Laplace's Equation — Question 6

Question 6 – Solution

Strategy. The logarithmic radial mode is admissible on an annulus, but its small-hole limit is not uniform near the shrinking inner boundary.

Step 1: Solve the radial equation. The equation $(ru_r)'=0$ gives $u=A\log r+B$. Writing $D=\log(R/a)>0$ and $\Delta T=T_i-T_o$, the boundary data give $$\boxed{u(r)=T_o+\Delta T\,\frac{\log(R/r)}{D}.}$$ Its radial Laplacian is zero and its two traces are correct. Any other continuous harmonic solution differs by a harmonic function with zero boundary data; the maximum principle on the annulus proves uniqueness.

Step 2: Respect the inner normal. Here $u_r=-\Delta T/(rD)$. At $r=R$ the outward normal points toward increasing $r$, whereas at $r=a$ it points toward decreasing $r$. Thus $$\boxed{F_{\mathrm{outer}}=\frac{2\pi\Delta T}{D},\qquad F_{\mathrm{inner}}=-\frac{2\pi\Delta T}{D},\qquad G=\frac{2\pi}{D}.}$$ The sum is zero. When $\Delta T>0$, heat enters the annulus at the inner boundary and leaves at the outer boundary.

Step 3: Integrate the energy. Direct radial integration yields $$\mathcal E=2\pi\int_a^R\frac{(\Delta T)^2}{r^2D^2}\,r\,dr =\boxed{\frac{2\pi(\Delta T)^2}{D}=G(\Delta T)^2}.$$ Equivalently, Green’s identity gives $\mathcal E=-T_iF_{\mathrm{inner}}-T_oF_{\mathrm{outer}}$, with the same sign and value.

Step 4: Separate local and uniform limits. As $a\downarrow 0$, $D\to\infty$. On each fixed $r_0\le r\le R$ the field tends uniformly to $T_o$, and $\mathcal E\to 0$. However $\sup_{a\le r\le R}|u-T_o|=|\Delta T|$ for every $a$. Also $\max|\nabla u|=|\Delta T|/[a\log(R/a)]\to\infty$ at the inner circle. The concentration near a shrinking boundary is compatible with vanishing integrated energy.