= Solution
Apply the <Laplacian> in <polar coordinates> twice to the proposed separated function. For any real exponent $m$,
$$
\nabla^2(r^m\cos2\theta)=(m^2-4)r^{m-2}\cos2\theta,
$$
so the <biharmonic equation> becomes
$$
\nabla^4(r^m\cos2\theta)=(m^2-4)((m-2)^2-4)r^{m-4}\cos2\theta=0.
$$
Its four distinct roots are $m=-2,0,2,4$. The <separated polar solutions of the biharmonic equation> therefore give the complete radial family in this angular sector,
$$
\psi=(Ar^{-2}+B+Cr^2+Dr^4)\cos2\theta.
$$
Equivalently, its radial equation is a fourth-order <Cauchy-Euler equation>; four distinct powers supply four <linearly independent> solutions, so no logarithmic terms are required.
\b[Interior.] Continuity with value zero at the origin removes both $r^{-2}\cos2\theta$ and $\cos2\theta$. The latter depends on the approach direction even though it is bounded. The two <boundary conditions> at unit radius then give $C+D=0$ and $2C+4D=1$. Consequently
$$
\boxed{\psi_{\mathrm{in}}(r,\theta)=\tfrac12(r^4-r^2)\cos2\theta,\qquad 0\leq r\leq1.}
$$
This extends smoothly through the origin: $r^2\cos2\theta=x^2-y^2$ and $r^4\cos2\theta=(x^2+y^2)(x^2-y^2)$ are <polynomials>. Its radial <derivative> at $r=1$ is $(2r^3-r)\cos2\theta=\cos2\theta$.
\b[Exterior.] The condition on the radial <derivative> at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos2\theta$ is permitted, since the required decay concerns the radial <derivative>. Now $A+B=0$ and $-2A=1$, giving
$$
\boxed{\psi_{\mathrm{out}}(r,\theta)=\tfrac12(1-r^{-2})\cos2\theta,\qquad r\geq1.}
$$
Indeed $\partial_r\psi_{\mathrm{out}}=r^{-3}\cos2\theta\to0$, and its value at unit radius is zero. These are particular interior and exterior solutions for the <biharmonic streamfunction>; no condition requiring $\psi$ itself to vanish at infinity was imposed.
Back to article page
