Appendix A: Proof of Theorem 2.1
Let \(\psi \) be new variable defined by \(\varphi (x,t) := e^{s\alpha (x,t)}\psi (x,t)\). It yields:
$$\begin{aligned} \displaystyle \varphi _t= & {} s\alpha _te^{s\alpha }\psi + e^{s\alpha }\psi _t, \quad \varphi _x = s\alpha _xe^{s\alpha } \psi + e^{s\alpha }\psi _x \,\, \text{ and } \\ \displaystyle \varphi _{xx}= & {} e^{s\alpha }\left( \psi _{xx} + s^2\alpha ^2_x\psi + 2s\alpha _x\psi _x + s\alpha _{xx}\psi \right) . \end{aligned}$$
The functions \(\alpha \) and \(\xi \) yield:
$$\begin{aligned} \displaystyle \alpha _x= & {} -\xi _x = -\lambda \alpha _{0,x}\xi , \quad \alpha _{xx} = -\lambda ^2\alpha _{0,x}^2\xi - \lambda \alpha _{0,xx}\xi , \\ \displaystyle \alpha _t= & {} -k\frac{T-2t}{(T-t)^{k+1} t^{k+1}}\left( e^{2\lambda \parallel \alpha _1\parallel _{\infty } } - e^{\lambda \alpha _1(x,t)} \right) - \lambda \alpha _{0,t}\xi , \\ \displaystyle \alpha _{x,t}= & {} k \lambda \xi \left( \frac{(T-2t)}{t(T-t)}\alpha _{0,x} -\frac{\alpha _{0,tx}}{k}-\lambda \frac{\alpha _{0,t} \alpha _{0,x}}{k}\right) \end{aligned}$$
and, for \(\lambda \) sufficiently large, one has:
$$\begin{aligned} \displaystyle |\alpha _x|\le & {} C\lambda \xi , \quad |\alpha _{xx}| \le C\lambda ^2\xi ,\\ \displaystyle |\alpha _t|\le & {} C\xi ^{1+\frac{1}{k}} + C\lambda \xi \le C \lambda \xi ^{\frac{3}{2}} \quad \text{ and } \\ |\alpha _{x,t}|\le & {} C \lambda \xi ^2. \end{aligned}$$
On the other hand, we have that \(\psi (x,0) \equiv 0\) in (0, L(0)) and \(\psi (x,T) \equiv 0\) in \((0,\overline{L}(T))\).
Moreover, replacing \(\varphi \) by \(e^{s\alpha }\psi \) in the PDE (2.2), it yields:
$$\begin{aligned}&e^{-2s\alpha }(s\alpha _t\psi + \psi _t) + e^{-2s\alpha }\Big [\beta \left( t\right) \psi _{xx} + \beta \left( t\right) s^2\alpha _{x}^2\psi + 2\beta \left( t\right) s\alpha _x\psi _x\\&\quad + \beta \left( t\right) s\alpha _{xx}\psi \Big ] + e^{s\alpha }(a\psi ) + e^{s\alpha }\Big [s\alpha _x(b\psi ) + (b\psi _x) \Big ] = F(x,t), \end{aligned}$$
therefore,
$$\begin{aligned}&\Big [\psi _t + 2\beta \left( t\right) s\alpha _x\psi _x \Big ] + \Big [\beta \left( t\right) \psi _{xx} + \beta \left( t \right) s^2\alpha _{x}^2\psi \Big ] \\&\quad = - e^{-s\alpha } F(x,t) - s\alpha _t\psi - \beta \left( t\right) s\alpha _{xx}\psi - a\psi - s\alpha _xb\psi - b\psi _x. \end{aligned}$$
Considering the following notation
$$\begin{aligned} \left\{ \begin{array}{l} U\psi := \psi _t + 2\beta \left( t\right) s\alpha _x\psi _x = (U\psi )_1 + (U\psi )_2\\ V\psi := \beta \left( t\right) \psi _{xx} + \beta \left( t\right) s^2\alpha _{x}^2\psi = (V\psi )_1 + (V\psi )_2\\ \end{array}\right. , \end{aligned}$$
it yields:
$$\begin{aligned} U\psi + V\psi = G(x,t), \end{aligned}$$
where
$$\begin{aligned} G(x,t)= & {} - e^{-s\alpha (x,t)} F(x,t) - s\alpha _t(x,t)\psi (x,t) - \beta (t)s\alpha _{xx}(x,t)\psi (x,t) \\&- a(x,t)\psi (x,t) - s\alpha _x(x,t)b(x,t)\psi (x,t) - b(x,t)\psi _x(x,t). \end{aligned}$$
Therefore,
$$\begin{aligned}&\displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})} \nonumber \\&\quad \displaystyle = \Vert e^{-s\alpha } F + s\alpha _t\psi + \beta s\alpha _{xx}\psi + a\psi + s\alpha _xb\psi + b\psi _x \Vert _{L^2(Q_{\overline{L}})}^2 \nonumber \\&\quad \displaystyle \le C\Big (\Vert e^{-2s\alpha }F\Vert _{L^2(Q_{\overline{L}})}^2 + \Vert s\alpha _t\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert \beta s\alpha _{xx}\psi \Vert ^2_{L^2(Q_{\overline{L}})}\Big .\nonumber \\&\qquad + \Big .\Vert a\psi \Vert ^2_{L^2(Q_{\overline{L}})} + \Vert s\alpha _xb\psi \Vert ^2_{L^2(Q_{\overline{L}})} + \Vert b\psi _x \Vert _{L^2(Q_{\overline{L}})}^2 \Big ) \nonumber \\&\quad \displaystyle \le C\Bigg (\iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2|\alpha _t|^2|\psi |^2dxdt \nonumber \\&\qquad + \iint _{Q_{\overline{L}}}\beta s^2|\alpha _{xx}|^2|\psi |^2dxdt \Big . \displaystyle + \Bigg . \iint _{Q_{\overline{L}}} |a|_{L^{\infty }(Q_{\overline{L}})}^2|\psi |^2dxdt \nonumber \\&\qquad + \iint _{Q_{\overline{L}}} |b|_{L^{\infty }(Q_{\overline{L}})}^2 s^2|\alpha _x|^2|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |b|_{L^{\infty }(Q_{\overline{L}})}^2|\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \displaystyle \le C\Bigg (\iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2\widetilde{C}\lambda ^2\xi ^3|\psi |^2dxdt \nonumber \\&\qquad + C_{\beta }\iint _{Q_{\overline{L}}} s^2\widetilde{C}\lambda ^4\xi ^2 |\psi |^2dxdt \Big . \displaystyle + \Bigg . |a|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} |\psi |^2dxdt \nonumber \\&\qquad + |b|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} s^2\lambda ^2|\alpha _{0,x}|^2\xi ^2|\psi |^2dxdt + |b|_{L^{\infty }(Q_{\overline{L}})}^2\iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \displaystyle \le C\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{Q_{\overline{L}}} s^2\lambda ^4\xi ^3|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg ). \nonumber \\ \end{aligned}$$
(4.3)
Let us compute \((U\psi ,V\psi )_{L^2(Q_{\overline{L}})}\) and use the fact that \(\displaystyle \iint _{Q_{\overline{L}}} = \int _0^T\int _0^{{\overline{L}}(t)}\).
$$\begin{aligned} I_1= & {} \left( (U\psi )_1,(V\psi )_1\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}\beta \psi _{t}\psi _{xx}dxdt \\= & {} \displaystyle \int _0^T\beta \left( \psi _t\psi _x \Big \vert _{0}^{{\overline{L}}(t)} - \int _{0}^{{\overline{L}}(t)}\psi _x\psi _{xt}ds \right) dt \\= & {} \displaystyle -\frac{1}{2}\int _0^T\beta \int _0^{{\overline{L}}(t)}\frac{d}{dt}[\psi _{x}]^2dxdt \\= & {} \displaystyle -\frac{1}{2}\int _0^T\left[ \beta \frac{d}{dt}\int _0^{{\overline{L}}(t)}(\psi _{x})^2dx - \beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)\right] dt\\= & {} \displaystyle \frac{1}{2}\int _0^T\int _{0}^{{\overline{L}}(t)}\beta '(\psi _{x})^2dxdt + \frac{1}{2}\int _{0}^{T}\beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)dt;\\ I_2= & {} \left( (U\psi )_1,(V\psi )_2\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}s^2\beta \psi _{t}\alpha _x^2\psi dxdt \\= & {} \displaystyle \int _0^T\int _0^{{\overline{L}}(t)}s^2\beta \psi _{t}\alpha _x^2\psi dxdt \\= & {} \displaystyle \frac{s^2}{2}\int _0^T\beta \int _0^{{\overline{L}}(t)}\frac{d}{dt}\left( \psi \right) ^2\alpha _x^2 dxdt \\= & {} \displaystyle -\frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}\left( \beta '\alpha ^2_x + 2\beta \alpha _x\alpha _{xt} \right) \psi ^2dxdt;\\ I_3= & {} \left( (U\psi )_2,(V\psi )_1\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} 2s\beta ^2\alpha _x\psi _x\psi _{xx} dxdt \\= & {} \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _x\frac{d}{dx}\left( \psi _x\right) ^2 dxdt \\= & {} \displaystyle -\int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _{xx}\psi _x^2 dxdt \displaystyle + \int _{0}^{T}s\psi _x^2({\overline{L}}(t),t)\beta ^2\alpha _x({\overline{L}}(t),t)dt\\&\displaystyle - \int _{0}^{T}s\psi _x^2(0,t)\beta ^2\alpha _x(0,t)dt;\\ I_4= & {} \left( (U\psi )_2,(V\psi )_2\right) _{L^2(Q_{\overline{L}})} = \displaystyle \int _0^T\int _0^{{\overline{L}}(t)} 2s^3\beta ^2\alpha _x^3\psi \psi _x dxdt \\= & {} \displaystyle s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _x^3\frac{d}{dx}(\psi )^2 dxdt = \displaystyle -3s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _{x}^2\alpha _{xx}\psi ^2 dxdt. \end{aligned}$$
Therefore,
$$\begin{aligned} \displaystyle (U\psi ,V\psi )_{L^2(Q_{\overline{L}})}= & {} \displaystyle -\int _0^T\int _0^{{\overline{L}}(t)} s\beta ^2\alpha _{xx}\psi _x^2 dxdt -3s^3\int _0^T\int _0^{{\overline{L}}(t)} \beta ^2\alpha _{x}^2\alpha _{xx}\psi ^2 dxdt\\&\displaystyle -\frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}2\beta \alpha _x\alpha _{xt}\psi ^2dxdt + \int _{0}^{T}s\psi _x^2({\overline{L}}(t),t)\beta ^2\alpha _x(L(t),t)dt\\&\displaystyle + \frac{1}{2}\int _{0}^{T}\beta \left[ \psi _x({\overline{L}}(t),t) \right] ^2{\overline{L}}'(t)dt - \int _{0}^{T}s\psi _x^2(0,t)\beta ^2\alpha _x(0,t)dt\\&\displaystyle + \frac{1}{2}\int _0^T\int _{0}^{{\overline{L}}(t)}\beta '(\psi _{x})^2dxdt - \frac{s^2}{2}\int _0^T\int _0^{{\overline{L}}(t)}\beta '\alpha ^2_x\psi ^2dxdt. \end{aligned}$$
Now, computing \(\displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})}\), we have:
$$\begin{aligned} (U\psi ,V\psi )_{L^2(Q_{\overline{L}})} \displaystyle= & {} I_1 + I_2 + I_3 + I_4 \nonumber \\ \displaystyle\ge & {} C\Bigg (\frac{1}{2}\int _0^T \beta \psi ^2_x({\overline{L}}(t),t){\overline{L}}'(t) dt + \frac{1}{2}\int _0^T\beta '\int _0^{{\overline{L}}(t)}\psi ^2_xdxdt\Bigg ) \nonumber \\&\displaystyle + \Bigg ( \frac{-s^2}{2}\iint _{Q_{\overline{L}}} \beta '\alpha _x^2\psi ^2dxdt - C_2s^2\lambda ^2\iint _{Q_{\overline{L}}} \xi ^3|\psi |^2dxdt \Bigg )\nonumber \\&\displaystyle + \Bigg (\frac{sC_3}{2}\iint _{Q_{\overline{L}}} \lambda ^2\xi |\psi _x|^2 dxdt \nonumber \\&- s\iint _{\omega _0\times (0,T)} \lambda ^2C_3\xi |\psi _x|^2dxdt + s\lambda C_3\int _0^T \left[ \xi ({\overline{L}}(t),t)\psi _x^2({\overline{L}}(t),t) \right. \nonumber \\&\displaystyle \left. +\xi (0,t)\psi _x^2(0,t) \right] dt \Bigg ) + \Bigg ( C_4\iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3\psi ^2dxdt\nonumber \\&- \overline{C}\iint _{\omega _0\times (0,T)} s^3\lambda ^4\xi ^3\psi ^2dxdt \Bigg ); \end{aligned}$$
(4.4)
$$\begin{aligned} \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 \displaystyle= & {} \iint _{Q_{\overline{L}}} |\psi _t + 2\beta s\alpha _x\psi _x|^2dxdt \displaystyle = \iint _{Q_{\overline{L}}} \big |\psi _t + 2\beta s(\lambda \alpha _{0,x}\xi )\psi _x\big |^2dxdt \nonumber \\ \displaystyle\ge & {} C\Bigg ( \iint _{Q_{\overline{L}}} |\psi _t|^2dxdt - \iint _{Q_{\overline{L}}}s^2\lambda ^2\xi ^2|\psi _x|^2dxdt\Bigg ) \nonumber \\ \displaystyle\ge & {} \widetilde{C} \Bigg ( \iint _{Q_{\overline{L}}} (s\xi )^{-1} |\psi _t|^2dxdt - \iint _{Q_{\overline{L}}}(s\xi )\lambda ^2|\psi _x|^2dxdt\Bigg ); \end{aligned}$$
(4.5)
$$\begin{aligned} \displaystyle \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 \displaystyle= & {} \iint _{Q_{\overline{L}}} |\beta \psi _{xx} + \beta s^2\alpha _x^2\psi |^2dxdt\nonumber \\ \displaystyle= & {} \iint _{Q_{\overline{L}}} \big |\beta \psi _{xx} - (-\beta s^2\lambda ^2 \alpha _{0,x}^2\xi ^2\psi )\big |^2dxdt \nonumber \\ \displaystyle\ge & {} C\Bigg ( \iint _{Q_{\overline{L}}} |\psi _{xx}|^2dxdt - \iint _{Q_{\overline{L}}} s^4\lambda ^4\xi ^4|\psi |^2dxdt\Bigg ) \nonumber \\ \displaystyle\ge & {} \widetilde{C} \Bigg ( \iint _{Q_{\overline{L}}} (s\xi )^{-1} |\psi _{xx}|^2dxdt - \iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3|\psi |^2dxdt\Bigg ). \end{aligned}$$
(4.6)
Combining the estimates (4.3) and (4.4), they yield:
$$\begin{aligned}&\displaystyle C\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt \Bigg . +\displaystyle \Bigg . \iint _{Q_{\overline{L}}} s^2\lambda ^4\xi ^3|\psi |^2dxdt + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt \Bigg )\nonumber \\&\quad \ge \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + 2(U\psi ,V\psi )_{L^2(Q_{\overline{L}})}\nonumber \\&\qquad \displaystyle \Vert U\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \Vert V\psi \Vert _{L^2(Q_{\overline{L}})}^2 + \bigg ( \int _0^T \beta \psi ^2_x({\overline{L}}(t),t){\overline{L}}'(t) dt\bigg .\nonumber \\&\qquad + \displaystyle \int _0^T\beta '\int _0^{{\overline{L}}(t)}\psi ^2_xdxdt - s^2\iint _{Q_{\overline{L}}} \beta '\alpha _x^2\psi ^2dxdt\nonumber \\&\qquad -\displaystyle s^2\lambda ^2\iint _{Q_{\overline{L}}} \xi ^3|\psi |^2dxdt + s\iint _{Q_{\overline{L}}} \lambda ^2\xi |\psi _x|^2 dxdt \nonumber \\&\qquad + \displaystyle s\iint _{\omega _0\times (0,T)} \lambda ^2\xi |\psi _x|^2dxdt + s\lambda \int _0^T \left[ \xi ({\overline{L}}(t),t)\psi _x^2({\overline{L}}(t),t)\right. \nonumber \\&\qquad - \Bigg . \displaystyle \left. \xi (0,t)\psi _x^2(0,t) \right] dt + \iint _{Q_{\overline{L}}} s^3\lambda ^4\xi ^3\psi ^2dxdt - \iint _{\omega _0\times (0,T)} s^3\lambda ^4\xi ^3\psi ^2dxdt\Bigg ). \nonumber \\ \end{aligned}$$
(4.7)
Also, given \(\varepsilon > 0\), we have:
$$\begin{aligned} \iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2|\psi _x|^2dxdt&\le \varepsilon \iint _{Q_{\overline{L}}}(s\xi )^{-1}|\psi _{xx}|^2 dxdt \nonumber \\&\quad + C_{\varepsilon }\iint _{\omega _0\times (0,T)}(s\xi )^3\lambda ^4|\psi |^2dxdt \end{aligned}$$
(4.8)
In fact,
$$\begin{aligned}&\displaystyle \Bigg |\iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2\psi _x^2dxdt\Bigg | = \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2(\xi \psi _x)_x\psi dxdt\Bigg |\\ \\&\quad \le \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2\xi _x\psi _x\psi dxdt\Bigg | + \displaystyle \Bigg |\iint _{\omega _0\times (0,T)} s\lambda ^2\xi \psi _{xx}\psi dxdt\Bigg |\\ \\&\quad \le \displaystyle C\Bigg |\iint _{\omega _0\times (0,T)} (s\xi )(\lambda \psi _x)(\lambda ^2\psi )dxdt\Bigg | + \Bigg |\iint _{\omega _0\times (0,T)} (s\xi )\psi _{xx}(\lambda ^2\psi )dxdt\Bigg |\\ \\&\quad \le \displaystyle C\Bigg (\varepsilon _1 \iint _{\omega _0\times (0,T)} (s\xi )|\lambda \psi _x|^2dxdt + \frac{1}{4\varepsilon _1}\iint _{\omega _0\times (0,T)} (s\xi )|\lambda ^2\psi |^2dxdt \Bigg )\\ \\&\qquad + \displaystyle \Bigg (\varepsilon _2 \iint _{\omega _0\times (0,T)} (s\xi )^{-1}|\psi _{xx}|^2dxdt + \frac{1}{4\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}|\lambda ^2\psi |^2dxdt \Bigg ). \end{aligned}$$
Therefore,
$$\begin{aligned} \displaystyle \iint _{\omega _0\times (0,T)} (s\xi )\lambda ^2|\psi _x|^2dxdt\le & {} \displaystyle \frac{\varepsilon _2 }{1-C\varepsilon _1} \iint _{\omega _0\times (0,T)} (s\xi )^{-1}|\psi _{xx}|^2dxdt \\&+ C_{\varepsilon _1\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}\lambda ^4|\psi |^2dxdt \\\le & {} \displaystyle \frac{\varepsilon _2 }{1-C\varepsilon _1} \iint _{Q_{\overline{L}}} (s\xi )^{-1}|\psi _{xx}|^2dxdt \\&+ C_{\varepsilon _1\varepsilon _2}\iint _{\omega _0\times (0,T)} (s\xi )^{3}\lambda ^4|\psi |^2dxdt . \end{aligned}$$
Combining the inequalities (4.5)–(4.8), they yield:
$$\begin{aligned}&\displaystyle \iint _{Q_{\overline{L}}} (s\xi )^{-1} \left[ |\psi _t|^2 + |\psi _{xx}|^2\right] dxdt + \iint _{Q_{\overline{L}}} \left[ (s\xi )\lambda ^2|\psi _x|^2 + (s\xi )^3\lambda ^4|\psi |^2\right] dxdt \\&\qquad +\displaystyle (s\lambda ) \int _{0}^{T}\left[ \xi ({\overline{L}}(t),t)|\psi _x({\overline{L}}(t),t)|^2 + \xi (0,t)|\psi _x(0,t)|^2 \right] dt \\&\qquad + \int _0^T\beta '\int _0^{{\overline{L}}(t)}|\psi _x|^2dxdt \\&\quad \le \displaystyle C \,\Bigg ( \iint _{Q_{\overline{L}}} e^{-2s\alpha }|F|^2dxdt + \iint _{\omega \times (0,T)} s^3\lambda ^4\xi ^3|\psi |^2dxdt\\&\qquad + \iint _{Q_{\overline{L}}} |\psi _x|^2dxdt + \iint _{Q_{\overline{L}}} \beta '(s\xi )^2\lambda ^2|\psi |^2dxdt \Bigg ). \end{aligned}$$
Therefore,
$$\begin{aligned}&\displaystyle \iint _{Q_{\overline{L}}} \left[ (s\xi )^{-1}(|\psi _t|^2 + |\psi _{xx}|^2 ) + (s\xi )\lambda ^2|\psi _x|^2 + (s\xi )^3\lambda ^4|\psi |^2 \right] dxdt \\&\qquad + s\lambda \int _{0}^T \left[ \xi ({\overline{L}}(t),t)|\psi _x({\overline{L}}(t),t)|^2 + \xi (0,t)|\psi _x(0,t)|^2 \right] dt \\&\quad \le C \left( \iint _{Q_{\overline{L}}} e^{-2s\alpha } |F|^2 dxdt + \iint _{\omega \times (0,T)} (s\xi )^3\lambda ^4|\psi |^2dxdt \right) . \end{aligned}$$
Coming back to the original variable \(\varphi \), we have (2.4).
Appendix B: Proof of Lemma 2.2 and 2.3
Proof of Lemma 2.2
(i) For all \(0 \le \theta < 1\), we have:
$$\begin{aligned} \frac{| \beta (t_1) - \beta (t_2) |}{ |t_1 - t_2|^{\theta } }= & {} \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \Big | \beta \left( \int _0^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx \right) - \beta \left( \int _0^{\overline{L}(t_2)} \overline{y}(x,t_2) \ dx \right) \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _0^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx - \int _0^{\overline{L}(t_2)} \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _{\overline{L}(t_2)}^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx + \int _0^{\overline{L}(t_2)} \overline{y}(x,t_1) - \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot ( | \overline{L}(t_1) - \overline{L}(t_2) | + |t_1 - t_2| ) \\\le & {} C\frac{|t_1 - t_2|}{ |t_1 - t_2|^{\theta } } < \infty . \end{aligned}$$
Then, \(\beta \in C^{\theta }([0,T])\), for all \(0 \le \theta < 1\).
(ii) For all \(0 \le \theta < 1\), one has:
$$\begin{aligned}&\frac{| a(x_1,t_1) - a(x_2,t_2) |}{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le \frac{ \displaystyle { \int _0^1 \Big | \frac{\partial {g}}{\partial {s}}( \lambda \overline{y}(x_1,t_1), \lambda \overline{y}_x(x_1,t_1)) - \frac{\partial {g}}{\partial {s}}( \lambda \overline{y}(x_2,t_2), \lambda \overline{y}_x(x_2,t_2)) \Big | \ d\lambda } }{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le C \cdot \frac{( |\overline{y}(x_1,t_1) - \overline{y}(x_2,t_2)| + |\overline{y}_x(x_1,t_1) - \overline{y}_x(x_2,t_2)| )}{ (|x_1-x_2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\qquad \le C \cdot \frac{ ( |x_1-x_2| + |t_1 - t_2| ) }{ ( |x_1-x_2| + |t_1 - t_2|^{1/2} )^{\theta } } < \infty . \end{aligned}$$
Then, \(a \in C_{x,t}^{\theta , \theta /2}(\overline{Q}_L)\), for all \(0 \le \theta < 1\). The proof is analogous to b(x, t).
(iii) For all \(0 \le \theta <1\), one has:
$$\begin{aligned} \frac{| \widetilde{a}(\xi _1, t_1) - \widetilde{a}(\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta }}= & {} \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }\\= & {} \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2)| }{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \cdot \\&\frac{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } , \end{aligned}$$
On the other hand, from (ii), we have
$$\begin{aligned} \displaystyle { \frac{| {a}( L(t_1)\xi _1, t_1) - {a}( L(t_2)\xi _2, t_2)| }{ (|L(t_1)\xi _1-L(t_2)\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } } \le C. \end{aligned}$$
Then, one has,
$$\begin{aligned} \frac{| \widetilde{a}(\xi _1, t_1) - \widetilde{a}(\xi _2, t_2) |}{(|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } }\le & {} C\cdot \frac{ (L(t_1)| \xi _1 - \xi _2 | + |\xi _2 | .|L(t_1) - L(t_2)| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \\&\le C_1\cdot \frac{ (| \xi _1 - \xi _2 | + |L(t_1) - L(t_2)| + |t_1 - t_2|^{1/2})^{\theta } }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\theta } } \\&< \infty , \quad \forall \theta \in [0,1). \end{aligned}$$
Then, \(\widetilde{a} \in C_{\xi ,t}^{\theta , \theta /2}(\overline{Q})\), for all \(0 \le \theta < 1\).
In a similar way, we obtain \(\widehat{b}(\xi ,t) = b(L(t)\xi ,t) \in C_{\xi ,t}^{\theta , \theta /2}(\overline{Q})\). Let us analyse the function \(\displaystyle { \overline{b}(\xi ,t) = \xi \frac{L'(t)}{L(t)} }\), for \(L\in C^{1+ \gamma /2}([0,T])\).
For all \(0 \le \gamma < 1\),
$$\begin{aligned}&\frac{| \overline{b}(\xi _1, t_1) - \overline{b}(\xi _2, t_2) |}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } }\\&\quad = \frac{1}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } \cdot \Big | \xi _1\frac{L'(t_1)}{L(t_1)} - \xi _2\frac{L'(t_2)}{L(t_2)} \Big | \\&\quad \le \frac{C}{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } \cdot \left( | \xi _1 - \xi _2| \cdot \Big | \frac{L'(t_1)}{L(t_1)} \Big | + |\xi _2| \cdot \Big | \frac{L'(t_1)}{L(t_1)} - \frac{L'(t_2)}{L(t_2)} \Big | \right) \\&\quad \le C_1 \cdot \frac{\left( \displaystyle { | \xi _1 - \xi _2| + |L'(t_1) - L'(t_2)|\frac{1}{|L(t_1)|} + |L'(t_2)| \cdot \frac{|L(t_2)-L(t_1)|}{|L(t_1)L(t_2)|} } \right) }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } }\\&\quad \le C_2 \cdot \frac{\left( \displaystyle { | \xi _1 - \xi _2| + |L'(t_1) - L'(t_2)| + |L(t_2)-L(t_1)| } \right) }{ (|\xi _1-\xi _2| + |t_1 - t_2|^{1/2})^{\gamma } } < \infty , \end{aligned}$$
since \(L\in C^{1+ \gamma /2}([0,T])\).
Then, \(\widetilde{b} \in C_{\xi ,t}^{\gamma , \gamma /2}(\overline{Q})\), for all \(0 \le \gamma < 1\). \(\square \)
Proof of Lemma 2.3
For \(0 \le \theta < 1\), one has:
$$\begin{aligned}&\frac{| L'(t_1) - L'(t_2) |}{ |t_1 - t_2|^{\theta } } \\&\quad = \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \left[ \displaystyle \beta \left( \int _{0}^{\overline{L}(t_1)} y(x,t_1)dx\right) y_x(\overline{L}(t_1),t_1) \right. \\&\qquad \left. - \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) y_x(\overline{L}(t_2),t_2) \right] \\&\quad =\frac{1}{ |t_1 - t_2|^{\theta } } \cdot {\displaystyle { \left[ \displaystyle { \beta \left( \int _{0}^{\overline{L}(t_1)} y(x,t_1)dx\right) - \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) } \right] y_x(\overline{L}(t_1),t_1) }} \\&\qquad + \displaystyle { \frac{1}{ |t_1 - t_2|^{\theta } } \cdot \displaystyle { \beta \left( \int _{0}^{\overline{L}(t_2)} y(x,t_2)dx\right) \cdot \left[ y_x(\overline{L}(t_1),t_1) - y_x(\overline{L}(t_2),t_2) \right] } } \\&\quad = I_1 + I_2 \end{aligned}$$
For \(0 \le \theta < 1\), we obtain the following estimates for \(I_1\) and \(I_2\):
$$\begin{aligned} |I_1|\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \displaystyle { {\displaystyle { \displaystyle { \int _{0}^{\overline{L}(t_1)} y(x,t_1) \ dx - \int _{0}^{\overline{L}(t_2)} y(x,t_2) \ dx } }} } \Big | \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot \Big | \int _{\overline{L}(t_2)}^{\overline{L}(t_1)} \overline{y}(x,t_1) \ dx + \int _0^{\overline{L}(t_2)} \overline{y}(x,t_1) - \overline{y}(x,t_2) \ dx \Big | \\\le & {} \frac{C_1}{ |t_1 - t_2|^{\theta } } \cdot |\overline{L}(t_1) - \overline{L}(t_2)| + c\int _0^{\overline{L}(t_2)} \frac{ |\overline{y}(x,t_1) - \overline{y}(x,t_2)| }{|t_1 - t_2|^{\theta }} \ dx \\\le & {} C_2 \left( \sup _{t_1, t_2\in [0,T]} \frac{|\overline{L}(t_1) - \overline{L}(t_2)|}{|t_1 - t_2|^{\theta }} + \sup _{t_1, t_2\in [0,T]} \frac{|\overline{y}(x,t_1) - \overline{y}(x,t_2)|}{|t_1 - t_2|^{\theta }} \right) < \infty , \end{aligned}$$
and
$$\begin{aligned} |I_2|\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot |y_x(\overline{L}(t_1),t_1) - y_x(\overline{L}(t_2),t_2)| \\\le & {} \frac{C}{ |t_1 - t_2|^{\theta } } \cdot ( |\overline{L}(t_1) - \overline{L}(t_2)| + |t_1 - t_2| ) < \infty . \end{aligned}$$
\(\square \)
