y)=-v(x,-y)[/tex]

[tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,y)}{\partial x}[/tex]

[tex]\dfrac{\partial U(x,y)}{\partial y} =\dfrac{\partial u(x,-y)}{\partial y} = - \dfrac{\partial u(x,y)}{\partial y} [/tex]

[tex]\dfrac{\partial V(x,y)}{\partial x} =\dfrac{\partial (-v(x,-y))}{\partial x} = -\dfrac{\partial v(x,y)}{\partial x} [/tex]

[tex]\dfrac{\partial V(x,y)}{\partial y} =\dfrac{\partial (-v(x,-y))}{\partial y} = \dfrac{\partial v(x,y)}{\partial y}[/tex]

Is apparently wrong, how would you do this Dick? Dont show me all, just show for one of them, then I try more myself.

thanks.

 

[Dick]

Is not wrong. Is right! I'll do one. Since du(x,y)/dy=-dv(x,y)/dx or du(x,-y)/dy=dv(x,-y)/dx (CR for f), from what you've sent, dU(x,y)/dy=du(x,-y)/dy=dv(x,-y)/dx=-dV(x,y)/dx which is the second CR for g.

 

[Dick]

Well, almost correct. Change things like
[tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,y)}{\partial x}[/tex]
to:
[tex]\dfrac{\partial U(x,y)}{\partial x} =\dfrac{\partial u(x,-y)}{\partial x} = \dfrac{\partial u(x,-y)}{\partial x}[/tex]

You can't just ignore the -y's, the U and V need them.

 

[malawi_glenn]

and for the thoughest one: ?

[tex]\dfrac{\partial V(x,y)}{\partial y} =\dfrac{\partial (-v(x,-y))}{\partial y} = (-1)(-1)\dfrac{\partial (-v(x,-y))}{\partial x}[/tex]

 

[Dick]

dV(x,y)/dy=d(-v(x,-y))/dy=-dv(x,-y)/dy=du(x,-y)/dx=dU(x,y)/dx. Since du(x,y)/dx=dv(x,y)/dy implies du(x,-y)/dx=-dv(x,-y)/dy.

 

[malawi_glenn]

hmm why is

dV(x,y)/dy=d(-v(x,-y))/dy=-dv(x,-y)/dy

?

I still think it is:

[tex]\dfrac{\partial}{\partial y}\left( V(h(x,y),g(x,y))\right) = \dfrac{\partial h(x,y)}{\partial y}\dfrac{\partial V}{\partial y}+ \dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}[/tex]

[tex]
V(h(x,y),g(x,y)) = -v(x,-y)[/tex]

[tex]
h(x,y) = x[/tex]

[tex]
g(x,y)=-y
[/tex]

gives

[tex]
\dfrac{\partial h(x,y)}{\partial y}\dfrac{\partial V}{\partial y}+
[/tex]

[tex]\dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}=
[/tex]

[tex]\dfrac{\partial g(x,y)}{\partial y}\dfrac{\partial V}{\partial y}=\\
[/tex]

[tex]
-1\dfrac{\partial V}{\partial y}= -1\dfrac{\partial (-v(h,g))}{\partial y}\\
[/tex]

[tex]
=(-1)(-1)\dfrac{\partial (v(x,-y))}{\partial y}\\
[/tex]

where am I wrong about this? Sorry for being noob :P

 

[malawi_glenn]

Since du(x,y)/dx=dv(x,y)/dy implies du(x,-y)/dx=-dv(x,-y)/dy

is due to:

d/dy(v(x,h(y)) = (dh/dy)*(dv(x,h)/dy) = -1(dv(x,h)/dy)

?

So why not chain rule when differentiating V(x,y)= -v(x,-y) ?

 

[malawi_glenn]

Never mind, my teacher will help me tomorrow. Thanks for all help!

 

[Dick]

hmm why is

dV(x,y)/dy=d(-v(x,-y))/dy=-dv(x,-y)/dy

?

This is not a chain rule. I'm just substituting equals for equals. V(x,y) is DEFINED to be -v(x,-y). Similarly using:

du(x,y)/dx=dv(x,y)/dy

I just SUBSTITUTE -y for y getting du(x,-y)/dx=dv(x,-y)/d(-y). If you want to appeal to the one-dimensional chain rule now dv(x,-y)/d(-y)=(dv(x,-y)/dy)*(dy/d(-y))=-dv(x,-y)/dy. Just a change of variable in y.
There is no use of the chain rule for partial derivatives here (which I don't think you are stating quite correctly). Does that help? I sympathize here, this can be confusing.

 

[malawi_glenn]

I think I got it now, will show my teacher my thougts and work tomorrow. Thanks for all the help guys!

see ya