Put $s=|\mathbf{r}-{\mathbf{r}}^{\prime }|$. Away from the source, the radial Laplacian in three dimensions givesThus $\boxed{\Delta G_0=0\text{ for }\mathbf{r}\ne {\mathbf{r}}^{\prime }}$. At the source it is a distribution, not a smooth function; its outward gradient flux through a small sphere is $4\pi s^2/(4\pi s^2)=1$, giving $\Delta G_0=\delta$.

Expand the solution and the forcing in the orthonormal Sturm-Liouville eigenfunction expansion. The forcing coefficients areEvery ${\lambda }_n$ is nonzero, so the solution coefficient is $b_n/{\lambda }_n$. ThusAs an independent check, solving $Y^{\prime \prime }+Y=-1$ directly with the two endpoint conditions givesIt satisfies both Dirichlet boundary conditions and the forced ordinary differential equation.

Use a logarithmic change of variable in a Sturm-Liouville problem: set $t=\log x$ and $Y(t)=y(e^t)$. The chain rule givesThe endpoint conditions become $Y(0)=Y(1)=0$. Multiplying $-Y^{\prime \prime }=(\lambda +1)Y$ by $Y^{\ast }$ and integrating shows that $\lambda +1>0$ for a nonzero solution. Put $\mu =\sqrt{\lambda +1}$. The first endpoint selects $Y=A\sin (\mu t)$ and the second requires $\mu =n\pi$, $n\ge 1$. HenceThese eigenfunctions are orthonormal in the specified weighted inner product, because $w(x)dx=dx/x=dt$ and

Factor out $x$ from the differential expression:Comparing with $\mathcal{L}y=w^{-1}[-(p{y}^{\prime }{)}^{\prime }+qy]$ givesMultiplying all three by the same positive constant would give the same Sturm-Liouville operator.

First, setting $f=g=y_n$ in the self-adjoint operator identity shows that ${\lambda }_n⟨y_n,y_n{⟩}_w$ is real, so every eigenvalue is real. For two eigenfunctions,Distinct eigenvalues therefore implyThis is orthogonality of Sturm-Liouville eigenfunctions.

The weight cancels the denominator in the Sturm-Liouville operator. Two integrations by parts, with real $p,q$, give the Lagrange identity for a Sturm-Liouville operatorThus the boundary form must vanish for every pair in the operator domain:Separated real Robin boundary conditions $A_{s}y(s)+B_{s}p(s)y^{\prime }(s)=0$ at $s=a,b$, with $(A_s,B_s)\ne (0,0)$, make each endpoint contribution vanish. This includes Dirichlet boundary conditions and Neumann boundary conditions. Coupled conditions can also work: periodic $y(a)=y(b)$ and $p(a)y^{\prime }(a)=p(b)y^{\prime }(b)$ cancel the two endpoint contributions. For the regular problem, imposing a complete pair of such self-adjoint boundary conditions gives a self-adjoint differential operator; the displayed identity is the boundary-form criterion.

Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the Fermat principle minimizes the length through a reflection point $P=(s,0)$:Its stationary equation isThe two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin {\theta }_i=\sin {\theta }_r$, where both angles are measured from the normal vector to the mirror and lie in $[0,\pi /2)$. Hence the law of reflection isFor uniqueness, differentiate once more:Thus $L$ is a strictly convex function, its derivative increases from $-2$ to $2$, and it has exactly one stationary point, its global minimum. Explicitly,Substitution gives $(s_{\ast }-a)/h_A=(b-s_{\ast })/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.

The Fermat principle states that the actual path of a light ray makes its travel time stationary under small admissible variations with fixed endpoints. Equivalently, its optical path length is stationary:where $v$ is the local light speed, $n$ is the refractive index, and the two integrals differ by the constant vacuum light speed. In a uniform medium this reduces to stationarity of geometric path length. The principle is a stationarity statement; for the planar reflection considered next the stationary path is also the unique minimum.

Apply the Cauchy-Hadamard theorem to the power series coefficients $a_n=[n/(n+\alpha ){]}^{n^2}$ for $n\ge 1$:For example, taking logarithms gives $-n\log (1+\alpha /n)\to -\alpha$. ThereforeAny convention for the isolated $n=0$ coefficient has no effect on the radius of convergence.

The denominator vanishes at $z=\pm i\sqrt{2}$, giving two simple poles. Their distances from the center are both $|1\mp i\sqrt{2}|=\sqrt{3}$. There are no nearer singularities, so the Taylor series has radius of convergence

For $0<t<1$, split the Green's function integral where the absolute value changes sign:The two terms inside the bracket are respectivelyAdding them yieldsThe affine term gives $x^{\prime \prime }-4x=t$, while both exponential terms are homogeneous solutions. The values at the joins are $x(0)=(3e^{-2}-1)/16$ and $x(1)=-(1+e^{-2})/16$, agreeing with parts (i) and (ii). Their first derivatives also match: $x^{\prime }(0)=(3e^{-2}-1)/8$ and $x^{\prime }(1)=(1+e^{-2})/8$. Thus the pieces introduce no unwanted Dirac delta distributions.

For $t>1$, every forcing point has $\tau <t$. Hence the Green's function givesSince $\int \tau e^{2\tau }d\tau =(\tau /2-1/4)e^{2\tau }$, the remaining integral is $(e^2+1)/4$. ThereforeThis decays at positive infinity.

For $t<0$, all forcing points satisfy $\tau >t$, so the Green's function formula becomesAn integration by parts gives $\int \tau e^{-2\tau }d\tau =-(\tau /2+1/4)e^{-2\tau }$. ConsequentlyThe exponential has the required decay as $t\to -\infty$.

Let $z={\omega }_1^2$, so ${\omega }_2^2=4z$. The sum and product of roots of the characteristic polynomial in (i) giveEliminating $z$ yieldsTherefore the octave tuning in a two-mass complete spring graph has two positive solutions:The two values are reciprocals. Both meet the requested frequency ratio; substituting either into the expressions in (i) gives ${\omega }_2/{\omega }_1=2$.

For two bars, after subtracting the gravitational equilibrium,A normal mode $\mathbf{\eta }=\mathbf{v}{e}^{i\omega t}$ requires the generalized eigenvalue problem $(K-{\omega }^{2}M)\mathbf{v}=0$. With $z={\omega }^2$,Thus, for $m_2>0$,Both squared angular frequencies are positive. A corresponding eigenvector is $(1,2-{\omega }_j^2{)}^T$, from the first row of the matrix equation.

For this real integral choose $0<\arg z<2\pi$, so the upper bank of the positive real axis has $z^{1/4}=x^{1/4}>0$. Use a counterclockwise keyhole contour with inner radius $\epsilon$ and outer radius $R$, enclosing $\pm i$. The upper-bank contribution tends to $J$. On the lower bank, $z^{-1/4}=e^{-i\pi /2}{x}^{-1/4}=-i{x}^{-1/4}$, and the reversed orientation makes its contribution $+iJ$.

The large arc is $O(R^{-5/4})$ because $g=O(R^{-9/4})$, while the small arc is $O({\epsilon }^{3/4})$. Both vanish, givingThe residues on this branch areThe residue theorem yieldsThereforeThe branch cut in (a) was arbitrary; here placing it on the integration axis makes the bank contributions directly proportional to the desired positive-real integral.

The factor $z^{-1/4}$ has algebraic branch points at zero and infinity. Near zero the leading behaviour is $z^{-1/4}$; with $w=1/z$, the behaviour near infinity is $g(1/w)\sim w^{9/4}$. These are four-sheeted branch points.

The polynomial denominator vanishes at $z=\pm i$, with nonzero derivative $2z$, and the fractional factor is finite and nonzero near either point on any chosen local branch. HenceA single-valued holomorphic branch is obtained by choosing a branch cut joining zero to infinity. Points along the chosen cut mark its discontinuity, rather than additional isolated poles.

Write $t=1-x$. The analytic solution has $y_1=a_0+O(t)$ with $a_0\ne 0$, so the integrand in the reduction of order formula satisfiesIntegrating with respect to $x$ gives $-a_0^{-2}\log (1-x)+O(1)$. Multiplication by $y_1$ therefore givesAny solution independent of $y_1$ has a nonzero coefficient of this second solution. HenceAn added multiple of $y_1$ stays bounded and cannot cancel this logarithm. On the real side $x>1$, the corresponding statement uses $\log |1-x|$.

For the normal-form coefficient $p(x)=-1/(1-x)$, Abel's identity gives the WronskianUse reduction of order: if $y_2=y_{1}v$, then $W=y_1^2{v}^{\prime }$. Absorbing a nonzero Wronskian constant into the second integration constant, the general solution isThis local representation applies on intervals where $y_1\ne 0$, in particular sufficiently near $x=1$ because $y_1(1)=a_0\ne 0$.

Put $t=1-x$ and $Y(t)=y(x)$. The differential equation becomesSubstitute the Frobenius solution $Y=t^{\sigma }{\sum }_{n\ge 0}{a}_n{t}^n$. The lowest power has coefficient ${\sigma }^2{a}_0$, soEquating successive coefficients gives, with $a_{-1}=0$,Thus $a_0\ne 0$ is the only free coefficient: $a_1=-\lambda a_0$, $a_2=({\lambda }^2-c^2)a_0/4$, and so on. Since every denominator is nonzero, there is exactly one independent pure Frobenius solution of this form. The repeated root supplies no second free coefficient; the other solution is logarithmic, as parts (iii) and (iv) demonstrate.