Recall from last lecture, conformal maps and Laplacian of harmonic functions.
If U is the concentration of something “in equilibrium”, that implies (somewhat) that \Delta U = 0. There are many solutions to this in general (constants, linear, etc) however we are often interested in boundary conditions.
Can we also study \frac{\partial U}{\partial t} = \alpha \Delta U? As the left hand side approaches 0, the Laplacian approaches 0 and the system approaches steady state. This has many physical applications.
Examples:
Note that these are radial functions around 0. But how badly do they behave?
Theorem. If f(z) = u(x,y) + iv(x,y) is analytic in \Omega \subseteq \mathbb C, then u and v are harmonic in \Omega.
Proof. Recall that if f is analytic then u and v have continuous partials of all orders and C/R holds. That is, u_x = v_y and u_y = -v_x. We can differentiate these and apply C/R again to get \begin{aligned} u_{xx} &= v_{yx} & u_{xy} &= -v_{xx} \\ u_{yx} &= v_{yy} & u_{yy} &= -v_{yx} \end{aligned} Since partials of all orders are continuous, by Clairaut’s theorem, u_{xy} = u_{yx} and v_{xy} = v_{yx}. Therefore, u_{xx} = v_{yx} = -u_{yy} and similarly for v, so \Delta u = 0 and \Delta v = 0. \square
Definition. If u and v are harmonic and satisfy C/R, then v is called a (not the) harmonic conjugate of u. Note that this is not symmetric.
Theorem. f = u+iv is analytic in \Omega if and only if v is a harmonic conjugate of u.
Proof. (\rightarrow) is done above. (\leftarrow) v is a harmonic conjugate so u and v are both harmonic and u, u_x, u_y, u_{xx}, u_{yy} all exist, are continuous and satisfy C/R throughout \Omega, f is analytic. \square
Example: Suppose v and w are harmonic conjugates of u. This means that u+iv and u+iw are both analytic. Applying C/R, \begin{aligned} u_x &= v_y = w_y, \quad \text{and}\quad u_y = -v_x = -w_x. \end{aligned} Integrating the derivatives of v and w wrt their partial variable, we get v = w + \phi(x) and v = w+\psi(y). Therefore, \phi(x) = \psi(y) which must be a constant. This means v = w+c. \circ
A similar procedure can be used to find a harmonic conjugate of a given harmonic function u.
Example: Find a harmonic conjugate of u(x,y) = y^3 - 3x^2y.
ui s a polynomial function of x and y so has continuous partials of all orders. Moreover, u_{xx}+u_{yy} = 0. Suppose v is a harmonic conjugate of u. C/R tells us u_x = v_y so v_y = -6xy. Integrating this wrt y gives us v = -3xy^2 + \phi(x). Using this in the second part of C/R, \begin{aligned} u_y &= -v_x \\ 3y^2 - 3x^2 &= 3y^2 - \phi'(x) \\ \phi'(x) &= 3x^2\\ \phi(x) &= x^3 + c \end{aligned} So, we can choose c =0 and v(x,y) = -3xy^2+x^3 is a harmonic conjugate of u. Note that in this example, u=\operatorname{Re}f and v=\operatorname{Im} f where f(z) = iz^3.