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For the nonconstant boundary data, use separation of variables in polar coordinates and put $\varphi =R(r)\cos 2\theta$. Laplace's equation reduces toThis Cauchy-Euler equation has independent solutions $r^2$ and $r^{-2}$. Unlike a disk containing the origin, the annulus permits both. The zero inner boundary value requires $R=A(r^2-a^4{r}^{-2})$, and the outer value requires $A=V{b}^2/(b^4-a^4)$. HenceSubstituting $r=a$ gives zero, while $r=b$ gives $V\cos 2\theta$. This is the Dirichlet problem on an annulus for one Fourier mode; no other angular mode is needed.

For the changed inner boundary value, use superposition to add a radial harmonic function that is $-V$ at $a$ and zero at $b$. The radial zero-mode equation is $(r{R}^{\prime }{)}^{\prime }=0$, giving $R=C+D\log r$. Imposing the two constant boundary values givesTherefore the second answer isThe logarithmic term is independent of $\theta$, so it changes only the angular average of the electric potential. Both answers are unique for these Dirichlet boundary conditions, by uniqueness for Laplace's equation.

Apply the Laplacian in polar coordinates twice to the proposed separated function. For any real exponent $m$,so the biharmonic equation becomesIts 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,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.

Interior. Continuity with value zero at the origin removes both $r^{-2}\cos 2\theta$ and $\cos 2\theta$. 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 $2C+4D=1$. ConsequentlyThis extends smoothly through the origin: $r^2\cos 2\theta =x^2-y^2$ and $r^4\cos 2\theta =(x^2+y^2)(x^2-y^2)$ are polynomials. Its radial derivative at $r=1$ is $(2r^3-r)\cos 2\theta =\cos 2\theta$.

Exterior. The condition on the radial derivative at infinity excludes the $r^2$ and $r^4$ terms. A nonzero $B\cos 2\theta$ is permitted, since the required decay concerns the radial derivative. Now $A+B=0$ and $-2A=1$, givingIndeed ${\partial }_r{\psi }_{\mathrm{out}}=r^{-3}\cos 2\theta \to 0$, 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.

The homogeneous Cauchy-Euler equation has powers $x^r$ with $r(r-1)+2r-2=(r-1)(r+2)=0$. Boundedness at zero selects $x$, and boundedness at infinity selects $x^{-2}$. Put $G=Ax$ for $x<\zeta$ and $G=B{x}^{-2}$ for $x>\zeta$. Continuity gives $B=A{\zeta }^3$. Since the operator is $L=(x^{2}d/dx{)}^{\prime }-2$, the required delta-function normalization is ${\zeta }^2[G_x{]}_{{\zeta }^{-}}^{{\zeta }^{+}}=1$. Thus $-3A{\zeta }^2=1$, giving the bounded Green kernel for a radial Euler operator:In the requested notation, $\boxed{a=1,b=-2,c=1,d=-2,k_{-}=k_{+}=-1/3}$. The two expressions agree at $x=\zeta$.

With ordinary measure $d\zeta$, the solution isFor bounded forcing these integrals converge and give a bounded solution. It is unique because no nonzero combination of $x,x^{-2}$ is bounded at both ends. For a more general forcing, convergence and boundedness of the displayed integral must be checked.

Put $\Phi (r,\theta )=R(r)Y(\theta )$ in Laplace's equation, multiply by $r^2/(RY)$, and separate the two variables:With $s=\cos \theta$, the angular equation becomes $(1-s^2)Y_{ss}-2s{Y}_s+\lambda Y=0$. Requiring a smooth axisymmetric harmonic function at both poles selects $\lambda =n(n+1)$ and the Legendre polynomial $Y=P_n(s)$, for nonnegative integers $n$. The radial Cauchy-Euler equation has solutions $r^n$ and $r^{-n-1}$. Superposition over the angular eigenfunctions therefore givesThis expansion assumes regularity on each whole angular sphere; without it, singular angular solutions would also be possible. The boundary data in the shell involve only the $n=0,1$ modes. Define the spherical-shell harmonic interpolation weightsEach $p_n$ equals one at the inner boundary and zero at the outer one, and $q_n$ does the reverse. Hence the steady temperature isEach term is a harmonic function, and substitution verifies both prescribed boundary values. Uniqueness follows from the maximum principle for harmonic functions.

The homogeneous Cauchy-Euler equation has solutions $x$ and $x^{-1}$. Choose $y_L=x$, which vanishes at zero, and $y_R=x-x^{-1}$, which vanishes at one. Their Wronskian is $W(s)=y_L(s)y_R^{\prime }(s)-y_L^{\prime }(s)y_R(s)=2/s$. For a Green function integrated against the unweighted measure $ds$, continuity at $x=s$ and the jump condition $s^2[G_x{]}_{s^{-}}^{s^{+}}=1$ giveIt satisfies $L_{x}G=\delta (x-s)$ for $L=x^2{\partial }_x^2+x{\partial }_x-1$. Therefore the candidate solution isThe leading coefficient vanishes at zero, so the endpoint condition needs care. If $f$ extends continuously to $[0,1]$, this expression has limit $-f(0)$ as $x\downarrow 0$. Indeed $x^{-1}{\int }_0^{x}f(s)ds\to f(0)$ and $x{\int }_x^{1}f(s)s^{-2}ds\to f(0)$, while the remaining terms vanish. Thus the continuous zero-endpoint problem in this class requires $f(0)=0$. With that compatibility condition the formula gives both boundary values and is unique: a difference of two bounded solutions is $Ax+B/x$, whose two endpoint conditions force both constants to vanish. For incompatible forcing such as $f\equiv 1$, the bounded solution with $u(1)=0$ is $x-1$, which cannot also vanish at zero. Pointwise vanishing of $G(x,s)$ for fixed $s$ does not justify interchanging the singular endpoint limit with its integral. This issue is treated in Green kernel and endpoint compatibility for a singular Euler equation.

For the supplied compatible forcing $f(s)=s$, the integrals are $x^2/2$ and $(1-x^2)/2+\log x$. Their polynomial pieces cancel, givingDirect differentiation gives $Lu=x$, and $x\log x\to 0$ at zero. Its derivative diverges there, which is allowed because the boundary condition requires the value, not a finite derivative.

Set $\Phi =R(r)P(\theta )$ and multiply the Laplace equation by $r^2/(RP)$. Separation of variables givesWith $\xi =\cos \theta$, the angular equation is $[(1-{\xi }^2)P_{\xi }{]}^{\prime }+\lambda P=0$. Regularity at both poles selects the Legendre polynomials $P_n$ and $\lambda =n(n+1)$. The radial Cauchy-Euler equation has solutions $r^n,r^{-n-1}$. Expanding the angular data in this complete orthogonal system therefore giveswith the usual convergence required of a harmonic expansion on the region considered. Angular Legendre functions singular at the poles are excluded.

For $0<a<b$, the outer data have expansion $b^2{\cos }^2\theta =b^2/3+(2b^2/3)P_2(\cos \theta )$, while the inner data are pure $P_0$. Thus only degrees zero and two are needed. A radial interpolation matching each pair of boundary values givesThe first term is a linear combination of $1,r^{-1}$ and takes values $a^2,b^2/3$ at the two radii. The second is a combination of $r^2,r^{-3}$ and takes values zero and $2b^2/3$. These checks establish both boundary conditions; each term is harmonic. The maximum principle for harmonic functions gives uniqueness. This is an example of axisymmetric harmonic interpolation between concentric spheres.

Trying $y=x^m$ in the Cauchy-Euler equation givesso the homogeneous solution isContinuity with zero value at the singular endpoint excludes the second branch. Use the left and right adapted solutions $\ell (x)=x$, $r(x)=x^{-4}-x$, whose Wronskian isThe Dirichlet Green kernel for an Euler operator with exponents one and minus four is thereforeIt vanishes at both endpoints, is continuous at $x=\xi$, and has first-derivative jumpThus $L_{x}G=\delta (x-\xi )$ with ordinary source measure $d\xi$. For bounded forcing the Green function representation isThe apparent singular prefactor in the first term is canceled by the integral's decay at zero. The only regular homogeneous solution vanishing at $1$ is zero, so this solution is unique in the stated regular class.

For the piecewise source, split the integrals at $1/2$. If $x<1/2$, the needed integrals are ${\int }_0^x{\xi }^{6}d\xi =x^7/7$ and ${\int }_x^{1/2}({\xi }^6-\xi )d\xi$. If $x>1/2$, only ${\int }_0^{1/2}{\xi }^{6}d\xi =1/896$ remains. Simplification yieldsAt the interface both expressions have value $-31/8960$ and derivative $129/4480$, so no unintended Dirac delta source appears. Their second derivatives have the allowed jump associated with the discontinuous forcing.

The homogeneous equation is a Cauchy-Euler equation. Trying $y=x^m$ gives the repeated indicial rootHence, for $x>0$,Both independent solutions diverge at the origin, so no nonzero homogeneous combination satisfies the zero-origin conditions by continuous extension.

For a source point $\xi >0$, finiteness on $0<x<\xi$ forces the Green function to vanish on that side. On $x>\xi$, continuity and the unit first-derivative jump requireUsing the homogeneous solutions gives the causal Green function for a repeated-root Euler operatorIndeed, the Wronskian of $x^{-1}$ and $x^{-1}\log x$ is $x^{-3}$, and direct differentiation gives $G_x({\xi }^{+},\xi )=1$. Since $G$ is continuous, its second derivative produces precisely a unit Dirac delta, with no derivative of a delta. For each positive source it is identically zero near the origin, so both required origin values vanish. The differential operator itself is interpreted on $x>0$; no value of its singular coefficients at zero is needed.

Integrating against the forcing yieldsThe last integral equals $1/16$, by integration by parts or differentiating ${\int }_0^1{s}^{p}ds=1/(p+1)$. ConsequentlySubstitution gives $y^{\prime \prime }+3y^{\prime }/x+y/x^2=(6+9+1)x/16=x$, while $y(0)=y^{\prime }(0)=0$. The divergent homogeneous solutions prove uniqueness in the regular class. No boundary condition at infinity was imposed or required.

The homogeneous equation is a Cauchy-Euler equation. With the power-law ansatz $y=x^r$ on $x>0$, its indicial equation isThus $r=(1-\alpha )/2$ is a repeated indicial root. The two independent solutions are $x^r$ and $x^r\log x$, givingTheir Wronskian, with $W=y_1{y}_2^{\prime }-y_1^{\prime }{y}_2$, is $x^{2r-1}=x^{-\alpha }\ne 0$, verifying linear independence.

A single-valued potential must be $2\pi$-periodic in $\theta$. For a separated solution $\Phi =F(r)H(\theta )$, the Laplace equation in polar coordinates becomesThe periodic angular eigenvalue problem has $\mu =n^2$, $n\ge 0$: its nonconstant eigenfunctions are $\cos n\theta ,\sin n\theta$. Negative $\mu$ gives exponentials that cannot be periodic, and $\mu =0$ gives only a constant angular factor. For $n\ge 1$, the radial Cauchy-Euler equation $r^2{F}^{\prime \prime }+r{F}^{\prime }-n^{2}F=0$ gives $F=A{r}^n+B{r}^{-n}$. At $n=0$ the two radial solutions are $1$ and $\log r$. Superposition therefore givesThe expansion is interpreted on an annulus where the harmonic function is defined; particular boundary conditions determine its constants and convergence region. Single-valuedness is essential: allowing a cut in the angular domain would also permit nonperiodic angular solutions.

For the harmonic extension in a circular sector, write a separated harmonic function as $F(r)H(\theta )$. The Laplace equation in polar coordinates yieldsThe angular Dirichlet boundary conditions select the Fourier sine basisIndeed, the angular equation has no nonzero solution with nonpositive ${\nu }^2$: multiplying $-H^{\prime \prime }={\nu }^{2}H$ by $H$ and integrating gives $\int (H^{\prime }{)}^2={\nu }^2\int H^2$. The radial Cauchy-Euler equation has solutions $r^{{\nu }_n}$ and $r^{-{\nu }_n}$. Boundedness at the vertex excludes the latter, so superposition gives

Write $Y(t)=y(e^t)$, so $t=\log x$. The chain rule givesSubstitution into the Cauchy-Euler equation therefore gives $Y^{\prime \prime }+Y^{\prime }-2Y=0$. Its characteristic equation factors as $(r-1)(r+2)=0$, so $Y=C_1{e}^t+C_2{e}^{-2t}$. Returning to $x$ gives

Write $Y(t)=y(e^t)$, so $x=e^t>0$. The chain rule givesHence $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$ and the Cauchy-Euler equation becomesIts characteristic equation is $r^2+3r+2=(r+1)(r+2)=0$. The distinct roots give $Y(t)=A{e}^{-t}+B{e}^{-2t}$. Returning to the original independent variable givesThe same form holds on any negative interval avoiding $0$, using $t=\log |x|$; the stated substitution covers the positive interval. At $x=0$ the differential equation is singular, so crossing it is a separate extension problem.

Resonance occurs when forcing overlaps a homogeneous mode, requiring multiplication of the usual particular ansatz by an extra power or logarithm.

For angular Fourier modes of integer order $n\ge 2$, the biharmonic equation in polar coordinates has radial basis $r^n,r^{-n},r^{n+2},r^{2-n}$. Apply the Laplacian twice to obtain the displayed indicial equation. For $n=0$, repeated roots give the radial basis $1,\log r,r^2,r^2\log r$; for $n=1$, the basis is $r^{-1},r,r\log r,r^3$. These are Cauchy-Euler equations. Regularity at the origin and boundary conditions select the admissible powers. For $n=2$, a constant radial factor still multiplies $\cos 2\theta$ or $\sin 2\theta$, so it is bounded but fails continuity at the origin.