Show: u = x+y2 is Not a Harmonic Function

[andrey21]

Show that the following is not a harmonic function:

u = x+y²

This is what I know:

du/dx = 1 this must be equal to dv/dy , therefore:

v = y

du/dy = 2y this must be equal to -dv/dx, therefore:

v = -2xy

Putting expressions together:

v = y-2xy.

Not sure where to go from here?

 

[Dick]

You don't need to compute the harmonic conjugate (the 'v') to check whether a function is harmonic. Just check whether the laplacian vanishes.

 

[andrey21]

So using the following:

d²u / dx² + d²u/dy²=0

Which for this example gicves:

2+ 0 = 2 so laplacian does not disappear!

 

[Dick]

So using the following:

d²u / dx² + d²u/dy²=0

Which for this example gicves:

2+ 0 = 2 so laplacian does not disappear!

Right. So u is not harmonic.

 

[andrey21]

Thanks Dick, I have a similar question which is giving me some problems:

I'm asked to show v = x+2y is part of a harmonic conjugate pair. What is harmonic conjugate pair?

 

[Dick]

Thanks Dick, I have a similar question which is giving me some problems:

I'm asked to show v = x+2y is part of a harmonic conjugate pair. What is harmonic conjugate pair?

Try to work it out using Cauchy-Riemann, like you were doing before.

 

[andrey21]

Ok so here's my attempt:

v=x+2y

dv/dy = 2 must equal du/dx therefore:

u = 2x

dv/dx = 1 which must equal -du/dy

u = -y

adding together:

u = 2x-y

Am I on the right track??

 

[Dick]

Ok so here's my attempt:

v=x+2y

dv/dy = 2 must equal du/dx therefore:

u = 2x

dv/dx = 1 which must equal -du/dy

u = -y

adding together:

u = 2x-y

Am I on the right track??

Sure. You might be a little more careful of keeping track of things. du/dx=2 means u=2x+f(y) where f(y) is any function of y. du/dy=(-1) means u=-y+g(x) where g(x) is any function of x. When you compare the two expressions you realize f(y)=(-y) and g(x)=2x.

 

[andrey21]

Ok so now I have:

u = 2x-y and v = x+2y

Which can be written as:

2x-y +xi +2yi

 

[Dick]

Ok so now I have:

u = 2x-y and v = x+2y

Which can be written as:

2x-y +xi +2yi

Ok, so u+iv is an analytic function. Is there more to the problem?

 

[andrey21]

I must find its harmonic conjugate and say what f(z) is.

 

[Dick]

I must find its harmonic conjugate and say what f(z) is.

I thought so. What kind of function does it look like f(z) is, linear, quadratic, exponential??

 

[andrey21]

Im really not sure, I'm pretty sure it isn't a quadratic.

 

[Dick]

Im really not sure, I'm pretty sure it isn't a quadratic.

Guess! What kind of function does it look like? What kind of powers of x and y do you see?

 

[andrey21]

exponential?

 

[Dick]

exponential?

Wrong. What kind of powers of x and y do you see in u+iv? Why would you say 'exponential'!?

 

[HallsofIvy]

You have f(x,y)= 2x-y + (x +2y)i; what part of that looks exponential?

 

[andrey21]

Im not to sure what you are asking:

I know that: (x+iy)² = x²-y²+2xyi

 

[Dick]

Im not to sure what you are asking:

I know that: (x+iy)² = x²-y²+2xyi

Ok, so in that example you've got z^2 on one side and powers like x^2 on the other. Is that a clue? What about your given function?

 

[andrey21]

So my function will be z on one side?

 

[Dick]

So my function will be z on one side?

The highest power of z you would expect to see in f(z) is first power, yes. But that doesn't mean that's all you can have. You could have constants as well. What's the most general form of a linear function in z?

 

[andrey21]

z = f(x,y) = c+mx+my

 

[Dick]

z = f(x,y) = c+mx+my

You want to find f(z). You already know f(x,y). What does a linear function of z look like??

 

[andrey21]

So will it just be c+mz

 

[Dick]

So will it just be c+mz

Sure. f(z) must be m*z+c where m and c are complex. Now try to figure out an m and c so that when you expand into real and imaginary parts it matches your u+iy.

 

[andrey21]

zi +2z

 

[Dick]

zi +2z

Sure, that's it.

 

[andrey21]

Thanks for your help Dick