Appendix A: Some technical proofs of results from Sect. 2
1.1 Proofs for Sect. 2.1
Proof of Proposition 2.1
Let S(x, t) denote the heat kernel, given by
$$\begin{aligned} S(x,t) := (2\sigma ^2 \pi t)^{-1/2}\exp \left\{ -\frac{x^2}{2\sigma ^2 t}\right\} . \end{aligned}$$
(A.1)
Using Duhamel’s principle and integration by parts, we have \(\mu = \mu _1 + \mu _2\), where
$$\begin{aligned} \begin{aligned} \mu _1(x,t)&=\int _0^t\int _0^\infty \big (S(x-y,t-s)-S(x+y,t-s)\big )\big ((b\mu )_y(y,s)+\nu _y(y,s)\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s\\&=\int _0^t\int _0^\infty \big (S_x(x-y,t-s)+S_x(x+y,t-s)\big )\big ((b\mu )(y,s)+\nu (y,s)\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \end{aligned} \end{aligned}$$
(A.2)
and
$$\begin{aligned} \mu _2(x,t) = \int _0^\infty \left( S(x-y,t) - S(x+y,t)\right) \mu _0(y){{\,\mathrm{d \!}\,}}y. \end{aligned}$$
(A.3)
Note that
$$\begin{aligned} \int _0^\infty \big |S_x(x,t)\big |{{\,\mathrm{d \!}\,}}x= & {} -\int _0^\infty S_x(x,t){{\,\mathrm{d \!}\,}}x=-\lim _{x\rightarrow \infty }\big (S(x,t)-S(0,t)\big )\nonumber \\= & {} (2\pi \sigma ^2 t)^{-1/2}. \end{aligned}$$
(A.4)
Call
$$\begin{aligned} c':=2(2\pi \sigma ^2)^{-1/2}, \end{aligned}$$
(A.5)
and let \(M>1\) to be determined later. Then
$$\begin{aligned} \begin{aligned}&e^{-Mt}\int _0^\infty |\mu _1(x,t)|{{\,\mathrm{d \!}\,}}x \\&\quad \le e^{-Mt}\int _0^t\int _0^\infty \int _0^\infty \big (|S_x(x-y,t-s)|\\&\quad +|S_x(x+y,t-s)|\big )\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|+|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \\&\quad \le c'e^{-Mt}\int _0^t(t-s)^{-1/2}\int _0^\infty \big (\left\| b\right\| _{L^\infty }|\mu (y,s)|+|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \\&\quad =c'\int _0^te^{-M(t-s)}(t-s)^{-1/2}\int _0^\infty e^{-Ms}\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|+|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s. \end{aligned} \end{aligned}$$
(A.6)
Here, the first inequality follows directly from (A.2); the second inequality follows from (A.4) and noting that \(\exp \big \{-y^2/(2\sigma ^2t)\big \}\le 1\) for all \(y\in (0,\infty )\). Define
$$\begin{aligned} B:=\sup _{0\le \tau \le T}e^{-M\tau }\int _0^\infty |\mu (x,\tau )|{{\,\mathrm{d \!}\,}}x. \end{aligned}$$
(A.7)
Then (A.6) implies
$$\begin{aligned} \begin{aligned} e^{-Mt}\int _0^\infty |\mu _1(x,t)|{{\,\mathrm{d \!}\,}}x&\le c'B\left\| b\right\| _{L^\infty }\int _0^te^{-M(t-s)}(t-s)^{-1/2}{{\,\mathrm{d \!}\,}}s\\&\quad +c'e^{-Mt}\left\| \nu \right\| _{L_t^\infty (L_x^1)}\int _0^t(t-s)^{-1/2}{{\,\mathrm{d \!}\,}}s \\&\le c'B\left\| b\right\| _{L^\infty }M^{-1/2}\sqrt{\pi }+2c'e^{-Mt}\left\| \nu \right\| _{L_t^\infty (L_x^1)}\sqrt{t}. \end{aligned} \end{aligned}$$
(A.8)
The second inequality follows from the substitutions and estimate below
$$\begin{aligned} \int _0^te^{-M(t-s)}(t-s)^{-1/2}{{\,\mathrm{d \!}\,}}s= & {} \int _0^te^{-Ms}s^{-1/2}{{\,\mathrm{d \!}\,}}s\nonumber \\= & {} M^{-1/2}\int _0^{Mt} e^{-s}s^{-1/2}{{\,\mathrm{d \!}\,}}s\le M^{-1/2}\sqrt{\pi }. \end{aligned}$$
(A.9)
Now, if \(M>1\), then \(2e^{-Mt}\sqrt{t}\le 1\) for all \(t>0\). Thus, (A.8) implies
$$\begin{aligned} e^{-Mt}\int _0^\infty |\mu _1(x,t)|{{\,\mathrm{d \!}\,}}x\le c'\left\| b\right\| _{L^\infty }\sqrt{\pi }M^{-1/2}B+c'\left\| \nu \right\| _{L_t^\infty (L_x^1)}. \end{aligned}$$
(A.10)
On the other hand, since \(S(\cdot ,t)\) is a probably density, we deduce
$$\begin{aligned} \int _0^\infty \left| \mu _2(x,t)\right| {{\,\mathrm{d \!}\,}}x \le \int _0^\infty \left| \mu _0(x)\right| {{\,\mathrm{d \!}\,}}x. \end{aligned}$$
(A.11)
Add \(e^{-Mt}\) times (A.11) to (A.10), take a supremum on both sides over all \(t\in [0,T]\), and we find that
$$\begin{aligned} B\le c'\left\| b\right\| _{L^\infty }\sqrt{\pi }M^{-1/2}B + \left\| \mu _0\right\| _{L^1} +c'\left\| \nu \right\| _{L_t^\infty (L_x^1)}. \end{aligned}$$
(A.12)
Choose \(M>1\) large enough so that \(M>(2c'\left\| b\right\| _{L^\infty }\sqrt{\pi })^2\). Consequently, we find that
$$\begin{aligned} B\le 2\left\| \mu _0\right\| _{L^1} + 2c'\left\| \nu \right\| _{L_t^\infty (L_x^1)}, \end{aligned}$$
(A.13)
which, given the definition of B in (A.7), establishes part (a).
For part (b) we argue similarly. First, note that
$$\begin{aligned} 2\int _0^\infty x|S_x(x,t)|{{\,\mathrm{d \!}\,}}x=-2\int _0^\infty xS_x(x,t){{\,\mathrm{d \!}\,}}x=2\int _0^\infty S(x,t){{\,\mathrm{d \!}\,}}t=1. \end{aligned}$$
(A.14)
Knowing this and using (A.4), we can estimate
$$\begin{aligned} \begin{aligned} \int _0^\infty x\big (|S_x(x-y,t-s)|&+|S(x+y,t-s)|\big ){{\,\mathrm{d \!}\,}}x \\&=\int _{-y}^\infty (x+y)\left| S_x(x,t-s)\right| {{\,\mathrm{d \!}\,}}x+\int _y^\infty (x-y)\left| S_x(x,t-s)\right| {{\,\mathrm{d \!}\,}}x \\&\le \int _{-\infty }^\infty (x+y)\left| S_x(x,t-s)\right| {{\,\mathrm{d \!}\,}}x+\int _0^\infty x\left| S_x(x,t-s)\right| {{\,\mathrm{d \!}\,}}x \\&=\frac{3}{2}+y\int _{-\infty }^\infty \left| S_x(x,t-s)\right| {{\,\mathrm{d \!}\,}}x \\&\le 2+c'y(t-s)^{-1/2}, \end{aligned} \end{aligned}$$
(A.15)
where \(c'\) is as in (A.5). Note that the second equality above follows from (A.14), and the last inequality from (A.4). We now use Fubini’s Theorem to estimate
$$\begin{aligned} \begin{aligned} \int _0^\infty&x|\mu _1(x,t)|{{\,\mathrm{d \!}\,}}x \\&\le \int _0^t\int _0^\infty \int _0^\infty x\big (|S_x(x-y,t-s)|+|S_x(x+y,t-s)|\big )\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|\\&\quad +|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \\&\le \int _0^t\int _0^\infty \big (2+c'y(t-s)^{-1/2}\big )\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|+|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s. \end{aligned} \end{aligned}$$
(A.16)
The second inequality follows directly from (A.15). Let \(M>1\) to be determined later and define
$$\begin{aligned} B':=\sup _{0\le t\le T}e^{-Mt}\int _0^\infty x|\mu (x,t)|{{\,\mathrm{d \!}\,}}x. \end{aligned}$$
(A.17)
We then estimate in a similar manner as in the proof of part (a)
$$\begin{aligned} \begin{aligned} e^{-Mt}\int _0^\infty x|\mu _2(x,t)|{{\,\mathrm{d \!}\,}}x&\le e^{-Mt}\int _0^t\int _0^\infty \big (2+c'y(t-s)^{-1/2}\big )\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|\\&\quad +|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \\&\le c'\int _0^te^{-M(t-s)}(t-s)^{-1/2}\int _0^\infty e^{-Ms}y\big (\left\| b\right\| _{L^\infty }|\mu (y,s)|\\&\quad +|\nu (y,s)|\big ){{\,\mathrm{d \!}\,}}y{{\,\mathrm{d \!}\,}}s \\&\quad +2te^{-Mt}\big (\left\| b\right\| _{L^\infty }\left\| \mu \right\| _{L_t^\infty (L_x^1)}+\left\| \nu \right\| _{L_t^\infty (L_x^1)}\big ). \end{aligned} \end{aligned}$$
(A.18)
In similar fashion, we estimate
$$\begin{aligned} \begin{aligned} \int _0^\infty x\left| \mu _2(x,t)\right| {{\,\mathrm{d \!}\,}}x&\le \int _0^\infty \int _{-y}^\infty (x+y)S(x,t) \left| \mu _0(y)\right| {{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}y\\&\quad + \int _0^\infty \int _{y}^\infty (x-y)S(x,t) \left| \mu _0(y)\right| {{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}y \\&\le 3\int _0^\infty xS(x,t){{\,\mathrm{d \!}\,}}x \int _0^\infty \left| \mu _0(y)\right| {{\,\mathrm{d \!}\,}}y + \int _0^\infty y\left| \mu _0(y)\right| {{\,\mathrm{d \!}\,}}y, \end{aligned} \end{aligned}$$
(A.19)
where we have also used that \(S(\cdot ,t)\) is an even function. We calculate (by a change of variables) that \(\int _0^\infty xS(x,t){{\,\mathrm{d \!}\,}}x = \sqrt{\frac{\sigma ^2 t}{2\pi }}\). Add together (A.18) and (A.19) to get
$$\begin{aligned}&e^{-Mt}\int _0^\infty x|\mu (x,t)|{{\,\mathrm{d \!}\,}}x \le c'\left\| b\right\| _{L^\infty }\sqrt{\pi }M^{-1/2}B'+ \left\| x\mu _0\right\| _{L^1} \nonumber \\&\quad + c'\left\| x\nu \right\| _{L_t^\infty (L_x^1)}+c_{b,\sigma }\left( \left\| \mu _0\right\| _{L^1} + \left\| \nu \right\| _{L_t^\infty (L_x^1)}\right) . \end{aligned}$$
(A.20)
Here, \(c_{b,\sigma }<\infty \) is a constant that depends only on \(\left\| b\right\| _{L^\infty }\) and \(\sigma \); its existence is guaranteed by part (a) of this proposition. Now, taking a supremum on both sides of (A.20) over all \(t\in [0,T]\), then choosing \(M>1\) large enough so that \(M>(2\left\| b\right\| _{L^\infty }c'\sqrt{\pi })^2\), we derive
$$\begin{aligned} B'\le 2\left( \left\| x\mu _0\right\| _{L^1} + c'\left\| x\nu \right\| _{L_t^\infty (L_x^1)}+c_{b,\sigma }\left( \left\| \mu _0\right\| _{L^1} + \left\| \nu \right\| _{L_t^\infty (L_x^1)}\right) \right) . \end{aligned}$$
(A.21)
Recalling the definition of \(B'\) in (A.17), the above estimate implies part (b) of the proposition. \(\square \)
Before proving Proposition 2.2, we present an abstract lemma.
Lemma A.1
Let \(A\ge 1\), \(B,\delta > 0\) be given constants. Suppose \(f,g:[0,\infty ) \rightarrow [0,\infty )\) are functions that satisfy
$$\begin{aligned} f(t_1) \le Af(t_0) + B\int _{t_0}^{t_1}(t_1-s)^{-1/2}\left( f(s) + g(s)\right) {{\,\mathrm{d \!}\,}}s \quad \forall 0 \le t_0 \le t_1 \le t_0 + \delta \end{aligned}$$
(A.22)
Then for any \(\lambda > \frac{1}{\delta }\ln (A)\), we have
$$\begin{aligned} \left( 1 - \frac{2\delta ^{1/2}B}{1 - Ae^{-\lambda \delta }}\right) \int _0^T e^{-\lambda t}f(t){{\,\mathrm{d \!}\,}}t \le \frac{A}{\kappa }f(0) + \frac{2\delta ^{1/2}B}{1 - Ae^{-\lambda \delta }}\int _0^T e^{-\lambda t}g(t) {{\,\mathrm{d \!}\,}}t.\nonumber \\ \end{aligned}$$
(A.23)
Proof
Set \(h(t) = f(t) + g(t)\), so that (A.22) reads simply
$$\begin{aligned} f(t_1) \le Af(t_0) + B\int _{t_0}^{t_1}(t_1-s)^{-1/2}h(s){{\,\mathrm{d \!}\,}}s \quad \text {for all }0 \le t_0 \le t_1 \le t_0 + \delta .\nonumber \\ \end{aligned}$$
(A.24)
For arbitrary \(t > 0\) let \(n = \left\lfloor \frac{t}{\delta } \right\rfloor \). Use (A.24) \(n+1\) times to get
$$\begin{aligned} f(t) \le A^{n+1}f(0) + \sum _{j=0}^n A^jB\int _{\left( t-(j+1)\delta \right) _+}^{t-j\delta }(t-j\delta -s)^{-1/2}h(s){{\,\mathrm{d \!}\,}}s, \end{aligned}$$
(A.25)
where \(s_+ := \max \{s,0\}\). Note that
$$\begin{aligned} t - (j+1)\delta < s \le t-j\delta \Rightarrow j = \left\lfloor \frac{t-s}{\delta } \right\rfloor , \end{aligned}$$
so we define \(\phi (s) = \left( s - \left\lfloor \frac{s}{\delta } \right\rfloor \delta \right) ^{-1/2}\). Then (A.25) implies
$$\begin{aligned} f(t)&\le A^{\frac{t}{\delta }+1}f(0) + \sum _{j=0}^n B \int _{\left( t-(j+1)\delta \right) _+}^{t-j\delta } A^{\frac{t-s}{\delta }}\phi (t-s)h(s){{\,\mathrm{d \!}\,}}s \\&= A^{\frac{t}{\delta }+1}f(0) + B\int _0^t A^{\frac{t-s}{\delta }}\phi (t-s)h(s){{\,\mathrm{d \!}\,}}s.\nonumber \end{aligned}$$
(A.26)
Let \(\lambda > \frac{1}{\delta }\ln (A)\) and set \(\kappa = \lambda - \frac{1}{\delta }\ln (A) > 0\). Multiply (A.26) by \(e^{-\lambda t}\), then integrate from 0 to T to get
$$\begin{aligned} \begin{aligned} \int _0^T e^{-\lambda t}f(t){{\,\mathrm{d \!}\,}}t&\le \frac{A}{\kappa }f(0) + B\int _0^T\int _0^t e^{-\kappa (t-s)}\phi (t-s)e^{-\lambda s}h(s){{\,\mathrm{d \!}\,}}s {{\,\mathrm{d \!}\,}}t\\&= \frac{A}{\kappa }f(0) + B\int _0^T\int _0^{T-s} e^{-\kappa t}\phi (t)e^{-\lambda s}h(s){{\,\mathrm{d \!}\,}}t {{\,\mathrm{d \!}\,}}s. \end{aligned} \end{aligned}$$
(A.27)
We now observe that
$$\begin{aligned} \begin{aligned} \int _0^\infty e^{-\kappa t}\phi (t){{\,\mathrm{d \!}\,}}t&= \sum _{n=0}^\infty \int _{n\delta }^{(n+1)\delta } e^{-\kappa t}(t-n\delta )^{-1/2}{{\,\mathrm{d \!}\,}}t\\&= \sum _{n=0}^\infty e^{-n\kappa \delta }\int _{0}^{\delta } e^{-\kappa t}t^{-1/2}{{\,\mathrm{d \!}\,}}t\\&\le \frac{1}{1 - e^{-\kappa \delta }}\int _0^\delta t^{-1/2}{{\,\mathrm{d \!}\,}}t\\&= \frac{2\delta ^{1/2}}{1 - e^{-\kappa \delta }} = \frac{2\delta ^{1/2}}{1 - Ae^{-\lambda \delta }} \end{aligned} \end{aligned}$$
(A.28)
Applying (A.28) to (A.27), we get
$$\begin{aligned} \int _0^T e^{-\lambda t}f(t){{\,\mathrm{d \!}\,}}t \le \frac{A}{\kappa }f(0) + \frac{2\delta ^{1/2}B}{1 - Ae^{-\lambda \delta }}\int _0^T e^{-\lambda s}\left( f(s) + g(s)\right) {{\,\mathrm{d \!}\,}}s, \end{aligned}$$
(A.29)
which implies (A.23). \(\square \)
We now apply Proposition A.1 to the Fokker–Planck equation.
Proof of Proposition 2.2
First we will treat \(\mu \) as a solution of the abstract Fokker–Planck equation (2.1) by identifying \(b:=F^{(0)}\) and \(\nu :=\Phi +\frac{1}{2}\big (G(\varepsilon )-w_x\big )m^{(0)}\). We start with the following formula: for every \(t_1 \ge t_0 \ge 0\),
$$\begin{aligned}&\mu (x,t_1) = \int _0^\infty \left( S(x-y,t_1-t_0) - S(x+y,t_1-t_0)\right) \mu (y,t_0){{\,\mathrm{d \!}\,}}y \nonumber \\&\quad + \int _{t_0}^{t_1}\int _0^\infty \left( {{\dfrac{\partial {^{}}S}{\partial {x^{}}}}}(x-y,t_1-s) + {{\dfrac{\partial {^{}}S}{\partial {x^{}}}}}(x+y,t_1-s)\right) \left( b(y,s)\mu (y,s) + \nu (y,s)\right) {{\,\mathrm{d \!}\,}}y {{\,\mathrm{d \!}\,}}s,\nonumber \\ \end{aligned}$$
(A.30)
Using the same calculations as in (A.6), we get
$$\begin{aligned}&\int _0^\infty \left| \mu (x,t_1)\right| {{\,\mathrm{d \!}\,}}x \le \int _0^\infty \left| \mu (y,t_0)\right| {{\,\mathrm{d \!}\,}}y\nonumber \\&\quad + c'\int _{t_0}^{t_1}\int _0^\infty (t_1-s)^{-1/2}\left( \left\| b\right\| _\infty \left| \mu (y,s)\right| + \left| \nu (y,s)\right| \right) {{\,\mathrm{d \!}\,}}y {{\,\mathrm{d \!}\,}}s, \end{aligned}$$
(A.31)
where \(c'\) is defined in (A.5). Let
$$\begin{aligned} f(t) := \left( \int _0^\infty \left| \mu (x,t)\right| {{\,\mathrm{d \!}\,}}x\right) ^2 \,\,\text { and }\,\, g(t) := \left( \int _0^\infty \left| \nu (x,t)\right| {{\,\mathrm{d \!}\,}}x\right) ^2, \end{aligned}$$
so we can write
$$\begin{aligned} f(t_1)^{1/2} \le f(t_0)^{1/2} + c\int _{t_0}^{t_1} (t_1-s)^{-1/2}\left( \left\| b\right\| _\infty f(s)^{1/2} + g(s)^{1/2}\right) {{\,\mathrm{d \!}\,}}s. \end{aligned}$$
(A.32)
By Hölder’s inequality, we get
$$\begin{aligned} f(t_1)^{1/2}&\le f(t_0)^{1/2} + \left\| b\right\| _\infty c\left( 2(t_1-t_0)^{1/2}\right) ^{1/2}\left( \int _{t_0}^{t_1} (t_1-s)^{-1/2}f(s){{\,\mathrm{d \!}\,}}s\right) ^{1/2} \\&\quad + c\left( 2(t_1-t_0)^{1/2}\right) ^{1/2}\left( \int _{t_0}^{t_1} (t_1-s)^{-1/2}g(s){{\,\mathrm{d \!}\,}}s\right) ^{1/2}.\nonumber \end{aligned}$$
(A.33)
Next, we square both sides and use the inequality \((a+b)^2 \le 2(a^2 + b^2)\) to get
$$\begin{aligned} f(t_1) \le 2f(t_0) + 4c(1+\left\| b\right\| _{\infty })(t_1-t_0)^{1/2}\int _{t_0}^{t_1} (t_1-s)^{-1/2}\left( f(s)+ g(s)\right) {{\,\mathrm{d \!}\,}}s.\nonumber \\ \end{aligned}$$
(A.34)
We now estimate
$$\begin{aligned} \int _0^\infty \left| [\right| 2]{\Phi +\frac{1}{2}\big (G(\varepsilon )-w_x\big )m^{(0)}}{{\,\mathrm{d \!}\,}}x\le & {} \left\| \Phi \right\| _{L_x^1}+2\int _0^\infty \left| w_xm^{(0)}\right| {{\,\mathrm{d \!}\,}}x\nonumber \\&+\int _0^\infty |u_x^{(0)}\mu |{{\,\mathrm{d \!}\,}}x, \end{aligned}$$
(A.35)
so that
$$\begin{aligned} g(t)&\le 3\left\| \Phi \right\| _{L_x^1}^2+12\Bigg (\int _0^\infty |w_xm^{(0)}|{{\,\mathrm{d \!}\,}}x\Bigg )^2+3\Bigg (\int _0^\infty |u_x^{(0)}\mu |{{\,\mathrm{d \!}\,}}x\Bigg )^2 \\&\le 3\left\| \Phi \right\| _{L_x^1}^2+12\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x+3\left\| u_x^{(0)}\right\| _{\infty }^2f(t).\nonumber \end{aligned}$$
(A.36)
The first inequality follows from the fact that \(\big (\sum _{k=1}^nx_k\big )^2\le n\sum _{k=1}^nx_k^2\). The second inequality follows from the Cauchy–Schwarz inequality and the fact that \(\int _0^\infty m^{(0)}{{\,\mathrm{d \!}\,}}x\le 1\). Now, define
$$\begin{aligned} h(t):=\frac{1}{1+3\left\| u_x^{(0)}\right\| _{\infty }^2}\Big (3\left\| \Phi \right\| _{L_t^\infty (L_x^1)}+12\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x\Big ). \end{aligned}$$
(A.37)
Since
$$\begin{aligned} 1 + \left\| b\right\| _{\infty }=1 + \left\| F^{(0)}\right\| _{\infty }\le 2\big (1+\left\| u_x^{(0)}\right\| _{\infty }\big ), \end{aligned}$$
we see that (A.36) and (A.34) yield
$$\begin{aligned} f(t_1)\le 2f(t_0)+24c\big (1+\left\| u_x^{(0)}\right\| _{\infty }\big )^3(t_1-t_0)^{1/2}\int _0^\infty (t_1-s)^{1/2}\big (f(s)+h(s)\big ),\nonumber \\ \end{aligned}$$
(A.38)
upon factoring out \(\big (1+3\left\| u_x^{(0)}\right\| _{\infty }^2\big )\), and estimating \(\big (1+3\left\| u_x^{(0)}\right\| _{\infty }^2\big )\le 3\big (1+\left\| u_x^{(0)}\right\| _\infty \big )^2\). From (A.38), we see that Proposition A.1 applies with \(A:=2\), and \(B:=24c\big (1+\left\| u_x^{(0)}\right\| _{\infty }\big )^3\delta ^{1/2}\).
Now, choose \(\delta >0\) small enough such that
$$\begin{aligned} 2\delta ^{1/2}B<\frac{1}{4}\iff \delta <\frac{1}{192c\big (1+\left\| u_x^{(0)}\right\| _\infty \big )^3}. \end{aligned}$$
(A.39)
Then choose \(\lambda >\frac{1}{\delta }\ln (2)\) large enough such that
$$\begin{aligned} 1-Ae^{-\lambda \delta }>\frac{1}{2}\iff \lambda>\frac{2}{\delta }\ln (2)\iff \lambda >C_0. \end{aligned}$$
(A.40)
As a result,
$$\begin{aligned} \frac{2\delta ^{1/2}B}{1-Ae^{-\lambda \delta }}<\frac{1}{2}, \end{aligned}$$
(A.41)
and by Proposition A.1 we obtain
$$\begin{aligned} \int _0^Te^{-\lambda t}\Bigg (\int _0^\infty |\mu |{{\,\mathrm{d \!}\,}}x\Bigg )^2{{\,\mathrm{d \!}\,}}t&\le \frac{3}{1+3\left\| u_x^{(0)}\right\| _\infty ^2}\int _0^Te^{-\lambda t}\left\| \Phi \right\| _{L_x^1}^2{{\,\mathrm{d \!}\,}}t\nonumber \\&\quad +\frac{12}{1+3\left\| u_x^{(0)}\right\| _{\infty }^2}\int _0^Te^{-\lambda t}\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}t\nonumber \\&\le \frac{3\left\| \Phi \right\| _{L_t^\infty (L_x^1)}^2}{C_0\big (1+3\left\| u_x^{(0)}\right\| _\infty ^2\big )}\nonumber \\&\quad +\frac{12}{1+3\left\| u_x^{(0)}\right\| _{\infty }^2}\int _0^Te^{-\lambda t}\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}t. \end{aligned}$$
(A.42)
The second inequality is obtained by the simple estimate
$$\begin{aligned} \int _0^Te^{-\lambda t}{{\,\mathrm{d \!}\,}}t=\frac{1}{\lambda }(1-e^{-\lambda T})<\frac{1}{\lambda }<\frac{1}{C_0}. \end{aligned}$$
(A.43)
Thus, (2.3) holds.
Now, in the case that T is finite, we can estimate for all \(\lambda \in (0,C_0]\),
$$\begin{aligned} e^{-\lambda t}=e^{(2C_0-\lambda )t}e^{-2C_0t}\le e^{(2C_0-\lambda )T}e^{-2C_0t}. \end{aligned}$$
(A.44)
Consequently, using (2.3) with \(\lambda =2C_0\), we obtain
$$\begin{aligned} \int _0^Te^{-\lambda t}\Bigg (\int _0^\infty |\mu |{{\,\mathrm{d \!}\,}}x\Bigg )^2{{\,\mathrm{d \!}\,}}t&\le e^{(2C_0-\lambda )T}\int _0^Te^{-2C_0 t}\Bigg (\int _0^\infty |\mu |{{\,\mathrm{d \!}\,}}x\Bigg )^2{{\,\mathrm{d \!}\,}}t\nonumber \\&\le e^{(2C_0-\lambda )T}\Bigg (\frac{C_1\left\| \Phi \right\| _{L_t^\infty (L_x^1)}^2}{C_0}+C_2\int _0^T e^{-2C_0 t}\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}t\Bigg )\nonumber \\&\le \frac{C_1\left\| \Phi \right\| _{L_t^\infty (L_x^1)}^2e^{2C_0T}}{C_0}+C_2e^{2C_0T}\int _0^T e^{-\lambda t}\int _0^\infty w_x^2m^{(0)}{{\,\mathrm{d \!}\,}}x{{\,\mathrm{d \!}\,}}t,\nonumber \\ \end{aligned}$$
(A.45)
whereby the last inequality follows, because \(e^{-2C_0t}\le e^{-\lambda t}\) for all \(t\in [0,T]\). This completes the proof. \(\square \)
1.2 An Abstract Hölder Estimate
In this subsection we will give an abstract result on Hölder regularity for a parabolic equation with Dirichlet boundary conditions and bounded coefficients. Our argument is in the spirit of [3, Lemma 3.2.2.], but we cover the case of an unbounded domain. The result stated below is also meant to allow for a possibly infinite time horizon, though in the present work we will not exploit this.
Lemma A.2
Fix \(0< \alpha < 1\). Let u be a solution of
$$\begin{aligned} {{\dfrac{\partial {^{}}u}{\partial {t^{}}}}} + \lambda u - \frac{\sigma ^2}{2}{{\dfrac{\partial {^{2}}u}{\partial {x^{2}}}}} + V(x,t){{\dfrac{\partial {^{}}u}{\partial {x^{}}}}} = F, \quad u(0,t) = 0, \quad u(x,0) = u_0(x) \end{aligned}$$
(A.46)
where \(\lambda \) is any positive constant, V and F are a bounded continuous functions, and \(u_0 \in {\mathcal {C}}^{1+\alpha }_\diamond ({\mathcal {D}})\) (i.e. \(u_0 \in {\mathcal {C}}^{1+\alpha }(\overline{{\mathcal {D}}})\) and \(u_0(0) = 0\)). Then
$$\begin{aligned} \left\| u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,T]\right) } + \left\| {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,T]\right) } \le C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \left( \left\| F\right\| _\infty + \left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}\right) ,\nonumber \\ \end{aligned}$$
(A.47)
where \(C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \) is independent of T.
As a corollary, we have
$$\begin{aligned} \left\| u\right\| _{{\mathcal {C}}^{\alpha /4}\left( [0,T];{\mathcal {C}}^{1+\alpha /2}(\overline{{\mathcal {D}}})\right) } \le C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \left( \left\| F\right\| _\infty + \left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}\right) . \end{aligned}$$
(A.48)
Proof
We will start by assuming \(u_0 = 0\). By a standard application of the maximum principle we have that \(\left\| u\right\| _0 \le \frac{1}{\lambda }\left\| F\right\| _0\). Let \(\phi (x)\) be a smooth function, and observe that
$$\begin{aligned} {\dfrac{\partial {^{}}(\phi u)}{\partial {t^{}}}} - \frac{\sigma ^2}{2}{\dfrac{\partial {^{2}}(\phi u)}{\partial {x^{2}}}} + \left( V + \frac{\sigma ^2\phi '}{\phi }\right) {\dfrac{\partial {^{}}(\phi u)}{\partial {x^{}}}} + \lambda (\phi u) = g(x,t), \end{aligned}$$
(A.49)
where
$$\begin{aligned} g(x,t) = \phi (x)F(x,t) + \left( \frac{\sigma ^2\left( \phi '(x)\right) ^2}{\phi (x)} - \frac{\sigma ^2}{2}\phi ''(x) + V(x,t)\phi '(x)\right) u(x,t).\nonumber \\ \end{aligned}$$
(A.50)
Fix a time \(\tau > 0\), and set \(v(x,t) = e^{\lambda (t-\tau )}\phi (x)u(x,t)\). Then (A.49) becomes
$$\begin{aligned} {\dfrac{\partial {^{}}v}{\partial {t^{}}}} - \frac{\sigma ^2}{2}{\dfrac{\partial {^{2}}v}{\partial {x^{2}}}} + \left( V + \frac{\sigma ^2\phi '}{\phi }\right) {\dfrac{\partial {^{}}v}{\partial {x^{}}}} = {\tilde{g}}, \end{aligned}$$
(A.51)
where
$$\begin{aligned} {\tilde{g}}(x,t) = e^{\lambda (t-\tau )}g(x,t). \end{aligned}$$
(A.52)
Fix any \(a > 0\) and let \(\phi (x) = \left( 1+(x-a)^2\right) ^{-1/2}\). Note that \(\phi \) satisfies the following properties:
$$\begin{aligned} \left| \frac{\phi '}{\phi }\right| \le 1, \ \left| \frac{\phi ''}{\phi }\right| \le 2, \quad \left( \int _{-\infty }^\infty \left| \phi (x)\right| ^p{{\,\mathrm{d \!}\,}}x\right) ^{1/p} = \left( \frac{2p}{p-1}\right) ^{1/p} \ \forall p > 1, \end{aligned}$$
(A.53)
and also \(\left( \int _0^\tau e^{p\lambda (t-\tau )}{{\,\mathrm{d \!}\,}}t\right) ^{1/p} \le (\lambda p)^{-1/p}\). It follows that
$$\begin{aligned} \left\| {\tilde{g}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}\le & {} C(p)\lambda ^{-1/p}\left( \left\| F\right\| _0 + \left( 1 +\left\| V\right\| _\infty \right) \left\| u\right\| _0\right) \nonumber \\\le & {} C(p)\lambda ^{-1/p}\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty \right) , \end{aligned}$$
(A.54)
where C(p) remains bounded as \(p\rightarrow \infty \).
Take p arbitrarily large. By the potential estimates in [26, Sect. IV.3] we have an estimate of the form
$$\begin{aligned} \left\| {{\dfrac{\partial {^{}}v}{\partial {t^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} + \left\| {{\dfrac{\partial {^{2}}v}{\partial {x^{2}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} \le C\left\| {\tilde{g}} - \left( V + \frac{2\phi '}{\phi }\right) {{\dfrac{\partial {^{}}v}{\partial {x^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}\nonumber \\ \le C(p)\lambda ^{-1/p}\left\| F\right\| _0\left( 1 + \left\| V\right\| _0\right) + C\left( 1+\left\| V\right\| _0\right) \left\| {{\dfrac{\partial {^{}}v}{\partial {x^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))},\nonumber \\ \end{aligned}$$
(A.55)
where the constants do not depend on \(\tau \). On the other hand we have
$$\begin{aligned} \left\| v\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} \le C(p)\lambda ^{-1/p}\left\| u\right\| _0 \le C(p)\lambda ^{-1-1/p}\left\| F\right\| _0. \end{aligned}$$
(A.56)
By interpolation (see e.g. [26, Lemma II.3.3]) we have
$$\begin{aligned} \left\| {{\dfrac{\partial {^{}}v}{\partial {x^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} \le \delta \left( \left\| {{\dfrac{\partial {^{}}v}{\partial {t^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} + \left\| {{\dfrac{\partial {^{2}}v}{\partial {x^{2}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}\right) + \frac{C}{\delta }\left\| v\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}\nonumber \\ \end{aligned}$$
(A.57)
for some constant C, where \(\delta > 0\) is sufficiently small. Choosing \(\delta > 0\) small enough, we deduce
$$\begin{aligned} \left\| {{\dfrac{\partial {^{}}v}{\partial {t^{}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))} + \left\| {{\dfrac{\partial {^{2}}v}{\partial {x^{2}}}}}\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}&\le C(p)\lambda ^{-1/p}\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty \right) \nonumber \\&\quad + C\left( 1+\left\| V\right\| _\infty ^2\right) \left\| v\right\| _{L^p({\mathcal {D}} \times (0,\tau ))}\nonumber \\&\le C(p)\lambda ^{-1/p}\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty \right) \nonumber \\&\quad + C\left( 1+C(p)\lambda ^{-1-1/p}\left\| F\right\| _0\left\| V\right\| _\infty ^2\right) \nonumber \\ \end{aligned}$$
(A.58)
Then by a Sobolev type embedding theorem [26, Lemma II.3.3] we have
$$\begin{aligned} \left\| v\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| {{\dfrac{\partial {^{}}v}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C(p,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C \end{aligned}$$
(A.59)
for \(\alpha = 1-\frac{3}{p}\) (assuming \(p > 3\)). We can rewrite this as
$$\begin{aligned}&\left\| e^{\lambda (\cdot - \tau )}\phi u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| e^{\lambda (\cdot - \tau )}\left( \phi ' u + \phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right) \right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}}\nonumber \\&\quad \le C(p,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C. \end{aligned}$$
(A.60)
Since \(t \mapsto e^{\lambda t}\) is locally Lipschitz with constant depending on \(\lambda \), we can write this as
$$\begin{aligned}&\left\| \phi u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}(\overline{{\mathcal {D}}} \times [\tau -1,\tau ])} + \left\| \phi ' u + \phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}(\overline{{\mathcal {D}}} \times [\tau -1,\tau ])}\nonumber \\&\quad \le C(p,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C(\lambda ). \end{aligned}$$
(A.61)
Using the fact that \(\frac{\phi '}{\phi }\) is bounded by 1 and is globally Lipschitz with \({{\,\mathrm{Lip}\,}}\left( \frac{\phi '}{\phi }\right) = 1\), we also have
$$\begin{aligned} \left\| \phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le \left\| \phi 'u+\phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| \phi 'u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le \left\| \phi 'u+\phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| \phi u\right\| _{{\mathcal {C}}^{\alpha }},\nonumber \\ \end{aligned}$$
(A.62)
hence
$$\begin{aligned}&\left\| \phi u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}(\overline{{\mathcal {D}}} \times [\tau -1,\tau ])} + \left\| \phi {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}(\overline{{\mathcal {D}}} \times [\tau -1,\tau ])}\nonumber \\&\quad \le C(p,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C(\lambda ). \end{aligned}$$
(A.63)
Now, since \(\frac{1}{\phi }\) is globally Lipschitz with \({{\,\mathrm{Lip}\,}}(1/\phi ) = 1\), and bounded on \([a-1,a+1]\) with an upper bound of 2, we deduce that
$$\begin{aligned}&\left\| u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( [a-1,a+1] \times [\tau -1,\tau ]\right) } + \left\| {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( [a-1,a+1] \times [\tau -1,\tau ]\right) }\nonumber \\&\quad \le C(p,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C(\lambda ). \end{aligned}$$
(A.64)
This estimate is independent of \(\tau \) and a. Letting \(\tau \) and a vary through all the positive integers, and since p can be determined through \(\alpha \), it follows that
$$\begin{aligned} \left\| u\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,\infty )\right) } + \left\| {{\dfrac{\partial {^{}}u}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,\infty )\right) } \le C(\alpha ,\lambda )\left\| F\right\| _0\left( 1 + \left\| V\right\| _\infty ^2\right) + C(\lambda ).\nonumber \\ \end{aligned}$$
(A.65)
We now remove the assumption that \(u_0 = 0\). Let w be the solution of
$$\begin{aligned} {\dfrac{\partial {^{}}w}{\partial {t^{}}}} - \frac{\sigma ^2}{2}{\dfrac{\partial {^{2}}w}{\partial {x^{2}}}} + \lambda w = 0, \quad w(x,0) = u_0(x). \end{aligned}$$
(A.66)
As \(\lambda > 0\), by the maximum principle, we have \(\left\| w\right\| _0 \le \left\| u_0\right\| _0\). Then \([w]_{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C\left\| u_0\right\| _{{\mathcal {C}}^\alpha }\) by [26, Theorem 10.1]; this estimate does not depend on time because of the global in time \(L^\infty \) bound. This establishes \(\left\| w\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C\left\| u_0\right\| _{{\mathcal {C}}^{\alpha }}\). We then take the derivative in x of Equation (A.66) and apply the same argument as above to \({{\dfrac{\partial {^{}}w}{\partial {x^{}}}}}\) to establish \(\left\| {{\dfrac{\partial {^{}}w}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C\left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}.\) Then let \({\hat{u}}\) be the solution of
$$\begin{aligned} {\dfrac{\partial {^{}}{\hat{u}}}{\partial {t^{}}}} - \frac{\sigma ^2}{2}{\dfrac{\partial {^{2}}{\hat{u}}}{\partial {x^{2}}}} + V(x,t){\dfrac{\partial {^{}}{\hat{u}}}{\partial {x^{}}}} + \lambda {\hat{u}} = F(x,t) - V(x,t){\dfrac{\partial {^{}}w}{\partial {x^{}}}} \end{aligned}$$
(A.67)
with zero initial conditions. Then by (A.65) we have
$$\begin{aligned} \begin{aligned} \left\| {\hat{u}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,\infty )\right) } + \left\| {{\dfrac{\partial {^{}}{\hat{u}}}{\partial {x^{}}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}\left( \overline{{\mathcal {D}}} \times [0,\infty )\right) }&\le C(\alpha ,\lambda ,\left\| V\right\| _\infty )\left\| F+V{{\dfrac{\partial {^{}}w}{\partial {x^{}}}}}\right\| _0\\&\le C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \left( \left\| F\right\| _0 + \left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}\right) . \end{aligned} \end{aligned}$$
(A.68)
As \(u = {\hat{u}} + w\) is the solution to (A.46), the claim is proved.
Finally, to prove (A.48), note that (A.47) immediately implies
$$\begin{aligned} \sup _{0\le t\le T} \left\| u(\cdot ,t)\right\| _{{\mathcal {C}}^{1+\alpha /2}(\overline{{\mathcal {D}}})}\le & {} \sup _{0\le t\le T} \left\| u(\cdot ,t)\right\| _{{\mathcal {C}}^{1+\alpha }(\overline{{\mathcal {D}}})}\nonumber \\\le & {} C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \left( \left\| F\right\| _\infty + \left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}\right) , \end{aligned}$$
(A.69)
and we also have
$$\begin{aligned} \begin{aligned} \sup _{t \ne s} \sup _{x \ne y}&\frac{\left| {\dfrac{\partial {^{}}u}{\partial {x^{}}}}(x,t) - {\dfrac{\partial {^{}}u}{\partial {x^{}}}}(x,s) - {\dfrac{\partial {^{}}u}{\partial {x^{}}}}(y,t) + {\dfrac{\partial {^{}}u}{\partial {x^{}}}}(y,s)\right| }{\left| t-s\right| ^{\alpha /4}\left| x-y\right| ^{\alpha /2}}\\&\le \sup _{t \ne s} \sup _{x \ne y}\frac{2\left\| {\dfrac{\partial {^{}}u}{\partial {x^{}}}}\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}}\min \left\{ \left| t-s\right| ^{\alpha /2},\left| x-y\right| ^{\alpha }\right\} }{\left| t-s\right| ^{\alpha /4}\left| x-y\right| ^{\alpha /2}} \\&\le C\left( \left\| V\right\| _\infty ,\alpha ,\lambda \right) \left( \left\| F\right\| _\infty + \left\| u_0\right\| _{{\mathcal {C}}^{1+\alpha }}\right) . \end{aligned} \end{aligned}$$
(A.70)
Combine (A.69) and (A.70) to get (A.48). \(\square \)
Appendix B: Nonlocal Existence Lemma
For this lemma, recall the definition \({\mathcal {X}} = {\mathcal {C}}^{\alpha ,\alpha /2} \cap L_t^\infty (L_x^1) \cap {\mathcal {C}}^{\alpha /2}\left( [0,T];({\mathcal {C}}^{1+2\alpha })^*\right) \) with norm
$$\begin{aligned} \left\| m\right\| _{{\mathcal {X}}} := \left\| m\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| m\right\| _{L_t^\infty (L_x^1)} + \left\| m\right\| _{{\mathcal {C}}^{\alpha /2}\left( [0,T];({\mathcal {C}}^{1+2\alpha })^*\right) }. \end{aligned}$$
(B.1)
Lemma B.1
Let \(\alpha \in (0,1/2)\) be such that \(u^{(0)},m^{(0)} \in {\mathcal {C}}^{2+2\alpha ,1+\alpha }\). Let \(m \in {\mathcal {X}}\), \(u_T \in {\mathcal {C}}^{2+\alpha }([0,\infty ))\), and \(\Psi \in {\mathcal {C}}^{\alpha ,\alpha /2}([0,\infty ) \times [0,T])\) be given. Consider the backward parabolic equation
$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{lll} (i)&{}\displaystyle u_t+\frac{\sigma ^2}{2}u_{xx}-ru + \lambda \Psi +\lambda F^{(0)}(\varepsilon )\big (G(u_x,m;\varepsilon )-u_x\big )=0, &{}0\le x< \infty ,\,\,0\le t\le T\\ (ii)&{}\displaystyle u(x,T)=u_T(x), &{}0\le x< \infty \\ (iii)&{}\displaystyle u(0,t)=0, &{}0\le t\le T, \end{array} \end{array}\right. } \end{aligned}$$
(B.2)
where G is defined as in (1.5). We assume the usual compatibility conditions of first order:
$$\begin{aligned} u_T(0) = 0, \quad \frac{\sigma ^2}{2}u_T''(0) + \lambda \Psi (0,T) - \lambda F^{(0)}(\epsilon )(0,0)u_T'(0) = 0. \end{aligned}$$
(B.3)
Then there exists a unique classical solution \(u \in {\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [0,T])\) to the boundary value problem (B.2). Moreover, the following estimate holds:
$$\begin{aligned} \left\| u\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} \le C\left( \left\| m\right\| _{{\mathcal {X}}} + \left\| \Psi \right\| _{{\mathcal {C}}^{\alpha /2,\alpha }} + \left\| u_T\right\| _{{\mathcal {C}}^{2+\alpha }}\right) \end{aligned}$$
(B.4)
where C depends only on the data.
Proof
Let \(u \in C^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [0,T])\) be given. Note that
$$\begin{aligned} \left\| G(u_x,m;\varepsilon )\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C\left( \left\| m\right\| _{{\mathcal {C}}^{\alpha /2}\left( [0,T];({\mathcal {C}}^{1+2\alpha })^*\right) } + \left\| u\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}}\right) , \end{aligned}$$
(B.5)
where C depends on the \({\mathcal {C}}^{2+2\alpha ,1+\alpha }\) norms of \(u^{(0)}\) and \(m^{(0)}\). By [26, Theorem IV.5.2], there exists a unique solution \(w \in C^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [0,T])\) the boundary value problem
$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{lll} (i)&{}\displaystyle w_t+\frac{\sigma ^2}{2}w_{xx}-rw+\lambda \Psi +\lambda F^{(0)}(\varepsilon )\big (G(u_x,m;\varepsilon )-w_x\big )=0, &{}0\le x< \infty ,\,\,0\le t\le T\\ (ii)&{}\displaystyle w(x,T)=u_T(x), &{}0\le x< \infty \\ (iii)&{}\displaystyle w(0,t)=0. &{}0\le t\le T, \end{array} \end{array}\right. } \end{aligned}$$
(B.6)
and it satisfies the estimate
$$\begin{aligned} \left\| w\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} \le C\left( \left\| \Psi \right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| G(u_x,m;\varepsilon )\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} + \left\| u_T\right\| _{{\mathcal {C}}^{2+\alpha }}\right) . \end{aligned}$$
(B.7)
We will denote \(w = F(u)\). Our goal it to show that F is a contraction on a suitably defined metric space, and use this to prove (B.2) has a solution.
Let \(u_1,u_2 \in {\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [0,T])\), then define \(u = u_1 - u_2\) and \(w = F(u_1) - F(u_2)\). Note that w satisfies
$$\begin{aligned} {\left\{ \begin{array}{ll} \begin{array}{lll} (i)&{}\displaystyle w_t+\frac{\sigma ^2}{2}w_{xx}-rw+\lambda F^{(0)}(\varepsilon )\big (G(u_x,0;\varepsilon )-w_x\big )=0, &{}0\le x< \infty ,\,\,0\le t\le T\\ (ii)&{}\displaystyle w(x,T)=0, &{}0\le x< \infty \\ (iii)&{}\displaystyle w(0,t)=0. &{}0\le t\le T. \end{array} \end{array}\right. } \end{aligned}$$
(B.8)
Recalling the definition of G in (1.5), we see that there is a constant \(C_1\), depending only on the data, such that
$$\begin{aligned} \left\| G(u_x,0;\varepsilon )\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le C_1\left\| u_x\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}}. \end{aligned}$$
(B.9)
Combining this with classical estimates from [26, Theorem IV.5.2], we see that there is some constant \(C_2\), depending only on the data, such that
$$\begin{aligned} \left\| w\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} \le C_2\left\| u_x\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}}. \end{aligned}$$
(B.10)
Using interpolation on Hölder spaces, we deduce that there exists a constant \(C_3\), depending only on \(C_2\), such that
$$\begin{aligned} C_2\left\| v_x\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le \frac{1}{4}\left\| v\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} + C_3\left\| v\right\| _\infty \quad \forall v \in {\mathcal {C}}^{2+\alpha ,1+\alpha /2}. \end{aligned}$$
(B.11)
Hence
$$\begin{aligned} \left\| w\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} \le \frac{1}{4}\left\| u\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}} + C_3\left\| u\right\| _\infty . \end{aligned}$$
(B.12)
Next, we define \({\tilde{w}}(x,t) = e^{r(T-t)}w \pm r^{-1}(e^{r(T-t)} - 1)C_1\left\| u_x\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}}\), which satisfies
$$\begin{aligned} \pm (-{\tilde{w}}_t - \frac{\sigma ^2}{2} {\tilde{w}}_{xx} - \lambda F^{(0)}(\varepsilon ){\tilde{w}}_x) \le 0. \end{aligned}$$
(B.13)
Apply the maximum principle to \({\tilde{w}}\) to deduce that
$$\begin{aligned} \left| w(x,t)\right| \le r^{-1}(e^{r(T-t)} - 1)C_1\left\| u_x\right\| _{{\mathcal {C}}^{\alpha ,\alpha /2}} \le (T-t)e^{rT}C_1\left\| u\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}}.\nonumber \\ \end{aligned}$$
(B.14)
We now apply all of these estimates only on the time interval \([T-\tau ,T]\) for some \(\tau > 0\). We deduce that
$$\begin{aligned}&\left\| w\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [T-\tau ,T])} + 2C_3\left\| w\right\| _{L^\infty ([0,\infty ) \times [T-\tau ,T])} \\&\quad \le \left( \frac{1}{4} + 2C_3\tau e^{rT}C_1\right) \left\| u\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [T-\tau ,T])} + C_3\left\| u\right\| _{L^\infty ([0,\infty ) \times [T-\tau ,T])}.\nonumber \end{aligned}$$
(B.15)
We now set \(\tau = \dfrac{1}{8C_3 e^{rT}C_1}\). Define \({\mathcal {Y}}_\tau \) to be the space \({\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [T-\tau ,T])\) endowed with the norm
$$\begin{aligned} \left\| w\right\| _{{\mathcal {Y}}_\tau } := \left\| w\right\| _{{\mathcal {C}}^{2+\alpha ,1+\alpha /2}([0,\infty ) \times [T-\tau ,T])} + 2C_3\left\| w\right\| _{L^\infty ([0,\infty ) \times [T-\tau ,T])}. \end{aligned}$$
(B.16)
Observe that \({\mathcal {Y}}_\tau \) is a Banach space. Moreover, by the above estimates, \(F:{\mathcal {Y}}_\tau \rightarrow {\mathcal {Y}}_\tau \) is a contraction, since \(\left\| F(u_1) - F(u_2)\right\| _{{\mathcal {Y}}_\tau } \le \frac{1}{2}\left\| u_1 - u_2\right\| _{{\mathcal {Y}}_\tau }\). Hence F has a unique fixed point u, which is a solution to (B.2) and satisfies the estimate (B.4) on the time interval \([T-\tau ,T]\). However, T is arbitrary. We can now partition the interval [0, T] into subintervals that are each at most \(\tau \) in length, i.e. \(0 = t_0< t_1< \cdots < t_N = T\) where \(t_{j+1} - t_j \le \tau \). Apply the same argument on each subinterval \([t_{j-1},t_j]\), replacing the final condition \(u_T(x)\) with \(w(x,t_j)\), for each j starting with N and going down to 1. (Cf. the proof of [10, Proposition 3.11].) In this way we obtain a solution u to Equation (B.2), which indeed satisfies (B.4). Uniqueness of this solution follows from uniqueness on each subinterval. \(\square \)
