Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a way to show that every harmonic function is the real part of a holomorphic function without using integration equations if later theorems are allowed also?

share|improve this question
2  
This amounts to solving the Cauchy–Riemann equations as a system of PDEs. Integration is probably the only method that works generically... –  Zhen Lin Feb 11 '12 at 12:38

1 Answer 1

up vote 7 down vote accepted

Is the following "integral free" enough?

Let $$h:\quad \Omega\to{\mathbb R}\ ,\qquad (x,y)\mapsto h(x,y)$$ be the given harmonic function which we assume to be $C^2$. Using $h$ we define the function $$f(x,y):= h_x(x,y)- i\ h_y(x,y)$$ with real part $u(x,y)=h_x(x,y)$ and imaginary part $v(x,y)=- h_y(x,y)$. As $h$ is harmonic one has $u_x\equiv v_y$, furthermore $u_y\equiv -v_x$ by equality of the mixed derivatives. So $f\colon \Omega\to{\mathbb C}$ is $C^1$ and satisfies the CR equations; therefore it is an analytic function of $z=x+iy\in\Omega$.

Assume that $\Omega$ is simply connected and chose a point $z_0\in\Omega$. Then by a standard theorem of complex analysis the function $$F(z)\ :=\ h(z_0)+\int_\gamma f(z)\ dz\ , \qquad \hbox{$\gamma\ $ a path from $z_0$ to $z$}\ ,$$ is an analytic primitive of $f$ in $\Omega$. Let $(x,y)\mapsto U(x,y)$ be the real part of $F$. Then by the CR equations, this time applied to $F$, we have $$U_x(z) -i U_y(z)=F'(z)=f(z)= h_x(z)-i h_y(z)\qquad(z\in\Omega)\ .$$ It follows that $$\nabla U(x,y)\equiv\nabla h(x,y)\qquad \bigl( (x,y)\in\Omega\bigr)\ ,$$ and as $U(x_0,y_0)=h(x_0,y_0)$ we conclude that in fact $U(x,y)\equiv h(x,y)$ in $\Omega$.

Note that we had to assume $\Omega$ simply connected. The function $h(x,y):=\log\sqrt{x^2+y^2}$ is harmonic in the punctured plane but is not the real part of an analytic function in this domain.

share|improve this answer
    
Thank you very much!! –  VVV Feb 11 '12 at 17:47
    
How we know that $log \sqrt{x^2+y^2}$ is not the real part of an analytic function? –  Idonknow Oct 6 '13 at 17:26
    
@Idonknow: This is one of the most basic facts of complex analysis. The analytic function you are looking for would be $f(z)=\log|z|+i{\rm Arg}(z)$, but this function is not well defined in all of ${\mathbb C}\setminus\{0\}$. –  Christian Blatter Oct 6 '13 at 19:01

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.