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Econ-46 · economics · Economics

Viscosity existence without the BSDE comparison structure

LLM status Solved unverified

Statement

SC-02 |  Stochastic control and dynamic optimization

Published question

Definitions and assumptions

Let $b:[0,T]\times\R^n\to\R^n$ and $\sigma:[0,T]\times\R^n\to\R^{n\times d}$ be jointly continuous, with \[ |b(t,0)|+\|\sigma(t,x)\|\leq K,\qquad |b(t,x)-b(t,x')|+\|\sigma(t,x)-\sigma(t,x')\|\leq K|x-x'|. \] Let $g:[0,T]\times\R^n\times\R\times\R^d\to\R$ and $h:\R^n\to\R$ be continuous. For constants $k\geq0$, $\alpha\in(1,2]$, $\alpha^*=\alpha/(\alpha-1)$ and $p\in[1,\alpha^*)$, assume, everywhere, \begin{align*} \operatorname{sgn}(y)g(t,x,y,z)&\leq k(1+|x|^p+|y|+|z|^\alpha),\\ |g(t,x,y,z)|+|h(x)|&\leq k\left(1+|x|^p+e^{k|y|^{2/\alpha^*}}+|z|^2\right). \end{align*} Set $\mathcal L\varphi=b\cdot\nabla\varphi+ \frac12\operatorname{tr}(\sigma\sigma^\top D^2\varphi)$.

Formal question

Does there exist a continuous viscosity solution of \[ \partial_tu+\mathcal Lu+ g(t,x,u,\sigma(t,x)^\top\nabla u)=0, \qquad u(T,x)=h(x)? \] Precisely, at an interior local maximum of $u-\varphi$, with $\varphi\in C^{1,2}$ touching $u$, require $\partial_t\varphi+\mathcal L\varphi+ g(t,x,u,\sigma^\top\nabla\varphi)\geq0$; at a local minimum require the reverse inequality. No convexity, concavity, or comparison condition from the remaining part of assumption (A2) is imposed.

Scope and status

The forward diffusion has a unique solution under the stated assumptions, including bounded diffusion coefficient. Existence of individual associated BSDE solutions does not by itself produce a continuous, consistently selected Markovian value function. The question concerns viscosity existence; uniqueness of the PDE solution is an additional issue.

References for the source of the problem

Fan, Hu and Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2, §5, Problem 5.2; §4.3, (A1), equations (4.8), (4.13)--(4.14), and Definition 4.36. Theorem 4.37 includes the extra comparison structure being removed here. https://arxiv.org/html/2309.06233v2#S5.

Common conventions: Source mathematical conventions

Well-defined objects. All probability spaces and measurable variables are specified or inherited from a local statistical model. Expectations used as finite targets require the stated integrability. A decision rule or estimator is measurable with respect to its available information; randomized rules may use an auxiliary independent random variable. Norms without a subscript are Euclidean in finite-dimensional spaces. An optimizer is requested only when attainment is assured; otherwise the question concerns the optimal value and $\varepsilon$-optimal admissible rules for every $\varepsilon>0$.

Identification. A structural model $m\in\mathcal M$ implies an observable law $P_m^{\rm obs}$ and target $\theta(m)$. The target is point identified on $\mathcal M$ if \[ P_{m_1}^{\rm obs}=P_{m_2}^{\rm obs} \quad\Longrightarrow\quad\theta(m_1)=\theta(m_2) \qquad(m_1,m_2\in\mathcal M). \] When this fails, the sharp identified set is \[ \Theta(P;\mathcal M)=\{\theta(m):m\in\mathcal M, P_m^{\rm obs}=P\}. \] Specifying an intervention or estimand does not establish identification. Positivity, exclusion, causal sufficiency, measurement restrictions, and transport assumptions are substantive local inputs, never automatic consequences of notation.

Minimax risk and nontrivial inference. For a statistical experiment with data law $P^{(n)}$, target $\theta(P)$, and nonnegative loss $\ell$, define \[ \mathcal R_n(\mathcal P,\ell) =\inf_{\widehat\theta_n}\sup_{P\in\mathcal P} \E_{P^{(n)}}\ell(\widehat\theta_n,\theta(P)). \] The infimum is over measurable estimators. A sharp rate is a sequence $r_n$ for which $0<c\leq C<\infty$, independent of $n$, satisfies $c r_n\leq\mathcal R_n\leq C r_n$ for all sufficiently large $n$. Squared-error parametric risk is of order $n^{-1}$; the corresponding root-mean-squared error is of order $n^{-1/2}$. These different conventions are distinguished locally. A uniform asymptotic confidence region $C_n$ must satisfy \[ \liminf_{n\to\infty}\inf_{P\in\mathcal P} \Pr_{P^{(n)}}\{\theta(P)\in C_n\}\geq 1-\alpha, \qquad 0<\alpha<1, \] together with an informative diameter or expected-width requirement. Coverage alone permits the vacuous whole parameter space. Testing questions similarly impose both size control and power against a declared separated alternative class.

Binary causal baseline. When an entry explicitly invokes this baseline, one observes independent copies of $O=(X,W,Y)$, with $W\in\{0,1\}$ and potential outcomes $Y(0),Y(1)$. Assume consistency $Y=Y(W)$, no interference, conditional exchangeability $(Y(0),Y(1))\perp W\mid X$, and overlap $\epsilon\leq e(X)\leq1-\epsilon$ almost surely for a fixed $0<\epsilon<1/2$, where \[ e(x)=\Pr(W=1\mid X=x),\qquad m_w(x)=\E[Y\mid W=w,X=x],\qquad \tau(x)=m_1(x)-m_0(x). \] These assumptions identify conditional mean effects almost everywhere under the covariate law. Pointwise or uniform effects require appropriate regularity and a specified support. Continuous treatment, longitudinal data, adaptive experiments, and network interference require their own assumptions in place of this baseline. Bounded outcomes, densities, and smoothness are additional assumptions only where stated.

Function classes. On $[0,1]^d$, $\mathcal H_d^s(L)$ is the isotropic Hölder ball of radius $L>0$: put $k=\lceil s\rceil-1$ and $\gamma=s-k\in(0,1]$, bound derivatives through order $k$, and require the order-$k$ derivatives to be $\gamma$-Hölder with constant at most $L$. Thus integer orders use a Lipschitz final derivative convention. The class $\operatorname{BV}_d(L)$ consists of functions $f\in L^1((0,1)^d)$ whose distributional derivative is a finite vector measure and for which $\|f\|_{L^1}+|Df|((0,1)^d)\leq L$. An $L^\infty$ bound or a particular pointwise version is imposed separately when needed. Sparse and neural-network classes require a declared dictionary or architecture and coefficient bounds.

An anisotropic Hölder class with declared block smoothness indices means coordinatewise Hölder regularity in each block, uniformly in the other blocks, with all corresponding derivative and Hölder bounds at most the stated radius. Derivative orders and the integer-order convention in each block are those just given. Mixed-derivative bounds are additional restrictions only when explicitly stated.

Decision and computational questions. Policy regret compares admissible feasible policies under the same law and constraints. In computational entries, finite data and thresholds have rational encodings; running time is measured in their total bit length. Real-valued equilibrium search uses the explicitly declared algebraic or FIXP encoding. An approximation ratio for maximization is written $\operatorname{OPT}/\operatorname{ALG}$ and for minimization as $\operatorname{ALG}/\operatorname{OPT}$, with zero optima and infeasible instances handled locally. Historical approximation bounds are not automatically current bounds.

Common conventions: BSDE conventions for SC-01--SC-06

Fix $T>0$ and an integer $d\geq1$. The probability space $(\Omega,\mathcal F,\Pr)$ is complete, $B$ is a standard $d$-dimensional Brownian motion, and $(\mathcal F_t)_{0\leq t\leq T}$ is its usual augmented natural filtration, with $\mathcal F=\mathcal F_T$. A scalar BSDE solution is a pair $(Y,Z)$, where $Y$ is continuous and adapted, $Z$ is predictable and $\R^d$-valued, and, almost surely, \[ \int_0^T\left(|g(s,Y_s,Z_s)|+|Z_s|^2\right)\,ds<\infty, \qquad Y_t=\xi+\int_t^T g(s,Y_s,Z_s)\,ds-\int_t^T Z_s\cdot dB_s. \] The identity holds simultaneously for every $t\in[0,T]$. Uniqueness means indistinguishability of $Y$ and equality of $Z$ for $dt\otimes\Pr$ almost every $(t,\omega)$. A process $U$ is of class (D) when $\{U_\tau:\tau\text{ is an }(\mathcal F_t)\text{-stopping time},\ 0\leq\tau\leq T\}$ is uniformly integrable. For $p>0$, $L^p$ denotes $\{\xi:\E|\xi|^p<\infty\}$; for $\mu>0$, $\exp(\mu L)$ denotes $\{\xi:\E e^{\mu|\xi|}<\infty\}$. These conventions remove ambiguity about filtration, stochastic integration, and integrability; they do not add an $L^2$ expectation requirement to the scalar solution definition. The conventions follow Fan, Hu and Tang, 1D nonlinear backward stochastic differential equations: a unified theory and applications, arXiv:2309.06233v2 (11 February 2026), §§1 and 1.3, https://arxiv.org/html/2309.06233v2.

Statement licence: CC-BY-4.0 · Source file

Contributions 3

Econ-46-A3 · Full solution claimed · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-46, version 2

C65-45: Viscosity existence without comparison structure

Problem ID: C65-45  Model: GPT 6 Astra Pro Version: 2

Revision provenance. This version incorporates the supplied Pro audit's scope annotation. The complete coefficient check, viscosity lemmas and nonexistence proof from version 1 are retained. No new general existence theorem or independent numerical audit is asserted. Outcome: FULL SOLUTION: negative resolution of continuous viscosity existence under the stated, possibly degenerate, diffusion assumptions.

Tools: The mathematical proof and literature record are carried forward from version 1, checked on 9 October 2026. The supplied audit adds a scope qualification; its coalition and matching scripts do not verify this PDE argument.

1) FORMAL RESTATEMENT

For every $T>0$, integers $n,d\ge1$, and continuous coefficients $b,\sigma,g,h$ satisfying the two growth inequalities and the spatial Lipschitz assumptions in the uploaded statement, is there a function $u\in C([0,T]\times{\mathbb{R}}^n)$ with terminal value $h$ that solves \[ u_t+b\cdot Du+\tfrac12\operatorname{tr}(\sigma\sigma^\top D^2u) +g(t,x,u,\sigma^\top Du)=0 \] in the stated viscosity sense? At a local maximum of $u-\varphi$ in $(0,T)\times{\mathbb{R}}^n$, the displayed expression, evaluated on $\varphi$ and $u$, must be nonnegative; at a local minimum it must be nonpositive. Tests are $C^{1,2}$. No growth restriction on the candidate continuous solution is added.

The constants satisfy $K>0$, $k\ge0$, $\alpha\in(1,2]$, $\alpha^*=\alpha/(\alpha-1)$, and $1\le p<\alpha^*$. In particular, $\sigma\equiv0$ is permitted. This is the decisive stress point. The other stress point is a continuous, non-Lipschitz dependence on $y$. We use ${\operatorname{sgn}}(0)=0$ and do not replace the question by a uniformly elliptic version.

Meaning of the resolution. The universal existence assertion is false because one admissible instance has no continuous viscosity solution. This is a counterexample/disproof, not an affirmative existence theorem for all coefficients in the class. Requiring uniform ellipticity, or adding a comparison or uniqueness condition in $y$, defines a different question; this proof does not decide those strengthened formulations.

2) QUICK LITERATURE/CONTEXT CHECK

The primary source is Problem 5.2 of Fan--Hu--Tang [FHT], with growth conditions (4.13)--(4.14). The PDF numbers its accompanying existence result Theorem 4.10; the linked HTML uses different theorem numbering. The PDF assumptions explicitly permit zero diffusion. The argument below is independent of that existence theorem and makes no claim of bibliographic priority.

3) ATTACK PLAN

Proof routes tested. A BSDE-based construction would need continuous selection of solutions as the initial state varies. Perron's method would need adequate continuous barriers or a comparison substitute. Neither property follows just from pointwise ODE existence.

Disproof routes. First set $b=\sigma=0$; then force opposite nonunique ODE branches by a sign-changing spatial parameter. Finally check that viscosity solutions cannot evade the ODE obstruction. This is the selected route.

Landscape and tools. This is a degenerate semilinear PDE and selection problem. The relevant tools are time reversal (to fix signs), viscosity penalization (to restrict to a spatial slice), a one-dimensional monotonicity test (to identify viscosity ODE solutions), an integral identity (to avoid an unproved uniqueness theorem), scalar differential inequalities (to force branch separation), and direct growth estimates (to verify every assumption).

4) WORK

Theorem (Explicit nonexistence example): Set \[ T=1,\quad n=d=1,\quad b=0,\quad\sigma=0,\quad h=0, \qquad g(t,x,y,z)=x+2{\operatorname{sgn}}(y)\sqrt{|y|}. \] These data satisfy every assumption in the question, with $K=1$, $k=2$, $\alpha=2$, $\alpha^*=2$, and $p=1$. Nevertheless, there is no continuous viscosity solution on $[0,1]\times{\mathbb{R}}$.
Proof[Verification of the coefficients]: The map $y\mapsto{\operatorname{sgn}}(y)\sqrt{|y|}$ is continuous, including at zero. All remaining coefficients are continuous. Both spatial Lipschitz inequalities for $b,\sigma$ hold with left side zero. Write $r=|y|$. For $y\ne0$, \[ {\operatorname{sgn}}(y)g\le |x|+2\sqrt r\le |x|+1+r \le2(1+|x|+r+|z|^2). \] At $y=0$ the one-sided left side is zero. Further, \[ |g|+|h|\le |x|+1+r \le2(1+|x|+e^{2r}+|z|^2). \] The first inequality uses $(\sqrt r-1)^2\ge0$, and the second follows, for example, from $e^{2r}\ge1+2r$. Thus both required growth inequalities hold globally. Also $p=1<2=\alpha^*$. ∎

We give the viscosity argument in detail; absence of a classical solution alone would not suffice.

Lemma (Restriction to spatial slices): Let $F:{\mathbb{R}}\times{\mathbb{R}}\to{\mathbb{R}}$ be continuous. Suppose $v$ is a continuous viscosity solution of $v_s-F(x,v)=0$ on $(0,1)\times{\mathbb{R}}$, with the conventional subsolution inequality $\varphi_s\le F(x,v)$ at a maximum. Then $w(s)=v(s,x_0)$ is a viscosity solution of $w'=F(x_0,w)$ for every $x_0$.
Proof: Let $w-\phi$ have a local maximum at $s_0\in(0,1)$. Replace $\phi$ by $\phi_\eta(s)=\phi(s)+\eta(s-s_0)^2$, $\eta>0$, to make this maximum strict on a sufficiently small closed time interval. Choose a compact spatial interval about $x_0$ and maximize \[ v(s,x)-\phi_\eta(s)-N|x-x_0|^2 \] on the resulting rectangle. Its maximum is at least its value at $(s_0,x_0)$. Boundedness on the rectangle implies $|x_N-x_0|\to0$. Every subsequential limit of $s_N$ maximizes $w-\phi_\eta$ on the time interval; strictness gives $s_N\to s_0$. The maximizing points are therefore interior for all sufficiently large $N$. Adding a constant to the penalized test makes it touch $v$ there. Since the PDE has no spatial derivatives, its subsolution inequality gives $\phi_\eta'(s_N)\le F(x_N,v(s_N,x_N))$. Continuity and passage to the limit give $\phi'(s_0)\le F(x_0,w(s_0))$. For a local minimum, minimize $v-\phi_\eta+N|x-x_0|^2$, strictifying in the opposite direction. This proves the supersolution inequality as well. ∎
Lemma (Continuous viscosity ODE solutions are classical): If a continuous $w:[0,1)\to{\mathbb{R}}$ solves $w'=F(x_0,w)$ in viscosity sense on $(0,1)$, then \[ w(s)=w(0)+\int_0^s F(x_0,w(r))\,dr. \]
Proof: Put $A(s)=\int_0^s F(x_0,w(r))dr$ and $r(s)=w(s)-A(s)$. The function $A$ is $C^1$. Adding $A$ to any test for $r$ shows that $r$ solves $r'=0$ in viscosity sense.

Here is a direct proof that a continuous viscosity subsolution of $r'\le0$ is nonincreasing. Suppose $r(b)>r(a)$ for $0<a<b<1$. Choose $\eta>0$ with $\eta(b-a)<r(b)-r(a)$, and choose $c\in(b,1)$. For sufficiently small $\delta>0$, \[ r(b)-\eta b-\frac\delta{c-b} >r(a)-\eta a-\frac\delta{c-a}. \] The function $r(t)-\eta t-\delta/(c-t)$ tends to $-\infty$ as $t\uparrow c$ and has a maximum on $[a,c)$ at some point in $(a,c)$. The corresponding touching test has derivative $\eta+\delta/(c-t)^2>0$, a contradiction. Thus $r$ is nonincreasing. Applying the same argument to $-r$, which is a subsolution when $r$ is a supersolution, makes $r$ nondecreasing. It is constant on $(0,1)$ and, by continuity, at zero too. This proves the integral identity without assuming local Lipschitz continuity of $F$. ∎

Proof[Nonexistence]: Assume a continuous viscosity solution $u$ exists and put $v(s,x)=u(1-s,x)$. Time reversal turns precisely the sign convention in the question into \[ v_s=F(x,v),\qquad F(x,y)=x+2{\operatorname{sgn}}(y)\sqrt{|y|},\qquad v(0,x)=0. \] Indeed a touching test $\phi(s,x)$ corresponds to $\phi(1-t,x)$ and its time derivative changes sign. The two lemmas show that each slice $w(s)=v(s,x)$ is a classical solution of this ODE, including its right derivative at zero.

Fix $x>0$. Since $w'(0)=x$, $w(s)>0$ for small positive $s$. It cannot first return to zero: on a positive interval $w'=x+2\sqrt w\ge x$, so integration from any positive point makes $w$ strictly increase. Consequently $w(s)>0$ for every $s\in(0,1)$. On this interval $q(s)=\sqrt{w(s)}$ satisfies \[ q'(s)=1+\frac{x}{2\sqrt{w(s)}}\ge1. \] Integrating from $\varepsilon$ to $s$ and letting $\varepsilon\downarrow0$ gives $q(s)\ge s$, hence $v(s,x)\ge s^2$ for every $x>0$. If $x<0$, then $-w$ solves the positive-parameter equation with parameter $-x$. Therefore $v(s,x)\le-s^2$ for every $x<0$.

For fixed $s\in(0,1)$, continuity at $x=0$ would imply simultaneously $v(s,0)\ge s^2$ and $v(s,0)\le-s^2$. Since $s^2>0$, this is impossible. ∎

5) VERIFICATION

Supplied scope audit. The supplied Pro corrections explicitly classify this as a negative continuous-viscosity-existence result when degenerate diffusion is allowed. That annotation agrees with the theorem: $\sigma\equiv0$ is admissible, and the contradiction concerns continuity across $x=0$. The supplied finite verification programs address coalition formation and matching only, and are not cited as a check of this proof.

Tiny cases. At $x=0$, the ODE has the solutions $0$, $s^2$, $-s^2$, and delayed variants; selecting any one cannot repair the limits from both spatial sides. At $T=0$ the contradiction disappears, as it must. The example needs only one space coordinate, one Brownian coordinate, and zero terminal data.

Computational check. For $x>0$, separation of variables yields \[ s=q-\frac{x}{2}\log(1+2q/x),\qquad q=\sqrt{v(s,x)}. \] Version 1 reported a separate scalar bracketing calculation. Its recorded values at $s=1/2$ and $x=10^{-1},10^{-2},10^{-3},10^{-6}$ are respectively $0.3975850$, $0.2738441$, $0.2534698$, and $0.2500069$ for $v$; the negative side is its negative. These figures are carried forward as a numerical illustration, not as a computation in the current audit bundle, which contains no PDE verifier. They are not used to prove the gap of at least $1/2$ between the two sides.

Adversarial checks. No uniqueness of the non-Lipschitz ODE was invoked. Spatial restriction was proved for viscosity tests rather than assumed. Only compact rectangles were used in penalization, so no bounds at spatial infinity are hidden. The absence of diffusion is authorized by the literal assumptions; imposing uniform ellipticity would be a substantive new problem, not a correction of a typo. The contradiction rules out every continuous solution, not merely polynomial-growth solutions or solutions obtained by a particular BSDE selection.

6) FINAL

{LABEL: FULL SOLUTION. SUBLABEL: COUNTEREXAMPLE/DISPROOF.}

The explicit data in the theorem satisfy all the uploaded hypotheses and admit no continuous viscosity solution. The complete proof consists of coefficient verification, the two viscosity lemmas, and the forced opposite inequalities $v(s,x)\ge s^2$ for $x>0$ and $v(s,x)\le-s^2$ for $x<0$. No claim is made here about a version that adds uniform ellipticity or a suitable one-sided uniqueness condition in $y$.

Completion Estimate

100%

This is a post hoc, heuristic estimate for the literal universal existence target. One admissible coefficient tuple is checked against every stated assumption, and the two viscosity lemmas prove that no continuous solution exists for it. The complete negative resolution does not imply an affirmative general existence theorem, a result under uniform ellipticity, or a novelty claim.

References

  • [FHT] Shengjun Fan, Ying Hu and Shanjian Tang. 1D nonlinear backward stochastic differential equations: a unified theory and applications. arXiv:2309.06233v2, 11 February 2026. In particular Problems 5.1--5.5, PDF pp. 32--33; for C65-45 see (A1), (4.13)--(4.14), PDF p. 31. https://arxiv.org/pdf/2309.06233v2.
  • [audit] User-supplied Pro audit bundle, Independent checks for the economics "solved" audit, supplied 9 October 2026: README.md and proposed_corrections.md, C65-45 scope annotation. Reviewed snapshot ccccd798864807c1f3fb8fe36727fe35ee875aeb. Unpublished mathematical scope audit; no finite PDE verification is claimed by the supplied programs.

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Econ-46-A2 · Full solution claimed · LLM-generated · gpt 6 astra pro ·

GPT 6 Astra Pro attempt on Econ-46

C65-45: Viscosity existence without comparison structure

Outcome: FULL SOLUTION: counterexample to the statement as written.

Tools: web browsing available; Python computation available. Literature checked on 9 October 2026.

1) FORMAL RESTATEMENT

For every $T>0$, integers $n,d\ge1$, and continuous coefficients $b,\sigma,g,h$ satisfying the two growth inequalities and the spatial Lipschitz assumptions in the uploaded statement, is there a function $u\in C([0,T]\times{\mathbb{R}}^n)$ with terminal value $h$ that solves \[ u_t+b\cdot Du+\tfrac12\operatorname{tr}(\sigma\sigma^\top D^2u) +g(t,x,u,\sigma^\top Du)=0 \] in the stated viscosity sense? At a local maximum of $u-\varphi$ in $(0,T)\times{\mathbb{R}}^n$, the displayed expression, evaluated on $\varphi$ and $u$, must be nonnegative; at a local minimum it must be nonpositive. Tests are $C^{1,2}$. No growth restriction on the candidate continuous solution is added.

The constants satisfy $K>0$, $k\ge0$, $\alpha\in(1,2]$, $\alpha^*=\alpha/(\alpha-1)$, and $1\le p<\alpha^*$. In particular, $\sigma\equiv0$ is permitted. This is the decisive stress point. The other stress point is a continuous, non-Lipschitz dependence on $y$. We use ${\operatorname{sgn}}(0)=0$ and do not replace the question by a uniformly elliptic version.

2) QUICK LITERATURE/CONTEXT CHECK

The primary source is Problem 5.2 of Fan--Hu--Tang [FHT], with growth conditions (4.13)--(4.14). The PDF numbers its accompanying existence result Theorem 4.10; the linked HTML uses different theorem numbering. The PDF assumptions explicitly permit zero diffusion. The argument below is independent of that existence theorem and makes no claim of bibliographic priority.

3) ATTACK PLAN

Proof routes tested. A BSDE-based construction would need continuous selection of solutions as the initial state varies. Perron's method would need adequate continuous barriers or a comparison substitute. Neither property follows just from pointwise ODE existence.

Disproof routes. First set $b=\sigma=0$; then force opposite nonunique ODE branches by a sign-changing spatial parameter. Finally check that viscosity solutions cannot evade the ODE obstruction. This is the selected route.

Landscape and tools. This is a degenerate semilinear PDE and selection problem. The relevant tools are time reversal (to fix signs), viscosity penalization (to restrict to a spatial slice), a one-dimensional monotonicity test (to identify viscosity ODE solutions), an integral identity (to avoid an unproved uniqueness theorem), scalar differential inequalities (to force branch separation), and direct growth estimates (to verify every assumption).

4) WORK

Theorem (Explicit nonexistence example): Set \[ T=1,\quad n=d=1,\quad b=0,\quad\sigma=0,\quad h=0, \qquad g(t,x,y,z)=x+2{\operatorname{sgn}}(y)\sqrt{|y|}. \] These data satisfy every assumption in the question, with $K=1$, $k=2$, $\alpha=2$, $\alpha^*=2$, and $p=1$. Nevertheless, there is no continuous viscosity solution on $[0,1]\times{\mathbb{R}}$.
Proof[Verification of the coefficients]: The map $y\mapsto{\operatorname{sgn}}(y)\sqrt{|y|}$ is continuous, including at zero. All remaining coefficients are continuous. Both spatial Lipschitz inequalities for $b,\sigma$ hold with left side zero. Write $r=|y|$. For $y\ne0$, \[ {\operatorname{sgn}}(y)g\le |x|+2\sqrt r\le |x|+1+r \le2(1+|x|+r+|z|^2). \] At $y=0$ the one-sided left side is zero. Further, \[ |g|+|h|\le |x|+1+r \le2(1+|x|+e^{2r}+|z|^2). \] The first inequality uses $(\sqrt r-1)^2\ge0$, and the second follows, for example, from $e^{2r}\ge1+2r$. Thus both required growth inequalities hold globally. Also $p=1<2=\alpha^*$. ∎

We give the viscosity argument in detail; absence of a classical solution alone would not suffice.

Lemma (Restriction to spatial slices): Let $F:{\mathbb{R}}\times{\mathbb{R}}\to{\mathbb{R}}$ be continuous. Suppose $v$ is a continuous viscosity solution of $v_s-F(x,v)=0$ on $(0,1)\times{\mathbb{R}}$, with the conventional subsolution inequality $\varphi_s\le F(x,v)$ at a maximum. Then $w(s)=v(s,x_0)$ is a viscosity solution of $w'=F(x_0,w)$ for every $x_0$.
Proof: Let $w-\phi$ have a local maximum at $s_0\in(0,1)$. Replace $\phi$ by $\phi_\eta(s)=\phi(s)+\eta(s-s_0)^2$, $\eta>0$, to make this maximum strict on a sufficiently small closed time interval. Choose a compact spatial interval about $x_0$ and maximize \[ v(s,x)-\phi_\eta(s)-N|x-x_0|^2 \] on the resulting rectangle. Its maximum is at least its value at $(s_0,x_0)$. Boundedness on the rectangle implies $|x_N-x_0|\to0$. Every subsequential limit of $s_N$ maximizes $w-\phi_\eta$ on the time interval; strictness gives $s_N\to s_0$. The maximizing points are therefore interior for all sufficiently large $N$. Adding a constant to the penalized test makes it touch $v$ there. Since the PDE has no spatial derivatives, its subsolution inequality gives $\phi_\eta'(s_N)\le F(x_N,v(s_N,x_N))$. Continuity and passage to the limit give $\phi'(s_0)\le F(x_0,w(s_0))$. For a local minimum, minimize $v-\phi_\eta+N|x-x_0|^2$, strictifying in the opposite direction. This proves the supersolution inequality as well. ∎
Lemma (Continuous viscosity ODE solutions are classical): If a continuous $w:[0,1)\to{\mathbb{R}}$ solves $w'=F(x_0,w)$ in viscosity sense on $(0,1)$, then \[ w(s)=w(0)+\int_0^s F(x_0,w(r))\,dr. \]
Proof: Put $A(s)=\int_0^s F(x_0,w(r))dr$ and $r(s)=w(s)-A(s)$. The function $A$ is $C^1$. Adding $A$ to any test for $r$ shows that $r$ solves $r'=0$ in viscosity sense.

Here is a direct proof that a continuous viscosity subsolution of $r'\le0$ is nonincreasing. Suppose $r(b)>r(a)$ for $0<a<b<1$. Choose $\eta>0$ with $\eta(b-a)<r(b)-r(a)$, and choose $c\in(b,1)$. For sufficiently small $\delta>0$, \[ r(b)-\eta b-\frac\delta{c-b} >r(a)-\eta a-\frac\delta{c-a}. \] The function $r(t)-\eta t-\delta/(c-t)$ tends to $-\infty$ as $t\uparrow c$ and has a maximum on $[a,c)$ at some point in $(a,c)$. The corresponding touching test has derivative $\eta+\delta/(c-t)^2>0$, a contradiction. Thus $r$ is nonincreasing. Applying the same argument to $-r$, which is a subsolution when $r$ is a supersolution, makes $r$ nondecreasing. It is constant on $(0,1)$ and, by continuity, at zero too. This proves the integral identity without assuming local Lipschitz continuity of $F$. ∎

Proof[Nonexistence]: Assume a continuous viscosity solution $u$ exists and put $v(s,x)=u(1-s,x)$. Time reversal turns precisely the sign convention in the question into \[ v_s=F(x,v),\qquad F(x,y)=x+2{\operatorname{sgn}}(y)\sqrt{|y|},\qquad v(0,x)=0. \] Indeed a touching test $\phi(s,x)$ corresponds to $\phi(1-t,x)$ and its time derivative changes sign. The two lemmas show that each slice $w(s)=v(s,x)$ is a classical solution of this ODE, including its right derivative at zero.

Fix $x>0$. Since $w'(0)=x$, $w(s)>0$ for small positive $s$. It cannot first return to zero: on a positive interval $w'=x+2\sqrt w\ge x$, so integration from any positive point makes $w$ strictly increase. Consequently $w(s)>0$ for every $s\in(0,1)$. On this interval $q(s)=\sqrt{w(s)}$ satisfies \[ q'(s)=1+\frac{x}{2\sqrt{w(s)}}\ge1. \] Integrating from $\varepsilon$ to $s$ and letting $\varepsilon\downarrow0$ gives $q(s)\ge s$, hence $v(s,x)\ge s^2$ for every $x>0$. If $x<0$, then $-w$ solves the positive-parameter equation with parameter $-x$. Therefore $v(s,x)\le-s^2$ for every $x<0$.

For fixed $s\in(0,1)$, continuity at $x=0$ would imply simultaneously $v(s,0)\ge s^2$ and $v(s,0)\le-s^2$. Since $s^2>0$, this is impossible. ∎

5) VERIFICATION

Tiny cases. At $x=0$, the ODE has the solutions $0$, $s^2$, $-s^2$, and delayed variants; selecting any one cannot repair the limits from both spatial sides. At $T=0$ the contradiction disappears, as it must. The example needs only one space coordinate, one Brownian coordinate, and zero terminal data.

Computational check. For $x>0$, separation of variables yields \[ s=q-\frac{x}{2}\log(1+2q/x),\qquad q=\sqrt{v(s,x)}. \] The supplied script solves this scalar equation by bracketing. At $s=1/2$ and $x=10^{-1},10^{-2},10^{-3},10^{-6}$ it gives respectively $0.3975850$, $0.2738441$, $0.2534698$, and $0.2500069$ for $v$; the negative side is its negative. These finite tests illustrate, but are not used to prove, the gap of at least $1/2$ between the two sides.

Adversarial checks. No uniqueness of the non-Lipschitz ODE was invoked. Spatial restriction was proved for viscosity tests rather than assumed. Only compact rectangles were used in penalization, so no bounds at spatial infinity are hidden. The absence of diffusion is authorized by the literal assumptions; imposing uniform ellipticity would be a substantive new problem, not a correction of a typo. The contradiction rules out every continuous solution, not merely polynomial-growth solutions or solutions obtained by a particular BSDE selection.

6) FINAL

{LABEL: FULL SOLUTION. SUBLABEL: COUNTEREXAMPLE/DISPROOF.}

The explicit data in the theorem satisfy all the uploaded hypotheses and admit no continuous viscosity solution. The complete proof consists of coefficient verification, the two viscosity lemmas, and the forced opposite inequalities $v(s,x)\ge s^2$ for $x>0$ and $v(s,x)\le-s^2$ for $x<0$. No claim is made here about a version that adds uniform ellipticity or a suitable one-sided uniqueness condition in $y$.

References

  • [FHT] Shengjun Fan, Ying Hu and Shanjian Tang. 1D nonlinear backward stochastic differential equations: a unified theory and applications. arXiv:2309.06233v2, 11 February 2026. In particular Problems 5.1--5.5, PDF pp. 32--33; for C65-45 see (A1), (4.13)--(4.14), PDF p. 31. https://arxiv.org/pdf/2309.06233v2.

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Econ-46-A1 · Full solution claimed · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Econ-46

An obstruction to continuous viscosity selection for a degenerate semilinear terminal problem

Problem ID: C65-45. LLM claim: negative resolution by counterexample for the stated degenerate-diffusion formulation; independent external review pending.

Target and result

The target asks for a continuous viscosity solution under the growth assumptions of [FHT, Problem 5.2], without its additional comparison structure. The diffusion is allowed to be degenerate. We give a negative example in that allowed class, with a bounded smooth terminal function. The obstruction is already a family of scalar ordinary differential equations: their solutions exist for every spatial parameter but cannot be selected continuously.

Our argument is self-contained, including the passage from a viscosity PDE solution to its spatial slices. The counterexample addresses the literal catalogue formulation. It makes no assertion about uniformly elliptic versions or priority of the example. Independent external review of this proposed negative resolution is pending.

A precise negative statement

Fix any $T>0$, take $n=d=1$, and set \begin{equation}\label{wsref-0-64-61-74-61} b(t,x)=0,\qquad \sigma(t,x)=0,\qquad h(x)=\tanh x,\qquad g(t,x,y,z)=F(y):=\begin{cases}\sqrt y,&y\geq0,\\-\sqrt{-y},&y<0.\end{cases} \end{equation} Here $F(0)=0$ and $F$ is continuous. The proposed PDE is \begin{equation}\label{wsref-0-70-64-65} u_t+F(u)=0\quad\hbox{on }(0,T)\times{\mathbb{R}},\qquad u(T,x)=\tanh x. \end{equation} Use exactly the backward viscosity convention of the target: a test function at an upper contact satisfies $\varphi_t+F(u)\geq0$, and a test function at a lower contact satisfies $\varphi_t+F(u)\leq0$.
Theorem: The coefficients (data) satisfy all the catalogue assumptions with $\alpha=2$, $\alpha^*=2$, $p=1$, $k=1$, and $K=1$. Nevertheless, for every $T>0$ there is no function $u\in C([0,T]\times{\mathbb{R}})$ that is a viscosity solution of (pde) in the stated convention.

We first prove two elementary viscosity facts needed to exclude a nonclassical continuous solution.

Lemma (Restriction to a spatial line): Let $J\subset{\mathbb{R}}$ be an open interval, $V\subset{\mathbb{R}}^n$ open, and $F:{\mathbb{R}}\to{\mathbb{R}}$ continuous. Suppose $w\in C(J\times V)$ satisfies $w_s-F(w)=0$ in viscosity sense, with $\varphi_s-F(w)\leq0$ at upper contacts and the reverse inequality at lower contacts. For every $x_0\in V$, the function $z(s)=w(s,x_0)$ is a viscosity solution of $z'-F(z)=0$ on $J$.
Proof: Suppose $z-\psi$ has a local maximum at $s_0$, where $\psi$ is $C^1$. Adding $(s-s_0)^4$ makes this maximum strict on a sufficiently small closed interval $I$ around $s_0$, without changing the derivative at $s_0$. Choose a closed spatial ball $B\subset V$ centered at $x_0$ and maximize \[ w(s,x)-\psi(s)-(s-s_0)^4-\varepsilon^{-1}|x-x_0|^2 \] over $I\times B$. Let $(s_\varepsilon,x_\varepsilon)$ be a maximizer. Comparison with $(s_0,x_0)$ and boundedness on the compact cylinder show $|x_\varepsilon-x_0|^2=O(\varepsilon)$. Any limit of $s_\varepsilon$ maximizes $z-\psi-(s-s_0)^4$, hence equals $s_0$. Both coordinates are therefore interior for small $\varepsilon$. Add a constant to the displayed smooth test function to obtain upper contact at the maximizing point. The PDE inequality gives \[ \psi'(s_\varepsilon)+4(s_\varepsilon-s_0)^3 -F(w(s_\varepsilon,x_\varepsilon))\leq0. \] Continuity and passage to the limit yield $\psi'(s_0)-F(z(s_0))\leq0$. For a lower contact, subtract $(s-s_0)^4$, minimize after adding $\varepsilon^{-1}|x-x_0|^2$, and use the lower-contact inequality. This proves both required inequalities. ∎
Lemma (A continuous viscosity ODE solution is classical): Let $J\subset{\mathbb{R}}$ be an open interval. If $z\in C(J)$ is a viscosity solution of $z'-F(z)=0$ for continuous $F$, then $z\in C^1(J)$ and $z'(s)=F(z(s))$ for every $s\in J$.
Proof: Fix $s_*\in J$ and put $A(s)=\int_{s_*}^sF(z(r))\,dr$. Then $A$ is $C^1$. Testing $z-A$ with $\psi$ is the same as testing $z$ with $\psi+A$; since $A'=F(z)$, the continuous function $q=z-A$ is a viscosity solution of $q'=0$.

For completeness, an upper-contact viscosity inequality $q'\leq0$ forces $q$ to be nonincreasing. If $a<b$ in $J$ and $q(b)>q(a)$, choose $0<c<(q(b)-q(a))/(b-a)$ and then $b'>b$ in $J$. For sufficiently small $\varepsilon>0$ the function \[ q(s)-cs-\frac{\varepsilon}{b'-s},\qquad a\leq s<b', \] has value at $b$ larger than at $a$. It tends to $-\infty$ at $b'$, so its maximum occurs at an interior point of $(a,b')$. The corresponding test derivative is $c+\varepsilon/(b'-s)^2>0$, a contradiction. Applying this conclusion to $-q$, using the lower-contact inequality for $q$, shows that $q$ is also nondecreasing. Thus $q$ is constant. Consequently $z=A+q$ is $C^1$ and has derivative $F(z)$. ∎

Proof[Proof of Theorem Theorem]: The conditions on $b$ and $\sigma$ are immediate. For every $y$, ${\operatorname{sgn}}(y)F(y)=\sqrt{|y|}\leq1+|y|$, independently of the convention for ${\operatorname{sgn}}(0)$. Moreover $|\tanh x|\leq1$ and $\sqrt{|y|}\leq e^{|y|}$, so \[ |g(t,x,y,z)|+|h(x)|\leq1+e^{|y|} \leq1+|x|+e^{|y|}+|z|^2. \] These are both required growth inequalities at the announced parameters; $p=1<\alpha^*=2$. In particular no unbounded terminal data are needed.

Suppose a continuous viscosity solution existed and define $w(s,x)=u(T-s,x)$ for $0\leq s\leq T$. The viscosity inequalities transform to those in Lemma Lemma (Restriction to a spatial line), so every spatial slice is a classical solution of $w_s=F(w)$ for $0<s<T$. Continuity at $s=0$, together with boundedness of the slice and continuity of $F$, gives its integral equation also from $s=0$.

For $x>0$ the initial value $\tanh x$ is positive. On every interval where the solution is positive its derivative is positive, so it cannot reach zero: at a hypothetical first zero its values before that time would be at least $\tanh x>0$. Thus \[ \frac{d}{ds}\sqrt{w(s,x)}=\frac12, \qquad w(s,x)=\left(\sqrt{\tanh x}+\frac{s}{2}\right)^2. \] For $x<0$, the same reasoning applied to $-w$ gives \[ w(s,x)=-\left(\sqrt{-\tanh x}+\frac{s}{2}\right)^2. \] At any fixed $s\in(0,T)$ the right and left spatial limits at zero are $s^2/4$ and $-s^2/4$. They differ. This contradicts continuity of $w(s,\cdot)$, proving the assertion. ∎

What the example does and does not establish

At $x=0$, the ODE has the zero solution and delayed positive and negative solutions; for example $\pm((s-a)^+)^2/4$ for $0\leq a\leq T$. Each associated deterministic BSDE is perfectly well-defined with $Z=0$. Indeed the forward state is constant, and for every fixed initial state these solution paths are bounded, hence satisfy every exponential class-(D) condition. Therefore individual BSDE existence does not repair the lack of a continuous spatial selection.

The source version was inspected on 9 October 2026: its assumption (A1) in Section 4.3 imposes continuity, a bounded diffusion coefficient, and spatial Lipschitz bounds, but no positive ellipticity lower bound. Problem 5.2 removes the remaining comparison assumption on $g$ while retaining its growth bounds. Our example exploits precisely these freedoms. Peano nonuniqueness itself is classical; recent related work [Peano] studies uniqueness under additional positivity conditions. We do not attribute the particular PDE counterexample or a priority claim to that paper.

The complete literal existence assertion is contradicted by Theorem Theorem. A possible repaired question would add uniform ellipticity, or an assumption that ensures continuous dependence of the scalar terminal problem on its terminal value. Neither repair is proved here. No numerical experiment or unproved stochastic representation is used in the counterexample.

References

Completion Estimate

100%

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