An optimal consumption problem in finite time with a constraint on the ruin probability

Abstract

In this paper, we investigate the following problem: For a given upper bound for the ruin probability, maximize the expected discounted consumption of an investor in finite time. The endowment of the agent is modeled by a Brownian motion with positive drift. We give an iterative algorithm for the solution of the problem, where in each step an unconstrained, but penalized problem is solved. For the discontinuous value function $V(t,x)$ of the penalized problem, we show that it is the unique viscosity solution of the corresponding Hamilton–Jacobi–Bellman equation. Moreover, we characterize the optimal strategy as a barrier strategy with continuous barrier function.

Appendix

1.1 9.1 Auxiliary results for Sect. 3

Proof of Lemma 3.1

In the following proof, we use the notation $X^{C}$ for the underlying process if the starting point is clear from the context.

(a) Let us assume first $0 \leq t < T,0<x$. With $u(t):= \max(t,\hat{t})$, define the stopping time

$$ \kappa:= \kappa_{1} \wedge\kappa_{2} \wedge T:= \inf \big\{ s \geq u(t) \big| X^{\hat{t},\hat{x},0}_{s} \geq X^{t,x,C}_{s+} \big\} \wedge\inf\big\{ s \geq\hat{t} \big| X^{\hat{t},\hat{x},0}_{s} =0 \big\} \wedge T. $$

We provide now the definition of the new strategy $\hat{C}$ as

$$\left\{ \textstyle\begin{array}{ll} \hat{C}_{s}:= \hat{C}_{\hat{t}},&\quad s \in[\hat{t},\kappa], \\ \Delta\hat{C}_{\kappa}:= -(X^{\hat{t},\hat{x},0}_{\kappa}- X^{t,x,C}_{\kappa+}),& \quad\kappa_{2}\wedge T > \kappa, \\ \hat{C}_{s}:= C_{s},&\quad s > \kappa, \kappa_{2}\wedge T > \kappa, \\ \hat{C}_{s}:= \hat{C}_{\hat{t}},&\quad s > \kappa, \kappa_{2}\wedge T \leq \kappa. \end{array}\displaystyle \right. $$

Hence, after time $\hat{t}$ we run our endowment without paying out anything, until the first moment our trajectory is above or equal to the level of the old trajectory. Then we jump down to this old trajectory and use the given strategy $C$.

Let $B^{(\hat{t}, \hat{x})}:= \{0 \leq-X^{\hat{C}}_{u(\hat{t})+} + X^{C}_{u(\hat{t})+} \leq|\Delta t|^{\frac{1}{3}}+\vert\Delta x \vert\} \cap\left\{ \tau =T\right\}$. By construction,

$$ \mathbf{P}\big[B^{(\hat{t}, \hat{x})} \big] \longrightarrow\mathbf{P}[\tau=T] \qquad \text{ as } (\Delta t,\Delta x) \to0. $$

(9.1)

We distinguish 3 cases.

Case 1: $\omega\in B_{1}:= \{ \omega\in B^{(\hat{t}, \hat{x})} | C_{T}-C_{u+}>0\}$. Let now $|\Delta t|^{\frac{1}{3}}+\vert\Delta x \vert\leq \Delta(\omega)$. Because $C$ is left-continuous, there exists a $\delta(\omega)>0$ with $C_{T-s}-C_{u+} \geq \frac{C_{T}-C_{u+}}{2}$ for $0 \leq s \leq\delta$. Hence, we arrive at

$$ \kappa(\omega) \leq T-\delta(\omega),\quad\Delta\leq \frac{C_{T}-C_{u+}}{2}. $$

(9.2)

Furthermore, since $\tau(\omega)=T$ for $\omega \in B$, we know that for all $\epsilon>0$,

$$ \inf_{u< s \leq T-\epsilon}\big(X^{C}_{u+}+\mu(s-u)+ \sigma(W_{s}-W_{u})-(C_{s}-C_{u+})\big)>0. $$

Choosing now $\epsilon$ as $\delta(\omega)/2$ gives

$$ \inf_{u< s \leq T-\delta/2}\big(X^{C}_{u+}+\mu(s-u)+ \sigma (W_{s}-W_{u})-(C_{s}-C_{u+})\big)=: \nu(\omega)>0, $$

which implies for $\Delta<\nu$ that

$$ \inf_{u< s \leq T-\delta/2}\big(X^{\hat{C}}_{u+}+\mu(s-u)+ \sigma (W_{s}-W_{u})-(C_{s}-C_{u+})\big)>0. $$

Hence,

$$ \inf_{u< s \leq T-\delta/2}\big(X^{\hat{C}}_{u+}+\mu(s-u)+ \sigma (W_{s}-W_{u})\big)>0. $$

Therefore (9.2) gives $X^{\hat{C}}_{\kappa+}>0$ for $\Delta$ small enough. Finally, we arrive at

$$ \hat{\tau}=\tau=T $$

for $(\Delta t, \Delta x)$ small enough on the set $B_{1}$.

Case 2: $\omega\in B_{2}:= \{ \omega\in B | C_{T}-C_{u+}=0, X^{C}_{u+}+\mu(T-u)+ \sigma(W_{T}-W_{u})>0\}$. In this case, we choose $\hat{C}_{s}=\hat{C}_{u+}$ for $s>u$ and get $X^{\hat{C}}_{T}\geq X^{C}_{T}-\Delta$, so that we have $\hat{\tau}=T$ for all $\omega$ in $B_{2}$, for $\Delta$ small enough.

Case 3: $\omega\in B_{3}:= \{ \omega\in B | C_{T}-C_{u+}=0, X^{C}_{u+}+\mu(T-u)+ \sigma(W_{T}-W_{u})=0\}$. We first note that $\mathbf{P}[B_{3}]=0$, and hence (9.1) gives

$$ \mathbf{P}[B_{1} \cup B_{2}] \longrightarrow\mathbf{P}[\tau=T] \qquad \text{ as } (\Delta t,\Delta x) \to0. $$

So by Cases 1 and 2, we conclude that

$$ \liminf_{(\Delta t,\Delta x) \to0}\mathbf{P}[\hat {\tau }=T] \geq\mathbf{P}[\tau=T]. $$

Finally, since $\mathbf{P}[\hat{\tau}=T] \leq \mathbf{P}[\tau=T]$ by construction, we get

$$ \lim_{(\Delta t,\Delta x) \to0}\mathbf{P}[\hat{\tau }=T] = \mathbf{P}[\tau=T]. $$

The cases $(t=T,x>0)$ and $(0\leq t < T,x=0)$ are trivial, hence part a is proved.

Let us finally remark that for $(t=T,x=0)$, part a is wrong since $\mathbf{P}[\tau=T]=1$. But if the point $(\hat{t},\hat{x})$ approaches $(T,0)$ in a flat enough way, say linearly, e.g., as $\hat{x}=T-\hat{t}$, then $\mathbf{P}[\hat{\tau}< T]$ is clearly not $o(1)$ for $\hat{t} \to T$.

(b) We first consider the case $0< t< T,0< x$. Take $\hat{C},u$ from part a and let $G:= \{X^{C}_{u+}-X^{\hat{C}}_{u+} \leq|\Delta t|^{\frac{1}{3}}+\vert\Delta x \vert\}$. Hence

$$ \mathbf{P}[G^{c}] \to0 $$

(9.3)

for $(\Delta t,\Delta x) \to0$, uniformly in $(t,x)$. We split the probability space $\varOmega$ and arrive at

$$ \Vert(\hat{\tau}-\tau)\Vert_{L^{2}}\leq\Vert (\hat{\tau }-\tau)\mathbf{1}_{G}\Vert_{L^{2}}+ \Vert(\hat{\tau}-\tau)\mathbf{1}_{G^{c}}\Vert_{L^{2}}. $$

The second term tends to zero by (9.3), and for the first term, it is sufficient that $\Vert\nu_{1}-\nu_{2}\Vert_{L^{2}} \to0$ for $\Delta x \searrow0$, uniformly in $x$, where we use

$$\begin{aligned} \nu_{1} :=& \inf\{t>0 |x+\mu t + \sigma W_{t} =0 \} \wedge T, \\ \nu_{2} :=& \inf\{t>0 |x-\Delta x+\mu t + \sigma W_{t} =0 \} \wedge T. \end{aligned}$$

But the above $L^{2}$-convergence is implied by Lemma 9.1, so that we get part b for $0< t< T,0< x$. The case $t=T$ is trivial, and for $t< T$, $x=0$, one can again use Lemma 9.1. □

Lemma 9.1

Let $X_{t}:= \Delta x+\mu t + \sigma W_{t}$, with small positive $\Delta x$, and let further $\tau := \inf\{t>0 | X_{t}=0\}\wedge T$. Then we have $\Vert\tau\Vert_{L^{2}} \leq\mathrm{const.}\Delta x$, where the constant depends only on the model parameters.

Proof

By (3.5.12) in [19], $\tau$ has the defective density

$$ f(t) = \frac{\Delta x}{\sqrt{2\pi t^{3}}}e^{-\frac {(\Delta x+\mu t)^{2}}{2t}}\mathbf{1}_{[0,T]}(t) $$

and a point mass $\mathbf{P}[\tau=T] = \int_{T}^{\infty}f(s)\, ds$. Hence, we get

$$ \mathbf{E}[\tau^{2}]= \int_{0}^{T} \frac{\Delta x}{\sqrt {2\pi }}t^{\frac{1}{2}}e^{-\frac{(\Delta x+\mu t)^{2}}{2t}}\,dt + T^{2} \int_{T}^{\infty}\frac{\Delta x}{\sqrt{2\pi t^{3}}}e^{-\frac{(\Delta x+\mu t)^{2}}{2t}}\,dt. $$

In the following, $D$ denotes a generic constant which may vary from place to place, and which depends only on the model parameters. So we get

$$\begin{aligned} \mathbf{E}[\tau^{2}] \leq&D \int_{0}^{T} \Delta x t^{\frac {1}{2}}e^{-\frac{(\Delta x+\mu t)^{2}}{2t}}\,dt + D \int_{T}^{\infty}\frac{\Delta x}{\sqrt{t^{3}}}e^{-\frac{(\Delta x+\mu t)^{2}}{2t}}\,dt \\ \leq& D \Delta x \bigg( (\Delta x)^{3} \int_{\frac{(\Delta x)^{2}}{2T}}^{\infty}w^{-\frac{5}{2}} e^{-w} \,dw + \frac{1}{\Delta x} \int^{\frac{(\Delta x)^{2}}{2T}}_{0} w^{-\frac{1}{2}} e^{-w} \,dw \bigg) \\ \leq& D \Delta x, \end{aligned}$$

which concludes the proof. □

1.2 9.2 Auxiliary results for Sect. 4

Proof of Lemma 4.2

We define our partition as

$$\begin{aligned} Q^{i,j}:= \bigg(\frac{(i-1)T}{M},\frac{iT}{M}\bigg)\times\bigg(\frac {j}{M},\frac{j+1}{M} \bigg) &\cup\bigg\{ \frac{iT}{M}\bigg\} \times\bigg[\frac{j}{M},\frac {j+1}{M}\bigg)\\ &\cup\bigg(\frac{(i-1)T}{M},\frac{iT}{M}\bigg]\times\bigg\{ \frac{j}{M} \bigg\} \end{aligned}$$

for $i=2,3,\dots,M$, $j \in\mathbf{N}_{0}$, and $Q^{1,j}:= [0,\frac{T}{M}] \times[\frac{j}{M}, \frac{j+1}{M})$ for $j \in\mathbf{N}_{0}$. Let $C^{(i,j)}$ be an admissible strategy for $X^{t_{i},x_{j}}$, $(t_{i},x_{j}):= (\frac{T(i-1)}{M},\frac{j}{M})$, $(i,j) \in {\mathcal{I}}$. By our definition of the half-open rectangles $Q^{i,j}$, this strategy is also admissible for all $(t,x) \in Q^{i,j}$, for all $(i,j)\in{\mathcal{I}}$. Similarly as in the proof of Proposition 3.2 (cf. (3.4) there), one finds that for all $\epsilon>0$, there exists an $M>0$ large enough such that

$$ | J_{n}(t,x,C^{(i,j)})- e^{-\beta t}x -J_{n}(t_{i},x_{j},C^{(i,j)})- e^{-\beta t_{i}}x_{j} | < \epsilon $$

(9.4)

for all $(t,x) \in Q^{i,j}$ and all $(i,j)\in{\mathcal{I}}$. Since $V_{n}(t,x)-e^{-\beta t}x$ is by Proposition 3.2 uniformly continuous, we have also

$$ | V_{n}(t,x)- e^{-\beta t}x -V_{n}(t_{i},x_{j})- e^{-\beta t_{i}}x_{j} | < \epsilon $$

(9.5)

for all $(t,x) \in Q^{i,j}$ and all $(i,j)\in{\mathcal{I}}$. We now choose $C^{(i,j)}$ such that one has

$$ J_{n}(t_{i},x_{j},C^{(i,j)})- e^{-\beta t_{i}}x_{j} \geq V_{n}(t_{i},x_{j})- e^{-\beta t_{i}}x_{j} -\epsilon. $$

(9.6)

By (9.4), (9.6) and (9.5), one concludes that

$$J_{n}(t,x,C^{(i,j)})- e^{-\beta t}x \geq V_{n}(t,x)- e^{-\beta t}x -3\epsilon $$

for all $(t,x) \in Q^{i,j}$ and all $(i,j)\in{\mathcal{I}}$. □

Proof of Lemma 4.5

Instead of the estimate (9.4) in the proof of Lemma 4.2, we claim that we have now

$$ J(t,x,C^{(i,j)})-e^{-\beta t}x -J(t_{i},x_{j},C^{(i,j)})+ e^{-\beta t_{i}}x_{j} > -\epsilon $$

(9.7)

for all $(t,x) \in Q^{i,j}$, all $(i,j)\in{\mathcal{I}}$ and $M$ large enough. Indeed, to get (9.4), we have used (3.4), and the point where the function $f_{n}$ entered in this estimate was via an estimate of $|\mathbf{E}[ f_{n}(\tau)-f_{n}(\hat{\tau})]|$. This is now replaced by $\mathbf{E}[f(\hat{\tau})-f(\tau)] \geq0$, by the definition of our partition and the points $(t_{i},x_{j})$. Moreover, we can replace, using the uniform upper semicontinuity of $W$ from Proposition 3.3(a), the estimate (9.5) by

$$ V_{n}(t_{i},x_{j})- e^{-\beta t_{i}}x_{j} - V_{n}(t,x)+ e^{-\beta t}x > -\epsilon $$

(9.8)

for all $(t,x) \in Q^{i,j}$, all $(i,j)\in{\mathcal{I}}$ and $M$ large enough. Finally, we can take (9.6) without alterations, and this gives, together with (9.7) and (9.8), our assertion. □

1.3 9.3 Auxiliary results for Sect. 5

Proof of Proposition 5.2

Since we know by Proposition 3.3 that the value function $V$ is upper semicontinuous, we have $V^{*}=V$. Now let $(\bar{t},\bar{x}) \in Q$ be a maximizer of $V-\phi$ with

$$V(\bar{t},\bar{x})= \phi( \bar{t},\bar{x}), $$

where $\phi\in C^{1,2}(Q)$. We have to show that

$$ \min\{-\mathcal{L}\phi(\bar{t},\bar {x}),-e^{-\beta \bar{t}}+\phi_{x}(\bar{t},\bar{x}) \} \leq0. $$

Let $(t_{m},x_{m})\to(\bar{t},\bar{x})$ for $m\to\infty$, with $V(t_{m},x_{m}) \to V(\bar{t},\bar{x})$. Moreover, let

$$ \gamma_{m}:= V(t_{m},x_{m})-\phi(t_{m},x_{m}) \to0 $$

for $m \to\infty$. We argue by contradiction. So assume that we have

$$\begin{aligned} -\mathcal{L}\phi(\bar{t},\bar{x})>0, \\ -e^{-\beta\bar{t}}+\phi_{x}(\bar{t},\bar{x}) >0. \end{aligned}$$

By continuity, this implies the existence of $\epsilon_{0},\eta>0$ with

$$\begin{aligned} -\mathcal{L}\phi(t,x) >&\epsilon_{0}>0, \\ -e^{-\beta t}+\phi_{x}(t,x) >& \epsilon_{0}>0, \end{aligned}$$

(9.9)

on a sphere $K((\bar{t},\bar{x});\eta) \subset Q$ with radius $\eta$.

Let $\theta_{m}$ be the exit time of $X^{(t_{m},x_{m})}$ from $K_{m}:= K((t_{m},x_{m});\eta/2)$. We have $K_{m} \subset K((\bar{t},\bar{x});\eta)$, where the inclusion holds for $m$ large enough. For $h>0$, $\theta_{m}+h$ is a stopping time, and we apply our DPP in Proposition 4.4(b) for this stopping time. Hence for all $\delta>0$, there exists an admissible strategy $C^{m}$ such that

$$ V(t_{m},x_{m}) - \delta\leq\mathbf{E}\bigg[\int_{t_{m}}^{\tau\wedge (\theta _{m}+h)} e^{-\beta s} dC^{m}_{s} +V\big(\tau\wedge(\theta_{m}+h), X^{(t_{m},x_{m})}_{\tau\wedge(\theta _{m}+h)}\big)\bigg]. $$

(9.10)

Let now $\lambda_{m}:= [X^{(t_{m},x_{m})}_{(\theta _{m})+},X^{(t_{m},x_{m})}_{(\theta_{m})}] \cap\partial K_{m}$, where $\partial K_{m}$ denotes the boundary of $K_{m}$. Using this definition, we have

$$\begin{aligned} &V\big(\tau\wedge(\theta_{m}+h), X^{(t_{m},x_{m})}_{\tau\wedge(\theta _{m}+h)}\big)-\phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}})\\ &=V\big(\tau\wedge(\theta_{m}+h), X^{(t_{m},x_{m})}_{\tau\wedge(\theta _{m}+h)}\big)-V(\theta_{m}, \lambda_{m})+ V(\theta_{m}, \lambda_{m})- \phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}}). \end{aligned}$$

Hence,

$$\begin{aligned} &\limsup_{h\to0} V\big(\tau\wedge(\theta_{m}+h), X^{(t_{m},x_{m})}_{\tau \wedge(\theta_{m}+h)}\big)- \phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}}) \\ &\leq V\big(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}+}\big)-V(\theta_{m}, \lambda_{m}) +V(\theta_{m}, \lambda_{m})- \phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}}) \\ &\leq -e^{-\beta\theta_{m}}(\lambda_{m}- X^{(t_{m},x_{m})}_{\theta_{m}+})+ \phi(\theta_{m}, \lambda_{m})- \phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}}) \\ &\leq -e^{-\beta\theta_{m}}(X^{(t_{m},x_{m})}_{\theta_{m}}- X^{(t_{m},x_{m})}_{\theta_{m}+}) -\epsilon_{0}(X^{(t_{m},x_{m})}_{\theta_{m}}-\lambda_{m}). \end{aligned}$$

(9.11)

For the first inequality, we have used the upper semicontinuity of $V$ from Proposition 3.3. The argument for the second one is exactly as in [12, Lemma VIII.3.1] (moreover we use $\phi\geq V$ to estimate the third term). Finally, for the third inequality we use

$$\begin{aligned} \phi(\theta_{m}, \lambda_{m})- \phi(\theta_{m}, X^{(t_{m},x_{m})}_{\theta_{m}}) \leq& \phi_{x}(\theta_{m},\hat{X}_{\theta_{m}})(\lambda _{m}-X^{(t_{m},x_{m})}_{\theta_{m}}) \\ \leq& -(e^{-\beta\theta_{m}}-\epsilon_{0})(-\lambda_{m}+X^{(t_{m},x_{m})}_{\theta_{m}}) \end{aligned}$$

with $\hat{X}_{\theta_{m}} \in[\lambda_{m},X_{\theta_{m}}]$, employing (9.9). For notational convenience, we now omit the upper index $(t_{m},x_{m})$ of the state process. Itô’s formula implies

$$\begin{aligned} \phi(\theta_{m}, X_{\theta_{m}}) =&\phi(t_{m}, x_{m})+ \int_{t_{m}}^{\theta_{m}} {\mathcal{L}}\phi(s,X_{s}) \,ds - \int_{t_{m}}^{\theta_{m}} \phi_{x}(s,X_{s})\, dC^{m,c}_{s} \\ &{}+ \sum_{s \in J_{X},t_{m} \leq s < \theta_{m}}\big(\phi(s,X_{s+})-\phi (s,X_{s})\big)+\mathcal{S} \\ \leq& \phi(t_{m}, x_{m})-\epsilon_{0}(\theta_{m}-t_{m}) - \int_{t_{m}}^{\theta_{m}} e^{-\beta s}\, dC^{m,c}_{s} \\ &{}- \epsilon_{0} \int_{t_{m}}^{\theta_{m}} \, dC^{m,c}_{s} +\sum_{s \in J_{X},t_{m} \leq s < \theta_{m}}\phi_{x}(s,\tilde {X}_{s})\Delta X_{s}+\mathcal{S} \\ \leq& \phi(t_{m}, x_{m})-\epsilon_{0}(\theta_{m}-t_{m}) - \int_{t_{m}}^{\theta_{m}} e^{-\beta s}\, dC^{m,c}_{s} \\ &{}- \epsilon_{0} \int_{t_{m}}^{\theta_{m}} \, dC^{m,c}_{s} +\sum_{s \in J_{X},t_{m} \leq s < \theta_{m}}(e^{-\beta s}+\epsilon_{0})\Delta X_{s}+\mathcal{S} \\ =& \phi(t_{m}, x_{m})-\epsilon_{0}(\theta_{m}-t_{m}) - \int_{t_{m}}^{\theta_{m}} e^{-\beta s}\, dC^{m}_{s} \\ &{}- \epsilon_{0} \int_{t_{m}}^{\theta_{m}} \, dC^{m}_{s}+\mathcal{S}. \end{aligned}$$

(9.12)

Here we have used (9.9), the notation $C^{m,c}$ for the continuous part of $C^{m}$, $J_{X}$ for the jump times of $X$, and $\tilde{X}_{s} \in[X_{s+},X_{s}]$. Moreover, $\mathcal{S}$ denotes a stochastic integral with respect to Brownian motion with an integrand integrable enough to guarantee that this term vanishes in the expectation. Hence, we do not specify this term explicitly. Combining (9.11) and (9.12) gives

$$\begin{aligned} &\limsup_{h\to0} V\big(\tau\wedge(\theta_{m}+h), X_{\tau\wedge (\theta_{m}+h)}\big)- \phi(t_{m},x_{m}) \\ &\leq-\epsilon_{0}(X_{\theta_{m}}-\lambda_{m}) -\epsilon_{0}(\theta_{m}-t_{m}) - \int_{[t_{m},\theta_{m}]} e^{-\beta s} dC^{m}_{s}- \epsilon_{0} \int_{t_{m}}^{\theta_{m}} dC^{m}_{s}+\mathcal{S}, \end{aligned}$$

(9.13)

where the first integral includes a possible jump at $\theta_{m}$, too. Applying $\limsup_{h\to0}$ in (9.10) gives

$$- \delta\leq\mathbf{E}\left[\int_{[t_{m},\theta_{m}]} e^{-\beta s}\, dC^{m}_{s} +\limsup_{h\to0}V\big(\tau\wedge(\theta_{m}+h), X_{\tau\wedge (\theta _{m}+h)}\big)\right]-V(t_{m},x_{m}). $$

Using (9.13) and taking the expected value, we end up with

$$\begin{aligned} -\delta\leq -\epsilon_{0}\mathbf{E}[X_{\theta_{m}}-\lambda_{m}] -\epsilon_{0}\mathbf{E}\left[\theta_{m}-t_{m}\right] -\epsilon_{0} \mathbf{E} \left[\int_{t_{m}}^{\theta_{m}} \, dC^{m}_{s}\right]-\gamma_{m}. \end{aligned}$$

(9.14)

Let $\hat{\theta}_{m}:= \inf\{s> t_{m} | X_{s} \notin\hat{K}_{m}:= K((t_{m},x_{m});\eta/4) \subset K_{m} \}$ and define the set $A_{m}:= \{ \theta_{m} =\hat{\theta}_{m}\}$, which means that if we leave $\hat{K}_{m}$ on $A_{m}$, then we leave also $K_{m}$. Obviously, we have $|X_{\theta_{m}}-\lambda_{m} | \geq\eta/4$ on the set $A_{m}$. We distinguish now several cases.

Case 1: $\mathbf{P}[A_{m}] \geq1/2$. In this case, we can give for the right-hand side of (9.14) the upper estimate $-\frac{\epsilon_{0} \eta}{8}-\gamma_{m}$.

Case 2: $\mathbf{P}[A_{m}^{c}] \geq1/2$. Set now $B_{m}:= \{ \int_{t_{m}}^{\theta_{m}} dC^{m}_{s} \geq\eta/10\}$ and distinguish two subcases:

Case 2.1: $\mathbf{P}[A_{m}^{c} \cap B_{m}] \geq1/4$. This is the case where we leave only the smaller sphere and not the larger one, and where the integral term of the right-hand side of (9.14) gives a significant contribution. We find $-\frac{\epsilon_{0} \eta}{40}-\gamma_{m}$ as an upper estimate for the right-hand side of (9.14).

Case 2.2: $\mathbf{P}[A_{m}^{c} \cap B_{m}^{c}] \geq1/4$. We define

$$\tilde{\theta}_{m}:= \inf\{s>t_{m} | x_{m}+\mu(s-t_{m})+\sigma(W_{s}-W_{t_{m}}) \notin\hat{K}_{m}\} $$

and observe that we have $\tilde{\theta}_{m} \leq\theta_{m}$ on $B^{c}_{m}$. Moreover, one has $\mathbf{E} [\tilde{\theta}_{m}-t_{m}] \geq D$ for some generic constant $D$ depending on $\eta,\mu$ and $\sigma$. Hence $\mathbf{E}[\theta_{m}-t_{m}] \geq D$, and for the upper bound for the right-hand side of (9.14), we find $-\epsilon_{0} D-\gamma_{m}$.

Therefore, we have in all cases

$$ \mbox{right-hand side of (9.14)} \leq -\epsilon_{0} D-\gamma_{m} $$

for all $m$. Hence, by (9.14), $-\delta\leq-\epsilon_{0} D-\gamma_{m}$ for all $m$, which is a contradiction for $m$ large and $\delta$ small enough. □

Proof of Proposition 5.4

We first transform the HJB equation to an equivalent one, where we use $V=e^{\lambda_{1} t+\lambda_{2} x}\tilde{V}$ for some real $\lambda_{i}$. This yields (we replace now for convenience $\tilde{V}$ again by $V$)

$$ \min\{-\tilde{\mathcal{L}}V,-e^{-\nu_{2} t- \nu_{3} x}+V_{x} +\nu_{3} V \}=0, $$

with $\tilde{\mathcal{L}}V:= \nu_{1} V+V_{t} + \tilde{\mu}V_{x} +\frac {\sigma ^{2}}{2}V_{xx}$, $\nu_{1}:= \lambda_{1} +\lambda_{2}\mu+\frac{\sigma^{2}}{2}\lambda_{2}^{2}$, $\tilde{\mu}:= \mu+ \sigma^{2}\lambda_{2}$, $\nu_{2}:= \beta+\lambda_{1}$, $\nu_{3}:= \lambda_{2}$. In the sequel, we choose $\lambda_{2}>0$ arbitrarily large and $\lambda_{1} := -\lambda_{2}^{3}$. All together, we have the choice of parameters

$$\begin{aligned} \nu_{1} < &0 \mbox{ and of the magnitude }-\lambda_{2}^{3}, \\ \nu_{2} < &0 \mbox{ and of the magnitude }-\lambda_{2}^{3}, \\ \nu_{3} >&0 \mbox{ and of the magnitude }\lambda_{2}, \\ \tilde{\mu} >&0 \mbox{ and of the magnitude }\lambda_{2}. \end{aligned}$$

(9.15)

Our proof goes by contradiction. So assume that

$$ K:= \sup_{(t,x)\in[0,T]\times\mathbf{R}_{+}}(u-v) (t,x)= \sup_{(t,x)\in[0,T)\times\mathcal{O}}(u-v)(t,x)>0 $$

for some open bounded set $\mathcal{O}$ in $\mathbf{R}_{+}$. Here we have used the method of penalized supersolution ($u,v$ increase only linearly) for the last equality above (see [24, Th. 4.4.3, Step 2 and Th. 4.4.4, Step 1]).

Let now $w:= 10K e^{-\nu_{3} x -\nu_{3} (T-t)}$ with $\nu_{3}>0$ from above, and let $U$ be a small open rectangle at $(T,0)$, i.e., $U:= \{ (t,x) | T- \frac{1}{\nu_{3}}< t <T, 0<x < \frac{1}{\nu_{3}} \}$. We want to show that

$$ u-v-w \leq0 $$

(9.16)

holds on $\bar{Q} \setminus U$. To do this, we first show that $v+w$ is a supersolution, too. Indeed, we get $-\tilde{\mathcal{L}}w =(-\nu_{1}-\nu_{3} +\tilde{\mu}\nu_{3} -\frac {\sigma ^{2}}{2}\nu_{3}^{2})w >0$. Moreover, one has $w_{x} +\nu_{3}w=0$, and both together imply

$$ \min\{-\tilde{\mathcal{L}}(v+w),-e^{-\nu_{2} t- \nu_{3} x}+(v+w)_{x} +\nu_{3} (v+w) \}\geq0, $$

hence the asserted supersolution property. Indeed, on the sets $\{(t,0) | 0 \leq t \leq T-\frac{1}{\nu_{3}} \}$ (resp., $\{ (T,x) | x \geq\frac{1}{\nu_{3}} \}$) we have $u-v-w =-w <0$. Moreover, on the rest of the boundary one gets $K-\frac{10}{e^{2}}K$ as an upper bound for $u-v-w$.

We now show (9.16) by contradiction. So assume

$$\begin{aligned} 0< M:= \sup_{\bar{Q} \setminus U} (u-v-w)= \max_{(\bar{Q} \setminus U)\cap L} (u-v-w)= \max_{((\bar{Q} \setminus U)\cap L)^{o}} (u-v-w), \end{aligned}$$

(9.17)

where $A^{o}$ denotes the interior of $A$ and $L$ is the compact set $L:= [0,T] \times[0,z]$ for some $z>0$. (Here we have again used the method of penalized supersolution.) We now apply the well-known dedoubling variable technique and define

$$\varPhi_{\epsilon}(t,s,x,y):= u(t,x)-v(s,y)-w(s,y)-\phi_{\epsilon}(t,s,x,y), $$

where $\phi_{\epsilon}(t,s,x,y):= \frac{(t-s)^{2}+(x-y)^{2}}{2\epsilon}$. As $\varPhi_{\epsilon}$ is continuous on $((\bar{Q} \setminus U)\cap L)^{2}$, let us denote its maximum by $M_{\epsilon}$ and its maximizer by $(t_{\epsilon},s_{\epsilon},x_{\epsilon},y_{\epsilon})$. By going to subsequences, we can assume that $(t_{\epsilon},s_{\epsilon},x_{\epsilon},y_{\epsilon})\to(\bar{t},\bar {s},\bar {x},\bar{y})$ holds for $\epsilon\to0$. As in [24, Theorem 4.4.4], we infer that

$$ \bar{t}=\bar{s}, \bar{x}=\bar{y}, \phi_{\epsilon}(t_{\epsilon},s_{\epsilon},x_{\epsilon},y_{\epsilon}) \to0, M_{\epsilon}\to M,(u-v-w)(\bar{t},\bar{x})=M, $$

with $(\bar{t},\bar{x}) \in((\bar{Q} \setminus U)\cap L)^{o}$, because of (9.17). Note that the form of the HJB equation does not enter into this argument.

As in [24], we introduce now the concept of a super(sub)jet $\mathcal{P}^{2,+}u(t_{0},x_{0})$ (resp., $\mathcal{P}^{2,-}v(t_{0},x_{0})$) of an upper (lower) semicontinuous function $u$ (resp., $v$) at a point $(t_{0},x_{0}) \in[0,T) \times\mathbf{R}$ as the set of elements $(q_{0},p_{0},K_{0})$ in $\mathbf{R}^{3}$ satisfying

$$u(t,x) \leq u(t_{0},x_{0}) +q_{0}(t-t_{0})+p_{0}(x-x_{0})+\frac{1}{2}K_{0}(x-x_{0})^{2} +o(|t-t_{0}|+ |x-x_{0}|^{2}) $$

(or

$$v(t,x) \geq v(t_{0},x_{0}) +q_{0}(t-t_{0})+p_{0}(x-x_{0})+\frac{1}{2}K_{0}(x-x_{0})^{2} +o(|t-t_{0}|+ |x-x_{0}|^{2}), $$

respectively). For technical reasons, one also needs the limiting super- and subjets; we define $\overline{\mathcal{P}}^{2,+}u(t,x)$ as the set of elements $(q,p,K) \in\mathbf{R}^{3}$ for which there exists a sequence $(t_{\epsilon},x_{\epsilon},q_{\epsilon},p_{\epsilon},K_{\epsilon})$ in $[0,T)\times\mathbf{R} \times\mathcal{P}^{2,+}u(t_{\epsilon},x_{\epsilon})$ satisfying

$$\big(t_{\epsilon},x_{\epsilon},u(t_{\epsilon},x_{\epsilon}),q_{\epsilon},p_{\epsilon},K_{\epsilon}\big)\longrightarrow \big(t,x,u(t,x),q,p,K\big). $$

The set $\overline{\mathcal{P}}^{2,-}v(t,x)$ is defined similarly.

Now, Ishii’s lemma in the form of [24], Remark 4.4.9, gives the existence of real numbers $K,N$ such that

$$\begin{aligned} \Big(\frac{1}{\epsilon}(t_{\epsilon}-s_{\epsilon}),\frac {1}{\epsilon}(x_{\epsilon}-y_{\epsilon}),K\Big) \in& \overline {\mathcal {P}}^{2,+}u(t_{\epsilon},x_{\epsilon}), \\ \Big(\frac{1}{\epsilon}(t_{\epsilon}-s_{\epsilon}),\frac{1}{\epsilon }(x_{\epsilon}-y_{\epsilon}),N\Big) \in& \overline{\mathcal {P}}^{2,-}(v+w)(s_{\epsilon},y_{\epsilon}), \end{aligned}$$

with $c^{2}K-d^{2} N \leq\frac{3}{\epsilon}(c-d)^{2}$ for any real numbers $c,d$. Hence setting $c=d$ gives

$$ K \leq N. $$

(9.18)

The superjet (resp., subjet) characterization of supersolutions (resp., subsolutions) provides

$$\begin{aligned} \min\Big\{ &\!-\nu_{1} u(t_{\epsilon},x_{\epsilon})-\frac{1}{\epsilon }(t_{\epsilon}-s_{\epsilon})-\frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon})-\frac{\sigma^{2}}{2}K, \\ &\!-e^{-\nu_{2} t_{\epsilon}-\nu_{3} x_{\epsilon}}+\frac{1}{\epsilon }(x_{\epsilon}-y_{\epsilon})+\nu_{3} u(t_{\epsilon},x_{\epsilon}) \Big\} \leq0, \\ \min\Big\{ &\!-\nu_{1} (v+w)(s_{\epsilon},y_{\epsilon})-\frac {1}{\epsilon }(t_{\epsilon}-s_{\epsilon})-\frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon}) -\frac{\sigma^{2}}{2}N, \\ &\!-e^{-\nu_{2} s_{\epsilon}-\nu_{3} y_{\epsilon}}+\frac{1}{\epsilon}(x_{\epsilon}-y_{\epsilon})+ \nu_{3} (v+w)(s_{\epsilon},y_{\epsilon}) \Big\} \geq0. \end{aligned}$$

(9.19)

Now (9.19) is equivalent to

$$\begin{aligned} \max\Big\{ &\!\nu_{1} u(t_{\epsilon},x_{\epsilon})+\frac{1}{\epsilon }(t_{\epsilon}-s_{\epsilon})+ \frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon})+\frac{\sigma^{2}}{2}K , \\ &\!e^{-\nu_{2} t_{\epsilon}-\nu_{3} x_{\epsilon}}-\frac{1}{\epsilon }(x_{\epsilon}-y_{\epsilon})-\nu_{3} u(t_{\epsilon},x_{\epsilon}) \Big\} \geq0, \\ \max\Big\{ &\!\nu_{1} (v+w)(s_{\epsilon},y_{\epsilon})+\frac {1}{\epsilon }(t_{\epsilon}-s_{\epsilon})+ \frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon})+\frac{\sigma ^{2}}{2}N, \\ &\!e^{-\nu_{2} s_{\epsilon}-\nu_{3} y_{\epsilon}}-\frac{1}{\epsilon}(x_{\epsilon}-y_{\epsilon})- \nu_{3} (v+w)(s_{\epsilon},y_{\epsilon}) \Big\} \leq0, \end{aligned}$$

(9.20)

and the second equation of (9.20) implies

$$\begin{aligned} \nu_{1} (v+w)(s_{\epsilon},y_{\epsilon})+\frac{1}{\epsilon}(t_{\epsilon}-s_{\epsilon}) +\frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon})+\frac {\sigma ^{2}}{2}N \leq& 0, \\ e^{-\nu_{2} s_{\epsilon}-\nu_{3} y_{\epsilon}}-\frac{1}{\epsilon}(x_{\epsilon}-y_{\epsilon})- \nu_{3} (v+w)(s_{\epsilon},y_{\epsilon}) \leq& 0. \end{aligned}$$

(9.21)

We distinguish two cases:

Case 1: $\nu_{1} u(t_{\epsilon},x_{\epsilon})+\frac{1}{\epsilon }(t_{\epsilon}-s_{\epsilon})+ \frac{\tilde{\mu}}{\epsilon}(x_{\epsilon}-y_{\epsilon})+\frac{\sigma^{2}}{2}K \geq0$ for some subsequence, again denoted by $\epsilon$ , tending to zero. The assumption, as well as the first equation of (9.21), implies the inequality $\nu_{1}(u(t_{\epsilon},x_{\epsilon})-(v+w)(s_{\epsilon},y_{\epsilon}))+\frac {\sigma^{2}}{2}(K-N)\geq0$, from which one gets for $\epsilon\to0$ that $\nu_{1} M+\frac{\sigma^{2}}{2}(K-N)\geq0$. By (9.15), (9.17) and (9.18), this is a contradiction.

Case 2: $e^{-\nu_{2} t_{\epsilon}-\nu_{3} x_{\epsilon}}-\frac {1}{\epsilon }(x_{\epsilon}-y_{\epsilon})-\nu_{3} u(t_{\epsilon},x_{\epsilon}) \geq0$ for some subsequence, again denoted by $\epsilon$ , tending to zero. The assumption, as well as the second equation of (9.21), implies the inequality $-\nu_{3}(u(t_{\epsilon},x_{\epsilon})-(v+w)(s_{\epsilon},y_{\epsilon}))+e^{-\nu _{2} t_{\epsilon}-\nu_{3} x_{\epsilon}}-e^{-\nu_{2} s_{\epsilon}-\nu_{3} y_{\epsilon}} \geq0$ and in the limit $-\nu_{3}M \geq0$, which is by (9.15) and (9.17) a contradiction.

Hence, our assumption of a positive supremum of $u-v-w$ on $\bar{Q} \setminus U$ is wrong. Therefore, we have $u-v-w \leq0$ on $\bar{Q} \setminus U$, and also on $\bar{Q} \setminus V$ if we define $V:= \{ (t,x) | T- \frac{1}{\sqrt{\nu_{3}}}< t <T, 0<x < \frac{1}{\sqrt{\nu_{3}}} \}$ (remember that $\nu_{3}$ is large). Since $w$ can be made arbitrarily small on $\bar{Q} \setminus V$ and as the same holds for $V$, we conclude

$$u-v \leq0 \quad\mbox{on } \bar{Q} \setminus\big( \{ (t,0) | t \in [0,T] \} \cup \{ (T,x) | x \in\mathbf{R}_{+} \} \big). $$

Because we have assumed $u=v$ on $( \{ (t,0) | t \in[0,T] \} \cup \{ (T,x) | x \in\mathbf{R}_{+} \} )$, we get finally $u \leq v$ on $\bar{Q}$. □

1.4 9.4 Auxiliary results for Sect. 6

Proof of Lemma 6.2

In an analogous manner as in [15, Lemma 3.1], one can show that the following construction of $H(x)$ is successful. We set $H(x):= \underline{H}(x) +\overline{H}(x)$, and give first the definitions of $\underline{H} (x)$ and $\overline{H}(x)$ for $x \in [0,\epsilon]$. They are

$$\underline{H}(x):= \bar{\gamma}+H_{1}x-\bigg(\hat{\mu}+\frac{\hat {\mu}^{2} \bar{\gamma}}{2}\bigg)H_{1}x^{2}+ \alpha\left(\frac{x^{7}}{840}-\frac{\epsilon}{240}x^{6}+\frac {\epsilon ^{2}}{240}x^{5} \right), $$

where $\alpha$ depends on the model parameters and has the asymptotic (for $\epsilon\to0$) behavior $\alpha\sim\frac{120 (2\hat{\mu}+\hat{\mu}^{2} \bar{\gamma })}{\epsilon^{5}}$. For $\overline{H}(x)$ one can take

$$\overline{H}(x)=x^{3}(x-\epsilon)^{3} (\alpha_{0}+\alpha_{1} x +\alpha_{2} x^{2} +\alpha_{3} x^{3}) $$

with

$$\begin{aligned} \alpha_{0} =&-\frac{a}{6\epsilon^{3}},\quad\alpha_{1}=-\frac {a}{2\epsilon ^{4}}-\frac{c}{24\epsilon^{3}}, \quad\alpha_{2}=\frac{3a}{2\epsilon^{5}}+\frac{b}{\epsilon^{5}} +\frac {c}{12\epsilon^{4}}-\frac{d}{24\epsilon^{4}},\\ \alpha_{3} =&-\frac{5a}{6\epsilon^{6}}-\frac{5b}{6\epsilon^{6}} -\frac {c}{24\epsilon^{5}}+\frac{d}{24\epsilon^{5}}, \end{aligned}$$

and

$$\begin{aligned} a :=& 3\hat{\mu}^{2} H_{1}+H_{3},\quad b:= 2\hat{\beta},\quad c:= -4\hat{\mu}H_{3}-4\hat{\mu}^{3} H_{1}+5 \hat{\mu}^{4}\bar{\gamma},\\ d :=& \frac{4\hat{\beta}}{\sigma^{2}}H_{4}-4\hat{\mu}\hat{\beta}. \end{aligned}$$

For negative $x$, we use the definition $H(-x)=-H(x)e^{2\hat{\mu}x}$, $x>0$, which stems from the fact that the transformed terminal gain function $h(x)=H(x)e^{\hat{\mu}x -\lambda T}$, with $\lambda:= \beta+\frac{\mu^{2}}{2\sigma^{2}}$, is assumed to be an odd function. □

We describe now the sequence of transformations of BVPs needed in Sect. 6. Our first lemma transforms the original problem into a new one, where we replace the former region $I$ by $I\cup\hat{I}$, with $\hat{I}$ the mirror image of $I$ with respect to the line $x=0$. Moreover, it shows that the correct inequalities assumed in region I (resp., II) in Theorem 6.5 (see Definition 6.4) follow if we assume $C^{4}$-regularity (except at the point $(T,0)$) up to and including the boundary.

Lemma 9.2

A function $V$ and a boundary function $b$ solving the system

$$ \left\{ \textstyle\begin{array}{l} V(T,x)=e^{-\beta T}H(x),\quad x \in[-\epsilon,\infty);\\ V_{t}+\mu V_{x}+\frac{\sigma^{2}}{2}V_{xx}=0,\quad (t,x) \in\overline {G_{1}}\setminus\{(T,0)\}; \\ V_{x} = e^{-\beta t},\quad x \geq b(t),0\leq t \leq T;\\ V(t,0)=0,\quad 0\leq t < T;\\ V,V_{t},V_{tt},V_{x},V_{xx},V_{xxx},V_{xxxx} \in C(\overline {G_{1}}\setminus\{(T,0)\}),\\ \quad V,V_{t},V_{x},V_{xx} \in C\left(\overline{G_{2}}\setminus\{(T,0)\} \right);\\ b(t)>0,\quad 0\leq t \leq T;\\ b(T)=\epsilon;\\ b \in C^{1}([0,T]);\\ b(t)+d(t) >0,\quad 0\leq t \leq T, \end{array}\displaystyle \right. $$

(9.22)

solve also (6.1) of Theorem 6.5.

Proof

What we have actually to show is that we have the respective inequalities

$$ V_{t}+\mu V_{x}+\frac{\sigma^{2}}{2}V_{xx}\leq0,\qquad (t,x) \in \mathit{II}, $$

(9.23)

and

$$ V_{x} \geq e^{-\beta t},\qquad (t,x) \in I \setminus\{(T,0)\}. $$

(9.24)

The proof of (9.23) is completely analogous to the proof of Lemma 3.2 in [15], with the only alteration that we have now $d(T) =H_{0}+\bar{\gamma}$, which implies

$$ d(t) = e^{\beta t}\int_{t}^{T}\left(\mu e^{-\beta s}-\beta e^{-\beta s}b(s)\right)\,ds+(H_{0}+\bar{\gamma})e^{\beta(t-T)}. $$

(9.25)

We proceed with inequality (9.24) and start with the calculation of the left-hand boundary condition of $V$, i.e., $V(t,-b(t))$. Using the right-hand boundary conditions $V(t,b(t))=e^{-\beta t}(b(t)+d(t))$, $V_{x}(t,b(t))=e^{-\beta t}$, $V_{xx}(t,b(t))=0$, the transformation $v(t,x)=V(t,x)e^{\hat{\mu}x -\frac{\mu ^{2}}{2\sigma^{2}}t}$ and the fact that $v$ is an odd function in $x$ give after some elementary algebra

$$\begin{aligned} V\big(t,-b(t)\big) =&-e^{-\beta t +2\hat{\mu }b(t)}\big(b(t)+d(t)\big), \\ V_{x}\big(t,-b(t)\big) =&e^{-\beta t +2\hat{\mu}b(t)}\Big(1+2\hat {\mu}\big(b(t)+d(t)\big)\Big), \\ V_{xx}\big(t,-b(t)\big) =&e^{-\beta t +2\hat{\mu}b(t)}\Big(-4\hat {\mu}-4\hat{\mu} ^{2}\big(b(t)+d(t)\big)\Big). \end{aligned}$$

For the transformed function $v$, one gets

$$\begin{aligned} v\big(t,\pm b(t)\big) =&\pm e^{-\lambda t +\hat {\mu} b(t)}\big(b(t)+d(t)\big) =: \pm g(t), \\ v_{x}\big(t,\pm b(t)\big) =& e^{-\lambda t +\hat{\mu}b(t)}\Big(1+\hat{\mu}\big(b(t)+d(t)\big)\Big) =: k(t), \\ v_{xx}\big(t,\pm b(t)\big) =&\pm e^{-\lambda t +\hat{\mu}b(t)}\Big(2 \hat{\mu} +\hat{\mu}^{2}\big(b(t)+d(t)\big)\Big) =: \pm m(t). \end{aligned}$$

Moreover, $v$ fulfils the boundary value problem

$$\begin{aligned} \hat{\mathcal{L}}v(t,x)&:= v_{t}+\frac{\sigma ^{2}}{2}v_{xx}=0, \\ v\big(t,\pm b(t)\big)&\phantom{:}=\pm g(t), \\ v(T,x)&\phantom{:}=e^{\hat{\mu}x -\lambda T}H(x) =: h(x). \end{aligned}$$

The function $h$ has a discontinuity at $x=0$, and using the Standing Assumption on $H$, one finds $h(0\pm)=\pm\tilde{\gamma}$. Subtracting out this jump, we define $\tilde{h}$ as $\tilde{h}(x):= h(x)-\tilde{\gamma} \mathrm{sgn}(x)$. Our definition of $H$ in the Standing Assumption is chosen in such a way that we have $h(-x)=-h(x)$ for $x \in[-\epsilon,\epsilon]$, $h^{\prime\prime}(0+)=0$ and $h^{\prime \prime \prime\prime}(0+)=0$. This implies that $\tilde{h}\in C^{4}([-\epsilon,\epsilon])$.

We now decompose the function $v$ into a discontinuous part fulfilling a Cauchy problem and a regular part fulfilling a BVP, i.e., $v =: v^{(1)}+v^{(2)}$ with

$$\begin{aligned} \hat{\mathcal{L}}v^{(1)} =&0, \\ v^{(1)}(T,x) =&\tilde{\gamma}\mathrm{sgn}(x), \end{aligned}$$

respectively,

$$\begin{aligned} \hat{\mathcal{L}}v^{(2)} =&0, \\ v^{(2)}\big(t,\pm b(t)\big) =&\pm g(t)-v^{(1)}\big(t,\pm b(t)\big), \\ v^{(2)}(T,x) =&\tilde{h}(x). \end{aligned}$$

(9.26)

Note that the function on the right-hand side of the second equation in (9.26) is in $C^{1}([0,T])$, since away from the point $(T,0)$, the function $v^{(1)}$ is smooth, and $g \in C^{1}([0,T])$ by the assumptions of our lemma. Since the boundary data and the final data are obviously compatible, we get by Gevrey’s result, see [15, Lemma 3.5], that $v^{(2)}_{x} \in C(\overline{G_{1}})$. Hence, $v^{(2)}_{x}$ fulfils the BVP

$$\begin{aligned} \hat{\mathcal{L}}v^{(2)}_{x} =&0, \\ v^{(2)}_{x}\big(t,\pm b(t)\big) =&k(t)-v^{(1)}_{x}\big(t,b(t)\big), \\ v^{(2)}_{x}(T,x) =&\tilde{h}'(x). \end{aligned}$$

(9.27)

Again the function on the right-hand side of the second equation in (9.27) is in $C^{1}([0,T])$, and since the boundary and final data are compatible, we conclude by Gevrey’s result that $v^{(2)}_{xx} \in C(\overline{G_{1}})$.

We now approximate the function $v^{(1)}$ by smooth functions $v^{(1,\delta)}$ which fulfil

$$\begin{aligned} \hat{\mathcal{L}}v^{(1,\delta)} =&0, \\ v^{(1,\delta)}(T,x) =&\phi^{(\delta)}(x), \end{aligned}$$

(9.28)

where $\phi^{(\delta)}\in C^{\infty}(\mathbf{R})$ fulfils $\phi ^{(\delta)}(x) \to\tilde{\gamma}\mathrm{sgn}(x)$ pointwise for $\delta\to0$, except at $x=0$. Moreover, we impose that $\phi^{(\delta)}$ are odd functions, fulfil $\phi^{(\delta)'}(x)\geq0$, $\phi^{(\delta)''}(x)\leq0$ for $x\geq0$, and $\phi^{(\delta)}(x):= \tilde{\gamma}\mathrm{sgn}(x)$ for $|x|\geq \delta/2$. The solution is well known and given by

$$v^{(1,\delta)}(t,x)=\frac{1}{\sqrt{2\pi(T-t)}\sigma}\int_{-\infty }^{\infty}\phi^{(\delta)}(\xi )e^{-\frac{(x-\xi)^{2}}{2\sigma^{2}(T-t)}}\, d\xi. $$

Obviously, one has

$$ v^{(1,\delta)}\longrightarrow v^{(1)},\quad v^{(1,\delta)}_{x} \longrightarrow v^{(1)}_{x}, \quad v^{(1,\delta)} _{xx} \longrightarrow v^{(1)}_{xx} $$

(9.29)

for $\delta\to0$, pointwise in $\overline{G_{1}} \setminus\{(T,0)\}$ and uniformly on each compact $K$ with $K \subset\overline{G_{1}} \setminus\{(T,0)\}$. Let $v^{(\delta)}:= v^{(1,\delta)}+v^{(2)}$, which gives for $v^{(\delta)}$ the system

$$\begin{aligned} \hat{\mathcal{L}}v^{(\delta)} =&0, \\ v^{(\delta)}\big(t,\pm b(t)\big) =&v^{(1,\delta)}\big(t,\pm b(t)\big)\pm g(t)-v^{(1)}\big(t,b(t)\big), \\ v^{(\delta)}(T,x) =&v^{(1,\delta)}(T,x)+\tilde{h}(x)=\phi^{(\delta )}(x)+h(x)-\tilde{\gamma}\mathrm{sgn}(x). \end{aligned}$$

(9.30)

Because of (9.29), we have

$$ v^{(\delta)}\longrightarrow v, \quad v^{(\delta)}_{x} \longrightarrow v_{x},\quad v^{(\delta)}_{xx} \longrightarrow v_{xx} $$

(9.31)

for $\delta\to0$, pointwise in $\overline{G_{1}} \setminus\{(T,0)\}$ and uniformly on each compact $K$ with $K \subset\overline{G_{1}} \setminus\{(T,0)\}$, or (with $V^{(\delta)}:= v^{(\delta)}e^{-\hat{\mu}x +\frac{\mu ^{2}}{\sigma^{2}}t}$)

$$ V^{(\delta)}\longrightarrow V, \quad V^{(\delta)}_{x} \longrightarrow V_{x}, \quad V^{(\delta)}_{xx} \longrightarrow V_{xx} $$

(9.32)

for $\delta\to0$, pointwise in $\overline{G_{1}} \setminus\{(T,0)\}$ and uniformly on each compact $K$ with $K \subset\overline{G_{1}} \setminus\{(T,0)\}$. Since $v^{(1,\delta)}\in C^{\infty}$, we have $v^{(\delta)}_{xx} \in C(\overline{G_{1}})$ and $v^{(\delta)}_{x}$ fulfils

$$\begin{aligned} \hat{\mathcal{L}}v^{(\delta)}_{x} =&0, \\ v^{(\delta)}_{x}\big(t,\pm b(t)\big) =&v^{(1,\delta)}_{x}\big(t,\pm b(t)\big) + k(t)-v^{(1)}_{x}\big(t,b(t)\big) =: k^{(\delta)}(t), \\ v^{(\delta)}_{x}(T,x) =&\phi^{(\delta)'}(x)+\tilde{h}'(x). \end{aligned}$$

(9.33)

Furthermore, we have $\tilde{h}'(x)=h'(x)$ for $x \neq0$, and $\tilde{h} '(0)=h'(0+)=h'(0-)$. Our Standing Assumption on $H$, the transformation formula between $H$ and $h$, as well as the oddness of $h$, implies $\tilde{h}'(x) >0$ for $x \in[-\epsilon,\epsilon]$, which in turn gives

$$ v^{(\delta)}_{x}(T,x) >0,\quad x\in[-\epsilon,\epsilon]. $$

(9.34)

Since we know that the convergence $k^{(\delta)}(t) \to k(t) >0$ for $\delta\to 0$ is uniform on $[0,T]$ by (9.31), we conclude that $k^{(\delta)}(t)>0$ on $[0,T]$ for $\delta$ small enough. The maximum principle implies by (9.33) and (9.34) that $v^{(\delta)}_{x}>0$ on $G_{1}$ for small enough $\delta$, especially $v^{(\delta)}_{x}(t,0) >0$ for $0\leq t \leq T$. Hence, transforming to $V^{(\delta)}$ gives by the oddness of $v^{(\delta)}$ that

$$ V^{(\delta)}_{xx}(t,0) < 0, \quad t\in[0,T], $$

(9.35)

for $\delta$ small enough. Defining $g^{(\delta)}$ and $m^{(\delta )}$ analogously to $k^{(\delta)}$ and exploiting (9.29) gives $g^{(\delta)}(t) \to g(t)$, $k^{(\delta)}(t) \to k(t)$, $m^{(\delta )}(t) \to m(t)$, uniformly on $[0,T]$, hence $V^{(\delta)}_{xx}(t,b(t)) \to V_{xx}(t,b(t))$, uniformly on $[0,T]$, i.e., there exists a positive function $\nu (\delta ) \to0$ with $\delta\to0$ such that

$$ \big| V^{(\delta)}_{xx}\big(t,b(t)\big) \big| \leq\nu(\delta) $$

(9.36)

for $t \in[0,T]$.

For the terminal value, we claim that

$$ V^{(\delta)}_{xx}(T,x) \leq0,\quad x \in[0,\epsilon]. $$

(9.37)

By our transformation law, this is equivalent to $(v^{(\delta )}_{xx}-2\hat{\mu}v^{(\delta)} _{x}+\hat{\mu}^{2}v^{(\delta)})(T,x) \leq0$ or, if we use our terminal condition (9.30), equivalent to

$$\phi^{(\delta)''}(x)+\tilde{h}''(x)-2\hat{\mu}\phi^{(\delta )'}(x)-2 \hat{\mu}\tilde{h}'(x)+\hat{\mu}^{2}\phi^{(\delta)}(x)+ \hat{\mu}^{2} \tilde{h}(x) \leq0. $$

Now, the sum of the second, fourth and $\hat{\mu}^{2} h$ from the sixth term are less than or equal to zero by our Standing Assumption (viii). Hence it remains to show that we have $\phi^{(\delta )''}(x)-2\hat{\mu}\phi^{(\delta)'} (x)+\hat{\mu}^{2}(\phi^{(\delta)}(x)-\tilde{\gamma}) \leq0$, which is true by our assumptions on the function $\phi^{(\delta)}$ formulated after (9.28). Therefore, (9.37) is true.

Since $V^{(\delta)}_{xx} \in C(\overline{G_{1}})$, the maximum principle, together with (9.35)–(9.37), implies $V^{(\delta)}_{xx}(t,x) \leq\nu(\delta)$ on $\overline{G_{1}}$. Employing (9.32) and going to the limit $\delta\to0$ gives $V_{xx}(t,x) \leq0$ on $\overline{G_{1}} \setminus\{(T,0)\}$, which by $V_{x}(t,b(t))=e^{-\beta t}$ finishes the proof. □

Since we want to apply the Green’s function method for the backward heat equation, we remove the drift of the problem by a simple exponential transformation, and we formulate the problem now for a new dependent variable $v$.

Lemma 9.3

(i) If a function $v= v(t,x)$ and a boundary function $b$ solve the system

$$ \left\{ \textstyle\begin{array}{l} v \textit{ is an odd function in x on }\overline{G_{1}}, \textit{except }v(T,0)=\tilde{\gamma};\\ v(T,x)=e^{\hat{\mu}x-\lambda T}H(x) =: h(x),\quad x \in[-\epsilon ,\infty );\\ v_{t}+\frac{\sigma^{2}}{2}v_{xx}=0,\quad (t,x) \in\overline {G_{1}}\setminus \{(T,0)\}; \\ v(t,x)=e^{-\lambda t+\hat{\mu}x}(x+d(t)),\quad x \geq b(t),0\leq t \leq T;\\ v,v_{t},v_{tt},v_{x},v_{xx},v_{xxx},v_{xxxx} \in C(\overline {G_{1}}\setminus\{(T,0)\}),\\ \quad v,v_{t},v_{x},v_{xx} \in C(\overline{G_{2}}\setminus\{(T,0)\});\\ b(t)>0,\quad 0\leq t \leq T;\\ b(T)=\epsilon;\\ b \in C^{1}([0,T]);\\ b(t)+d(t) >0,\quad 0\leq t \leq T, \end{array}\displaystyle \right. $$

(9.38)

then $V$ defined by $V(t,x):= e^{-\hat{\mu}x +\frac{\mu^{2}}{2\sigma ^{2}}t}v(t,x)$ and $b$ solve system (9.22).

(ii) The transformed terminal condition $h$ fulfils

$$\textstyle\begin{array}{ll} h(x) \in C^{4}([-\epsilon,\epsilon]\setminus\{0\}) \cap C^{2}([-\epsilon ,\infty)\setminus\{0\})\textit{, with existing }h^{(4)}(0\pm);\\ h(0)=h(0+)=\tilde{\gamma},h(0-)=-\tilde{\gamma}, h(-x)=-h(x) \textit{ for }x>0, \\ \quad h^{\prime\prime}(0\pm)=h^{\prime\prime\prime\prime}(0\pm )=0;\\ -h(-\epsilon)=h(\epsilon)=e^{-\lambda T +\hat{\mu}b(T)}\big(b(T)+d(T)\big);\\ h^{\prime}(-\epsilon)=h^{\prime}(\epsilon)=e^{-\lambda T +\hat{\mu }b(T)}\Big(1+\hat{\mu}\big(b(T)+d(T)\big)\Big);\\ -h^{\prime\prime}(-\epsilon)=h^{\prime\prime}(\epsilon )=e^{-\lambda T +\hat{\mu}b(T)}\Big(2\hat{\mu}+\hat{\mu}^{2}\big(b(T)+d(T)\big)\Big);\\ h^{\prime\prime\prime}(-\epsilon+)=h^{\prime\prime\prime }(\epsilon -)=e^{-\lambda T +\hat{\mu}b(T)}\Big(2 \hat{\beta}+3\hat{\mu }^{2}+\hat{\mu}^{3}\big(b(T)+d(T)\big)\Big);\\ -h^{(4)}(-\epsilon+)=h^{(4)}(\epsilon-)\\ \quad =e^{-\lambda T +\hat{\mu}b(T)} \bigg(\dfrac{4\hat{\beta}}{\sigma^{2}}H_{4} +4\hat{\mu}\hat{\beta }+4\hat{\mu}^{3}+\hat{\mu}^{4}\big(b(T)+d(T)\big)\bigg). \end{array} $$

Proof

The proof consists of simple calculations. Note that the asserted values of $h$ and its derivatives at $\epsilon$ are the same as in [15], with the only difference that we now have a different value for $d(T)$. □

Remark 9.4

Note that in comparison to system (9.38) in [15, Lemma 3.3], we have skipped the fifth condition in (9.38) there. It was used there to prove the fifth property in (9.22) which we do not have any more in our new system (9.22).

In the following lemma, we define a function we use repeatedly in the sequel.

Lemma 9.5

Let $F$ be the solution of the Cauchy problem

$$\begin{aligned} \hat{\mathcal{L}} F(t,x) =&0, \\ F(T,x) =&\tilde{\gamma}\mathrm{sgn}(x). \end{aligned}$$

Then $F$ has the representation

$$F(t,x)=\frac{\tilde{\gamma}}{\sqrt{2 \pi(T-t)}\sigma} \int _{-\infty}^{\infty}\mathrm{sgn}(\xi) e^{-\frac{(x-\xi)^{2}}{2 \sigma^{2} (T-t)}}\,d\xi, $$

and $\frac{\partial^{k+\ell}}{\partial t^{k} \partial x^{\ell}}F(t,x)$ is continuous on $\overline{G_{1}} \setminus\{(T,0)\}$ (resp., uniformly continuous on each compact set $K \subset\overline{G_{1}} \setminus\{(T,0)\}$), for $k,\,\ell\in\mathbf{N}_{0}$. Moreover,

$$\hat{\mathcal{L}}\biggl( \frac{\partial^{k+\ell}}{\partial t^{k} \partial x^{\ell}}F(t,x)\biggr)=0 $$

on $\overline{G_{1}} \setminus\{(T,0)\}$. Finally, we note that

$$F_{x}(t,x)= \frac{\tilde{\gamma}}{\sigma} \sqrt{\frac{2}{\pi (T-t)}}e^{-\frac {x^{2}}{2\sigma^{2} (T-t)}} $$

is the fundamental solution of our backward heat equation.

By subtracting the function $F$ from $v$, we get a regularized version of $v$, namely $\tilde{v}(t,x):= v(t,x)-F(t,x)$. In the sequel, we consider $\tilde{v}$ and derive a sequence of BVPs for it. The proof of the following lemma follows obviously from Lemma 9.3.

Lemma 9.6

(i) If a function $\tilde{v}$ and a boundary function $b$ solve the system

$$ \left\{ \textstyle\begin{array}{l} \tilde{v}(t,x) \textit{ is an odd function in x on }\overline{G_{1}};\\ \tilde{v}(T,x)=h(x)-\tilde{\gamma}\mathrm{sgn}(x) =: \tilde {h}(x),\quad x \in[-\epsilon,\infty );\\ \tilde{v}_{t}+\frac{\sigma^{2}}{2}v_{xx}=0,\quad (t,x) \in\overline {G_{1}}; \\ \tilde{v}(t,x)=e^{-\lambda t+\hat{\mu}x}(x+d(t))-F(t,x),\quad x \geq b(t),0\leq t \leq T;\\ \tilde{v}, \tilde{v}_{t},\tilde{v}_{tt},\tilde{v}_{x},\tilde {v}_{xx},\tilde{v}_{xxx},\tilde{v}_{xxxx} \in C(\overline{G_{1}}), \tilde{v},\tilde{v}_{t},\tilde{v}_{x},\tilde{v}_{xx} \in C(\overline {G_{2}});\\ b(t)>0,\quad 0\leq t \leq T;\\ b(T)=\epsilon;\\ b \in C^{1}([0,T]);\\ b(t)+d(t) >0,\quad 0\leq t \leq T, \end{array}\displaystyle \right. $$

(9.39)

then $v$, defined by $v(t,x):= \tilde{v}(t,x)+F(t,x)$ and $b$ solve (9.38).

(ii) The transformed terminal condition $\tilde{h}$ fulfils

$$\textstyle\begin{array}{ll} \tilde{h}(x) \in C^{4}\big([-\epsilon,\epsilon]\cap C^{2}([-\epsilon ,\infty)\big);\\ \tilde{h}(x) \textit{ is an odd function on }[-\epsilon,\epsilon];\\ -\tilde{h}(-\epsilon)=\tilde{h}(\epsilon)=e^{-\lambda T +\hat{\mu }b(T)}\big(b(T)+d(T)\big)-\tilde{\gamma};\\ h^{\prime}(-\epsilon)=h^{\prime}(\epsilon)=e^{-\lambda T +\hat{\mu }b(T)}\Big(1+\hat{\mu}\big(b(T)+d(T)\big)\Big);\\ -h^{\prime\prime}(-\epsilon)=h^{\prime\prime}(\epsilon )=e^{-\lambda T +\hat{\mu}b(T)}\Big(2\hat{\mu}+\hat{\mu}^{2}\big(b(T)+d(T)\big)\Big);\\ h^{\prime\prime\prime}(-\epsilon+)=h^{\prime\prime\prime }(\epsilon -)=e^{-\lambda T +\hat{\mu}b(T)}\Big(2 \hat{\beta}+3\hat{\mu }^{2}+\hat{\mu}^{3}\big(b(T)+d(T)\big)\Big);\\ -h^{(4)}(-\epsilon+)=h^{(4)}(\epsilon-)\\ \quad =e^{-\lambda T +\hat{\mu}b(T)} \bigg(\dfrac{4\hat{\beta}}{\sigma^{2}}H_{4} +4\hat{\mu}\hat{\beta }+4\hat{\mu}^{3}+\hat{\mu}^{4}\big(b(T)+d(T)\big)\bigg). \end{array} $$

Our next result shows that if we choose proper boundary conditions for $\tilde{v}$, then $\tilde{v}$ has automatically the smoothness asserted in Lemma 9.6.

Lemma 9.7

If a function $\tilde{v}$ and a boundary function $b$ solve the system

$$ \left\{ \textstyle\begin{array}{l} \tilde{v}(T,x)=\tilde{h}(x),\quad x \in[-\epsilon,\infty);\\ \tilde{v}_{t}+\frac{\sigma^{2}}{2}\tilde{v}_{xx}=0,\quad (t,x) \in \overline{G_{1}}; \\ \tilde{v}\big(t,\pm b(t)\big)=\pm e^{-\lambda t+\hat{\mu}b(t)}\big(b(t)+d(t)\big)\mp F\big(t,b(t)\big),\quad 0\leq t \leq T;\\ \tilde{v}_{x}\big(t,\pm b(t)\big)=e^{-\lambda t+\hat{\mu}b(t)}\Big(1 + \hat{\mu}\big(b(t)+d(t)\big)\Big)-F_{x}\big(t,b(t)\big),\quad 0\leq t \leq T;\\ \tilde{v}(t,x)=e^{\hat{\mu}x-\lambda t}\big(x+d(t)\big)-F(t,x),\quad (t,x) \in \mathit{II};\\ b(t)>0,\quad 0\leq t \leq T;\\ b(T)=\epsilon;\\ b \in C^{1}([0,T]);\\ b(t)+d(t) >0,\quad 0\leq t \leq T, \end{array}\displaystyle \right. $$

(9.40)

then $\tilde{v}$ and $b$ solve (9.39), especially the smoothness conditions formulated there. Moreover, we have the explicit representations

$$\begin{aligned} \tilde{v}_{xx}\big(t,\pm b(t)\big) &=\pm e^{-\lambda t +\hat{\mu}b(t)}\Big(2\hat{\mu}+\hat{\mu }^{2}\big(b(t)+d(t)\big)\Big)\mp F_{xx}\big(t,b(t)\big) \\ &=: \pm\tilde{A}(t), \\ \tilde{v}_{xxx}\big(t, \pm b(t)\mp\big) &= e^{-\lambda t +\hat{\mu}b(t)}\Big(2 \hat{\beta} +3\hat{\mu}^{2}+\hat{\mu}^{3}\big(b(t)+d(t)\big)\Big)-F_{xxx}\big(t,b(t)\big) \\ &=: \tilde{B}(t), \\ \tilde{v}^{(\epsilon)}_{xxxx}\big(t, \pm b(t)\mp\big) &=\pm e^{-\lambda t +\hat{\mu} b(t)}\bigg(\frac{4\hat{\beta}}{\sigma^{2}}b'(t) +4\hat{\mu}\hat {\beta} +4\hat{\mu}^{3}+\hat{\mu}^{4}\big(b(t)+d(t)\big)\bigg) \\ &\phantom{=:}\mp F_{xxxx}\big(t,b(t)\big) \\ &=: \pm\tilde{D}(t). \end{aligned}$$

(9.41)

Proof

The proof is analogous to the proof of Lemma 3.4 in [15]. We only have to modify the boundary conditions, using our function $F(t,x)$. □

We now start to derive similar systems for the derivatives of $\tilde{v}$.

Lemma 9.8

If a function $\tilde{u}$ and a boundary function $b$ solve the system

$$ \left\{ \textstyle\begin{array}{l} \tilde{u}(T,x)=\tilde{h}'(x),\quad x \in[-\epsilon,\infty);\\ \tilde{u}_{t}+\frac{\sigma^{2}}{2}u_{xx}=0,\quad (t,x) \in\overline {G_{1}}; \\ \tilde{u}\big(t,\pm b(t)\big)=e^{-\lambda t +\hat{\mu}b(t)}\Big(1+\hat{\mu}\big(b(t)+d(t)\big)\Big) -F_{x}\big(t,b(t)\big)=: \tilde{R}(t),\\ \quad 0\leq t \leq T;\\ \tilde{u}_{x}\big(t,\pm b(t)\big)=\pm\tilde{A}(t),\quad 0\leq t \leq T;\\ \tilde{u}(t,x)=e^{\hat{\mu}x-\lambda t}\Big(1+\hat{\mu}\big(x+d(t)\big)\Big)-F_{x}(t,x),\quad (t,x) \in \mathit{II};\\ \tilde{u}_{xx} \in C(\overline{G_{1}}); \\ b(t)>0,\quad 0\leq t \leq T;\\ b(T)=\epsilon;\\ b \in C^{1}([0,T]);\\ b(t)+d(t) >0,\quad 0\leq t \leq T, \end{array}\displaystyle \right. $$

(9.42)

then $\tilde{v}$ defined by $\tilde{v}(t,x):= -\tilde{M}(t)+\int_{-b(t)}^{x} \tilde{u}(t,z) \, dz$, where we define the function $\tilde{M}(t):= e^{\hat{\mu}b(t)-\lambda t}\left (b(t)+d(t)\right)-F(t,b(t))$, and $b$ solve (9.40) of Lemma  9.7.

Proof

The proof is analogous to the proof of Lemma 3.7 in [15]. We just have to use the relation $\hat{\mathcal{L}}F(t,b(t))=0$ at one instance. □

Proof of Lemma 6.6

Again, the proof is analogous to the proof of Lemma 3.8 in [15]. This time we just have to use the relations $\hat{\mathcal{L}}F_{x}(t,b(t))=0$ and $F_{xx}(T,b(T))=0$. □

Proof of Lemma 6.10, part a

The proof is identical to the proof of Lemma 5.1 in [15], since the only point where the form of the terminal function $H$ is used is inequality (46) there, where one has to estimate $\mathbf{E} [H^{2}(X_{T})]$. So exchanging the old terminal function by the new one gives just a different constant. □

Proof of Lemma 6.10, part b

Let $(r_{n})$ be a sequence such that $r_{n} \downarrow T-\eta$ and define $\tilde{V}(t,x):= e^{-\hat{\mu}x +\frac{\mu^{2}}{2\sigma^{2}}t} \tilde{v}(t,x)$. Then one has $\mathcal{L}\tilde{V}_{xt}=0$ in $\overline{G}_{1}$ and after some elementary transformations,

$$\begin{aligned} \tilde{V}_{xt}\big(t,b(t)-\big) =&-\beta e^{-\beta t}+ e^{-\hat{\mu }b(t)+\frac {\mu^{2}}{2\sigma^{2}}t}\sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big), \\ \tilde{V}_{xt}\big(t,-b(t)+\big) =&(-\beta-2 \mu\hat{\mu}) e^{-\beta t+2\hat{\mu} b(t)} \\ &{}+e^{+\hat{\mu}b(t)+\frac{\mu^{2}}{2\sigma^{2}}t}\sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big), \end{aligned}$$

(9.43)

where $F^{(i)}$ denotes the $i$th $x$-derivative of $F$ and $c_{i}$ some generic, model-dependent constants which may vary from place to place, and where $c_{0}$ in the second equation of (9.43) fulfils $c_{0} <0$. Define now $\hat{V}:= \tilde{V}_{xt}-\rho$ for some positive constant $\rho$ to be chosen later. This gives for $\hat{V}$ the system

$$\begin{aligned} \hat{V}(T,x) =&K(x), \\ \hat{V}\big(t,b(t)-\big) =&-\beta e^{-\beta t}+ e^{-\hat{\mu }b(t)+\frac{\mu ^{2}}{2\sigma^{2}}t}\sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big)-\rho, \\ \hat{V}\big(t,-b(t)+\big) =&(-\beta-2 \mu\hat{\mu}) e^{-\beta t+2\hat{\mu} b(t)} +e^{+\hat{\mu}b(t)+\frac{\mu^{2}}{2\sigma^{2}}t}\sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big)-\rho. \end{aligned}$$

Since we know by Lemma 6.10(a) that $b(t) \geq C$ for some generic ($\epsilon$-dependent) constant $C$, one concludes that $|\sum_{i=0}^{3} c_{i}F^{(i)}(t,b(t))| \leq C$. Note also that for $b(t)$ large, $|F^{(i)}(t,b(t))|$, $i=1,2,3$, becomes small. Hence we can choose a model-dependent positive constant $\rho$ such that we have by the maximum principle $\hat{V} <0$ on $\overline{G}_{1}$. Introducing $w:= \frac{\hat {V}}{-\beta e^{-\beta t}}$ gives $w>0$ on $\overline{G}_{1}$, on the one hand, and the system

$$\begin{aligned} w(T,x) =&\frac{K(x)}{-\beta e^{-\beta T}}, \\ w\big(t,b(t)\big) =&1+ \sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big)e^{-\hat{\mu} b(t)+\lambda t}+\frac{\rho}{\beta} e^{\beta t}, \\ w\big(t,-b(t)\big) =&e^{2 \hat{\mu}b(t)}\left(1+\frac{2\mu ^{2}}{\beta\sigma ^{2}}\right) +\sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(t)\big)e^{+\hat{\mu}b(t)+\lambda t}+\frac {\rho}{\beta} e^{\beta t}=: M(t), \\ \mathcal{L}w =&\beta w, \end{aligned}$$

(9.44)

with $c_{0} >0$ in the third equation of (9.44), on the other hand. Since we know that $w$ is positive, it fulfils the maximum principle.

Moreover, $M \in C^{1}((T-\eta,T])$ and $M^{*}(t):= \sup_{s \in[t,T]} M(s)$ is absolutely continuous with respect to Lebesgue measure on $[r_{n},T]$ for all $n$. Now let us define the set $I:= \{ t \in [r_{n},T] | M^{*}(t)=M(t), M(t) \geq\varGamma\}$ for some positive constant $\varGamma> \Vert\frac{K(x)}{-\beta e^{-\beta T}}\Vert_{C([-\epsilon,\epsilon])}$. We also choose $\varGamma$ large enough such that $t \in I$ implies $w(t,-b(t))=M^{*}(t)>w(s,b(s))$, for all $s \in[t,T]$. By the maximum principle, this implies $w_{x}(t,-b(t)) \leq0$, or $\hat {V}_{x}(t,-b(t)+) \geq0$, or $\tilde{V}_{xxt}(t,-b(t)+)\geq0$, for $t \in I$. Using the boundary condition for $\tilde{V}_{xxt}$, one arrives at

$$2\hat{\beta}b'(t)+4\mu\hat{\beta}+\frac{4 \mu^{3}}{\sigma^{4}}+ e^{-\hat{\mu}b(t)+\lambda t}\sum_{i=0}^{4} c_{i}F^{(i)}\big(t,b(t)\big) \geq0,\quad t \in I, $$

or

$$ b'(t) \geq-C,\quad t \in I, $$

for a generic positive ($\epsilon$-dependent) constant $C$.

We set now $C^{+}:= 1+\frac{2\mu^{2}}{\beta\sigma^{2}}$, view the functions $M$ and $b$ as functions of the backward time $\nu := T-t$ and calculate the derivative of $M$ for $\nu$ with $t(\nu) \in I$. Note that since with $\varGamma$ large, also $M(\nu)$ becomes large, we can infer that $b(\nu)$ becomes large, hence $|F^{(i)}(t,b(\nu))|$, $i=1,2,3$, become small, whereas $F^{(0)}(t,b(\nu ))$ is approximately $\tilde{\gamma}>0$. We get

$$\begin{aligned} M'(\nu) =& 2 \hat{\mu}b'(\nu) C^{+} e^{2 \hat{\mu}b(\nu)}+ \hat{\mu}e^{ \hat{\mu}b(\nu)+\lambda t} b'(\nu) \sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(\nu)\big)\\ &{}-\lambda e^{ \hat{\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(\nu)\big)+ e^{ \hat{\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F^{(i+1)}\big(t,b(\nu)\big)b'(\nu)\\ &{}-e^{ \hat{\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F_{t}^{(i+1)}\big(t,b(\nu )\big)- \rho e^{\beta t}\\ \leq& b'(\nu) \bigg( 2 \hat{\mu}C^{+} e^{2 \hat{\mu}b(\nu)}+ \hat{\mu}e^{ \hat {\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F^{(i)}\big(t,b(\nu)\big)\\ &{} +e^{ \hat{\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F^{(i+1)}\big(t,b(\nu )\big)\bigg) -e^{ \hat{\mu}b(\nu)+\lambda t} \sum_{i=0}^{3} c_{i}F_{t}^{(i+1)}\big(t,b(\nu)\big). \end{aligned}$$

By the explicit formula for $M$ in the third equation of (9.44), the fact that $M(\nu)$ and $b(\nu)$ are large, $c_{0} >0$, $F^{(0)}(t,b(\nu))>0$ and $| F^{(i)}(t,b(\nu))|$, $i=1,2,3$, are small, we have $M(\nu) > \frac{C^{+} e^{2 \hat{\mu}b(\nu)}}{2}> -e^{ \hat {\mu}b(\nu )+\lambda t} \sum_{i=0}^{3} c_{i}F_{t}^{(i+1)}(t,b(\nu))$. Hence we can estimate the right-hand side above by $b'(\nu) ( \dots )+M(\nu)$. Moreover, we have

$$\hat{\mu}C^{+} e^{2 \hat{\mu}b(\nu)} \leq(\dots)\leq3 \hat{\mu }C^{+} e^{2 \hat{\mu}b(\nu)}, $$

which gives us in the case where $b'(\nu) \geq0$ the upper estimate

$$3 b'(\nu) \hat{\mu}C^{+} e^{2 \hat{\mu}b(\nu)}+M(\nu) \leq C e^{2 \hat{\mu}b(\nu)} +M(\mu) \leq C M(\nu), $$

for some generic positive constant $C$. If $b'(\nu) <0$, we get the upper estimate $M(\nu)$. Summing up, we have, for $t(\nu) \in I$,

$$ M^{\prime}(\nu) \leq C M(\nu). $$

This is obviously sufficient for $M(t) \leq C$, $t \in[r_{n},T]$, uniformly in $n$. By the formula for $M(t)$, the same holds true for $b(t)$. □

Proof of Lemma 6.10, part c

Let $(t_{n})$ be a decreasing sequence in the interval $(T-\eta,T]$ such that $b'(t_{n})$ is a maximizer of $b'$ in the interval $[t_{n},T]$. If the sequence $(b'(t_{n}))$ is bounded, there is nothing to show. So let us assume that $b'(t_{n})$ becomes large with increasing $n$. We fix a large $n$ and define $\overline{b}:= b'(t_{n})$. The goal is to construct an upper bound for $\overline{b}$ independent of $n$.

Let $t_{0}:= t_{n}+\frac{1}{\sqrt{\overline{b}}}$. We further define $w$ as the solution of the boundary value problem

$$\begin{aligned} \left\{ \textstyle\begin{array}{l} w(t_{0},x)=-\beta e^{-\beta t_{0}},\quad x \in(-\infty,-b(t_{0})];\\ w(t_{0},x)=\max\{V_{xt}(t_{0},x),-\beta e^{-\beta t_{0}}\},\quad x \in (-b(t_{0}),b(t_{0})];\\ w(t_{0},x)=-\beta e^{-\beta t_{0}},\quad x \in(b(t_{0}),x_{0}];\\ w(t,x)=-\beta e^{-\beta t},\quad x_{0}-x=\overline{b}(t_{0}-t),t \in [t_{n},t_{0}];\\ w_{t}+\mu w_{x}+\frac{\sigma^{2}}{2}w_{xx}=0,\quad x_{0}-x > \overline {b}(t_{0}-t),t \in (t_{n},t_{0}), \end{array}\displaystyle \right. \end{aligned}$$

where $x_{0}$ solves $\frac{-b(t_{n})+x_{0}}{t_{0}-t_{n}}=\overline{b}$. So we have a boundary value problem in a wedge with small ($\overline{b}$ is large) angle. The maximum principle for $w$ implies

$$\begin{aligned} w\big(t,-b(t)\big) \geq&-\beta e^{-\beta t},\\ w\big(t,b(t)\big) \geq&-\beta e^{-\beta t} \end{aligned}$$

for $t \in[t_{n},t_{0}]$. The maximum principle applied for the function $w-V_{xt}$ on the set $G_{0}:= \{ (t,x)| -b(t) \leq x \leq b(t),t_{n} \leq t\leq t_{0} \}$ gives $w-V_{xt} \geq0$ on $G_{0}$ (for the boundary values of $V_{xt}$ see [15, Eq. (51)]). Using the fact that $w$ and $V_{xt}$ agree at $(t_{n},b(t_{n}))$, we conclude $\left(w_{x}-V_{xtx} \right)(t_{n},b(t_{n})) \leq 0$, or

$$\begin{aligned} w_{x}\big(t_{n},b(t_{n})\big) \leq V_{xtx}\big(t_{n},b(t_{n})\big). \end{aligned}$$

(9.45)

Now we split $w$ into three functions

$$\begin{aligned} w=w^{(1)}+w^{(2)}+w^{(3)} \end{aligned}$$

(9.46)

and define $w^{(1)}:= -\beta e^{-\beta t_{0}}$. For $w^{(2)}$ and $w^{(3)}$, we define proper BVPs such that (9.46) holds. The function $w^{(2)}$ is defined as the solution of the BVP

$$\begin{aligned} \textstyle\begin{array}{l} w^{(2)}(t_{0},x)=0,\quad x \in(-\infty,x_{0}];\\ w^{(2)}(t,x)=-\beta e^{-\beta t}+\beta e^{-\beta t_{0}},\quad x_{0}-x=\overline{b} (t_{0}-t),t \in[t_{n},t_{0}];\\ w^{(2)}_{t}+\mu w^{(2)}_{x}+\dfrac{\sigma^{2}}{2}w^{(2)}_{xx}=0,\quad x_{0}-x > \overline{b} (t_{0}-t),t \in(t_{n},t_{0}). \end{array}\displaystyle \end{aligned}$$

The proof of

$$\begin{aligned} w^{(2)}_{x}\big(t_{n},b(t_{n})\big) \geq -c\sqrt{\overline{b}} \end{aligned}$$

(9.47)

is exactly as in [15, Lemma 5.3], and we are left with the problem for $w^{(3)}$, where the new terminal condition enters. We define $w^{(3)}$ as solution of the BVP

$$\begin{aligned} \textstyle\begin{array}{l} w^{(3)}(t_{0},x)=0,\quad x \in(-\infty,-b(t_{0})];\\ w^{(3)}(t_{0},x)=\max\{V_{xt}(t_{0},x),-\beta e^{-\beta t_{0}}\} +\beta e^{-\beta t_{0}},\quad x \in(-b(t_{0}),b(t_{0})];\\ w^{(3)}(t_{0},x)=0,\quad x \in(b(t_{0}),x_{0}];\\ w^{(3)}(t,x)=0,\quad x_{0}-x=\overline{b}(t_{0}-t),t \in[t_{n},t_{0}];\\ w^{(3)}_{t}+\mu w^{(3)}_{x}+\dfrac{\sigma^{2}}{2}w^{(3)}_{xx}=0,\quad x_{0}-x > \overline{b} (t_{0}-t),t \in(t_{n},t_{0}). \end{array}\displaystyle \end{aligned}$$

Let now

$$ M > \max\Big\{ \Vert b\Vert_{C((T-\eta,T])},\sup_{x \in [-b(t_{0}),b(t_{0})]} \big( \max\{V_{xt}(t_{0},x),-\beta e^{-\beta t_{0}}\}+\beta e^{-\beta t_{0}}\big) \Big\} . $$

(9.48)

This constant depends only on the quantities mentioned in the formulation of the lemma. (We show in the next section, Lemma 7.3(a), that for a special sequence $\epsilon_{k} \to0$, $M$ depends only on the model parameters.) Define now $w^{(4)}$ as the solution of

$$\begin{aligned} \textstyle\begin{array}{l} w^{(4)}(t_{0},x) =M \mathbf{1}_{[-M,M]}(x),\quad x \in(-\infty,x_{0}];\\ w^{(4)}(t,x)=0,\quad x_{0}-x=\overline{b}(t_{0}-t),t \in[t_{n},t_{0}];\\ w^{(4)}_{t}+\mu w^{(4)}_{x}+\dfrac{\sigma^{2}}{2}w^{(4)}_{xx}=0,\quad x_{0}-x > \overline{b} (t_{0}-t),t \in(t_{n},t_{0}). \end{array}\displaystyle \end{aligned}$$

The maximum principle implies $w^{(4)}\geq w^{(3)}$, and as

$$w^{(4)}\bigl(t_{n},b(t_{n})\bigr) =w^{(3)}\bigl(t_{n},b(t_{n})\bigr), $$

one concludes that

$$ w^{(4)}_{x}\big(t_{n},b(t_{n})\big) \leq w^{(3)}_{x}\big(t_{n},b(t_{n})\big). $$

(9.49)

Introducing the new coordinates

$$\begin{aligned} z=-(x-x_{0})-\overline{b}(t_{0}-t),\qquad \tau=t_{0}-t, \end{aligned}$$

and the new dependent variable $u:= w^{(4)}e^{\frac{\tilde{b}}{\sigma ^{2}}z+\frac{\tilde{b}^{2}}{2\sigma^{2}}\tau}$, where $\tilde{b}:= \overline{b}-\mu$, gives for $u(\tau,z)$ the system

$$\begin{aligned} \textstyle\begin{array}{l} u(\tau=0,z)=M e^{\frac{\tilde{b}}{\sigma^{2}}z} \mathbf{1} _{[x_{0}-M,x_{0}+M]}(z),\quad z \in[0,\infty);\\ u(\tau,z=0)=0,\quad \tau\in(0,\tau_{n});\\ u_{\tau}=\dfrac{\sigma^{2}}{2}u_{zz},\quad z>0,\tau\in(0,\tau_{n}), \end{array}\displaystyle \end{aligned}$$

which has the solution $u(\tau,z)= \frac{M}{\sqrt{2\pi\tau}\sigma } \int _{x_{0}-M}^{x_{0}+M} e^{\frac{\tilde{b}}{\sigma^{2}}z} (e^{-\frac{(z-\xi)^{2}}{2\sigma^{2} \tau}}- e^{-\frac{(z+\xi)^{2}}{2\sigma^{2} \tau}})\, d\xi$. By elementary calculation, one gets the value of $u_{z}(\tau,0)$, and transforming back to the original variables gives

$$w^{(4)}_{x}\big(t_{n},b(t_{n})\big)=-\sqrt{\frac{2}{\pi}}\frac {M\overline{b}^{1/4}}{\sigma } \bigg(e^{-\frac{(x_{0}-M)^{2}\sqrt{\overline{b}}}{2\sigma^{2}}}- e^{-\frac{(x_{0}+M)^{2}\sqrt{\overline{b}}}{2\sigma^{2}}}\bigg)e^{-\frac {\tilde {b}^{2}}{2\sigma^{2}\sqrt{\overline{b}}}}, $$

hence

$$\begin{aligned} w^{(4)}_{x}\big(t_{n},b(t_{n})\big)\geq-c \end{aligned}$$

for a generic positive constant $c$, depending on the model parameters and $M$. We conclude by (9.49) and (9.47)–(9.45) that $V_{xtx}(t_{n},b(t_{n})) \geq-c\sqrt{\overline{b}}-c$ holds, i.e., by the boundary condition for $V_{xtx}$,

$$ -2 \hat{\beta}e^{-\beta t_{n}} \overline{b}\geq -c\sqrt{\overline{b}}-c, $$

which implies immediately $\overline{b}\leq c$. □

Proof of Lemma 6.10, part d

The proof works analogously to the proof of Lemma 5.4 in [15]. Indeed, instead of the integral equation (11) there, we work with our new integral equation (6.2), i.e., we have to replace $B', D$ and $h$ by $\tilde{B}', \tilde{D}$ and $\tilde{h}$. By this replacement, we generate a new term in (77), stemming from $\frac{d}{dt}\left(-F_{xxx}(t,b(t))\right)$. But the new term can be incorporated in $-c+cg'(T-s)$ there (with an $\epsilon$-dependent constant $c$). The replacement of $D$ by $\tilde{D}$ in (78) leads to a term which can be incorporated in $cb(T-s)+c$ there. Hence, (79) and (80) are true as well. (81) can be dealt with as (78) before, and finally, the replacement of $h$ by $\tilde{h}$ does not generate alterations at all. □

1.5 9.5 Auxiliary results for Sect. 7

Proof of Lemma 7.3 a

Checking the proof of Lemma 6.10(c), one sees that we have to check only that the constant $M$ defined in (9.48) is independent of $k$. Sufficient for this is $\sup_{x \in[-b^{(k)}(t_{0}),b^{(k)}(t_{0})]} V^{(k)}_{xt}(t,x) \leq C$, for a generic $k$-independent constant $C$ depending on $t_{0}$ and the model parameters. By $\mathcal{L}V^{(k)}_{x}=0$, this is equivalent to $\inf_{x \in[-b^{(k)} (t_{0}),b^{(k)}(t_{0})]} V^{(k)}_{xxx}(t,x) \geq-C$ or, after transforming to the $v$-variable, $\inf_{x \in[-b^{(k)}(t_{0}),b^{(k)}(t_{0})]} v^{(k)}_{xxx}(t,x) \geq-C$. By definition we have $v^{(k)}=: \tilde{v}^{(k)}+F(t,x)$, which implies that it is sufficient that

$$ \inf_{x \in[-b^{(k)}(t_{0}),b^{(k)}(t_{0})]} \tilde{v}^{(k)}_{xxx}(t,x) \geq-C, $$

(9.50)

for a generic $k$-independent constant $C$ depending on $t_{0}$ and the model parameters. We recall (see Lemma 9.7) that $\tilde{v}^{(k)}_{xxx}$ solves the system

$$\begin{aligned} \tilde{v}^{(k)}_{xxx}(T,x) =& \tilde{h}^{(k)\prime\prime\prime}(x),\quad x \in[-\epsilon _{k},\epsilon_{k}]; \\ \hat{\mathcal{L}}\tilde{v}^{(k)}_{xxx} =& 0,\quad (t,x) \in G_{1}^{(k)}; \\ \tilde{v}^{(k)}_{xxx}\big(t, \pm b^{(k)}(t)\mp\big) =& e^{-\lambda t +\hat{\mu}b^{(k)}(t)}\Big(2 \hat{\beta}+3\hat{\mu }^{2}+\hat{\mu}^{3}\big(b^{(k)}(t)+d^{(k)} (t)\big)\Big) \\ & -F_{xxx}\big(t,b^{(k)}(t)\big)\\ =:& \tilde{B}^{(k)}(t). \end{aligned}$$

We now assert the following:

Claim 1: $\Vert\tilde{B}^{(k)}(t)\Vert_{L^{p}(0,T)} \leq C$ , uniformly in $k$ , for all $1 \leq p < \infty$. For the first term of $\tilde{B}^{(k)}$, this is obvious since the $b^{(k)}$ are uniformly bounded, and for $F_{xxx}(t,b^{(k)}(t))$, this follows from the explicit formula

$$F_{xxx}(t,x)=\frac{C}{(T-t)^{5/2}}\big(x^{2}-\sigma^{2}(T-t)\big)e^{-\frac {x^{2}}{2\sigma^{2}(T-t)}} $$

and

Claim 2: Let $0 <\delta_{0} <\sqrt{3}$ be an arbitrary fixed number. Then there exists $\Delta>0$ independent of $k$ such that for $t\in[T-\Delta,T]$ ,

$$ \inf_{k} \frac{b^{(k)}(t)}{\sigma\sqrt{-(T-t)\ln(T-t)}} \geq\delta_{0}. $$

We prove Claim 2 by contradiction. So let $f^{(k)}(\nu):= \frac {b^{(k)}(\nu )}{g(\nu)}$ with $\nu:= T-t$ and $g(\nu):= \sigma\sqrt{-\nu\ln{\nu}}$. Furthermore, let $\underline{f}(\nu):= \inf_{k} f^{(k)}(\nu)$ and assume that there exists a sequence $\nu_{\ell}\to0$ such that $\underline{f}(\nu_{\ell}) < \delta_{0}$ for all $\ell\in\mathbf{N}$. Hence there exist numbers $K(\ell)$ such that

$$ f^{(k)}(\nu_{\ell}) \leq\delta_{1} $$

for some $\delta_{0} < \delta_{1} < \sqrt{3}$ and for all $k \geq K(\ell)$.

Note that we use alternatingly in the following the physical time $t$ and the backward running time $\nu$, which is a slight abuse of notation. Since the barrier strategy with barrier function $b^{(k)}$ is the optimal strategy, using the notation $V_{k}$ for the value function with terminal gain $H^{(k)}$ and using $\hat{H}^{(k)}(x):= H^{(k)}(x)-\bar {\gamma}$ gives for all $k \geq K(\ell)$

$$\begin{aligned} V_{k}\big(t_{\ell}, \sqrt{3}g(\nu_{\ell})\big) =& \mathbf{E}\bigg[ \int_{t_{\ell}+}^{\tau^{(k)}_{\ell}} e^{-\beta s}\,dC^{(k)}_{s} +e^{-\beta\tau ^{(k)}_{\ell}}\hat{H}^{(k)}\big(X_{\tau^{(k)}_{\ell}}\big) +\gamma\mathbf{1}_{\{\tau^{(k)}_{\ell}=T\}} \bigg] \\ &{}+ e^{-\beta t_{\ell}}\big(\sqrt{3}g(\nu_{\ell})-b^{(k)}(\nu_{\ell})\big) \\ =& \frac{\mu}{\beta}\big(e^{-\beta t_{\ell}}-\mathbf{E}\big[e^{-\beta \tau^{(k)}_{\ell}}\big]\big) \\ &{}- \mathbf{E}\bigg[ e^{-\beta s }X_{s}\bigg|^{s=\tau^{(k)}_{\ell}}_{s=t_{\ell}+}+ \beta\int _{t_{\ell}+}^{\tau^{(k)}_{\ell}} X_{s} e^{-\beta s}\,ds \bigg] \\ &{}+ \mathbf{E}\big[e^{-\beta\tau^{(k)}_{\ell}}\hat{H}^{(k)}\big(X_{\tau ^{(k)}_{\ell}}\big)\big]+ \gamma\mathbf{P}[\tau^{(k)}_{\ell}=T] \\ &{}+e^{-\beta t_{\ell}}\big(\sqrt{3}g(\nu_{\ell})-b^{(k)}(\nu_{\ell})\big) \\ \leq& \frac{\mu}{\beta}\big(e^{-\beta t_{\ell}}-e^{-\beta T}\big)- \mathbf{E}\big[e^{-\beta\tau^{(k)}_{\ell}}X_{\tau^{(k)}_{\ell}}\big] \\ &{}+ \mathbf{E}\big[e^{-\beta\tau^{(k)}_{\ell}}\hat{H}^{(k)}\big(X_{\tau ^{(k)}_{\ell}}\big)\big]+ \gamma\mathbf{P}[\tau^{(k)}_{\ell}=T]+e^{-\beta t_{\ell}}\sqrt{3}g(\nu _{\ell}) \\ \leq& e^{-\beta T}\big(\mu\nu_{\ell}+O(\nu_{\ell}^{2})\big)+2 \epsilon _{k}+\gamma \mathbf{P}[\tau^{(k)}_{\ell}=T]+e^{-\beta t_{\ell}}\sqrt{3}g(\nu_{\ell}) \end{aligned}$$

(9.51)

Here we have used $dX_{s}=\mu ds+\sigma dW_{s} -dC^{(k)}_{s}$ and integration by parts in the first equality, $\tau^{(k)}_{\ell}\leq T$ and the non-negativity of $X$ in the first inequality, and finally $\hat{H}^{(k)}(X_{\tau^{(k)}_{\ell}})-X_{\tau^{(k)}_{\ell}} \leq 2 \epsilon_{k}$ in the last inequality. Note that the $O$ above is uniform in $k$. We are now looking for an upper estimate of $\mathbf{P}[\tau^{(k)}_{\ell}=T]$ and denote by $\hat{\tau}^{(k)}_{\ell}$ the ruin time if we do not consume anything after the lump sum at $t_{\ell}$. Clearly, $\mathbf{P}[\tau^{(k)}_{\ell}=T] \leq\mathbf{P}[\hat{\tau}^{(k)}_{\ell}=T]$, and we estimate the latter probability by

$$\begin{aligned} \mathbf{P}[\hat{\tau}^{(k)}_{\ell}=T] =&\mathbf{P}\Big[ \inf_{0 \leq s \leq\nu_{\ell}}(\mu s + \sigma W_{s} )\geq -b^{(k)}(\nu_{\ell})\Big] \\ \leq& \mbox{erf}\big(r(\nu_{\ell})\big)\leq1-\frac{e^{-r^{2}(\nu_{\ell})}}{2\sqrt {\pi}r(\nu_{\ell})}, \end{aligned}$$

(9.52)

with $r(\nu_{\ell}):= \frac{\mu}{\sigma}\sqrt{\frac{\nu_{\ell}}{2}}+\delta_{1} \frac{g(\nu_{\ell})}{\sigma\sqrt{2\nu_{\ell}}}$. Here we used for the first inequality standard results (see, e.g., (2.8.3) in [19]), and the second holds for $r(\nu_{\ell}) \geq1/2$, which we assume without loss of generality. Noting that $r(\nu_{\ell}) \leq\delta_{2} \sqrt{\frac{-\ln\nu_{\ell}}{2}}$ for some $\delta_{2}$ with $\delta_{0} < \delta_{1} < \delta_{2} <\sqrt{3}$, we end up, using (9.51) and (9.52), with

$$ V_{k}\big(t_{\ell}, \sqrt{3}g(\nu_{\ell})\big)\leq e^{-\beta T}\big(\mu\nu_{\ell}+O(\nu_{\ell}^{2})\big)+2 \epsilon _{k}+e^{-\beta t_{\ell}}\sqrt{3}g(\nu_{\ell}) +\gamma-C\nu_{\ell}^{\delta_{3}} $$

(9.53)

for a constant $\delta_{3} < \frac{3}{2}$ arbitrarily close to $3/2$, for all $\ell$ and for all $k \geq K(\ell)$, with the $O$ uniform in $k$.

We now consider a barrier strategy with barrier $\sqrt{3}g(\nu)$ for $0 \leq\nu\leq\nu_{\ell}$ and denote it by $C^{\sqrt{3}}$. We get for the target functional, using integration by parts as before,

$$\begin{aligned} J\big(t_{\ell},\sqrt{3}g(\nu_{\ell}),C^{\sqrt{3}}\big) =& \frac{\mu}{\beta}(e^{-\beta t_{\ell}}-\mathbf{E}[e^{-\beta\tau_{\ell}}]) +e^{-\beta t_{\ell}}\sqrt{3}g(\nu_{\ell}) \\ &{}- \mathbf{E}\left[\beta\int_{t_{\ell}}^{\tau_{\ell}} X_{s} e^{-\beta s}\, ds \right] +\gamma\mathbf{P}[\tau_{\ell}=T]. \end{aligned}$$

(9.54)

For the first term, we use $-\mathbf{E}[e^{-\beta\tau_{\ell}}] \geq -e^{-\beta T} \mathbf{P}[\tau_{\ell}=T]-e^{-\beta t_{\ell}} \mathbf{P}[\tau_{\ell}< T]$ and arrive at

$$ \frac{\mu}{\beta}(e^{-\beta t_{\ell}}-\mathbf{E}[e^{-\beta \tau_{\ell}}])\geq e^{-\beta T}\big( \mu\nu_{\ell}+O(\nu_{\ell}^{2})-C\nu_{\ell}\mathbf{P}[\tau _{\ell}< T]\big) $$

for some generic positive constant $C$. We now use from Lemma 9.9 below the upper estimate $\mathbf{P}[\tau_{\ell}< T] \leq C \nu_{\ell}^{\frac {3}{2}}\sqrt{-\ln\nu_{\ell}}$ for $\ell$ large enough, say $\ell\geq L$. All together, and observing that in the last integral of (9.54), $X_{s} \leq\sqrt{3}g(\nu)$ holds, we conclude that

$$ J\big(t_{\ell},\sqrt{3}g(\nu_{\ell}),C^{\sqrt {3}}\big) \geq e^{-\beta T} \mu\nu_{\ell}-C \nu_{\ell}^{\frac{3}{2}}\sqrt{-\ln\nu_{\ell}} +e^{-\beta t_{\ell}}\sqrt{3}g(\nu_{\ell})+\gamma $$

for all $\ell\geq L$. Since the target functional should be less than or equal to the value function, this gives via (9.53)

$$0 \leq2 \epsilon_{k} -C \nu_{\ell}^{\delta_{3}} $$

for all $\ell\geq\bar{L}$ and $k \geq K(\ell)$. Letting $k$ tend to infinity, this is an obvious contradiction, proving Claim 2, and therefore also Claim 1.

One checks that $\Vert\tilde{h}^{(k)\prime\prime\prime}(x)\Vert _{L^{1}(-\epsilon_{k},\epsilon_{k})}$ is uniformly bounded, and the fact that $\tilde{B}^{(k)}$ is uniformly bounded above implies that $\Vert\tilde{v}^{(k)}_{xxx+}\Vert_{L^{1}(-b^{(k)}(t),b^{(k)}(t))}$ is uniformly bounded in $t \in[0,T]$ and $k$. Hence, in order to show

Claim 3: $\Vert\tilde{v}^{(k)}_{xxx}\Vert _{L^{1}(-b^{(k)}(t),b^{(k)}(t))}$ is uniformly bounded in $t \in[0,T]$ and $k$ , it suffices to show that $\int_{t}^{T} (\frac{d}{ds} \int_{-b^{(k)}(s)}^{b^{(k)} (s)} \tilde{v}^{(k)}_{xxx}(s,x)\,dx)\,ds$ is uniformly bounded in $t \in[0,T]$ and $k$. We calculate

$$\begin{aligned} \frac{d}{ds} \int_{-b^{(k)}(s)}^{b^{(k)}(s)} \tilde {v}^{(k)}_{xxx}(s,x)\,dx &=\frac{d}{ds} \Big( \tilde{v}^{(k)}_{xx}\big(s,b^{(k)}(s)\big) - \tilde{v}^{(k)}_{xx}\big(s,-b^{(k)}(s)\big)\Big) \\ &=2 \tilde{A}^{(k)\prime}(s)=-\sigma^{2} \tilde{D}^{(k)}(s)+2 b^{(k)\prime}(s) \tilde{B}^{(k)}(s) \\ &=:-\sigma^{2} \Big( D^{(k)}(s)-F_{xxxx}\big(s,b^{(k)}(s)\big)\Big) \\ &\phantom{=:}{} +2 b^{(k)\prime}(s) \Big( B^{(k)}(s) -F_{xxx}\big(s,b^{(k)}(s)\big)\Big) \\ &=\big(-\sigma^{2} D^{(k)}(s)+2 b^{(k)\prime}(s)B^{(k)}(s)\big) \\ &\phantom{=:}{} +\Big( \sigma^{2} F_{xxxx}\big(s,b^{(k)}(s)\big)- 2b^{(k)\prime} (s)F_{xxx}\big(s,b^{(k)}(s)\big)\Big) \\ &=: J_{1}(s)+J_{2}(s), \end{aligned}$$

(9.55)

where the first equality holds because $\tilde{v}^{(k)}\in C(\overline {G_{1}^{(k)}})$, the second follows from the first equation in (9.41), the third follows from (9.41), the fact that $F_{xx}$ solves the backward heat equation (away from $(T,0)$) and $d^{(k)\prime}(s) = \beta( b^{(k)}(s)+d^{(k)}(s) ) -\mu$, which is itself implied by (9.25). Finally, the fourth equality defines $B^{(k)}$ (resp., $D^{(k)}$) as the boundary values of $\tilde{v}^{(k)}_{xxx}$ (resp., $\tilde{v}^{(k)}_{xxxx}$), but without the $F$-term.

Using now the backward heat equation for $F_{xx}$, we get for the time derivative that $\frac {d}{ds}(2F_{xx}(s,b^{(k)}(s)))=-J_{2}(s)$. Hence, we can estimate, with a generic constant $C$,

$$ \left|\int_{t}^{T} J_{2}(s) \, ds \right| = \big| 2 F_{xx}\big(T,b^{(k)}(T)\big)-2 F_{xx}\big(t,b^{(k)}(t)\big) \big| \leq C $$

because of Claim 2 and the explicit formula for $F_{xx}(t,x)=Ce^{-\frac{x^{2}}{2\sigma^{2}(T-t)}}\frac{x}{(T-t)^{3/2}}$.

Concerning $J_{1}$, one calculates

$$\begin{aligned} &-\sigma^{2} \tilde{D}^{(k)}(s)+2 b^{(k)\prime}(s)\tilde {B}^{(k)}(s) \\ &= e^{-\lambda s +\hat{\mu}b^{(k)}(s)}\Big( -C-C\big(b^{(k)}(s)+d^{(k)}(s)\big) \\ &\phantom{=}{}+ Cb^{(k)\prime}(s)+ Cb^{(k)\prime}(s) b^{(k)}(s)+Cb^{(k)\prime}d^{(k)}(s)\Big) \\ &=: J_{3}(s)+ J_{4}(s)+J_{5}(s)+J_{6}(s)+J_{7}(s) \end{aligned}$$

with some generic positive constants $C$. We claim that

$$ \left| \int_{t}^{T} J_{i}(s)\, ds \right| \leq C $$

(9.56)

for some positive constant $C$, for $i=3,4,\dots,7$, uniformly in $t \in[0,T]$ and $k$. We show this only for $J_{6}$, because the other cases work analogously, or are even simpler. One calculates

$$\begin{aligned} e^{-\lambda s +\hat{\mu}b^{(k)}(s)}b^{(k)\prime}(s) b^{(k)}(s) =& C \frac{d}{ds}\big(e^{-\lambda s +\hat{\mu}b^{(k)}(s)}b^{(k)}(s)\big) \\ &{}+Ce^{-\lambda s +\hat{\mu}b^{(k)}(s)}b^{(k)}(s) -Ce^{-\lambda s +\hat{\mu}b^{(k)}(s)}b^{(k)\prime}(s). \end{aligned}$$

From this, (9.56) for $i=6$ is immediate. (For the estimate of the last term on the right-hand side, one can use the result for $J_{5}$, which can obviously be proved without using the case $i=6$). Equation (9.55) and bound (9.56) provide Claim 3.

Let now $t_{0}$ be given with $0 \leq t_{0} < T$. Set $t_{1}:= t_{0}+ \frac {T-t_{0}}{2}$ and consider the BVP for $\tilde{v}^{(k)}_{xxx}$ on the time interval $[0,t_{1}]$. By Claim 2, one has obviously $\tilde{B}^{(k)}(t) \geq-C(t_{0})$ on $[0,t_{1}]$, with the constant independent of $k$. Moreover, Claim 3 provides the estimate $\Vert\tilde{v}^{(k)} _{xxx}(t_{1},x)\Vert_{L^{1}(-b^{(k)}(t_{1}),b^{(k)}(t_{1}))} \leq C$. Both together give $\tilde{v}^{(k)}_{xxx}(t,x) \geq-C(t_{0})$, for $t\in[0,t_{0}]$ and $x \in[-b^{(k)}(t),b^{(k)}(t)]$, with the constant independent of $k$, which concludes our proof (see (9.50)). □

Lemma 9.9

Let $b(t):= \sigma\sqrt{-3 (T-t) \ln(T-t)}$, $t>T-1$, and let $X_{s}$, $s \in[t,T]$, be the process $x+\mu(s-t)+\sigma(W_{s}-W_{t})$ reflected on the barrier $b(t)$. Then we have

$$ \mathbf{P}\Big[ \inf_{s \in[t,T]}X_{s} \geq0 \Big| X_{t}=b(t) \Big] \leq K(t) \sim1 - C \frac{(T-t)^{\frac{3}{2}}}{\sqrt{-\ln(T-t)}} $$

(9.57)

and

$$ \mathbf{P}\Big[ \inf_{s \in[t,T]}X_{s} \geq0 \Big| X_{t}=b(t) \Big] \geq L(t) \sim1 - C (T-t)^{\frac{3}{2}}\sqrt{-\ln(T-t)}, $$

(9.58)

for some positive constant $C$ and for $t\to T$.

Proof

In the following, we use the abbreviation $\nu=T-t$ for the time to maturity $T$ and $\mathbf{P}_{t}[\,\cdot\,]$ as shortcut for $\mathbf{P} [\,\cdot\,| X_{t}=b(t)]$. We start with the proof of (9.57).

One simply estimates the survival probability of our reflected process $X$ by the survival probability of an agent who consumes nothing, which is evidently larger. So

$$\begin{aligned} \mathbf{P}_{t} \Big[ \inf_{s \in[t,T]}X_{s} \geq0 \Big] \leq& \mathbf{P}\Big[\inf_{s \in[0,\nu]} \sigma\sqrt{-3 \nu\ln\nu} +\mu s+ \sigma W_{s} \geq0 \Big] \\ \leq& 1- \mathbf{P}\Big[\sup_{s \in[0,\nu]}W_{s} \geq\sqrt{-3 \nu\ln\nu }+\frac {\mu}{\sigma} \nu\Big] \\ =& 1-\sqrt{\frac{2}{\pi}}\int_{\frac{\mu}{\sigma}\sqrt{\nu}+\sqrt {-3\ln\nu }}^{\infty}e^{-\frac{z^{2}}{2}} \,dz \\ \sim& 1 - \sqrt{\frac{2}{\pi}} \frac{\nu^{\frac{3}{2}}}{\sqrt{-3\ln \nu}} \end{aligned}$$

for $\nu\to0$. Here we have used in the last equality a well-known formula for the distribution of the supremum of Brownian motion (see, e.g., (2.8.4) in [19]), and for the last asymptotic equivalence the asymptotic behavior of the standard normal distribution.

In order to prove (9.58), we first note that setting $\mu =0$ makes the probability in question certainly smaller; so we stick in the proof to this case. Let $\hat{X}$ be our process $X$, but not stopped at the ruin time $\tau$. Hence, $\hat{X}_{T} {\in} (-\infty,0]$. Define $D {:=} \{ \tau {<} T \}$ and consider on the set $D$ the stopping time $\tau_{1} {:=} \inf\{ t>\tau| \hat{X}_{t} =\pm b(t)\}$. Set

$$D_{1}:= \{ \hat{X}_{\tau_{1}} =b(\tau_{1}) \}, \quad D_{2}:= \{ \hat {X}_{\tau_{1}} =-b(\tau_{1}) \}, $$

which implies by the obvious symmetry $\mathbf{P}_{t}[D_{1}|D]=\mathbf{P}_{t}[D_{2}|D]=1/2$. Since the process $\hat {X}_{t}$ is not reflected on the lower branch $-b(t)$, it is clear that the probability $\mathbf{P}_{t} [\hat{X}_{T} <0 | \hat{X}_{\tau _{1}}=-b(\tau_{1})]$ is very high by the proof of (9.57), and in any case we have

$$\mathbf{P}_{t} [\hat{X}_{T} < 0 \big| \hat{X}_{\tau_{1}}=-b(\tau_{1})]>\frac{1}{2} $$

for $\nu$ small enough, which implies

$$ \mathbf{P}_{t} [\hat{X}_{T} < 0 ] \geq\mathbf{P}_{t} [\hat{X}_{T} < 0 | D_{2} ]\mathbf{P}_{t}[D_{2}] >\frac {\mathbf{P}_{t}[D]}{4}. $$

(9.59)

The process $\hat{X}$ is considered in [7], and if one has $\mathbf{P} _{t}[\hat{X}_{T}=0]>0$, the authors call this phenomenon a heat atom. Moreover, they show that this occurs if and only if the reflecting boundary is a so-called upper function of Brownian motion. In the second part of the proof of their Theorem 2.2, the authors show (in our notation) that $\mathbf{P}_{t}[\hat{X}_{T}=b(T)] \geq\mathbf{P}[A]>0$, where the set $A$ is defined as

$$\begin{aligned} A :=& \{ b(t)+\sigma(W_{T}-W_{t})>b(T)=0 \} \\ &{}\cap\{ \sigma W_{T-s}-\sigma W_{T} < b(T-s)-b(T), \text{ for all } s \in(0,T-t) \} \\ =:& A_{1} \cap A_{2}. \end{aligned}$$

Now, we obviously have $\mathbf{P}[A_{1}] \sim1-C\frac{\nu^{3/2}}{\sqrt {-\ln \nu}}$ for $\nu\to0$, and by time reversal and the classical considerations concerning the Kolmogorov test as in [18, Sect. 1.8], we get $\mathbf{P}[A_{2}] \geq1-\int_{0}^{\nu}\frac{b(s)}{\sigma\sqrt{2\pi s^{3}}}e^{-\frac{b(s)^{2}}{2\sigma^{2} s}}\, ds \sim1-C(-\ln\nu)\nu^{3/2}$. Therefore, we obtain $\mathbf{P}[A] \geq L_{1}(\nu) \sim1-C(-\ln \nu)\nu ^{3/2}$, and finally, using (9.59),

$$\begin{aligned} \mathbf{P}_{t}\Big[ \inf_{s \in[t,T]}X_{s} \geq0 \Big] =&1-\mathbf{P}_{t}[D] \geq1-4\mathbf{P}_{t}[\hat{X}_{T}< 0] \\ \geq& -3+4\mathbf{P}[A] \geq L(\nu) \sim 1 - C \nu^{\frac{3}{2}}\sqrt{-\ln\nu}, \end{aligned}$$

which concludes the proof. □