Real part of a analytic function

[Pouyan]

Homework Statement

Show that xu_x + yu_y is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂²φ/∂x²) + (∂²φ/∂y²) =0

The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=u_x+xu_xx+yu_xy =(∂v/∂y)
(∂u/∂y)=xu_xy+u_y+yu_yy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!

 

[Pouyan]

If I say there is a function F(z)=U(x,y)+iV(x,y)
and
U(x,y) = x (∂u/∂x) + y(∂u/∂y)
Can I use Cauchy-Riemann to find V(x,y)?

Is this right ?!

 

[Pouyan]

Please can somebody help me ?!

 

[Mark44]

Homework Statement

Show that xu_x + yu_y is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

In the second equation, it should be (∂u/∂y)=-(∂v/∂x)

I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂²φ/∂x²) + (∂²φ/∂y²) =0

The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=u_x+xu_xx+yu_xy =(∂v/∂y)
(∂u/∂y)=xu_xy+u_y+yu_yy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!

For the first part, I don't think you are. The statement to prove is: If u(x, y) is the real part of an analytic function f(z), then Re(f(z)) = xu_x + yu_y.

 

[Pouyan]

In the second equation, it should be (∂u/∂y)=-(∂v/∂x)

For the first part, I don't think you are. The statement to prove is: If u(x, y) is the real part of an analytic function f(z), then Re(f(z)) = xu_x + yu_y.

OK if I say f(z)= U(x,y) + i V(x,y), then U(x,y) = x (∂u/∂x) + y(∂u/∂y) ?! (U is a new function and is not the same like u)
But how can I use it ?! It's not easy to use Cauchy-Riemann to find V

 

[Mark44]

OK if I say f(z)= U(x,y) + i V(x,y), then U(x,y) = x (∂u/∂x) + y(∂u/∂y) ?! (U is a new function and is not the same like u)

What you're calling U above is, I believe, the same as u in ∂u/∂x. If they are the same, you should use just one letter and be consistent.
The situation as I see it is that f(z) = f(x + iy) = u(x, y) + iv(x, y).

If I'm wrong in this assumption, what's the meaning of u?

But how can I use it ?! It's not easy to use Cauchy-Riemann to find V

 

[Mark44]

Show that xu_x + yu_y is the real part of an analytic function if u(x,y) is.

Is this the exact wording of the problem?

It suggests to me that if f(z) = f(x + iy) = u(x, y) + iv(x, y), then u(x, y) = xu_x(x, y) + yu_y(x, y). I don't see why that would be true, but I have to admit it's been a long time since I took a class in complex analysis.

 

[Pouyan]

OK somebody told me F(z) = U(x,y)+iV(x,y) and U(x,y) =xu_x + yu_y
But if I use Cauchy -Riemann , it's going to be very hard to find V(x,y) ... I can come with more post and show why

 

[Pouyan]

Is this the exact wording of the problem?

It suggests to me that if f(z) = f(x + iy) = u(x, y) + iv(x, y), then u(x, y) = xu_x(x, y) + yu_y(x, y). I don't see why that would be true, but I have to admit it's been a long time since I took a class in complex analysis.

OK thank you anyway

 

[Pouyan]

∂V/∂x = (-∂u/∂y) - x(∂²/∂x∂y) -y(∂²/∂y²)

∂V/∂y = y(∂u/∂x) +x(∂²u/∂x²)+y(∂²u/∂x∂y)

But I don't know how I can integrate to find V

 

[Mark44]

∂V/∂x = (-∂u/∂y) - x(∂²/∂x∂y) -y(∂²/∂y²)
∂V/∂y = y(∂u/∂x) +x(∂²u/∂x²)+y(∂²u/∂x∂y)

These parts in the first line above don't make sense:

x(∂²/∂x∂y)
y(∂²/∂y²)

This -- ∂²/∂x∂y -- and the other one are operators. You have to indicate what you are taking the partial of, like you did in the second line of the first quote above.

But I don't know how I can integrate to find V

 

[Ray Vickson]

Homework Statement

Show that xu_x + yu_y is the real part of an analytic function if u(x,y) is.
To which analytic function is the real part of u = Re (f(z))?

Homework Equations

What I know about analytic functions is Cauchy-Riemann condition
(∂u/∂x) =(∂v/∂y) and (∂y/∂y)=-(∂v/∂x)

I know actually Harmonic functions and Laplace equation (2-dim) but I don't know if I need it here:
(∂²φ/∂x²) + (∂²φ/∂y²) =0

The Attempt at a Solution

[/B]I say that there is a analytic function : f(z)=u(x,y)+iv(x,y)
(∂u/∂x)=u_x+xu_xx+yu_xy =(∂v/∂y)
(∂u/∂y)=xu_xy+u_y+yu_yy=-(∂v/∂x)
But further , should I integrate to find v(x,y) ?!

Am I in right path ?!

Basically, you start with some analytic $f(z) = f(x+iy) = u(x,y) + i v(x,y)$ and are then asked to show that there is an analytic $F(z)$ such that $F(x+iy) = [x u_x + y u_y]+ i V(x,y)$ for some $V$. Apply Cauchy-Riemann to the pair $U(x,y) = x u_x(x,y) + y u_y(x,y)$ and $V(x,y)$, so see what can be said about $V$ in terms of $u$ and $v$. In particular, can you show that an appropriate $V$ actually exists?

 

[Pouyan]

Basically, you start with some analytic $f(z) = f(x+iy) = u(x,y) + i v(x,y)$ and are then asked to show that there is an analytic $F(z)$ such that $F(x+iy) = [x u_x + y u_y]+ i V(x,y)$ for some $V$. Apply Cauchy-Riemann to the pair $U(x,y) = x u_x(x,y) + y u_y(x,y)$ and $V(x,y)$, so see what can be said about $V$ in terms of $u$ and $v$. In particular, can you show that an appropriate $V$ actually exists?

Cauchy-Riemann : ∂U/∂x = ∂V/∂y and ∂U/∂y = - ∂V/∂x

∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

But I don't know how I find V in this equation!

I know that the U(x,y) is harmonic (∂²U/∂x²) +(∂²U/∂y²) =0
But that gives me nothing to find V!

 

[Mark44]

Cauchy-Riemann : ∂U/∂x = ∂V/∂y and ∂U/∂y = - ∂V/∂x

∂V/∂x = (-∂u/∂y) - x(∂2/∂x∂y) -y(∂2/∂y2)

∂V/∂y = y(∂u/∂x) +x(∂2u/∂x2)+y(∂2u/∂x∂y)

But I don't know how I find V in this equation!

You could integrate both sides of the first equation above with respect to x, and integrate the second equation with respect to y.Keep in mind that when you integrate V_x with respect to x, you don't end up with an arbitrary constant, you end up with an arbitrary function of y alone, say g(y). Similar idea when you integrate the second equation with respect to y -- you end up with an arbitrary function of x alone, say f(x).

This will give you two "views" of V, which you can set equal. I believe that's what Ray was suggesting.

I know that the U(x,y) is harmonic (∂²U/∂x²) +(∂²U/∂y²) =0
But that gives me nothing to find V!