# Harmonic Functions

1. Aug 16, 2011

### Gh0stZA

Hi everyone,
if I could get a push in the right direction with this question I'd really appreciate it:

2. Aug 16, 2011

### HallsofIvy

To be harmonic, a function must satisfy
$$\nabla u= \frac{\partial^2 u}{\partial x^2}+ \frac{\partial^2 u}{\partial y^2}$$

$$\frac{\partial e^u}{\partial x}= \frac{\partial u}{\partial x}e^u$$
$$\frac{\partial^2 e^u}{\partial x^2}= \frac{\partial u}{\partial x^2}e^u+ \left(\partial u}{\partial x}\right)^2 e^u$$

Do the same with the y variable and add.

3. Aug 17, 2011

### Gh0stZA

Thank you.

So we know:
$$e^u \left( \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} \right) = \Delta u$$
and
$$\Delta u = \frac{\partial ^2 u}{\partial x^2} u + \frac{\partial ^2 u}{\partial y^2} u = 0 + 0 = 0$$

So we can say:
$$e^u \left( \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} \right) = 0$$

While $$e^u$$ is not zero, how can I prove that it's a constant from here on?

4. Aug 17, 2011

### Gh0stZA

I think I made a mistake.

$$\frac{\partial ^2 u}{\partial x^2} u + \frac{\partial ^2 u}{\partial y^2} u = \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2}$$ is it not?

$$e^u \left( \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} \right) = \left( \frac{\partial ^2 u}{\partial x^2} + \frac{\partial ^2 u}{\partial y^2} \right)$$

and

$$e^u = 1 \Rightarrow u = 0$$ which is a constant.

5. Aug 17, 2011

### jackmell

I think you're having problems with this Ghost. Look, I fixed Hall's Latex here:

$$\frac{\partial e^u}{\partial x}= \frac{\partial u}{\partial x}e^u$$
$$\frac{\partial^2 e^u}{\partial x^2}= \frac{\partial^2 u}{\partial x^2}e^u+ \left(\frac{\partial u}{\partial x}\right)^2 e^u$$

See, that's the second partial with respect to x. Now you compute the second partial with respect to y. Do the first partial, then using the chain-rule, get the second partial. Now add up the two sets. Since u is harmonic, the u_xx+u_yy term will drop out since it's zero right. Now, what must the remaining part of that sum be if it is to be harmonic, that is, if it is to equal zero? Keep in mind e^u is never zero for any value of u.

6. Aug 17, 2011

### Gh0stZA

Okay so we have:
$$\frac{\partial^2 e^u}{\partial x^2}= \frac{\partial^2 u}{\partial x^2}e^u+ \left(\frac{\partial u}{\partial x}\right)^2 e^u$$
$$\frac{\partial^2 e^u}{\partial y^2}= \frac{\partial^2 u}{\partial y^2}e^u+ \left(\frac{\partial u}{\partial y}\right)^2 e^u$$

$$\frac{\partial^2 e^u}{\partial x^2} + \frac{\partial^2 e^u}{\partial y^2} = \frac{\partial^2 u}{\partial x^2}e^u + \frac{\partial^2 u}{\partial y^2}e^u + \left(\frac{\partial u}{\partial x}\right)^2 e^u + \left(\frac{\partial u}{\partial y}\right)^2 e^u$$

$$\frac{\partial^2 e^u}{\partial x^2} + \frac{\partial^2 e^u}{\partial y^2} = e^u (0) + \left(\frac{\partial u}{\partial x}\right)^2 e^u + \left(\frac{\partial u}{\partial y}\right)^2 e^u$$

$$\frac{\partial^2 e^u}{\partial x^2} + \frac{\partial^2 e^u}{\partial y^2} = \left(\frac{\partial u}{\partial x}\right)^2 e^u + \left(\frac{\partial u}{\partial y}\right)^2 e^u$$

$$e^u \left( \frac{\partial^2 }{\partial x^2} + \frac{\partial^2}{\partial y^2} \right) = e^u \left( \left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 \right)$$

Am I right up to here?

7. Aug 17, 2011

### jackmell

Don't do the last step. So just consider the next to last expression written as:

$$\frac{\partial^2 e^u}{\partial x^2} + \frac{\partial^2 e^u}{\partial y^2} = e^u\left[\left(\frac{\partial u}{\partial x}\right)^2 + \left(\frac{\partial u}{\partial y}\right)^2 \right]$$

Now, in order for the function e^u to be harmonic, then the sum of the double partials on the left have to be zero right. So, if that has to be zero, then the right then has to be zero, but e^u is never zero. So that means, in order for e^u to be harmonic, that sum in the brackets has to be zero. But the squares mean that it's always positive or zero. But in order for it to be zero, then each first partial has to be zero. Well, there you go. What kind of function has it's first derivative zero?

8. Aug 18, 2011

### Gh0stZA

Thank you, I get it now. :)