Nonuniqueness for a BSDE with generator |z| sin |z|

\begin{abstract} We construct two solutions with the same terminal value for the scalar Brownian BSDE with generator $g(z)=|z|\sin|z|$. Both solutions have integrable supremum powers of every finite order. This gives a negative answer to Fan, Hu and Tang's Problem 5.3. \end{abstract}

Proof: We use Itô's formula, stopped martingale identities and the Burkholder--Davis--Gundy inequality in their usual continuous-martingale forms; see [RY, Chapters IV and IX].

1. A positive process with an integrable inverse-square clock.

Let \begin{equation}\label{wsref-0-62-65-73-73-65-6c} dR_t=dB_t+R_t^{-1}\,dt,\qquad R_0=1, \qquad A_t=\int_0^tR_s^{-2}\,ds. \end{equation} The stopped Picard construction gives a strong solution up to exit from $(0,\infty)$, since the coefficients are locally Lipschitz there. For $n\ge2$, let $\sigma_n$ be its first exit from $(1/n,n)$ and put $\tau_n=T\wedge\sigma_n$. Itô's formula gives \[ d(R_t^{-1})=-R_t^{-2}\,dB_t, \qquad d(R_t^2)=2R_t\,dB_t+3\,dt. \] The stopped integrands are bounded, so $\mathbb E R_{\tau_n}^{-1}=1$ and $\mathbb E R_{\tau_n}^{2}=1+3\mathbb E\tau_n\le1+3T$. Consequently \[ \mathbb P(\sigma_n\le T)\le \frac1n+\frac{1+3T}{n^2}. \] Thus $R$ exists throughout $[0,T]$ and stays strictly positive.

Another application of Itô's formula shows that \[ L_t=R_t^{-1/2}\exp(A_t/8),\qquad dL_t=-\frac{L_t}{2R_t}\,dB_t. \] In particular $\mathbb E L_{\tau_n}=1$. Cauchy--Schwarz and the preceding second-moment bound yield \[ \mathbb E e^{A_{\tau_n}/16} =\mathbb E\left[L_{\tau_n}^{1/2}R_{\tau_n}^{1/4}\right] \le(\mathbb E R_{\tau_n}^{1/2})^{1/2} \le(1+3T)^{1/8}. \] Fatou's lemma therefore gives \begin{equation}\label{wsref-0-63-6c-6f-63-6b} \mathbb E e^{A_T/16}\le(1+3T)^{1/8}<\infty. \end{equation} In particular $A_T$ has moments of every finite order.

2. A nonzero gap that vanishes at the terminal time.

Let $\varphi$ and $\Phi$ denote the standard normal density and distribution function. For $0\le h\le T$, $r>0$, define \begin{equation}\label{wsref-0-75} u(h,r)=\frac1{2\sqrt T}\int_0^h\Phi(-r/\sqrt s)\,ds, \qquad u(0,r)=0. \end{equation} Differentiating for $h,r>0$ gives \[ u_h=\frac1{2\sqrt T}\Phi(-r/\sqrt h),\qquad u_{rr}=\frac1{\sqrt T}\Phi(-r/\sqrt h),\qquad u_r=-\frac1{2\sqrt T}\int_0^h \frac{\varphi(r/\sqrt s)}{\sqrt s}\,ds. \] For the second identity, integrate $\partial_s\Phi(-r/\sqrt s) =r\varphi(r/\sqrt s)/(2s^{3/2})$. These differentiations are justified by exponential decay as $s\downarrow0$. Thus $u_h=\tfrac12u_{rr}$, $-1\le u_r\le0$, and $0\le u\le\sqrt T/4$; the derivative bound follows from $\varphi\le1$ and $h\le T$.

Set \[ D_t=u(T-t,R_t),\qquad q_t=u_r(T-t,R_t), \qquad q_T=0. \] The processes are continuous, $D_T=0$, and $D_t>0$ for $t<T$. Using (bessel) and the heat equation gives \begin{equation}\label{wsref-0-67-61-70} dD_t=\frac{q_t}{R_t}\,dt+q_t\,dB_t. \end{equation} The identity extends to $T$ by continuity: $q$ is bounded and every path of $R$ has a positive minimum on $[0,T]$.

3. Realizing the required secant of the generator.

For $q\in[-1,0]$, write \[ s(q)=\begin{cases}\sin(q/2)/(q/2),&q\ne0,\\1,&q=0,\end{cases} \qquad s_0=2\sin(1/2). \] Then $0<s_0\le s(q)\le1$ and $\cos(q/2)>0$. For $r>0$ put $k(r)=\lceil(2\pi s_0r)^{-1}\rceil$. On the interval \[ I_r=[2\pi k(r)+\pi/2,\,(2k(r)+1)\pi] \] the function \[ F(a,q)=\sin a\cos(q/2)+a\cos a\,s(q) \] is strictly decreasing, because \[ F_a=(\cos(q/2)+s(q))\cos a-a\,s(q)\sin a<0. \] Its left endpoint value is $\cos(q/2)>0$, and its right endpoint value is at most $-1/r$. There is therefore a unique $a=a(r,q)\in I_r$ with $F(a,q)=-1/r$. This choice is Borel: on each Borel set where $k(r)$ is constant, the strictly monotone inverse depends continuously on $(r,q)$. It satisfies \begin{equation}\label{wsref-0-61-62-6f-75-6e-64} 0<a(r,q)\le \frac1{s_0r}+3\pi. \end{equation} Both $a+q/2$ and $a-q/2$ are positive. The addition formula for sine now gives the exact identity \begin{equation}\label{wsref-0-73-65-63-61-6e-74} g(a+q/2)-g(a-q/2)=qF(a,q)=-q/r. \end{equation} The formula also holds at $q=0$.

Define predictable processes \[ Z_t^\pm=a(R_t,q_t)\pm q_t/2, \qquad Y_t^-=-\int_0^t g(Z_s^-)\,ds+\int_0^t Z_s^-\,dB_s, \qquad Y_t^+=Y_t^-+D_t. \] By (abound), $|Z_t^\pm|\le C(1+R_t^{-1})$ for a fixed constant $C$. Hence (clock) implies \[ \mathbb E\left(\int_0^T|Z_t^\pm|^2\,dt\right)^{p/2}<\infty \qquad(1\le p<\infty). \] Since $|g(z)|\le|z|$, Cauchy--Schwarz controls the drift integral, and the Burkholder--Davis--Gundy inequality controls the stochastic integral. Thus $\mathbb E\sup_{t\le T}|Y_t^-|^p<\infty$ for every finite $p\ge1$. The same holds for $Y^+$ because $D$ is bounded.

Finally, (gap) and (secant) give \[ dY_t^+=-g(Z_t^-)\,dt+Z_t^-\,dB_t +\frac{q_t}{R_t}\,dt+q_t\,dB_t =-g(Z_t^+)\,dt+Z_t^+\,dB_t. \] Taking $\xi=Y_T^-=Y_T^+$ proves (bsde) for both pairs. Their initial values differ by $u(T,1)>0$. For each $p\ge1$, the integrable variable $\sup_{t\le T}|Y_t^\pm|^p$ dominates all stopped powers, establishing the required class-(D) condition. ∎