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For a harmonic function between $\tau =a$ and $\tau =b$, $0<a<b$, in bipolar coordinates, suppose $\Phi (\sigma ,a)=0$ and $\Phi (\sigma ,b)=F(\sigma )$. Write the periodic boundary values as a Fourier seriesSeparation of variables givesEach radial factor solves $M^{\prime \prime }-n^{2}M=0$ and has endpoint values zero and one; the mean mode solves $M^{\prime \prime }=0$. Sufficiently regular boundary data give the required Fourier series solution. Uniqueness follows from the maximum principle for a harmonic function with zero Dirichlet boundary conditions on both circles. The formula follows the coordinate values; $\tau =b$ is the geometrically smaller circle.

A solution exists when at least one object satisfies all the stated conditions. In first-order logic, this is expressed by existential quantification. Uniqueness additionally requires that any two solutions are equal.

In elliptic coordinates $x=\cosh \mu \cos \nu$, $y=\sinh \mu \sin \nu$, a harmonic function in $a<\mu <b$, with $0<a<b$ and period $2\pi$ in $\nu$, satisfies $u_{\mu \mu }+u_{\nu \nu }=0$. Prescribe $u(b,\nu )=f(\nu )$ and $u_{\mu }(a,\nu )=g(\nu )$. Expand both boundary functions in Fourier series, writing their constant terms as $f_0,g_0$. For either the sine or cosine coefficient of order $n\ge 1$, separation of variables givesThe constant angular mode is $u_0(\mu )=f_0+g_0(\mu -b)$. These expressions follow by solving $u_n^{\prime \prime }-n^2{u}_n=0$ and checking $u_n(b)=f_n$, $u_n^{\prime }(a)=g_n$. Their Fourier series gives the solution for sufficiently regular boundary data.

The phrase normal derivative needs care: the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$, so the outward unit normal derivative on the inner ellipse is ${\partial }_{n}u=-h^{-1}{u}_{\mu }$. Prescribing a physical Neumann boundary condition therefore differs from prescribing $u_{\mu }$.

For uniqueness, subtract two solutions to obtain $v$. Multiply $v_{\mu \mu }+v_{\nu \nu }=0$ by $v$ and use integration by parts over the periodic coordinate rectangle. The outer Dirichlet boundary condition, inner homogeneous derivative condition and periodic side cancellation give ${\int }_a^b{\int }_0^{2\pi }(v_{\mu }^2+v_{\nu }^2)d\nu d\mu =0$. Thus $v$ is constant, and the outer condition makes it zero.

The only nonzero angular Fourier mode is $\cos (2\sigma )$. Its radial equation is $T^{\prime \prime }-4T=0$. A solution vanishing at $\tau =a$ is proportional to $\sinh [2(\tau -a)]$; making its value one at $b$ fixes the constant. The Dirichlet problem between bipolar coordinate circles therefore givesIts second derivatives in $\sigma$ and $\tau$ cancel, and the endpoint values are respectively zero and $\cos (2\sigma )$. The difference of any two solutions would be a harmonic function with zero Dirichlet boundary conditions, so the maximum principle proves uniqueness.

Geometric wording. In the printed bipolar coordinates, the circle radius $1/\sinh \tau$ decreases with $\tau$. For $0<a<b$, the $b$ circle is contained inside the $a$ circle. Thus the source’s “inner” and “outer” labels are interchanged geometrically. The solution above follows the explicit coordinate boundary values, which unambiguously specify the problem.

Write $Y={\sum }_{n\ge 0}{b}_n{y}_n$ in the complete Sturm-Liouville eigenfunction expansion. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The Sturm-Liouville operator has the symmetry identitybecause both functions obey the homogeneous Dirichlet boundary conditions. Consequentlyso that $({\lambda }_m-\alpha )b_m=a_m$. The stipulated exclusion of every eigenvalue makes all denominators nonzero, and givesThis is the resolvent operator applied to $f$ in the weighted space. The same projection shows uniqueness: a homogeneous solution would have $({\lambda }_m-\alpha )b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.

Let $d=b-a$. In the mixed Dirichlet-Neumann problem in an elliptic annulus, the angular Fourier modes separate. The outer data excite the $\cos \nu$ mode, while the inner derivative excites $\cos 2\nu$.

For the first mode, a radial factor with zero derivative at $\mu =a$ is $\cosh (\mu -a)$. Normalizing its outer value gives $F_1(\mu )=\cosh (\mu -a)/\cosh d$. For the second mode, a radial factor vanishing at $\mu =b$ is $\sinh [2(b-\mu )]$. Its derivative at $a$ must be $-1$, fixing the coefficient to $1/[2\cosh (2d)]$. ThereforeEach term satisfies Laplace's equation; at $\mu =b$ the second term vanishes and the first is $\cos \nu$. At $\mu =a$, differentiating gives $0-\cos 2\nu$, as required. All other modes, including the constant mode, vanish by the homogeneous mixed conditions. Alternatively, integration by parts applied to the difference of two solutions gives a zero integral of $|\nabla \Phi {|}^2$, proving uniqueness.

Boundary convention. The calculation follows the explicitly printed coordinate derivative ${\partial }_{\mu }\Phi$. In elliptic coordinates, the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$ and the outward unit normal at the inner boundary points toward decreasing $\mu$. A physical normal derivative there would be $-h^{-1}{\partial }_{\mu }\Phi$. Thus the phrase “normal derivative” in the source must be understood in its displayed coordinate sense; imposing the same numerical data on a unit-normal derivative would be a different problem.

Write the equation as $(L+\beta )y=-\cos x/x$ with the positive Sturm-Liouville operator from part (a). A Sturm-Liouville eigenfunction expansion therefore gives $a_n=-f_n/(4n^2+\beta )$, where weighted orthogonal projection givesHere the product-to-sum formulas reduce the integral to half the sum of the integrals of $\sin ((2n+1)x)$ and $\sin ((2n-1)x)$. ConsequentlyAt $x=0$, use the continuous limits of the terms. The coefficients are $O(n^{-3})$, so the resulting endpoint series converges. Completeness is the ordinary Fourier sine series completeness after $v=xy$. For an independent check, solving $v^{\prime \prime }-\beta v=\cos x$ with zero endpoint values givesIts numerator vanishes at zero, and it has the same orthogonal projection coefficients. Positivity of $L+\beta$ proves uniqueness.

Set $u=\cos \theta$. The boundary data simplify to $u+2u^2=\frac{2}{3}{P}_0(u)+P_1(u)+\frac{4}{3}{P}_2(u)$. The axisymmetric harmonic extension from a sphere keeps only decaying radial powers outside and regular powers inside. HenceandEach Legendre polynomial coefficient agrees at $r=a$. The exterior solution tends to zero, and the interior solution is finite at $r=0$. The maximum principle for harmonic functions proves uniqueness in these respective classes.

Take the trace first: the antisymmetric tensor has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the scalar, symmetric-traceless and axial tensor decomposition:The Kronecker delta satisfies ${\delta }_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its uniqueness.