Determine whether functions are harmonic

[Shackleford]

Homework Statement

Determine whether or not the following functions are harmonic:

u = z + \bar{z}

u = 2z\bar{z}

Homework Equations

z = u(x,y) + v(x,y)i

\bar{z} = u(x,y) - v(x,y)i

A function is harmonic if Δu = 0.

The Attempt at a Solution

Δu = Δz +Δ \bar{z} = u_{xx} + v_{xx} + u_{yy} + v_{yy} + u_{xx} - v_{xx} + u_{yy} + -v_{yy} = 2u_{xx} + 2u_{yy}<br /> <br />

u = 2z\bar{z} = 2[u(x,y) + v(x,y)i][u(x,y) - v(x,y)i] = 2[u^2(x,y) - v^2(x,y)]

Δu = 2[2uu_{xx} + 2uu_{yy} - 2vv_{xx} - 2vv_{yy}]<br /> <br />

 

[Dick]

Homework Statement

Determine whether or not the following functions are harmonic:

u = z + \bar{z}

u = 2z\bar{z}

Homework Equations

z = u(x,y) + v(x,y)i

\bar{z} = u(x,y) - v(x,y)i

A function is harmonic if Δu = 0.

The Attempt at a Solution

Δu = Δz +Δ \bar{z} = u_{xx} + v_{xx} + u_{yy} + v_{yy} + u_{xx} - v_{xx} + u_{yy} + -v_{yy} = 2u_{xx} + 2u_{yy}<br /> <br />

u = 2z\bar{z} = 2[u(x,y) + v(x,y)i][u(x,y) - v(x,y)i] = 2[u^2(x,y) - v^2(x,y)]

Δu = 2[2uu_{xx} + 2uu_{yy} - 2vv_{xx} - 2vv_{yy}]<br /> <br />

I don't see why you are struggling with this. $z=x+iy$. If $u=z+\bar{z}$ then $u(x,y)=2x$. Is that harmonic?

 

[Shackleford]

I don't see why you are struggling with this. $z=x+iy$. If $u=z+\bar{z}$ then $u(x,y)=2x$. Is that harmonic?

I wanted to use the more general case. To be honest, I just wanted to check my work.

If it's not zero, then it's not harmonic.

 

[Dick]

I wanted to use the more general case. To be honest, I just wanted to check my work.

If it's not zero, then it's not harmonic.

I'm not sure what you are saying here. Don't do the general case. Just do these two special cases. What about those?

 

[Shackleford]

I'm not sure what you are saying here. Don't do the general case. Just do these two special cases. What about those?

Sorry. It was my mistake. For some reason I wanted to generalize to a function f(z).

Here, the first is harmonic. Δ(2x) = 0 and Δ(2x²+2y²) = 4 + 4 = 8.

 

[Dick]

$(x+iy)(x-iy)$ is not equal to $x^2-y^2$.

 

[Shackleford]

$(x+iy)(x-iy)$ is not equal to $x^2-y^2$.

Corrected.

 

[Dick]

Sorry. It was my mistake. For some reason I wanted to generalize to a function f(z).

Here, the first is harmonic. Δ(2x) = 0 and Δ(2x²+2y²) = 4 + 4 = 8.

That's better.

 

[Shackleford]

That's better.

Thanks again.