Lecture 24
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.
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.
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. \[u_x = v_y\quad\text{and}\quad u_y=-v_x\qquad (1)\]
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 \[\Delta u = 0 \quad \text{and}\quad \Delta v = 0.\]
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. \(\square\)
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 \(v = w+ \phi(x).\)
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 \(u\).
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 \[v = -3xy^2 + \phi(x).\]
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 \( v = \Im(\,f)\).