Power-Law Boundary Singularities for a Planar Riccati Equation: A Counterexample to the Literal Formulation of Maz'ya's Problem 54

1. Introduction and Precise Formulation

Let $\mathrm{\Omega }\subset {\mathbb{R}}^2$ be a bounded domain with smooth boundary, let $O\in \partial \mathrm{\Omega }$, and consider

$$\Delta u+\alpha (x,y)u_x^2+\beta (x,y)u_y^2=0\mathrm{in}\mathrm{\Omega },$$

(1.1)

subject to

$$u=0\mathrm{on}\partial \mathrm{\Omega }\setminus \left\{O\right\}.$$

(1.2)

Problem 54 in [3] assumes that $\alpha$ and $\beta$ are smooth and positive in $\mathrm{\Omega }$ and states:

Prove that either $u=0$ on $\partial \mathrm{\Omega }$, or u has a logarithmic singularity at O and depends upon an arbitrary constant.

Because (1.2) has already prescribed the trace on the punctured boundary, we interpret the first alternative as the regular branch in which the trace extends across O with value zero. The counterexample below concerns the second alternative and is independent of any stronger interpretation of the regular branch.

The point of the present note is the distinction between two types of assumptions:

(i)

interior regularity and pointwise positivity,

$$\alpha ,\beta \in C^{\infty }\left(\mathrm{\Omega }\right),\alpha \left(x\right),\beta \left(x\right)>0(x\in \mathrm{\Omega }),$$

(ii)

boundary regularity and quantitative nondegeneracy near O, for example

$$\alpha ,\beta \in C(\overline{\mathrm{\Omega }}\cap B_r\left(O\right)),min\{\alpha \left(O\right),\beta \left(O\right)\}>0.$$

The published wording explicitly gives the first type but not the second. The two are not equivalent: a smooth positive function in the open domain may tend to zero, oscillate, or fail to have a limit at the excluded boundary point. The construction below exploits precisely this freedom.

We use the following minimal test for logarithmic behavior. Let ${\nu }_{\mathrm{in}}\left(O\right)$ denote the inward unit normal.

Definition 1.1

(Logarithmic profile). A function u has a standard logarithmic profile at O if there is a finite nonzero constant A such that, in every fixed closed nontangential region at O,

$$u\left(x\right)=Alog{\displaystyle \frac{1}{|x-O|}}+O\left(1\right)(x\to O).$$

(1.3)

In particular, (1.3) implies the necessary normal asymptotic

$$u(O+t{\nu }_{\mathrm{in}}\left(O\right))=Alog{\displaystyle \frac{1}{t}}+O\left(1\right)(t\downarrow 0).$$

(1.4)

Thus it is enough to construct a solution for which

$${\displaystyle \frac{u(O+t{\nu }_{\mathrm{in}}\left(O\right))}{log(1/t)}}⟶+\infty .$$

In fact, the examples below have the stronger nontangential behavior $u\left(x\right)\asymp {|x-O|}^{-\gamma }$.

A formal asymptotic construction for a broader Riccati-type equation near a corner was developed by Maz’ya and Slutskii [4]; related Neumann asymptotics were obtained by Kozlov and Maz’ya [1]. Those works provide important context, but their coefficient hypotheses are substantially more structured than mere positivity at interior points. We return to this issue in Section 5.

Throughout, a classical solution means

$$u\in C^2\left(\mathrm{\Omega }\right)\cap C\left(\overline{\mathrm{\Omega }}\setminus \left\{O\right\}\right).$$

The notation $f\asymp g$ means that $c^{-1}g\le f\le cg$ with a constant $c\ge 1$ independent of the limiting variable.

2. A Harmonic-Composition Identity

Lemma 2.1

(Composition identity). Let $\mathrm{\Omega }\subset {\mathbb{R}}^2$ be a domain, let $V\in C^{\infty }\left(\mathrm{\Omega }\right)$ be positive and harmonic, and let $F\in C^{\infty }\left(I\right)$ on an interval I containing $V\left(\mathrm{\Omega }\right)$. Assume

$$F^{\prime }\left(s\right)>0,F^{\prime \prime }\left(s\right)<0(s\in I).$$

Define

$$u=F\left(V\right),q\left(x\right)=-{\displaystyle \frac{F^{″}\left(V\left(x\right)\right)}{F^{\prime }{\left(V\left(x\right)\right)}^2}}.$$

(2.1)

Then $q\in C^{\infty }\left(\mathrm{\Omega }\right)$, $q>0$, and

$$\Delta u+{q\left(x\right)|\nabla u|}^2=0\mathit{in}\mathrm{\Omega }.$$

(2.2)

If V extends continuously to $\partial \mathrm{\Omega }\setminus \left\{O\right\}$ with $V=1$ there and $F\left(1\right)=0$, then u satisfies (1.2).

Proof. 

The chain rule and $\Delta V=0$ give

$$\nabla u=F^{\prime }\left(V\right)\nabla V,\Delta u=F^{\prime \prime }\left(V\right){|\nabla V|}^2.$$

Therefore

$${q|\nabla u|}^2=-{\displaystyle \frac{F^{″}\left(V\right)}{F^{\prime }{\left(V\right)}^2}}{F}^{\prime }{\left(V\right)}^2{|\nabla V|}^2=-F^{\prime \prime }\left(V\right){|\nabla V|}^2=-\Delta u.$$

The sign, smoothness, and boundary assertions follow directly. □

Two choices of F distinguish the logarithmic and power-law regimes. First,

$$F\left(s\right)=logs⟹-{\displaystyle \frac{F^{″}\left(s\right)}{F^{\prime }{\left(s\right)}^2}}=1.$$

(2.3)

Thus the logarithm is generated by a constant, nondegenerate coefficient. Second, for $0<\gamma <1$,

$$F_{\gamma }\left(s\right)=s^{\gamma }-1⟹-{\displaystyle \frac{F_{\gamma }^{″}\left(s\right)}{F_{\gamma }^{\prime }{\left(s\right)}^2}}={\displaystyle \frac{1-\gamma }{\gamma }}{s}^{-\gamma }.$$

(2.4)

The latter coefficient remains positive for every finite s but tends to zero as $s\to \infty$.

Remark 2.2.

Any increasing concave function F with $F\left(1\right)=0$ generates a positive coefficient through (2.1). Consequently, interior smoothness and pointwise positivity alone allow many singular scales. A rigidity statement for the singularity must use quantitative information on the coefficient as the singular point is approached.

3. Poisson-Kernel Construction

Let

$$\delta \left(x\right)=\mathrm{dist}(x,\partial \mathrm{\Omega }).$$

For a bounded $C^2$ domain in ${\mathbb{R}}^2$, the Poisson kernel satisfies the two-sided estimate

$$C^{-1}{\displaystyle \frac{\delta \left(x\right)}{{|x-z|}^2}}\le P_{\mathrm{\Omega }}(x,z)\le C{\displaystyle \frac{\delta \left(x\right)}{{|x-z|}^2}},x\in \mathrm{\Omega },z\in \partial \mathrm{\Omega },$$

(3.1)

with a domain-dependent constant C; see [2, Theorem 1]. Standard potential theory also gives that, for fixed z, $P_{\mathrm{\Omega }}(·,z)$ is positive and harmonic in $\mathrm{\Omega }$ and extends continuously with zero trace to $\partial \mathrm{\Omega }\setminus \left\{z\right\}$.

For $M>1$ and $r>0$, set

$${\mathrm{\Gamma }}_M(O,r)=\left\{x\in \mathrm{\Omega }:0<|x-O|<r,|x-O|\le M\delta \left(x\right)\right\}.$$

(3.2)

On this nontangential region, (3.1) implies

$$P_{\mathrm{\Omega }}(x,O)\asymp {|x-O|}^{-1}(x\to O,x\in {\mathrm{\Gamma }}_M(O,r)).$$

(3.3)

Theorem 3.1

(Power singularities in every smooth planar domain). Let $\mathrm{\Omega }\subset {\mathbb{R}}^2$ be a bounded $C^{\infty }$ domain and let $O\in \partial \mathrm{\Omega }$. Fix $c>0$ and $\gamma \in (0,1)$, and define

$$\begin{array}{cccc}\hfill V\left(x\right)& =1+c{P}_{\mathrm{\Omega }}(x,O),\hfill & \hfill u\left(x\right)& =V{\left(x\right)}^{\gamma }-1,\hfill \\ \hfill \alpha \left(x\right)& =\beta \left(x\right)={\displaystyle \frac{1-\gamma }{\gamma }}V{\left(x\right)}^{-\gamma }.\hfill \end{array}$$

(3.4)

Then:

(a)

$\alpha =\beta \in C^{\infty }\left(\mathrm{\Omega }\right)$ and

$$0<\alpha \left(x\right)=\beta \left(x\right)\le {\displaystyle \frac{1-\gamma }{\gamma }}(x\in \mathrm{\Omega });$$

(3.5)

(b)

u is a positive classical solution of (1.1)–(1.2);

(c)

for every fixed $M>1$ there is $r_M>0$ such that

$$u\left(x\right)\asymp {|x-O|}^{-\gamma }(x\to O,x\in {\mathrm{\Gamma }}_M(O,r_M));$$

(3.6)

(d)

in particular,

$${\displaystyle \frac{u(O+t{\nu }_{\mathrm{in}}\left(O\right))}{log(1/t)}}⟶+\infty (t\downarrow 0),$$

(3.7)

and ${inf}_{\mathrm{\Omega }}\alpha ={inf}_{\mathrm{\Omega }}\beta =0$.

Proof. 

The function V is positive and harmonic in $\mathrm{\Omega }$, extends continuously to $\partial \mathrm{\Omega }\setminus \left\{O\right\}$, and equals 1 there. Applying Theorem 2.1 with $F=F_{\gamma }$ from (2.4) proves the equation and the boundary condition. Since $V\ge 1$, (3.5) follows immediately.

Fix $M>1$. By (3.3), $P_{\mathrm{\Omega }}(x,O)\asymp {|x-O|}^{-1}$ in ${\mathrm{\Gamma }}_M(O,r)$, and hence $P_{\mathrm{\Omega }}(x,O)\to \infty$ there. After reducing r, the additive constant 1 in $V=1+c{P}_{\mathrm{\Omega }}$ is negligible, so

$$V\left(x\right)\asymp {|x-O|}^{-1}.$$

Raising this relation to the power $\gamma$ gives (3.6). The inward normal belongs to every sufficiently wide nontangential region and satisfies $|O+t{\nu }_{\mathrm{in}}\left(O\right)-O|=t$, yielding

$$u(O+t{\nu }_{\mathrm{in}}\left(O\right))\asymp t^{-\gamma }.$$

This proves (3.7), since every positive power dominates $log(1/t)$. Finally, along the inward normal $V\to \infty$, so (3.4) gives $\alpha =\beta \to 0$ and therefore the infima are zero. □

Corollary 3.2

(Literal formulation of Problem 54). Under the assumptions explicitly stated in Problem 54–namely, smoothness and strict positivity of α and β in the open domain–the proposed logarithmic classification is false.

Proof. 

The coefficients and solution in Theorem 3.1 satisfy the stated interior regularity, positivity, and punctured-boundary condition, whereas (3.7) contradicts the necessary logarithmic behavior (1.4). □

4. A Fully Explicit Tangent-Disk Counterexample

The general construction can be made completely elementary. Let

$$D=\left\{(x,y)\in {\mathbb{R}}^2:{(x-1)}^2+y^2<1\right\},O=(0,0).$$

(4.1)

The disk is tangent to the y-axis at O. Define

$$V(x,y)={\displaystyle \frac{2x}{x^2+y^2}}.$$

(4.2)

Because $V=2Re(1/(x+iy\left)\right)$, it is harmonic in D. The disk inequality is equivalent to

$$x^2+y^2<2x,$$

so $V>1$ in D. On $\partial D\setminus \left\{O\right\}$, one has $x^2+y^2=2x$ and therefore $V=1$.

Theorem 4.1

(Explicit counterexample). Fix $\gamma \in (0,1)$ and define

$$u(x,y)={\left({\displaystyle \frac{2x}{x^2+y^2}}\right)}^{\gamma }-1,$$

(4.3)

$$\alpha (x,y)=\beta (x,y)={\displaystyle \frac{1-\gamma }{\gamma }}{\left({\displaystyle \frac{2x}{x^2+y^2}}\right)}^{-\gamma }.$$

(4.4)

Then:

(a)

$\alpha ,\beta \in C^{\infty }\left(D\right)$ and

$$0<\alpha (x,y)=\beta (x,y)<{\displaystyle \frac{1-\gamma }{\gamma }};$$

(b)

$u\in C^{\infty }\left(D\right)\cap C(\overline{D}\setminus \left\{O\right\})$ and $u=0$ on $\partial D\setminus \left\{O\right\}$;

(c)

u solves

$$\Delta u+\alpha u_x^2+\beta u_y^2=0\mathit{in}D;$$

(d)

along the inward normal $y=0$, $x\downarrow 0$,

$$u(x,0)={\left({\displaystyle \frac{2}{x}}\right)}^{\gamma }-1\sim 2^{\gamma }{x}^{-\gamma },$$

(4.5)

and hence the singularity is not logarithmic;

(e)

the coefficient is globally bounded in D but has neither a positive lower bound nor a limit at O.

Proof. 

The first two assertions follow from $V>1$ in D and $V=1$ on the regular part of the boundary. Direct differentiation gives

$$\Delta u=\gamma (\gamma -1)V^{\gamma -2}{|\nabla V|}^2{,|\nabla u|}^2={\gamma }^2{V}^{2\gamma -2}{|\nabla V|}^2.$$

Using (4.4),

$$\alpha u_x^2+\beta u_y^2={\displaystyle \frac{1-\gamma }{\gamma }}{V}^{-\gamma }{|\nabla u|}^2=\gamma (1-\gamma )V^{\gamma -2}{|\nabla V|}^2=-\Delta u.$$

This proves the equation. Formula (4.5) follows from $V(x,0)=2/x$ and yields

$${\displaystyle \frac{u(x,0)}{log(1/x)}}⟶+\infty .$$

It remains to examine the coefficient at O. Along the normal,

$$\alpha (x,0)={\displaystyle \frac{1-\gamma }{\gamma }}{\left({\displaystyle \frac{x}{2}}\right)}^{\gamma }⟶0.$$

(4.6)

For any fixed $L>1$, consider the upper arc

$${\mathrm{\Gamma }}_L=\left\{(x,y):x^2+y^2={\displaystyle \frac{2x}{L}},y>0\right\}.$$

(4.7)

Its points lie in D, approach O, and satisfy $V=L$. Hence

$$\alpha {|}_{{\mathrm{\Gamma }}_L}={\displaystyle \frac{1-\gamma }{\gamma }}{L}^{-\gamma }>0.$$

(4.8)

The two approaches (4.6) and (4.8) give different limiting values, so $\alpha$ and $\beta$ have no limit at O. In particular, their infimum in D is zero. □

5. Scope, Comparison with the Formal Theory, and a Corrected Formulation

The counterexample establishes exactly the following negative statement:

Interior $C^{\infty }$ regularity and pointwise positivity of the coefficients do not force a logarithmic isolated boundary singularity.

It does not disprove a version of the problem in which the coefficients possess positive boundary limits or satisfy a uniform ellipticity-type lower bound near O.

The distinction is consistent with the formal theory in [4]. That paper treats a more general operator whose second-order and quadratic-gradient coefficients are symmetric positive-definite matrices. The coefficients are assumed to approach limiting functions as the solution tends to $+\infty$, and the limiting matrices are evaluated at the corner. Under those structural hypotheses, the constructed formal profile has the form

$$U(r,\theta )=\mathrm{\Lambda }log{\displaystyle \frac{1}{r}}+C+Z\left(\theta \right)+\mathrm{boundary}\text{-}\mathrm{layer}\mathrm{terms},$$

(5.1)

where C is arbitrary and $\mathrm{\Lambda }>0$ is selected by an angular problem. This is not a classification theorem for all solutions of (1.1), but it shows that the cited logarithmic mechanism is tied to well-defined, positive limiting quadratic forms. In the diagonal equation (1.1), smooth positive extension of $\alpha$ and $\beta$ to O would provide such a limiting form. The coefficient in Theorem 4.1 deliberately lacks that property.

The same point is visible directly within the harmonic-composition ansatz.

Proposition 5.1

(Asymptotic nondegeneracy forces logarithmic composition). Let $F\in C^2\left([s_0,\infty )\right)$ satisfy $F^{\prime }>0$ and $F^{″}<0$, and set

$$Q\left(s\right)=-{\displaystyle \frac{F^{″}\left(s\right)}{F^{\prime }{\left(s\right)}^2}}.$$

If

$$Q\left(s\right)⟶q_0>0(s\to \infty ),$$

then

$${\displaystyle \frac{F\left(s\right)}{logs}}⟶{\displaystyle \frac{1}{q_0}}(s\to \infty ).$$

(5.2)

If $Q\equiv q_0$, then

$$F\left(s\right)={\displaystyle \frac{1}{q_0}}log(q_{0}s+b)+d$$

(5.3)

for suitable constants $b,d$.

Proof. 

Let $H=1/F^{\prime }$. Then

$$H^{\prime }\left(s\right)=-{\displaystyle \frac{F^{″}\left(s\right)}{F^{\prime }{\left(s\right)}^2}}=Q\left(s\right).$$

The assumption $Q\left(s\right)\to q_0$ implies $H\left(s\right)=q_{0}s+o\left(s\right)$, and hence

$$F^{\prime }\left(s\right)={\displaystyle \frac{1}{q_{0}s}}\left(1+o\left(1\right)\right).$$

Integration gives (5.2). If $Q\equiv q_0$, then $H\left(s\right)=q_{0}s+b$, and a second integration gives (5.3). □

Remark 5.2

(What the counterexample does and does not require). The example proves that some quantitative boundary regularity and nondegeneracy assumption is necessary for the proposed logarithmic classification. Natural candidate hypotheses include

$$\alpha ,\beta \in C(\overline{\mathrm{\Omega }}\cap B_r\left(O\right)),\alpha \left(O\right)>0,\beta \left(O\right)>0,$$

(5.4)

or, more weakly,

$$0<c_0\le \alpha \left(x\right),\beta \left(x\right)\le c_1(x\in \mathrm{\Omega }\cap B_r\left(O\right)).$$

(5.5)

No sufficiency claim for (5.4) or (5.5) is made here. Establishing a complete classification under such strengthened assumptions is a separate problem.

6. Conclusion

For every $\gamma \in (0,1)$, the explicit formulas (4.3)–(4.4) give a classical solution of the planar Riccati equation with zero Dirichlet trace away from one boundary point, with coefficients that are smooth, strictly positive, and globally bounded in the open domain, and with power-law boundary blow-up. The ratio of the solution to $log(1/|x-O\left|\right)$ diverges along the inward normal, so the singularity cannot be logarithmic in the standard sense.

Consequently, the literal formulation of Problem 54 is false unless the stated interior smoothness and positivity are understood to include additional boundary regularity and nondegeneracy not written in the problem. The result should therefore be read as a correction of the hypotheses, not as a resolution of the strengthened nondegenerate classification problem.