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For $Ly=(x^2{y}^{\prime }{)}^{\prime }-2y$ on $x>0$, choose solutions bounded at zero and infinity. The homogeneous Cauchy-Euler equation has basis $x,x^{-2}$. Joining the left-regular and right-regular solutions continuously and imposing $s^2[G_x{]}_{s^{-}}^{s^{+}}=1$ gives the displayed Green's function with ordinary integration measure $ds$. For bounded forcing,The integrals converge and give the unique bounded solution: no nonzero homogeneous combination is bounded at both ends. If $0\le f\le M$, negativity of the kernel gives $-M/2\le y\le 0$. Differentiation cancels the two moving-endpoint terms, leavingwhich implies $|y^{\prime }(x)|\le M/(3x)$. For constant forcing $f=M$, the bounded solution is $-M/2$; the endpoint conditions are boundedness, rather than zero endpoint values.

A Cauchy–Euler equation has powers of the independent variable matched to derivative order and is solved using power laws or a logarithmic change of variable.

The Euler differential operator differentiates in logarithmic position: if $r=\log x$ and $V(r)=v(e^r)$, then $Dv=V^{\prime }$. In particular $D(x^n)=n{x}^n$, so a polynomial $p(D)$ acts on the power law $x^n$ by multiplication by $p(n)$. This converts a Cauchy-Euler equation into a constant-coefficient linear differential equation. Repeated roots give additional logarithmic factors, as ordinary constant-coefficient equations give polynomial factors multiplying exponentials.

For $0<x<1$, the homogeneous Cauchy-Euler equation has solutions $x,x^{-1}$. The left-vanishing solution $x$ and right-vanishing solution $x-x^{-1}$ give the unweighted Green function $G(x,s)=x(1-s^{-2})/2$ for $x\le s$ and $(x-x^{-1})/2$ for $x\ge s$. It is continuous at $x=s$ and its first derivative has jump $1/s^2$, so $L_{x}G=\delta (x-s)$.

For continuous forcing $f$ on $[0,1]$, its integral isThis is bounded at zero and vanishes at one, but $u(0^{+})=-f(0)$. The first singular term uses the average $x^{-1}{\int }_0^{x}f\to f(0)$. In the other, the substitution $t=x/s$ gives $x{\int }_x^{1}f(s)s^{-2}ds={\int }_x^{1}f(x/t)dt\to f(0)$ by dominated convergence. Thus the two zero endpoint values require $f(0)=0$. When compatible, the solution is unique among bounded solutions because no nonzero combination of $x,x^{-1}$ meets both homogeneous endpoint conditions. This example shows why pointwise endpoint behavior of a Green function does not itself justify passing a limit through its integral.

The homogeneous Cauchy-Euler equation $Ly=0$ has fundamental solutions $x,x^{-1}$ on $x>0$, with Wronskian $-2/x$. Requiring decay at both endpoints gives the unweighted Green's functionRequiring zero initial value and derivative at zero instead gives the forward kernelBoth kernels are continuous and have first-derivative jump one at $x=\xi$, so $L_{x}G=\delta (x-\xi )$. Away from that point they solve the homogeneous equation. Integrating against the forcing with ordinary measure $d\xi$ yields the respective solutions whenever the integral and endpoint limits converge. Their difference is $G_0-G_D=x/2$, independent of $\xi$, so the solutions differ by the homogeneous term $(x/2)\int f(\xi )d\xi$ for integrable forcing. For example, the two choices for an indicator forcing on $(0,1)$ differ by $x/2$. Specifying a differential operator without its endpoint conditions does not specify its Green kernel.

An indicial equation is the algebraic equation for the leading exponent of a power-law or Frobenius solution near a regular singular point.

For a Cauchy-Euler equation, logarithmic coordinates turn scale changes in $x>0$ into translations in $t$. With $Y(t)=y(e^t)$, the chain rule gives $x{y}^{\prime }=Y^{\prime }$ and $x^2{y}^{\prime \prime }=Y^{\prime \prime }-Y^{\prime }$. Consequently the Sturm-Liouville operator $-x^2{y}^{\prime \prime }-x{y}^{\prime }-cy$ becomes $-Y^{\prime \prime }-cY$. Its natural weighted inner product has measure $dx/x=dt$, so this change of variable also preserves the norm when the transformed problem uses ordinary integration in $t$. On $1<x<e^{\ell }$ with Dirichlet boundary conditions, the orthonormal eigenfunctions are $\sqrt{2/\ell }\sin (n\pi \log x/\ell )$ and the eigenvalues are $(n\pi /\ell {)}^2-c$, for $n\ge 1$.

Introduce the Euler differential operator $D=x{\partial }_x$. The fourth-order spatial expression is $D^{4}u$. The moving boundary has similarity coordinate $\xi =1$, so its two conditions giveHence $\alpha =0$ and $v^{\prime }(1)=1$. With $D_{\xi }=\xi {\partial }_{\xi }$, substitution in the equation yieldsThis is a Cauchy-Euler equation; putting $r=\log \xi$ turns it into the constant-coefficient equation $V^{(4)}-\beta V^{\prime }=0$. Its characteristic polynomial is $n(n^3-\beta )$.

If $\beta >0$, write $a={\beta }^{1/3}$. The roots are $0,a,-a/2\pm i\sqrt{3}a/2$. Decay excludes the constant and growing modes, leavingThe boundary conditions force $A=0$ and $B=2/(\sqrt{3}a)$. If $\beta <0$, writing $\beta =-a^3$ leaves only one decaying mode, $C{\xi }^{-a}$; $v(1)=0$ then forces the zero function, which fails the derivative condition. If $\beta =0$, the solutions are polynomials of degree at most three in $\log \xi$, and none can both decay and satisfy $v^{\prime }(1)=1$.

Consequently all real self-similar solutions are the one-parameter familyThe decaying profile for a fourth-order dilation equation oscillates in logarithmic distance. No positivity condition on $u$ was imposed, so these sign changes are permitted.

On an interval with $x>0$, use the power-law ansatz $y=x^r$. Substitution gives $r(r-1)+2r+1=r^2+r+1=0$, whose roots are $r=-1/2\pm i\sqrt{3}/2$. Taking real and imaginary parts of $x^r$ givesThese two real solutions are linearly independent: their Wronskian is $(\sqrt{3}/2)x^{-2}\ne 0$. They therefore span the solution space of the second-order linear ordinary differential equation on that interval.

For $x<0$, replace $x^{-1/2}$ by $|x{|}^{-1/2}$ and $\log x$ by $\log |x|$, with independent real constants on the negative interval. This follows equally by the change of variable $t=\log |x|$, which turns the Cauchy-Euler equation into $d^{2}y/d{t}^2+dy/dt+y=0$. The coefficient of $y^{\prime \prime }$ vanishes at zero, so the general formula is stated separately on either side. Every nonzero mode has an unbounded oscillatory envelope as $x\to 0$; a continuous extension through zero requires the solution to vanish identically on both sides.

The useful change of independent variable in a second-order ODE is $t=\log x$, followed by $y(x)=xv(t)$. The chain rule givesThus the eigenvalue equation becomes $v_{tt}=\lambda v$ on $0<t<1$, and the endpoint values become $v(0)=v(1)=0$. At the specified eigenvalues,The first endpoint forces $A=0$; the second holds for every positive integer $k$. Choosing a convenient normalization yieldsEach nonzero constant multiple is the same eigenfunction up to normalization. This logarithmic conjugacy of a repeated-root Euler operator turns the Cauchy-Euler equation into the familiar Fourier sine series problem on a unit interval.

Work first on $x>0$. For $b=0$, the power-law ansatz $y=x^r$ gives the indicial equation $r(r-1)+r-a^2=r^2-a^2=0$. Therefore the Cauchy-Euler equation has general solutionFor $a=0$ the repeated indicial root supplies the logarithm rather than two copies of the constant solution. Alternatively, the change of variables $t=\log x$ turns the differential equation into $Y^{\prime \prime }-a^{2}Y=0$. On $x<0$ the corresponding real formulas use $|x|$ and $\log |x|$, with independent constants on the two sides of zero.

For general $b$, normalize the second-order linear differential equation asThe coefficient $P$ is singular at zero, but $xP(x)=1$ and $x^{2}Q(x)=-a^2+b{x}^{\beta }$ are analytic functions there because $\beta$ is a positive integer. The regular singular point criterion for a second-order equation therefore proves that zero is a regular singular point for every allowed $a,b,\beta$. The indicial equation remains $r^2-a^2=0$.

For a putative second Frobenius solution with the smaller indicial root $-a$, set $y_{-}=x^{-a}{\sum }_{n\ge 0}{d}_n{x}^n$, with $d_0=1$ and $d_j=0$ for $j<0$. Its recurrence relation isThe Frobenius case classification concerns the gap $2a$ between the indicial roots. In the course convention, Case IIIa is the obstructed recurrence requiring a Logarithmic Frobenius solution; Case IIIb permits two pure Frobenius series.

If $a>0$ and $2a\notin \mathbb{Z}$, this is Case I: no denominator vanishes. There are two linearly independent pure Frobenius solutions, withTheir different leading powers establish linear independence.

If $a=0$, this is Case II, with a repeated indicial root. There is only one pure Frobenius series with a nonzero leading coefficient. To see the second solution's form, reduction of order givesHere $y_{+}(t)=1+O(t^{\beta })$, so the integrand is $t^{-1}+O(t^{\beta -1})$. ConsequentlyThe logarithm has a nonzero coefficient and cannot be removed by adding a multiple of $y_{+}$.

If $N=2a$ is a positive integer, the only potentially obstructed coefficient is $n=N$. Below that index, $d_n=0$ unless $\beta$ divides $n$. At the exceptional index the equation readsIf $N=m\beta$ for some positive integer $m$, all coefficients $d_0,d_{\beta },\dots ,d_{(m-1)\beta }$ are nonzero because $b\ne 0$. The exceptional equation is inconsistent. This is Case IIIa, and the independent second solution has the formThe Frobenius resonance for an Euler equation with monomial potential explains why divisibility, rather than the integer gap alone, decides this outcome. For example, with $d_0=1$, applying the original differential operator to $y_{+}\log x$ gives $2x{y}_{+}^{\prime }$. Cancellation of the coefficient of $x^a$ requires $K=-b{d}_{N-\beta }/(2a)\ne 0$.

If $\beta$ does not divide $N$, then $d_{N-\beta }=0$; this includes $N<\beta$ through the negative-index convention. The exceptional equation is $0=0$, so $d_N$ is free. This is Case IIIb: set $d_N=0$ to select an independent pure Frobenius series starting with $x^{-a}$. Subsequent coefficients again follow the recurrence relation. Changing $d_N$ simply adds a multiple of $y_{+}$, the undetermined coefficient at Frobenius resonance.

The complete answer is thereforeIn particular, for $\beta =1$ every positive integer gap gives Case IIIa. For $\beta =2$, positive integer $a$ gives Case IIIa, while $a=1/2,3/2,5/2,\dots$ gives Case IIIb; the remaining positive $a$ give Case I. The condition $b\ne 0$ matters: at $b=0$ the two Cauchy-Euler equation powers already give independent solutions whenever $a>0$.

The power-law ansatz $y=x^r$ givesSince $ϵ>0$, the equation roots $1$ and $-1/ϵ$ are distinct. The Cauchy-Euler equation therefore has general solutionFor $C_1=C_2=1$, the first derivative and second derivative areThe strictly convex function diverges to $+\infty$ as $x\to 0^{+}$, passes through $(1,2)$, and has a unique lowest point atAs $x\to \infty$, $y-x\to 0$, so the line $y=x$ is an asymptote. When $0<ϵ<1$, the lowest point is to the right of one; at $ϵ=1$ it is at one, and for $ϵ>1$ it is to the left of one.

The ratio of the two terms isThus $x^{-1/ϵ}$ is larger for $0<x<1$, the terms are equal at $x=1$, and $x$ is larger for $x>1$. In the stronger asymptotic sense as $ϵ\to 0^{+}$, $x^{-1/ϵ}$ dominates at each fixed $x<1$, and $x$ dominates at each fixed $x>1$. Their crossover has $|\log x|=O(ϵ)$, equivalently $x=1+O(ϵ)$, as described by the power-law crossover in a singular perturbation. In that region $x=1+ϵX$ with bounded $X$ gives $y\sim 1+e^{-X}$.

Substitute the proposed regular perturbation $y=y_0+ϵy_1+\cdots$ into the differential equation. The leading and subsequent equations areConsequently $y_0=Cx$, and in fact each successive coefficient has the form $y_j=C_{j}x$, because the preceding second derivative is zero. The leading approximation is thereforeIt has only one adjustable constant and generally cannot impose two independent boundary conditions. Those assumed in the question require a nonzero $x^{-1/ϵ}$ coefficient, which a power series in $ϵ$ cannot recover: at fixed $x>1$ it is exponentially small, and at fixed $x<1$ it grows without bound as $ϵ\to 0^{+}$. Thus the regular perturbation describes the specified solution only in the region to the right of the crossover, away from $x=1$; it is not a uniform approximation on $(0,\infty )$.

Set $t=\log x$, so $0<t<\pi$, and write $y(x)=x^{-2}w(t)$. Direct differentiation of this Cauchy-Euler equation givesThe eigenvalue equation is therefore $-w^{\prime \prime }=(\lambda -3)w$ with $w(0)=w(\pi )=0$. For $\lambda -3\le 0$, the linear/hyperbolic solutions cannot satisfy both endpoint conditions except trivially. For $\lambda -3>0$, the sine solution vanishes at both endpoints exactly when its frequency is a positive integer. ThusEach associated eigenspace is one-dimensional.

The displayed differential operator is not self-adjoint for the unweighted inner product. Let $A=x^2{\partial }_x^2+5x{\partial }_x+1$. Integration by parts on compactly supported smooth functions gives its formal adjointwhich differs from $A$. One can see an explicit failure of symmetry without boundary terms: take a nonzero real compactly supported smooth $f$ in the interval and $g=xf$. ThenBoth test functions obey the Dirichlet endpoint conditions. Changing the sign of $A$ for the eigenproblem does not repair this failure. A different, weighted inner product with weight $x^3$ does give the Sturm-Liouville form $x^{3}Ay=(x^5{y}^{\prime }{)}^{\prime }+x^{3}y$, explaining why a real discrete spectrum is nevertheless natural.

For the Cauchy-Euler equation $x^2{y}^{\prime \prime }-x{y}^{\prime }+y=0$, the power trial $y=x^r$ on $x>0$ gives $(r-1{)}^2=0$. The repeated root supplies the logarithmic second solution. On either half-line,One may also obtain this by setting $y=xu$, which reduces the equation to $(x{u}^{\prime }{)}^{\prime }=0$. Constants on opposite sides of the singular point need not agree.

Assume $r>0$, put $b=S^{\star }\in (0,K)$ and let ${\tau }_b=\inf \{u\ge 0:S_u\le b\}$. In the continuation region $S>b$, the time-independent Black-Scholes equation is a Cauchy-Euler equation. Substituting $V=S^m$ givesSet $\gamma =2r/{\sigma }^2>0$. The two powers are $S$ and $S^{-\gamma }$. The boundary condition $V(S)\to 0$ as $S\to \infty$ removes the growing solution; value matching at $b$ sets the coefficient. Hence the fixed-trigger value isFor $0\le S\le b$ the policy exercises immediately, giving $V_b(S)=(K-S{)}^n$. The quoted continuation expression alone does not apply below the exercise boundary. Equivalently, the discounted boundary-hitting representation gives ${\mathbb{E}}_S[e^{-r{\tau }_b};{\tau }_b<\infty ]=(S/b{)}^{-\gamma }$, which multiplies the exercise financial payoff.

Use $\kappa =(\mu -r)/\sigma$ and ${\gamma }_M$ from Question 1, and take the finite-value case ${\gamma }_M>0$. After retirement the value is $V_M(w)={\gamma }_M^{-R}{w}^{1-R}/(1-R)$. Working for an infinitesimal extra interval while using the retired portfolio and consumption produces a gain $\epsilon V_M^{\prime }(w)-\lambda$. Its zero isThis proves the instantaneous indifference level appearing in the question. For the stated irreversible optimal stopping problem it is generally a lower bound on the retirement boundary, rather than the boundary itself: retaining the opportunity to keep working has value. The following explicit wealth-variable Legendre dual calculation exhibits the correction. In particular, imposing continuity of $V^{\prime \prime }$ at retirement would discard that option value; smooth fit requires continuity of $V^{\prime }$.

Assume the usual borrowing-against-income solvency condition: $w>-\epsilon /r$ while working, with nonnegative wealth after retirement, and take $r>0$, $\rho >0$, $\kappa \ne 0$. The question does not specify a tighter borrowing constraint. Let $J(z)={\sup }_w\{V(w)-zw\}$, so $w=-J^{\prime }(z)$ and $V=J-z{J}^{\prime }$. The working Hamilton-Jacobi-Bellman equation becomes the linear Cauchy-Euler equationThe retired dual is $J_M(z)=R{z}^{1-1/R}/[(1-R){\gamma }_M]$. Let $m<0$ be the negative root ofThe other root exceeds one. Its growing term is excluded by the solvency asymptotic as $z\to \infty$. Therefore the working dual has the formRetirement corresponds to $z\le z_{\ast }$, where $J=J_M$. Value matching and smooth fit giveConsequently the retirement boundary with an income option isUsing the root equation, $z_{\ast }/(\lambda /\epsilon )=1+{\kappa }^{2}m/(2\rho )$, which lies strictly between zero and one. Thus $\overline{w}>w_{\mathrm{myopic}}$ for a nonzero risk premium. For example, $r=\rho =\epsilon =\lambda =\kappa =1$ and $R=2$ give $m=-1$, ${\gamma }_M=9/8$, andThese parameters give an explicit counterexample to identifying the printed level with the irreversible retirement boundary.

For $-\epsilon /r<w<\overline{w}$, find the unique $z>z_{\ast }$ fromThis right side decreases strictly from $\overline{w}$ to $-\epsilon /r$, since $J^{\prime \prime }>0$. Recover the value and controls byFor $w\ge \overline{w}$, retire immediately and use the Merton controls. Along the working strategy, $z_t$ satisfies $d{z}_t=(\rho -r)z_{t}dt-\kappa z_{t}d{W}_t$, and the optimal retirement time is its first hitting time of $z_{\ast }$, equivalently wealth's first hitting time of $\overline{w}$.

To check the optimal stopping inequalities, write $h=J-J_M$. It satisfies $h(z_{\ast })=h^{\prime }(z_{\ast })=0$ and $h^{\prime \prime }(z)=Dm(m-1)z^{m-2}>0$ for $z>z_{\ast }$. Hence continuation has positive option value there. On $z\le z_{\ast }$ the stopping candidate has working residual $\epsilon z-\lambda \le 0$. The matched dual is continuously differentiable, so the Itô formula introduces no boundary local-time term. Localization, the state-price budget, and the appropriate infinite-horizon transversality justify verification.

If nonnegative financial wealth is imposed during employment as an additional constraint, the working dual also has an upper boundary, and that extra constraint must be included in the free-boundary problem. It cannot be recovered by assuming the printed myopic equality.

For a nonzero separable solution of the Laplace equation in polar coordinates, write $T(r,\theta )=R(r)\Theta (\theta )$ with $r>0$. Multiplying the partial differential equation by $r^2$ and dividing where $R\Theta \ne 0$ giveswhere $\lambda$ is a real separation constant. The resulting ordinary differential equations areThe equations continue across zeros of the factors by continuity. The radial equation is a Cauchy-Euler equation; putting $s=\log r$ changes it to $d^{2}R/d{s}^2-\lambda R=0$.

For positive separation constant, write $\lambda =k^2$ with $k>0$. The radial factors are powers, while the angular factors are sine and cosine:For zero separation constant, each transformed equation has zero second derivative, givingFor negative separation constant, write $\lambda =-k^2$ with $k>0$. The radial factors oscillate in $\log r$, and the angular factors are exponential functions:These exhaust the nonzero real separated solutions before imposing angular periodicity or regularity at the origin. The zero solution is also allowed. Since Laplace's equation is linear, sums of these separated solutions are solutions wherever the sums and required derivatives converge appropriately.

Put $u(r,\theta )=R(r)\Theta (\theta )$ in Laplace's equation and multiply by $r^2/(R\Theta )$ on intervals where the factors are nonzero. Separation of variables givesand henceThe angular Dirichlet boundary conditions are $\Theta (0)=\Theta (\pi /2)=0$. Multiplication of the angular equation by $\Theta$, followed by integration by parts, givesFor a nonzero angular eigenfunction this forces $\lambda >0$: equality would make $\Theta$ constant and then zero. Write $\lambda ={\mu }^2$. The first angular condition removes the cosine term, and the second requires $\sin (\mu \pi /2)=0$. ThusThe radial Cauchy-Euler equation has the two solutions $r^{2n}$ and $r^{-2n}$. By superposition, the general separated series is thereforeThe coefficients are arbitrary subject to convergence with enough partial derivatives on compact subsets to give a harmonic function and to the prescribed zero values on the two open radial edges. Completeness of the sine eigenfunctions gives the usual series representation for solutions smooth up to those edges on closed radial subintervals. There is no constant or logarithmic zero mode because its angular factor would be linear in $\theta$ and would vanish at both ends only if it were zero.

The negative powers must be retained: the stated domain omits $r=0$, and no boundedness at that vertex is prescribed. For example, $r^{-2}\sin (2\theta )$ is a harmonic function satisfying both edge conditions. An additional boundedness condition as $r\to 0$ would remove all $B_n$; an additional value on the circular arc would then determine the remaining $A_n$. With only the conditions in the paper, neither restriction is available and the solution is not unique.

For $x=e^z>0$, let $Y(z)=y(e^z)$. The chain rule givesMultiplying each original Cauchy-Euler equation by $x^2$ therefore givesandThe exponential change of variable has made both equations have constant coefficients.