# Functions of Complex Variables

1. Jan 22, 2010

### alchemistoff

1. The problem statement, all variables and given/known data
{Q 6.2.2 from Arfken "Mathematical Methods for Physicists"}
Having shown that the real part $$u(x,y)$$ and imaginary part $$v(x,y)$$ of an analytic function $$w(z)$$ each satisfy Laplace's equation, show that $$u(x,y)$$ and $$v(x,y)$$ cannot have either a maximum or a minimum in the interior of any region in which $$w(z)$$ is analytic. (They can have saddle points)

2. Relevant equations
Cauchy-Riemann (CR) relations for analyticity of the function $$u_x=v_y$$ and $$u_y=-v_x$$ where subscript stands for partial differentiation with respect to that variable.

$$\nabla^2u=0$$ and $$\nabla^2v=0$$ (it follows from CR relations and proves that analytic function satisfies Laplace's equation)

3. The attempt at a solution

The local minimum/maximum points are to satisfy $$u_x=0$$ and $$u_y=0$$
and
$$M=u_{xx}u_{yy}-(u_{xy})^2>0$$

$$\nabla^2u=u_{xx}+u_{yy}=0\therefore u_{xx}=-u_{yy}$$

$$M=-u_{yy}^2-u_{xy}^2\leq0$$

...and it looks like totally wrong direction...

2. Jan 22, 2010

### payumooli

i think you proved it, for a function of two variables to have a min or max
$$-u_{yy}^2-u_{xy}^2 > 0$$

but for a saddle point
$$-u_{yy}^2-u_{xy}^2 < 0$$
since the square of two real numbers is always positive, the condition will give only a number <= 0.
if it is = 0 you cannot conclude anything
if it is <0 it is a saddle point

so u(x,y) and v(x,y) cannot have either a maximum or a minimum in the interior of any region in which w(z) is analytic.

3. Jan 22, 2010

### snipez90

Wait, the square of a real number is non-negative, so haven't you completely ignored the case where you can have a saddle point or a local extremum?

4. Jan 22, 2010

### alchemistoff

Precisely, the fact that $$M$$ can be zero doesn't allow proof to be completed. The only way to complete it is to show that for points where $$u_x=0$$ and $$u_y=0$$ $$M$$ cannot be zero.

or there is probably different approach which I cannot see...

5. Jan 22, 2010

### Count Iblis

Well, what does Gauss's divergence theorem yield for the integral of the Laplacian of U over a small area containing a hypothetical maximum/minimum of U?

6. Jan 23, 2010

### alchemistoff

From Gaussian theorem we have

$$\int _V\nabla^2 u \, dV=\int_S \nabla u \cdot n\, dS$$
$$0=\int_S \nabla u \cdot n\, dS$$​

I intuitively see that zero flux of $$\nabla u$$ implies that $$u$$ cannot have minimum or maximum but do not fully grasp it.

If it is to be true, then in the enclosed region where $$u$$ is minimum/maximum $$\nabla u$$ must be all negative/positive but cannot see why this is true.

7. Jan 23, 2010

### Count Iblis

You have to make your intuition more explicit. Intuitively we see a maximum that looks like a mountain top, so we see a gradient that points away from the mountain top. You then have to think about how to translate that to pure math. What is clear is that you cannot see this effect when expanding U to first order around the maximum, you have to go to second order.

So, you can make the following plan. Let's try to use the fact that U is analytic. U has a series expansion and you can write U around its maximum as a quadratic plus an error term. This will capture the above picture correctly. If you now use Gauss' theorem and substitute in the flux term involving nabla U the second order expansion plus error term, you should be able to prove that the flux term has to be strictly positive if U has a maximum or strictly negative if U has a minimum.

8. Jan 23, 2010

Thanks.