A Appendix: Proof of Theorem 1
It is sufficient to prove Theorem 1 assuming that \(b_0=0\).
Let v be the solution to (6) (where \(v_T\in L^2(0,1)\) and \(g\in L^2(Q)\)). For any \(s\ge s_0>0\), we set \(z=e^{-s\sigma }v\). By a density argument, we can assume without loss of generality that v is regular enough. We have
$$\begin{aligned} v_t=e^{s\sigma }(s\sigma _tz+z_t), \ \ (x^\alpha v_x)_x=e^{s\sigma }[s^2\sigma _x^2x^\alpha z+2s\sigma _xx^\alpha z_x+s(\sigma _xx^\alpha )_xz+(x^\alpha z_x)_x] \end{aligned}$$
and, consequently,
$$\begin{aligned} P^+z+P^-z=e^{-s\sigma }g, \end{aligned}$$
where \(P^+z=s\sigma _tz+s^2x^\alpha \sigma _x^2z+(x^\alpha z_x)_x\) and \(P^-z=z_t+s(x^\alpha \sigma _x)_xz+2sx^\alpha \sigma _xz_x.\) This gives
$$\begin{aligned} \Vert e^{-s\sigma }g\Vert ^2=\Vert P^+z\Vert ^2+\Vert P^-z\Vert ^2+2(\!( P^+z,P^-z)\!). \end{aligned}$$
(42)
We have that \((\!( P^+z,P^-z)\!)=I_1+\cdots +I_4\), where
$$\begin{aligned} I_1= & {} \left( \!\left( s\sigma _tz+s^2\sigma _x^2x^\alpha z+\left( x^\alpha z_x\right) _x,z_t\right) \!\right) \\ I_2= & {} s^2\left( \!\left( \sigma _tz,\left( x^\alpha \sigma _x\right) _xz+2\sigma _x x^\alpha z_x\right) \!\right) \\ I_3= & {} s^3\left( \!\left( \sigma _x^2x^\alpha z, \left( x^\alpha \sigma _x\right) _xz+2\sigma _x x^\alpha z_x\right) \!\right) \\ I_4= & {} s\left( \!\left( \, \left( x^\alpha z_x\right) _x, \left( x^\alpha \sigma _x\right) _xz+2\sigma _x x^\alpha z_x\right) \!\right) . \end{aligned}$$
The next step is to compute \(I_1,\ I_2, \ I_3 \) and \(I_4\). To this purpose, we will use that \(z=z_x=0\) at \(t=0\) and \(t=T\) and, also,
$$\begin{aligned} \displaystyle \int \!\!\!\!\int _Q(x^\alpha z_x)_xz_t\,\mathrm{d}x\,\mathrm{d}t=0. \end{aligned}$$
After integrating by parts, we deduce easily that
$$\begin{aligned} I_1= & {} -\frac{s}{2}\int \!\!\!\!\int _Q|z|^2(\sigma _{tt}+2s\sigma _x\sigma _{xt}x^\alpha )\,\mathrm{d}x\,\mathrm{d}t\\ I_2= & {} -s^2\int \!\!\!\!\int _Qx^\alpha \sigma _x\sigma _{xt}|z|^2\mathrm{d}x\,\mathrm{d}t;\\ I_3= & {} -s^3\int \!\!\!\!\int _Qx^\alpha \sigma _x(x^\alpha \sigma _x^2)_x|z|^2\,\mathrm{d}x\,\mathrm{d}t;\\ I_4= & {} -s\int \!\!\!\!\int _Q\left( x^\alpha \sigma _x\right) _{xx}x^\alpha zz_x\mathrm{d}x\,\mathrm{d}t-2s\int \!\!\!\!\int _Q\left( x^\alpha \sigma _x\right) _xx^\alpha |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\\&+\,s\alpha \int \!\!\!\!\int _Q\sigma _xx^{2\alpha -1}|z_x|^2\,\mathrm{d}x\,\mathrm{d}t +\,\left. s\int _0^T\sigma _xx^{2\alpha }|z_x|^2\mathrm{d}t\right| _{x=0}^{x=1}. \end{aligned}$$
Consequently,
$$\begin{aligned} (\!( P^+z,P^-z)\!)\!= & {} \!\!\left. s\!\int _0^T\!\!\sigma _xx^{2\alpha }|z_x|^2\,\mathrm{d}t\right| _{x=0}^{x=1}\!-s^3\!\!\int \!\!\!\!\int _Q\!x^\alpha \sigma _x(x^\alpha \sigma _x^2)_x|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -2s\!\int \!\!\!\!\int _Q\!(x^\alpha \sigma _x)_xx^\alpha |z_x|^2\,\mathrm{d}x\,\mathrm{d}t -2s^2\int \!\!\!\!\int _Q\!\!\sigma _x\sigma _{xt}x^\alpha |z|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad +\alpha s\int \!\!\!\!\int _Q\!\sigma _xx^{2\alpha -1}|z_x|^2\,\mathrm{d}x\,\mathrm{d}t-s\int \!\!\!\!\int _Q\!\!(x^\alpha \sigma _x)_{xx}x^\alpha zz_x\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -\frac{s}{2}\int \!\!\!\!\int _Q\!\!\sigma _{tt}|z|^2\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(43)
Using the definitions of the functions \(\sigma \) and \(\xi \), we see that
$$\begin{aligned} \sigma _x\sigma _{xt}x^\alpha= & {} \lambda ^2x^\alpha |\eta '|^2\xi \xi _t\\ x^\alpha \sigma _x(x^\alpha \sigma _x^2)_x= & {} -\,\lambda ^3x^\alpha \eta '(x^\alpha |\eta '|^2)_x\xi ^3-\,2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3\\ (x^\alpha \sigma _x)_xx^\alpha= & {} -\,\lambda x^\alpha (x^\alpha \eta ')_x\xi -\,\lambda ^2x^{2\alpha }|\eta '|^2\xi \\ \sigma _xx^{2\alpha -\,1}= & {} -\,\lambda x^{2\alpha -\,1}\eta '\xi \\ (x^\alpha \sigma _x)_{xx}x^\alpha= & {} -\,\lambda x^\alpha (x^\alpha \eta ')_{xx}\xi -\,2\lambda ^2x^\alpha \eta '(x^\alpha \eta ')_x\xi -\,\lambda ^2x^{2\alpha }\eta '\eta ''\xi -\,\lambda ^3x^{2\alpha }(\eta ')^3\xi \\ \sigma _xx^{2\alpha }= & {} -\,\lambda x^{2\alpha }\eta '\xi . \end{aligned}$$
With this information, we will now estimate each term in the right-hand side of (43). For \(s_0\) and \(\lambda _0\) large enough, we have
$$\begin{aligned} \left. s\int _0^T\!\!\sigma _xx^{2\alpha }|z_x|^2\,\mathrm{d}t\right| _{x=0}^{x=1}\ge & {} 0 , \end{aligned}$$
(44)
$$\begin{aligned} -\,s^3\int \!\!\!\!\int _Q\!\!x^\alpha \sigma _x(x^\alpha \sigma _x^2)_x|z|^2\,\mathrm{d}x\,\mathrm{d}t\ge & {} C\left( s^3\lambda ^4\int \!\!\!\!\int _Q\!\!x^{2\alpha }|\eta '|^4\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\right. \nonumber \\&\left. +\,s^3\lambda ^3\int \!\!\!\!\int _{Q_0}\!\!\!x^{2-\,\alpha }\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) \nonumber \\&\quad \!\!\!\!\!-\,Cs^3\lambda ^3\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t, \end{aligned}$$
(45)
$$\begin{aligned} -\,2s\int \!\!\!\!\int _Qx^\alpha (x^\alpha \sigma _x)_x|z_x|^2\,\mathrm{d}x\,\mathrm{d}t\ge & {} Cs\lambda ^2\int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^2\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\!\!\!+\,2s\lambda \int \!\!\!\!\int _{Q_0}\!\!\!x^\alpha \xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t-\,Cs\lambda \!\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t, \nonumber \\ \end{aligned}$$
(46)
$$\begin{aligned} -\,2s^2\int \!\!\!\!\int _Qx^\alpha \sigma _x\sigma _{xt}|z|^2\,\mathrm{d}x\,\mathrm{d}t\!\!\ge & {} \!\! -\,Cs^2\lambda ^2\left( \int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^4\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t \right. \nonumber \\&\!\left. +\int \!\!\!\!\int _{Q_0}\!\!\!x^{2-\,\alpha }\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t +\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) \nonumber \\ \end{aligned}$$
(47)
and
$$\begin{aligned} \alpha s\int \!\!\!\!\int _Qx^{2\alpha -\,1}\sigma _x|z_x|^2\,\mathrm{d}x\,\mathrm{d}t \ge -\,\alpha s\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!x^\alpha \xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t-\,C s\lambda \!\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\!\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t,\nonumber \\ \end{aligned}$$
(48)
where C only depends on \(a_0\), T and \(\alpha \). On the other hand,
$$\begin{aligned} -\,s\!\int \int _Q\!x^\alpha (x^\alpha \sigma _x)_{xx}zz_x\mathrm{d}x\,\mathrm{d}t= & {} s\lambda \!\int \int _Q\!x^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\mathrm{d}x\,\mathrm{d}t\nonumber \\&+s\lambda ^2\!\int \!\!\!\!\int _Q\!x^{2\alpha }\eta '\eta ''\xi zz_x\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad +\,2s\lambda ^2\!\int \!\!\!\!\int _Q\!x^\alpha \eta '(x^\alpha \eta ')_x\xi zz_x\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad +\,s\lambda ^3\!\int \!\!\!\!\int _Q\!x^{2\alpha }(\eta ')^3\xi zz_x\mathrm{d}x\,\mathrm{d}t . \end{aligned}$$
(49)
In order to estimate this last term, we note that
$$\begin{aligned} s\lambda \int \!\!\!\!\int _Qx^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} -\,Cs\lambda \int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( \xi ^3|z|^2+\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t, \end{aligned}$$
(50)
$$\begin{aligned} s\lambda ^2\int \!\!\!\!\int _Qx^{2\alpha }\eta '\eta ''\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} -\,C\int \!\!\!\!\int _Q\left( s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-\,C\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^2\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-\,C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^2\lambda ^3\xi ^3 |z|^2+\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t, \end{aligned}$$
(51)
$$\begin{aligned} 2s\lambda ^2\int \!\!\!\!\int _Qx^\alpha \eta '(x^\alpha \eta ')_x\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} -\,C\int \!\!\!\!\int _Q\left( s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-\,C\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^2\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-\,C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^2\lambda ^3\xi ^3 |z|^2+\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \end{aligned}$$
(52)
and
$$\begin{aligned} s\lambda ^3\!\int \!\!\!\!\int _Q\!x^{2\alpha }(\eta ')^3\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\!\!\ge & {} \!\!-\,C\int \!\!\!\!\int _Q\left( \lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2+s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3 |z|^2\right) \,\mathrm{d}x\,\mathrm{d}t.\nonumber \\ \end{aligned}$$
(53)
By (49)–(53), we conclude that
$$\begin{aligned}&-\,s\int \!\!\!\!\int _Qx^\alpha (x^\alpha \sigma _x)_{xx}zz_x\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad \ge -\,C\int \!\!\!\!\int _Q\left( s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad -\,C\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^2\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\qquad -\,C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^2\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(54)
From (43)–(48) and (54), we conclude that
$$\begin{aligned} (\!( P^+z,P^-z)\!)\ge & {} C\left( \int \!\!\!\!\int _Q\left( s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \right. \\&\left. +\int _{Q_0}\!\!\left( s^3\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+s\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) \nonumber \\&-\,C\left( \int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t+s\int \!\!\!\!\int _Q\xi ^{3/2}|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) . \end{aligned}$$
Using (42), we find that
$$\begin{aligned}&\Vert P^+z\Vert ^2+\Vert P^-\,z\Vert ^2+\int \!\!\!\!\int _Q\left( s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad +\int \!\!\!\!\int _{Q_0}\!\!\left( s^3\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+s\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-\,s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t+s\int \!\!\!\!\int _Q\xi ^{3/2}|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) .\nonumber \\ \end{aligned}$$
(55)
Furthermore, we see that
$$\begin{aligned} s^3\lambda ^3\int \!\!\!\!\int _Q\xi ^3x^{2-\,\alpha }|z|^2\,\mathrm{d}x\,\mathrm{d}t\le & {} Cs^3\lambda ^3\left( \int \!\!\!\!\int _{Q_0\!}\xi ^3x^{2-\,\alpha }|z|^2\,\mathrm{d}x\,\mathrm{d}t\right. \\&\left. +\int \!\!\!\!\int _Q\xi ^3x^{2\alpha }|\eta '|^4|z|^2\,\mathrm{d}x\,\mathrm{d}t+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) \end{aligned}$$
and
$$\begin{aligned} s\lambda \int \!\!\!\!\int _Q\xi x^\alpha |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\le & {} Cs\lambda \left( \int \!\!\!\!\int _{Q_0}\!\xi x^\alpha |z_x|^2\,\mathrm{d}x\,\mathrm{d}t+\int \!\!\!\!\int _Q\xi x^{2\alpha }|\eta '|^2|z_x|^2\,\mathrm{d}x\,\mathrm{d}t\right. \\&\left. +\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\right) . \end{aligned}$$
Hence, from (55), we obtain:
$$\begin{aligned}&\Vert P^+z\Vert ^2+\Vert P^-\,z\Vert ^2+\int \!\!\!\!\int _Q\left( s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad +\int \!\!\!\!\int _Q\!\!\left( s^3\lambda ^3x^{2-\,\alpha }\xi ^3|z|^2+s\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-\,s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t+s\int \!\!\!\!\int _Q\xi ^{3/2}|z|^2\,\mathrm{d}x\,\mathrm{d}t\right) .\nonumber \\ \end{aligned}$$
(56)
Let us denote by L(z) all the terms in the left-hand side of (56). For instance, assume that \(\alpha \ne 1\). Using Young and Hardy’s Inequality we deduce that
$$\begin{aligned} s^2\lambda ^2\int \!\!\!\!\int _Q\xi ^2|z|^2\,\mathrm{d}x\,\mathrm{d}t\le & {} s^3\lambda ^3\int \!\!\!\!\int _Q\xi ^3x^{2-\alpha }|z|^2\,\mathrm{d}x\,\mathrm{d}t+s\lambda \int \!\!\!\!\int _Qx^{\alpha -2}|\xi ^{1/2}z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\\le & {} s^3\lambda ^3\int \!\!\!\!\int _Q\xi ^3x^{2-\alpha }|z|^2\,\mathrm{d}x\,\mathrm{d}t+s\lambda \int \!\!\!\!\int _Qx^\alpha |(\xi ^{1/2}z)_x|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\\le & {} CL(z). \end{aligned}$$
(57)
Now, let us assume that \(\alpha =1\). Using Hölder, Young and Hardy inequalities, we see that
$$\begin{aligned} s^{3/2}\lambda ^{7/4}\int \!\!\!\!\int _Q\xi ^{3/2}|z|^2\,\mathrm{d}x\,\mathrm{d}t= & {} \int \!\!\!\!\int _{Q_0}(s^3\lambda ^4x^2\xi ^3|z|^2)^{1/4}.(s\lambda x^{-2/3}\xi |z|^2)^{3/4}\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&+s^{3/2}\lambda ^{7/4}\int _0^T\!\!\!\int _{a_0}^1\xi ^{3/2}|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\\le & {} \quad Cs^3\lambda ^4\int \!\!\!\!\int _Q\xi ^3x^{2\alpha }|\eta '|^4|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&+s\lambda \int \!\!\!\!\int _Qx^{-\,2+4/3}(\xi ^{1/2}z)^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\\le & {} CL(z). \end{aligned}$$
(58)
From (56), (57) and (58), we deduce that
$$\begin{aligned}&\Vert P^+z\Vert ^2+\Vert P^-z\Vert ^2+\int \!\!\!\!\int _Q\left( s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad +\int \!\!\!\!\int _Q\!\!\left( s^3\lambda ^3x^{2-\alpha }\xi ^3|z|^2+s^2\lambda ^2\gamma _1(\lambda )\xi ^2\gamma _2(s\xi )|z|^2+s\lambda x^\alpha \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) . \end{aligned}$$
(59)
Now, we will work to include the terms with a first-order time derivative and second-order spatial derivatives in the left-hand side. Using the estimate (59) and the definitions of \(P^-z\) and \(P^+z\), we have
$$\begin{aligned}&s^{-1}\gamma _1(\lambda )\int \!\!\!\!\int _Q\xi ^{-1}|z_t|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) \end{aligned}$$
(60)
and
$$\begin{aligned}&s^{-1}\gamma _1(\lambda )\!\!\int \!\!\!\!\int _Q\xi ^{-1}|(x^\alpha z_x)_x|^2\,\mathrm{d}x\,\mathrm{d}t \le C\left( \Vert e^{-s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) \!.\nonumber \\ \end{aligned}$$
(61)
By (59), (60) and (61), it is now clear that
$$\begin{aligned}&\int \!\!\!\!\int _Q\left[ s^{-1}\gamma (\lambda )\xi ^{-1}\left( |z_t|^2+|(x^\alpha z_x)_x|^2\right) +s\lambda x^\alpha \xi |z_x|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right] \,\mathrm{d}x\,\mathrm{d}t \\&\qquad +\int \!\!\!\!\int _Q\left( s^2\lambda ^2\gamma _1(\lambda )\xi ^2\gamma _2(s\xi )|z|^2+s^3\lambda ^3x^{2-\alpha }\xi ^3|z|^2+s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2\right) \,\mathrm{d}x\,\mathrm{d}t\\&\quad \le C\left( \Vert e^{-s\sigma }g\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) . \end{aligned}$$
Coming back to the original variable v and using the estimate
$$\begin{aligned} \int _{\omega _{0T}}\!\!\!\!e^{-2s\sigma }\xi |v_x|^2\mathrm{d}x\,\mathrm{d}t\le & {} Cs^2\!\lambda ^2\int \!\!\!\!\int _{\omega _T}\!\!\!\!e^{-2s\sigma }\xi ^3 |v|^2\mathrm{d}x\,\mathrm{d}t\nonumber \\&+\,Cs^{-2}\lambda ^{-2}\int \!\!\!\!\int _Qe^{-2s\sigma }\xi ^{-1}|(x^\alpha v_x)_x|^2\,\mathrm{d}x\,\mathrm{d}t, \end{aligned}$$
we find the desired inequality (20).\(\Box \)
B Appendix: Proof of Theorem 2
Again, we will assume that \(b_0=0\). Let us set \(g_0:=g-b_1v_x\). Arguing as before, we can deduce that
$$\begin{aligned} (\!( P^+z,P^-z)\!)= & {} s\!\left. \int _0^T\!\sigma _xx^{2\alpha }|z_x|^2\,\mathrm{d}t\right| _{x=0}^{x=1}-s^3\!\int \!\!\!\!\int _Qx^\alpha \sigma _x(x^\alpha \sigma _x^2)_x|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-2s\!\int \!\!\!\!\int _Q\!(x^\alpha \sigma _x)_xx^\alpha |z_x|^2\,\mathrm{d}x\,\mathrm{d}t-2s^2\!\int \!\!\!\!\int _Q\!\sigma _x\sigma _{xt}x^\alpha |z|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&+\alpha s\!\int \!\!\!\!\int _Q\!\sigma _xx^{2\alpha -1}|z_x|^2\,\mathrm{d}x\,\mathrm{d}t-\frac{s}{2}\int \!\!\!\!\int _Q\sigma _{tt}|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-s\int \!\!\!\!\int _Q(x^\alpha \sigma _x)_{xx}x^\alpha zz_x\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(62)
Moreover, estimating the terms on the right-hand side of (62), we obtain that
$$\begin{aligned} (\!( P^+z,P^-z)\!)\ge & {} C\left[ s^3\lambda ^4\int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^4\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t+s^3\lambda ^3\int \!\!\!\!\int _{Q_0}\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\right. \nonumber \\&\left. s\lambda ^2\int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^2\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\right] \nonumber \\&+\frac{1}{3}(2-\alpha )s\lambda \int \!\!\!\!\int _{Q_0}\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&-C\left[ s^3\lambda ^3\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t+s\lambda \int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\right] \nonumber \\&-s\int \!\!\!\!\int _Q(x^\alpha \sigma _x)_{xx}x^\alpha zz_x\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(63)
Now, we can estimate the last term in the right-hand side of (63). It is just here where the restriction \(\alpha <1/2\) is required. We have
$$\begin{aligned} -s\!\int \!\!\!\!\int _Qx^\alpha (x^\alpha \sigma _x)_{xx}zz_x\,\mathrm{d}x\,\mathrm{d}t= & {} s\lambda ^2\!\!\int \!\!\!\!\int _Qx^{2\alpha }\eta '\eta ''\xi zz_x\,\mathrm{d}x\,\mathrm{d}t+s\lambda ^3\!\!\int \!\!\!\!\int _Qx^{2\alpha }(\eta ')^3\xi zz_x\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad +2s\lambda ^2\!\!\int \!\!\!\!\int _Qx^\alpha \eta '(x^\alpha \eta ')_x\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad +s\lambda \!\int \!\!\!\!\int _Qx^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t \nonumber \\\ge & {} -C\int \!\!\!\!\int _Q\left( s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad -C\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^2\lambda ^3\xi ^3|z|^2+\lambda x^{(4\alpha -2)/3}\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad -C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^2\lambda ^3\xi ^3 |z|^2+\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad + s\lambda \!\int \!\!\!\!\int _Qx^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(64)
In the previous case we had \((x^\alpha \eta ')_{xx}=0\) in \((0,1)\backslash \omega _0\). Here, this is lost and, therefore, in order to estimate the last term in the right-hand side of (64), we have to work differently.
Using integration by parts, we have that
$$\begin{aligned}&s\lambda \!\int \!\!\!\!\int _Q\!x^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad =-\int \!\!\!\!\int _Q\frac{s\lambda }{2}\xi |z|^2\left[ \lambda x^\alpha \eta '(x^\alpha \eta ')_{xx}+[x^\alpha (x^\alpha \eta ')_{xx}]_x\right] \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad \ge Cs\lambda ^2\!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(2\alpha -4)/3}\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t-Cs\lambda ^2\!\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\qquad -\,Cs\lambda ^2\!\int \!\!\!\!\int _{(b_0,1)\times (0,T)}x^{(2\alpha -4)/3}\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad -\frac{s\lambda }{2}\int \!\!\!\!\int _Q\xi |z|^2[x^\alpha (x^\alpha \eta ')_{xx}]_x\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(65)
In view of Proposition 2, we have
$$\begin{aligned}&-\frac{s\lambda }{2}\!\int \!\!\!\!\int _Q[x^\alpha (x^\alpha \eta ')_{xx}]_x\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \ge \ -\frac{2}{3}s\lambda \frac{(1+\alpha )(2-\alpha )}{5-4\alpha }\int \!\!\!\!\int _{Q_0}\!\!x^{(4\alpha -2)/3}(\xi ^{1/2}z)_x^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad -\,Cs\lambda \int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3 |z|^2\,\mathrm{d}x\,\mathrm{d}t-\,Cs\lambda \int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^4\xi ^3 |z|^2\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(66)
Furthermore,
$$\begin{aligned}&\frac{2}{3}s\lambda \frac{(1+\alpha )(2-\alpha )}{5-4\alpha }\int \!\!\!\!\int _{Q_{0}}x^{(4\alpha -2)/3}(\xi ^{1/2}z)_{x}^{2}\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad =-m(\alpha )s\lambda \int \!\!\!\!\int _{Q_{0}}x^{(4\alpha -2)/3}\left[ \xi |z_{x}|^{2}+\lambda x^{(1-2\alpha )/3}\xi zz_{x}\!+\!\frac{\lambda ^{2}}{4}x^{(2-4\alpha )/3}\xi |z|^{2}\right] \,\mathrm{d}x\,\mathrm{d}t,\nonumber \\ \end{aligned}$$
(67)
where \(m(\alpha ):=\frac{2}{3}(1+\alpha )(2-\alpha )(5-4\alpha )^{-1}\) and
$$\begin{aligned} -Cs\lambda ^2\!\!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(2\alpha -1)/3}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} -C\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!\left( \lambda x^{(4\alpha -2)/3}\xi |z_x|^2+s^2\lambda ^3\xi ^3|z|^2\right) \,\mathrm{d}x\,\mathrm{d}t.\nonumber \\ \end{aligned}$$
(68)
Then from (66), (67) and (68), we get
$$\begin{aligned} -\frac{s\lambda }{2}\int \!\!\!\!\int _Q[x^\alpha (x^\alpha \eta ')_{xx}]_x\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t\ge & {} \ -m(\alpha )s\lambda \int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-Cs^2\lambda ^3\!\!\int \!\!\!\!\int _{Q_0}\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-Cs\lambda \!\int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^4\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-C\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&-Cs\lambda \!\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(69)
Hence, from (69) and (65) we deduce that
$$\begin{aligned} s\lambda \!\int \!\!\!\!\int _Qx^\alpha (x^\alpha \eta ')_{xx}\xi zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} Cs\lambda ^2\!\!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(2\alpha -4)/3}\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t-Cs^2\lambda ^3\!\!\int \!\!\!\!\int _{Q_0}\!\!\!\xi ^3|z|^2\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -m(\alpha )s\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad -C\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -Cs\lambda ^2\!\!\int \!\!\!\!\int _Qx^{2\alpha }|\eta '|^4\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -Cs\lambda ^2\!\!\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\xi ^3|z|^2\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(70)
Note that the new strategy makes appear an additional term in the right-hand side of (70).
From (64) and (70), we conclude that
$$\begin{aligned} -s\!\int \!\!\!\!\int _Qx^\alpha (x^\alpha \sigma _x)_{xx}zz_x\,\mathrm{d}x\,\mathrm{d}t\ge & {} -C\int \!\!\!\!\int _Q\left( \lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2+s^2\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad +Cs\lambda ^2\!\!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(2\alpha -4)/3}\xi |z|^2\,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^2\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -C\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^2\lambda ^3\xi ^3|z|^2+\lambda x^{(4\alpha -2)/3}\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\nonumber \\&\quad -m(\alpha )s\lambda \!\int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\,\mathrm{d}x\,\mathrm{d}t. \end{aligned}$$
(71)
It is important to take care of the constants accompanying the term
$$\begin{aligned} s\lambda \int \!\!\!\!\int _{Q_0}\!\!\!x^{(4\alpha -2)/3}\xi |z_x|^2\mathrm{d}x\,\mathrm{d}t \end{aligned}$$
in the estimates (63) and (71). The sum of these constants is
$$\begin{aligned} \frac{(2-\alpha )(1-2\alpha )}{5-4\alpha }, \end{aligned}$$
that is only positive for \(\alpha \in [0,1/2)\).
From (62), (63) and (71), we deduce that
$$\begin{aligned}&\Vert P^+z\Vert ^2+\Vert P^-z\Vert ^2+\int \!\!\!\!\int _Q\left( s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|z|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\ \ \ \ \ \ \nonumber \\&\quad +\int \!\!\!\!\int _{Q_0}\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda ^2x^{(2\alpha -4)/3}\xi |z|^2+s\lambda x^{(4\alpha -2)/3}\xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-s\sigma }g_0\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right) \,\mathrm{d}x\,\mathrm{d}t\right) .\ \end{aligned}$$
Arguing as in Appendix A, we can replace the integral in \(Q_0\) in the left-hand side by integrals in Q. Moreover, thanks to the additional term in the left-hand side of this inequality, we can incorporate terms with time derivatives and second-order spatial derivatives more easily and also return to the variable v:
$$\begin{aligned}&\int \!\!\!\!\int _Qe^{-2s\sigma }\left[ s^{-1}\xi ^{-1}\left( |v_t|^2+|(x^\alpha v_x)_x|^2\right) +s\lambda x^{(4\alpha -2)/3}\xi |v_x|^2+s\lambda ^2x^{2\alpha }|\eta '|^2\xi |v_x|^2\right] \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\qquad +\int \!\!\!\!\int _Qe^{-2s\sigma }|v|^2\left[ s\lambda ^2x^{(2\alpha -4)/3}\xi +s^3\lambda ^3\xi ^3+s^3\lambda ^4x^{2\alpha }|\eta '|^4\xi ^3|v|^2\right] \,\mathrm{d}x\,\mathrm{d}t \nonumber \\&\quad \le C\left( \Vert e^{-s\sigma }(g-b_1v_x)\Vert ^2+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!e^{-2s\sigma }\left[ s^3\lambda ^3\xi ^3|z|^2+s\lambda \xi |z_x|^2\right] \,\mathrm{d}x\,\mathrm{d}t\right) . \end{aligned}$$
(72)
Since the power of x in the local term with first-order spatial derivatives is negative, we deduce that (72) remains true with \((g-b_1v_x)\) replaced by g. Finally, to eliminate the term with derivatives in the right-hand side, it suffices to work as in the case considered in Appendix A. \(\Box \)
C Appendix: Proof of Theorem 3
Again we will prove Theorem 3 when \(B=0\). In view of the presence of Robin conditions in (12), we will perform another change of variables:
$$\begin{aligned} v(y,t)=e^{-\eta (y)}w(y,t). \end{aligned}$$
Now, (12) becomes
$$\begin{aligned} \left\{ \begin{array} [c]{lll} v_{t}+v_{yy}=g_{0} &{} \text{ in } &{} Q^{\prime },\\ \left( v_{x}-\frac{1}{2}v\right) (0,t)=0\ \ \text{ and }\ \ \displaystyle \lim _{y\rightarrow \infty }\left( v_{y}-\frac{1}{2}v\right) (y,t)=0 &{} \text{ on } &{} (0,T),\\ v(x,T)=v_{T}(x) &{} \text{ in } &{} (0,1), \end{array} \right. \end{aligned}$$
(73)
where \(g_0:=e^{-\eta }F-(\eta ''+|\eta '|^2)v-2\eta 'v_y\).
Let v be a solution to (73). For any \(s\ge s_0>0\), we set \(z=e^{-s\sigma }v\) and \(\tilde{z}=e^{-s\tilde{\sigma }}v\). We have that z and \(\tilde{z}\) satisfy the following initial, final and boundary conditions:
$$\begin{aligned}&z=\tilde{z}=z_y=\tilde{z}_y=0 \ \ \text{ at } \ \ t=0 \ \ \text{ and }\ \ t=T, \\&\left( z_y+\left( s\lambda \xi -\frac{1}{2}\right) z\right) (0,t)=0 \ \ \ \text{ and } \ \ \displaystyle \lim _{y\rightarrow +\,\infty }z_y(y,t)=\lim _{y\rightarrow +\,\infty }z(y,t)=0\ \ \text{ on } \ \ [0,T], \\&\left( \tilde{z}_y+\left( -s\lambda \tilde{\xi }-\frac{1}{2}\right) \tilde{z}\right) (0,t)=0\ \ \ \text{ and } \ \ \displaystyle \lim _{y\rightarrow +\,\infty }\tilde{z}_y(y,t)=\lim _{y\rightarrow +\,\infty }\tilde{z}(y,t)=0 \ \ \text{ on } \ \ [0,T]. \end{aligned}$$
Again, we assume that v is regular enough. We have:
$$\begin{aligned} v_t=e^{s\sigma }[s\sigma _tz+z_t], \ \ \ \ v_{yy}=e^{s\sigma }[z_{yy}+s^2\sigma _y^2z+2s\sigma _yz_y+s\sigma _{yy}z] \end{aligned}$$
and, consequently,
$$\begin{aligned} M_1z+M_2z=g_1, \end{aligned}$$
with \(M_1z:=-2s\lambda ^2|\eta '|^2\xi z-2s\lambda |\eta '|\xi z_y+z_t\), \(M_2z:=s^2\lambda ^2|\eta '|^2\xi ^2z+z_{yy}+s\sigma _t z\) and \(g_1:=e^{-s\sigma }g_0+s\lambda \eta ''\xi z-s\lambda ^2|\eta '|^2\xi z\). This gives
$$\begin{aligned} \Vert M_1z\Vert ^2+\Vert M_2z\Vert ^2+2(\!( M_1z,M_2z)\!)= & {} \Vert g_1\Vert ^2. \end{aligned}$$
(74)
After some work, we can deduce that
$$\begin{aligned} (\!( M_1z,M_2z)\!)\ge & {} C\int \!\!\!\!\int _{Q'}\left( s^3\lambda ^4\xi ^3|z|^2+s\lambda ^2\xi |z_x|^2\right) \,\mathrm{d}y\,\mathrm{d}t\nonumber \\&-C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\!\left( s^3\lambda ^4\xi ^3|z|^2+s\lambda ^2\xi |z_y|^2\right) \,\mathrm{d}y\,\mathrm{d}t\nonumber \\&+\left. \int _0^T\left[ s^3\lambda ^3(\eta ')^3\xi ^3|z|^2-s\lambda ^3(\eta ')^3\xi |z|^2+s^2\lambda \eta '\xi \sigma _t|z|^2\right] \,\mathrm{d}t\right| _{y=0}\nonumber \\&+\left. \int _0^T\left( s\lambda \eta '\xi |z_y|^2+2s\lambda ^2|\eta '|^2\xi zz_y+zz_{yt}\right) \,\mathrm{d}t\right| _{y=0}. \end{aligned}$$
(75)
Working similarly with the function \(\tilde{z}\), we obtain
$$\begin{aligned} \tilde{M}_1\tilde{z}+\tilde{M}_2\tilde{z}=\tilde{g}_1 \end{aligned}$$
with \(\tilde{M}_1\tilde{z}:=\tilde{I}_{11}+\tilde{I}_{12}+\tilde{I}_{13}:=-2s\lambda ^2|\eta '|^2\tilde{\xi }\tilde{z}+2s\lambda \eta '\tilde{\xi }\tilde{z}_y+\tilde{z}_t\), \(\tilde{M}_2\tilde{z}:=\tilde{I}_{21}+\tilde{I}_{22}+\tilde{I}_{23}:=s^2\lambda ^2|\eta '|^2\tilde{\xi }^2\tilde{z}+\tilde{z}_{yy}+s\tilde{\sigma }_t\tilde{z}\) and \(\tilde{g}_1:=e^{-s\tilde{\sigma }}g_0-s\lambda \eta ''\tilde{\xi }\tilde{z}-s\lambda ^2|\eta '|^2\tilde{\xi }\tilde{z}\). This gives:
$$\begin{aligned} \Vert \tilde{g}_1\Vert ^2\!= & {} \!\Vert \tilde{M}_1\tilde{z}\Vert ^2+\Vert \tilde{M}_2\tilde{z}\Vert ^2+2(\!(\tilde{M}_1\tilde{z},\tilde{M}_2\tilde{z})\!) \end{aligned}$$
(76)
and
$$\begin{aligned} (\!(\tilde{M}_1\tilde{z},\tilde{M}_2\tilde{z})\!)\ge & {} C\int \!\!\!\!\int _{Q'}s^3\lambda ^4\tilde{\xi }^3|\tilde{z}|^2+s\lambda ^2\tilde{\xi }|\tilde{z}_y|^2\,\mathrm{d}y\,\mathrm{d}t\nonumber \\&-C\int \!\!\!\!\int _{\omega _{0T}}s^3\lambda ^4\tilde{\xi }^3|\tilde{z}|^2+s\lambda ^2\tilde{\xi }|\tilde{z}_y|^2\,\mathrm{d}y\,\mathrm{d}t\nonumber \\&+\left. \int _0^T[-s^3\lambda ^3(\eta ')^3\tilde{\xi }^3\tilde{z}^2+s\lambda ^3(\eta ')^3\tilde{\xi } \tilde{z}^2-s^2\lambda \eta '\tilde{\xi }\tilde{\sigma }_t\tilde{z}^2]\,\mathrm{d}t\right| _{y=0} \nonumber \\&+\left. \int _0^T\left( -s\lambda \eta '\tilde{\xi } \tilde{z}_y^2+2s\lambda ^2|\eta '|^2\tilde{\xi } \tilde{z}\tilde{z}_y+\tilde{z}\tilde{z}_{yt}\,\mathrm{d}t\right) \right| _{y=0}. \end{aligned}$$
(77)
Note that \(z=\tilde{z}\), \(\xi =\tilde{\xi }\), \(\sigma =\tilde{\sigma }\) and \(\sigma _t=\tilde{\sigma }_t\) for \(y=0\). Hence, from (75) and (77), we find that
$$\begin{aligned}&(\!( M_1z,M_2z)\!)+(\!(\tilde{M}_1\tilde{z},\tilde{M}_2\tilde{z})\!)\nonumber \\&\quad \ge C\int \!\!\!\!\int _{Q'}\left[ s^3\lambda ^4(\xi ^3|z|^2+\tilde{\xi }^3|\tilde{z}|^2)+s\lambda ^2(\xi |z_y|^2+\tilde{\xi }|\tilde{z}_y|^2)\right] \,\mathrm{d}y\,\mathrm{d}t\nonumber \\&\qquad -\,C\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\left[ s^3\lambda ^4(\xi ^3|z|^2+\tilde{\xi }^3|\tilde{z}|^2)\right. \nonumber \\&\qquad \left. +s\lambda ^2(\xi |z_y|^2+\tilde{\xi }|\tilde{z}_y|^2)\right] \,\mathrm{d}y\,\mathrm{d}t\nonumber \\&\qquad +\,\!\!\left. \int _0^T\!\!\left[ 2s\lambda ^2|\eta '|^2(\tilde{\xi }\tilde{z}\tilde{z}_y+\xi zz_y)-s\lambda \eta '(\tilde{\xi }\tilde{z}_y^2-\xi z_y^2)\right. \right. \nonumber \\&\qquad \left. \left. +\,\tilde{z}\tilde{z}_{ty}+zz_{ty}\right] \,\mathrm{d}t\right| _{y=0}\!\!\!. \end{aligned}$$
(78)
On the other hand, in view of the boundary conditions satisfied by z and \(\tilde{z}\), we see that
$$\begin{aligned} \left. \int _0^T[zz_{yt}+\tilde{z}\tilde{z}_{yt}]\mathrm{d}t\right| _{y=0}= & {} \left. \int _0^T\left[ s\lambda \xi _t(|\tilde{z}|^2-|z|^2)+\frac{1}{4}(|\tilde{z}|^2+|z|^2)_t\right. \right. \\&\left. \left. +\frac{1}{2}s\lambda \xi (|\tilde{z}|^2-|z|^2)_t\right] \mathrm{d}t\right| _{y=0}= 0,\\ \left. 2s\lambda ^2\!\!\int _0^T|\eta '|^2\xi [zz_y+\tilde{z}\tilde{z}_y]\,\mathrm{d}t\right| _{y=0}= & {} \left. 2s\lambda ^2\int _0^T|\eta '|^2\xi |z|^2\,\mathrm{d}t\right| _{y=0}\ge 0 \end{aligned}$$
and
$$\begin{aligned} -\left. s\lambda \int _0^T\eta '\xi [\tilde{z}_y^2-z_y^2]\,\mathrm{d}t\right| _{y=0}= & {} \left. 2s^2\lambda ^2\int _0^T\xi ^2 |z|^2\,\mathrm{d}t\right| _{y=0}\ge 0. \end{aligned}$$
As a consequence, from (74), (76) and (78), we deduce that
$$\begin{aligned}&\displaystyle \sum _{i=1}^{2}(\Vert M_{i}z\Vert ^{2}+\Vert \tilde{M}_{i}\tilde{z}\Vert ^{2})+\int \!\!\!\!\int _{Q^{\prime }}\left[ s\lambda ^{2}(\xi |z_{y} |^{2}+\tilde{\xi }|\tilde{z}_{y}|^{2})+s^{2}\lambda ^{4}(\xi ^{3}|z|^{2} +\tilde{\xi }^{3}|\tilde{z}|^{2})\right] \,\mathrm{d}y\,\mathrm{d}t\nonumber \\&\quad \le C\left( \Vert g_{1}\Vert ^{2}+\Vert \tilde{g}_{1}\Vert ^{2}+\int \!\!\!\!\int _{\omega _{0T}}\!\!\!\left[ s\lambda ^{2}(\xi |z_{y}|^{2}+\tilde{\xi }|\tilde{z}_{y}|^{2})+s^{3}\lambda ^{4}(\xi ^{3}|z|^{2}+\tilde{\xi } ^{3}|\tilde{z}|^{2})\right] \,\mathrm{d}y\,\mathrm{d}t\right) .\nonumber \\ \end{aligned}$$
(79)
Using (79) and the definitions of \(g_1\), \(\tilde{g}_1\), \(M_iz\) and \(\tilde{M}_i\tilde{z}\), we see that, for s and \(\lambda \) large enough,
$$\begin{aligned}&\int \!\!\!\!\int _{Q^{\prime }}\left[ s^{-1}\left[ \xi ^{-1}(|z_{t} |^{2}+|z_{yy}|^{2})+\tilde{\xi }^{-1}(|\tilde{z}_{t}|^{2}+|\tilde{z}_{yy} |^{2})\right] \right. \\&\qquad \left. +s\lambda ^{2}(\xi |z_{y}|^{2}+\tilde{\xi }|\tilde{z}_{y} |^{2})+\,s^{2}\lambda ^{4}(\xi ^{3}|z|^{2}+\tilde{\xi }^{3}|\tilde{z}|^{2})\right] \,\mathrm{d}y\,\mathrm{d}t\\&\quad \le C\left( \Vert \varrho ^{1/2}g_{0}\Vert ^{2}+\int \!\!\!\!\int _{\omega _{0T} }\!\!\!\left[ s\lambda ^{2}(\xi |z_{y}|^{2}+\tilde{\xi }|\tilde{z}_{y} |^{2})+s^{3}\lambda ^{4}(\xi ^{3}|z|^{2}+\tilde{\xi }^{3}|\tilde{z}|^{2})\right] \,\mathrm{d}y\,\mathrm{d}t\right) . \end{aligned}$$
From classical arguments, we can eliminate the terms with derivatives in the right-hand side of the last inequality and find that
$$\begin{aligned}&\int \!\!\!\!\int _{Q^{\prime }}\left[ s^{-1}\left[ \xi ^{-1}(|z_{t} |^{2}+|z_{yy}|^{2})+\tilde{\xi }^{-1}(|\tilde{z}_{t}|^{2}+|\tilde{z}_{yy} |^{2})\right] +s\lambda ^{2}(\xi |z_{y}|^{2}+\tilde{\xi }|\tilde{z}_{y} |^{2})\right. \\&\qquad \left. +\,s^{2}\lambda ^{4}(\xi ^{3}|z|^{2}+\tilde{\xi }^{3}|\tilde{z}|^{2})\right] \,\mathrm{d}y\,\mathrm{d}t\\&\quad \le C\left( \Vert \varrho ^{1/2}g_{0}\Vert ^{2}+s^{3}\lambda ^{4}\int \!\!\!\!\int _{\omega _{0T}}\left( \xi ^{3}|z|^{2}+\tilde{\xi }^{3}|\tilde{z}|^{2}\right) \,\mathrm{d}y\,\mathrm{d}t\right) . \end{aligned}$$
We then conclude the proof coming back to the original variable w and using the definition of \(g_0\) and the fact that \(\xi \le C\tilde{\xi }\le C\xi \). \(\Box \)
