# MATH3401

Lecture 24

### Harmonic functions

Last time:

$$U =$$ concentration of something "in equilibrium"

$$\Rightarrow$$ $$\quad\Delta U = 0$$

Can also study $$\dfrac{\partial U}{\partial t} = \alpha \nabla U\quad (*)$$

If $$U$$ is the concentration of a chemical, $$(*)$$ is Fick's Law of chemical diffusion;

If $$U$$ is temperature, $$(*)$$ is Fourier's Law of heat conduction;

If $$U$$ is the electric potential, $$(*)$$ is Ohm's Law of electric conduction.

### Harmonic functions

Examples

On $$\R^2\setminus \left\{(0,0)\right\}$$ $f(x,y) = \ln \big(||(x,y)||\big) = \ln \left(\sqrt{x^2+y^2}\right)$ is harmonic. 📝

On $$\R^n\setminus \left\{\mathbf 0 \right\}$$, $$n\geq 3$$ $f(\mathbf x) = \frac{1}{||\mathbf x||^{n-2}} = \frac{1}{\left(x_1^2 + \cdots + x_n^2\right)^{\frac{n-2}{2}}}$ is harmonic.

### Harmonic functions

Theorem: If $$f(z) = u(x,y) + i v(x,y)$$ is analytic in $$\Omega\subseteq \C$$, then $$u,v$$ are harmonic in $$\Omega$$.

Proof: We know that: $$f$$ is analytic $$\Rightarrow$$ $$u$$ and $$v$$ have continuous partials of all orders and C/R hold, i. e. $u_x = v_y\quad\text{and}\quad u_y=-v_x\qquad (1)$

### Harmonic functions

Proof (cont): $$\partial_x$$ of $$(1)$$ implies $u_{xx} = v_{yx}\quad\text{and}\quad u_{yx}=-v_{xx}.\qquad (2)$

$$\partial_y$$ of $$(1)$$ implies $u_{xy} = v_{yy}\quad\text{and}\quad u_{yy}=-v_{yx}.\qquad (3)$

Since all partial of all orders are continuous $u_{xy} = u_{yx}\quad\text{and}\quad v_{xy}=v_{yx}.$

So $$(2)$$ and $$(3)$$ imply $\Delta u = 0 \quad \text{and}\quad \Delta v = 0.$

### Harmonic functions

Definition: If $$u$$ and $$v$$ are harmonic and satisfy C/R, then $$v$$ is called
a harmonic conjugate of $$u$$.

Theorem: $$f= u+iv$$ is analytic in $$\Omega$$ $$\iff$$ $$v$$ is harmonic conjugate of $$u$$.

Proof: $$(\Rightarrow)$$ done!

$$(\Leftarrow)$$ $$v$$ is harmonic conjugate of $$u$$ says that $$u$$ and $$v$$ are both harmonic, $$u, u_x, u_y, v_x, v_y$$ exist and are continuous and C/R hold throughout $$\Omega$$. Then $$f=u+iv$$ is analytic. $$\square$$

### Harmonic functions

Now suppose $$v$$ and $$w$$ are harmonic conjugates of $$u$$. Then $u+iw \quad \text{and}\quad u+iv$ are both analytic. C/R$$_{\text{I}}$$ implies $u_x=v_y=w_y\qquad (**)$

Integrate $$(**)$$ with respect of $$y$$. Then $$v = w+ \phi(x).$$

### Harmonic functions

C/R$$_{\text{II}}$$ implies $u_y=-v_x=-w_x\qquad (\heartsuit)$

Integrate $$(\heartsuit)$$ with respect of $$x$$ $$\implies$$ $$v = w+ \psi(y).$$

Thus $$\phi(x) = \psi (y)$$, which hence must be a constant, i. e. $v= w+c.$

Use a similar procedure to find harmonic conjugate of a given harmonic function $$u$$.

### Harmonic functions

Example 1: $$u(x,y) = y^3 - 3x^2y$$. Find a harmonic conjugate.

Solution: $$u$$ is a polynomial function in $$x$$ and $$y$$, so has continuous partials of all orders.

Further $$u_{xx} + u_{yy}= 0$$.      📝

Let $$v$$ be a harmonic conjugate of $$u$$. C/R$$_{\text{I}}$$ implies $$u_x = v_y$$

i. e. $$v_y = -6xy$$. Integrate with respect of $$y$$ implies $v = -3xy^2 + \phi(x).$

### Harmonic functions

Solution (cont): C/R$$_{\text{II}}$$ implies $$u_y = -v_x$$.

$$\implies 3y^2 - 3x^2 = 3y^2 -\phi'(x)$$

$$\implies \phi'(x) = 3x^2$$

$$\implies \phi(x) = x^3+C$$

So $$(C=0)$$ $$v = -3xy^2+x^3$$ is a harmonic conjugate of $$u$$.

Note: For $$f(z)=iz^3$$, $$u = \Re(\,f)$$ and $$v = \Im(\,f)$$.