Laplace's Equation — Question 5

Question 5 – Solution

Strategy. Regular disk modes preserve the mean, damp angular oscillations and diagonalize the boundary energy calculation.

Step 1: Retain only regular radial factors. For mode $n\ge 1$, separation gives $r^n$ and $r^{-n}$; the mean mode gives a constant and $\log r$. The singular factors cannot occur in a harmonic function on the full disk. Thus $$\boxed{u=2+r\cos\theta+\tfrac 12r^2\cos(2\theta) =2+x+\tfrac 12(x^2-y^2).}$$ The Cartesian polynomial verifies $\Delta u=1-1=0$ even at the center, and setting $r=1$ gives the prescribed data.

Step 2: Find the sharp range. Put $c=\cos\theta$. Then $g=3/2+c+c^2=(c+1/2)^2+5/4$, with $-1\le c\le 1$. The minimum $5/4$ occurs at $\theta=2\pi/3,4\pi/3$; the maximum $7/2$ occurs at $\theta=0$ modulo $2\pi$. Consequently $$\boxed{\tfrac 54<u(x,y)<\tfrac 72\quad\text{inside the disk}.}$$ The inequalities are strict by the strong maximum principle. The constants cannot be improved over the whole interior, since approaching the stated boundary points approaches those values.

Step 3: Identify the preserved mean. Both cosine terms have zero angular integral, so $$\frac 1{2\pi}\int_0^{2\pi}u(r,\theta)\,d\theta=2=u(0) \qquad(0<r\le 1).$$ This explicitly realizes the mean-value property. The center equals the boundary average, although most boundary points and most interior points have different temperatures.

Step 4: Compute flux and energy. At $r=1$, the outward derivative is $u_r=\cos\theta+\cos(2\theta)$. Its integral is zero, hence so is the total conductive flux $-\int u_r\,d\theta$. Green’s identity and trigonometric orthogonality give $$\int_{\mathrm{disk}}|\nabla u|^2\,dA =\int_0^{2\pi}g(\theta)u_r(1,\theta)\,d\theta =\pi+\tfrac 12\pi=\boxed{\tfrac 32\pi}.$$ Zero net heat flux does not imply zero gradient or zero energy: heat enters through some boundary portions and leaves through others.