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log|z| is harmonic on the punctured plane

Example

On $C ∖ {0}$ , the function

$u (z) = \log ∣ z ∣ = \frac{1}{2} \log (x^{2} + y^{2})$

is harmonic.

Facts & Assumptions

Given: The function $u (x, y) = \frac{1}{2} \log (x^{2} + y^{2})$ on $(x, y) \neq (0, 0)$ .

[L1]

The real logarithm satisfies $(\log t)^{'} = 1 / t$ for $t > 0$ (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

Verification

technique · direct

1.1

Differentiating with [L1] gives $u_{x} = \frac{x}{x^{2} + y^{2}}, u_{y} = \frac{y}{x^{2} + y^{2}} .$

L1algebra

2.1

Differentiating once more gives $u_{x x} = \frac{y^{2} - x^{2}}{(x^{2} + y^{2})^{2}}, u_{y y} = \frac{x^{2} - y^{2}}{(x^{2} + y^{2})^{2}} .$

step 1.1algebra

3.1

The sum in step 2.1 is $0$ , so $u_{x x} + u_{y y} = 0$ on $C ∖ {0}$ ; by Plane harmonic functions, $u$ is harmonic there.

step 2.1∎

Depends on

Used by

log|z| has no global harmonic conjugate on C{0 Counterexample

Dependency tree · two levels

9 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jeremy Orloff, MIT 18.04 Topic 5: Introduction to Harmonic Functions (standard reference, not scraped)