Laplace's Equation — Question 7

Question 7 – Solution

Strategy. Test flux compatibility before separation; then divide each nonconstant disk mode by its normal-derivative multiplier.

Step 1: Test the solvability condition. The divergence theorem gives $$0=\int_{\mathrm{disk}}\Delta u\,dA =\int_{\partial{\mathrm{disk}}}\partial_nu\,ds.$$ The prescribed derivative integrates to $2\pi$, since both oscillatory terms integrate to zero. It violates a necessary condition, so the original problem has no classical solution.

Step 2: Correct the mean and construct the field. The corrected integral is $2\pi(1-c)$, so the only constant correction is $c=1$. Since $\partial_r(r^n)=n$ at $r=1$, the regular solution is $$u=3r\cos\theta+2r^2\sin(2\theta)+C=3x+4xy+C.$$ The nonconstant terms have zero area average by symmetry. Thus normalization gives $C=0$, and $$\boxed{u(x,y)=3x+4xy.}$$ Its Cartesian Laplacian vanishes, and its outward derivative is exactly $3\cos\theta+4\sin(2\theta)$.

Step 3: Establish uniqueness up to a constant. For two solutions, their difference $w$ has $\Delta w=0$ and $\partial_nw=0$. Green’s identity gives $\int|\nabla w|^2=\int_{\partial{\mathrm{disk}}}w\partial_nw=0$. Hence $\nabla w=0$, so $w$ is constant on the connected disk. The zero-area-mean condition forces that constant to vanish. Without a normalization, every constant translate is another solution and these are all possibilities.

Step 4: Check energy in two coordinate systems. The boundary trace is $3\cos\theta+2\sin(2\theta)$. Orthogonality gives $$\mathcal E=\int_0^{2\pi} [3\cos\theta+2\sin(2\theta)][3\cos\theta+4\sin(2\theta)]\,d\theta =\boxed{17\pi}.$$ Independently, $|\nabla u|^2=(3+4y)^2+16x^2=9+24y+16r^2$. Its disk integral is $9\pi+0+16(\pi/2)=17\pi$. The corrected normal derivative integrates to zero, so its negative, the conductive flux, has zero total as required.