Linear Inviscid Damping and Enhanced Dissipation for Monotone Shear Flows

Abstract

In this paper, we study the linearized Navier–Stokes system around monotone shear flows in a finite channel with non-slip boundary condition. We prove that if the flow is linearly stable for the Euler equations, then it is also linearly stable for the Navier–Stokes equations at high Reynolds number. More importantly, we establish the inviscid damping and enhanced dissipation estimates for the linearized Navier–Stokes system, which may be crucial for nonlinear stability. One of the key ingredients is the resolvent estimates of the linearized operator. For this, we develop the compactness method and establish some sharper estimates for the boundary layer corrector.

Appendices

Appendix A

Lemma A.1

If $G(y)\in H^{1}(\mathbb {R})$ with compact support, then it holds that

$$\begin{aligned}&\int _{0}^{+\infty }\left( \int _{-\infty }^{+\infty } G(y)\textrm{e}^{-\textrm{i}ty}\textrm{d}y \right) \textrm{d}t =\pi G(0) -p.v.\int _{-\infty }^{\infty }\frac{\textrm{i}G(y)}{y}\textrm{d}y. \end{aligned}$$

Proof

Observe that

$$\begin{aligned}&\int _{0}^{+\infty }\left( \int _{-\infty }^{+\infty } G(y)\textrm{e}^{-\textrm{i}ty}\textrm{d}y \right) \textrm{d}t\\&\quad =\int _{0}^{+\infty }\left( \int _{-\infty }^{+\infty } G(y) \bigg (\dfrac{\textrm{e}^{\textrm{i}ty}+\textrm{e}^{-\textrm{i}ty}}{2}- \dfrac{\textrm{e}^{\textrm{i}ty}-\textrm{e}^{-\textrm{i}ty}}{2} \bigg )\textrm{d}y \right) \textrm{d}t\\&\quad =\dfrac{1}{2}\int _{-\infty }^{+\infty }\left( \int _{-\infty }^{+\infty }G(y)\textrm{e}^{\textrm{i}ty}\textrm{d}y \right) \textrm{d}t -\dfrac{1}{2}\int _{0}^{+\infty }\left( \int _{-\infty }^{+\infty }(G(y)-G(-y))\textrm{e}^{\textrm{i}ty}\textrm{d}y \right) \textrm{d}t\\&\quad = I-II. \end{aligned}$$

By the Fourier inversion formula, we have $I=\pi G(0)$. For the second term, we have

$$\begin{aligned}&II =\lim _{R\rightarrow +\infty }\dfrac{1}{2} \int _{-\infty }^{+\infty } \left( \int _{0}^{R}\textrm{e}^{\textrm{i}ty} \textrm{d}t\right) (G(y)-G(-y))\textrm{d}y, \end{aligned}$$

where

$$\begin{aligned}&\int _{0}^{R}\textrm{e}^{\textrm{i}ty} \textrm{d}t= \dfrac{\textrm{e}^{\textrm{i}tR}-1}{\textrm{i}y}. \end{aligned}$$

Then we infer that

$$\begin{aligned} II-p.v.\int _{-\infty }^{+\infty }\dfrac{\textrm{i}G(y)}{y}\textrm{d}y&= II-\int _{-\infty }^{+\infty } \dfrac{\textrm{i}(G(y)-G(-y))}{2y}\textrm{d}y\\&=\lim _{R\rightarrow +\infty }\int _{-\infty }^{+\infty }\textrm{e}^{\textrm{i}tR} \dfrac{G(y)-G(-y)}{2\textrm{i}y}\textrm{d}y=0, \end{aligned}$$

where in the last step we used the Riemann-Lebesgue lemma. $\square$

Lemma A.2

If $F(y)\in L^2([0,1])$, $v_L\in C^1([0,1])$, $\textbf{Im}\ v_L\le 0,$ $\textbf{Re}\ v_L'\ge c>0,$ $s\in (0,1)$, then it holds that

$$\begin{aligned}&\int _{0}^{+\infty }\left| \int _{0}^{1} F(y)\textrm{e}^{-\textrm{i}tv_L(y)}\textrm{d}y \right| ^2t^{-s}\textrm{d}t \le C\int _{0}^{1}|F(y)|^2\textrm{d}y. \end{aligned}$$

Here C depends only on c and s.

Proof

We write

$$\begin{aligned}&\left| \int _{0}^{1} F(y)\textrm{e}^{-\textrm{i}tv_L(y)}\textrm{d}y \right| ^2=\int _{0}^{1}\int _{0}^{1} F(y)\overline{F(z)}\textrm{e}^{-\textrm{i}t(v_L(y)-\overline{v_L(z)})}\textrm{d}y\textrm{d}z. \end{aligned}$$

Then for fixed $\epsilon >0$, we get by the Fubini theorem that

$$\begin{aligned}&\int _{0}^{+\infty }\left| \int _{0}^{1} F(y)\textrm{e}^{-\textrm{i}tv_L(y)}\textrm{d}y \right| ^2\textrm{e}^{-\epsilon t}t^{-s}\textrm{d}t\\&\quad =\int _{0}^{1}\int _{0}^{1} F(y)\overline{F(z)}\int _{0}^{+\infty }\textrm{e}^{-\textrm{i}t(v_L(y)-\overline{v_L(z)})-\epsilon t}t^{-s}\textrm{d}t\textrm{d}y\textrm{d}z. \end{aligned}$$

Here we used $\textbf{Re}[-\textrm{i}(v_L(y)-\overline{v_L(z)})]=\textbf{Im}\ v_L(y)+\textbf{Im}\ v_L(z)\le 0.$ Note that

$$\begin{aligned} \int _{0}^{+\infty }\textrm{e}^{-\textrm{i}t(v_L(y)-\overline{v_L(z)})-\epsilon t}t^{-s}\textrm{d}t=\Gamma (1-s)\big [\textrm{i}(v_L(y)-\overline{v_L(z)})+\epsilon \big ]^{s-1}. \end{aligned}$$

Thus, we deduce that

$$\begin{aligned}&\int _{0}^{+\infty }\left| \int _{0}^{1} F(y)\textrm{e}^{-\textrm{i}tv_L(y)}\textrm{d}y \right| ^2\textrm{e}^{-\epsilon t}t^{-s}\textrm{d}t\\&\quad =\int _{0}^{1}\int _{0}^{1} F(y)\overline{F(z)}\Gamma (1-s)[\textrm{i}(v_L(y)-\overline{v_L(z)})+\epsilon ]^{s-1}\textrm{d}y\textrm{d}z\\&\quad \le \Gamma (1-s)\int _{0}^{1}\int _{0}^{1} |F(y)||{F(z)}||\textrm{i}(v_L(y)-\overline{v_L(z)})+\epsilon |^{s-1}\textrm{d}y\textrm{d}z\\&\quad \le \frac{\Gamma (1-s)}{2}\int _{0}^{1}\int _{0}^{1} (|F(y)|^2+|{F(z)}|^2)(c|y-z|)^{s-1}\textrm{d}y\textrm{d}z\\&\quad =\Gamma (1-s)\int _{0}^{1}|F(y)|^2\int _{0}^{1} (c|y-z|)^{s-1}\textrm{d}z\textrm{d}y\\&\quad =\Gamma (1-s)c^{s-1}\int _{0}^{1}|F(y)|^2\frac{y^s+(1-y)^s}{s}\textrm{d}y\\&\quad \le \frac{2\Gamma (1-s)c^{s-1}}{s}\int _{0}^{1}|F(y)|^2\textrm{d}y. \end{aligned}$$

Here we used that(as $\textbf{Re}\ v_L'\ge c>0$)

$$\begin{aligned} |\textrm{i}(v_L(y)-\overline{v_L(z)})+\epsilon |\ge&|\textbf{Im} [\textrm{i}(v_L(y)-\overline{v_L(z)})+\epsilon ]|=|\textbf{Re} (v_L(y)-\overline{v_L(z)})|\\=&\left| \int _z^y\textbf{Re} v_L'(t)\textrm{d}t\right| \ge c|y-z|. \end{aligned}$$

Now the result follows by letting $\epsilon \rightarrow 0+$. $\square$

Appendix B: Airy Function

We assume $\alpha >0$ in this section. Let Ai(y) by the Airy function, which is a nontrivial solution of $f''-yf=0$. We denote

$$\begin{aligned} A_0(z)&=\int _{\textrm{e}^{\textrm{i}\frac{\pi }{6}}z}^{\infty }Ai(t)\textrm{d}t= \textrm{e}^{\textrm{i}\pi /6}\int _{z}^{\infty }Ai(\textrm{e}^{\textrm{i}\pi /6}t)\textrm{d}t.,\\ \omega (z,x)&=\frac{A_0(z+x)}{A_{0}(z)}=\exp \Big (\int _{0}^{x}\frac{A'_0(z+t)}{A_0(z+t)}dt\Big ). \end{aligned}$$

The following two lemmas come from [14].

Lemma B.1

There exist $c>0$ and $\delta _0>0$ such that for ${\textbf {Im}}(z)\le \delta _0$,

$$\begin{aligned}&\left| \frac{A_0'(z)}{A_0(z)}\right| \lesssim 1+|z|^{\frac{1}{2}}, \end{aligned}$$

and for $\textbf{Im}z\le \delta _0$ and $x\ge 0$,

$$\begin{aligned} |\omega (z,x)|\le e^{-c(|z|^\frac{1}{2}x+x^\frac{3}{2})}. \end{aligned}$$

Lemma B.2

There exist $\delta _1>0$, $k_0>1$ and $c\in (0,1)$ such that if $L\ge k_0$, $\alpha \ge 1$, $\textbf{Im}z\le \delta _1-\alpha ^2/L^2$, then

$$\begin{aligned}&\left| \sinh (\alpha )- \frac{\alpha }{L}\int _{0}^{L}\cosh \big (\alpha -\frac{\alpha t}{L}\big )\omega (z,t)\textrm{d}t\right| \ge c\sinh (\alpha )\\&\quad \ge 2\left| \sinh (\alpha )\omega (z,L)- \frac{\alpha }{L}\int _{0}^{L}\cosh (\frac{\alpha t}{L})\omega (z,t)\textrm{d}t\right| . \end{aligned}$$

We denote

$$\begin{aligned} L_0&=|\alpha u'(0)/\nu |^{\frac{1}{3}},\quad d_0=(u(0)-\lambda -\textrm{i}\nu \alpha )/(u'(0)),\\ L_1&=|\alpha u'(1)/\nu |^{\frac{1}{3}},\quad d_1=(u(1)-u'(1)-\lambda -\textrm{i}\nu \alpha )/(u'(1)),\\ W_{a,1}(y)&=Ai(\textrm{e}^{\textrm{i}\frac{\pi }{6}}L_0(y+d_0)),\quad W_{a,2}(y)=Ai(\textrm{e}^{\textrm{i}\frac{5\pi }{6}}L_1(y+d_1)). \end{aligned}$$

Let $\Psi _{a,i}(i=1,2)$ solve

$$\begin{aligned} (\partial _y^2-\alpha ^2)\Psi _{a,i}=W_{a,i},\quad \Psi _{a,i}(0)=\Psi _{a,i}(1)=0. \end{aligned}$$

Lemma B.3

There exist $c>0, C>0$ independent of $L_0, d_0$ such that

$$\begin{aligned}&\dfrac{|W_{a,1}(y)|}{\big | A_0(L_0d_0)\big |} \le C(1+|L_0d_0|)^{\frac{1}{2}}\textrm{e}^{-{(c/2)}L_0y(1+|L_0d_0|)^{\frac{1}{2}}},\\&\quad \left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}+ \left\| \dfrac{(\partial _y,\alpha )\Psi _{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}\le CL_0^{-\frac{3}{2}}(1+|L_0d_0|)^{-\frac{1}{4}},\\&\quad \left\| \dfrac{y^2W_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}\le CL_0^{-\frac{5}{2}}(1+|L_0d_0|)^{-\frac{3}{4}},\\&\quad \left\| \dfrac{\Psi _{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}\le CL_0^{-2}|\alpha |^{-\frac{1}{2}}(1+|L_0d_0|)^{-\frac{1}{2}}, \end{aligned}$$

and

$$\begin{aligned}&\dfrac{|W_{a,2}(y)|}{\big | A_0(-L_1(\bar{d}_1+1)\big |} \le C(1+|L_1(d_1+1)|)^{\frac{1}{2}}\textrm{e}^{-(c/2)L_1y(1+|L_1(d_1+1)|)^{\frac{1}{2}}},\\&\quad \left\| \dfrac{(1-y)W_{a,2}(y)}{ A_0(-L_1(\bar{d}_1+1))} \right\| _{L^2}+ \left\| \dfrac{(\partial _y,\alpha )\Psi _{a,2}(y)}{ A_0(-L_1(\bar{d}_1+1))} \right\| _{L^2}\le CL_1^{-\frac{3}{2}}(1+|L_1(d_1+1)|)^{-\frac{1}{4}},\\&\quad \left\| \dfrac{(1-y)^2W_{a,2}(y)}{ A_0(-L_1(\bar{d}_1+1))} \right\| _{L^2}\le CL_1^{-\frac{5}{2}}(1+|L_1(d_1+1)|)^{-\frac{3}{4}},\\&\quad \left\| \dfrac{\Psi _{a,2}(y)}{ A_0(-L_1(\bar{d}_1+1))} \right\| _{L^2}\le CL_1^{-2}|\alpha |^{-\frac{1}{2}}(1+|L_1(d_1+1)|)^{-\frac{1}{2}}. \end{aligned}$$

Proof

By Lemma B.1, we have

$$\begin{aligned} \dfrac{|W_{a,1}(y)|}{|A_0(L_0d_0)|}&=\left| \dfrac{A'_0(L_0(d_0+y))}{A_0(L_0d_0)}\right| =\left| \dfrac{A_0'(L_0(d_0+y))}{A_0(L_0(d_0+y))}\right| \left| \dfrac{A_0(L_0(d_0+y))}{A_0(L_0d_0)}\right| \\&\le C\big (1+|L_0d_0|+L_0y\big )^{\frac{1}{2}}\textrm{e}^{-cL_0y(1+|L_0d_0|^{\frac{1}{2}})}\\&\le C(1+|L_0d_0|)^{\frac{1}{2}}\textrm{e}^{-{(c/2)}L_0y(1+|L_0d_0|)^{\frac{1}{2}}}, \end{aligned}$$

which gives

$$\begin{aligned} \left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^1}&\le CL_0^{-2}(1+|L_0d_0|)^{-\frac{1}{2}},\\ \left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}&\le CL_0^{-\frac{3}{2}}(1+|L_0d_0|)^{-\frac{1}{4}},\\ \left\| \dfrac{y^2W_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}&\le CL_0^{-\frac{5}{2}}(1+|L_0d_0|)^{-\frac{3}{4}}. \end{aligned}$$

By the Hardy’s inequality, we get

$$\begin{aligned} \Vert (\partial _y,\alpha )\Psi _{a,1}\Vert _{L^2}^2&=\big |\langle yW_{a,1},\Psi _{a,1}/y \rangle \big |\le \big \Vert yW_{a,1}\big \Vert _{L^2}\big \Vert \Psi _{a,1}/y\big \Vert _{L^2}\\ {}&\le 2\big \Vert yW_{a,1}\big \Vert _{L^2}\Vert (\partial _y,\alpha )\Psi _{a,1}\Vert _{L^2}. \end{aligned}$$

Then $\Vert (\partial _y,\alpha )\Psi _{a,1}\Vert _{L^2}\le 2\big \Vert yW_{a,1}\big \Vert _{L^2}$ and

$$\begin{aligned}&\left\| \dfrac{(\partial _y,\alpha )\Psi _{a,1}(y)}{A_0(L_0d_0)} \right\| _{L^2}+\left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}\le 3\left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^2}\le CL_0^{-\frac{3}{2}}(1+|L_0d_0|)^{-\frac{1}{4}}. \end{aligned}$$

Let $\varphi _1$ solve $(\partial _y^2-\alpha ^2)\varphi _1=\Psi _{a,1},\ \varphi _1(0)=\varphi _1(1)=0$. Then we have

$$\begin{aligned} \Vert \Psi _{a,1}\Vert _{L^2}^2=\alpha ^4\Vert \varphi \Vert _{L^2}^2+2\alpha ^2\Vert \partial _y\varphi \Vert _{L^2}^2+\Vert \partial ^2_y\varphi \Vert _{L^2}^2. \end{aligned}$$

As $\varphi _1(y)=\int _{0}^{y}\varphi _1'(z)dz$ for $y\in [0,1]$, we conclude

$$\begin{aligned} \left\| {\varphi _1}/{y}\right\| _{L^\infty } \le&\Vert \partial _y\varphi _1\Vert _{L^\infty }\le C\Vert \partial _y\varphi _1\Vert _{L^2}^{\frac{1}{2}} \Vert \partial ^2_y\varphi _1\Vert _{L^2}^{\frac{1}{2}} \le C|\alpha |^{-\frac{1}{2}}\Vert \Psi _{a,1}\Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert \Psi _{a,1}\Vert ^2_{L^2}&=\left| \big \langle \varphi _1,W_{a,1} \big \rangle \right| \le \left\| {\varphi _1}/y\right\| _{L^\infty }\Vert yW_{a,1}\Vert _{L^1}\\&\le C|\alpha |^{-\frac{1}{2}}\Vert \Psi _{a,1}\Vert _{L^2}\Vert yW_{a,1}\Vert _{L^1}. \end{aligned}$$

Then $\Vert \Psi _{a,1}\Vert _{L^2}\le C|\alpha |^{-\frac{1}{2}}\Vert yW_{a,1}\Vert _{L^1}$ and

$$\begin{aligned}&\left\| \dfrac{\Psi _{a,1}(y)}{A_0(L_0d_0)} \right\| _{L^2}\le C|\alpha |^{-\frac{1}{2}}\left\| \dfrac{yW_{a,1}(y)}{ A_0(L_0d_0)} \right\| _{L^1}\le CL_0^{-2}|\alpha |^{-\frac{1}{2}}(1+|L_0d_0|)^{-\frac{1}{2}}. \end{aligned}$$

The proof for $W_{a,2}$ and $\Psi _{a,2}$ is similar. $\square$

Lemma B.4

Let $\psi$ solve $(\partial _y^2-\alpha ^2)\psi =w,\ \psi |_{y=0,1}=0$. Then it holds that

$$\begin{aligned}&\left[ \tanh (\alpha y)+\tanh \big (\alpha (1-y)\big )\right] \partial _y\psi (y)\\ {}&\quad = \int _{0}^{y}w(z)\dfrac{\sinh (\alpha z)}{\cosh (\alpha y)}\textrm{d}z-\int _{y}^{1}w(z)\dfrac{\sinh \big (\alpha (1-z)\big )}{\cosh \big (\alpha (1-y)\big )}\textrm{d}z. \end{aligned}$$

In particular, we have

$$\begin{aligned}&\partial _y\psi (0)=-\int _{0}^{1}w(z) \dfrac{\sinh \big (\alpha (1-z)\big )}{\sinh (\alpha )}\textrm{d}z,\quad \partial _y\psi (1)=\int _{0}^{1}w(z)\dfrac{\sinh (\alpha z)}{\sinh (\alpha )}\textrm{d}z. \end{aligned}$$

Proof

We get by integration by parts that

$$\begin{aligned}&\int _{0}^{y}w(z)\sinh \big (\alpha z\big )\textrm{d}z=\int _{0}^{y} \big [(\partial _z^2-\alpha ^2)\psi (z)\big ]\sinh \big (\alpha z\big )\textrm{d}z\\&\quad = \int _{0}^{y}\psi (z)\big [(\partial _z^2-\alpha ^2)\sinh \big (\alpha z\big )\big ]\textrm{d}z +(\partial _z\psi )\sinh \big (\alpha z\big )\big |^{z=y}_{z=0}-\psi \big (\partial _z\sinh \big (\alpha z\big )\big )\big |^{z=y}_{z=0}\\&\quad =\partial _y\psi (y) \sinh (\alpha y)-\alpha \psi (y)\cosh (\alpha y), \end{aligned}$$

and

$$\begin{aligned}&\int _{y}^{1}w(z)\sinh \big (\alpha (1-z)\big )\textrm{d}z= \int _{y}^{1}\big [(\partial _z^2-\alpha ^2)\psi (z)\big ] \sinh \big (\alpha (1-z)\big )\textrm{d}z\\&\quad =\int _{y}^{1}\psi (z)\big [(\partial _z^2-\alpha ^2)\sinh \big (\alpha (1-z)\big )\big ] +(\partial _z\psi )\sinh \big (\alpha (1-z)\big )\big |^{z=1}_{z=y} \\&\qquad -\psi \big (\partial _z\sinh \big (\alpha (1-z)\big )\big )\big |^{z=1}_{z=y}\\&\quad =-\partial _y\psi (y)\sinh \big (\alpha (1-y)\big )- \alpha \psi (y)\cosh \big (\alpha (1-y)\big ). \end{aligned}$$

Then we conclude that

$$\begin{aligned}&\left[ \tanh (\alpha y)+\tanh \big (\alpha (1-y)\big )\right] \partial _y\psi (y) = \int _{0}^{y}w(z)\dfrac{\sinh (\alpha z)}{\cosh (\alpha y)}\textrm{d}z-\int _{y}^{1}w(z)\dfrac{\sinh \big (\alpha (1-z)\big )}{\cosh \big (\alpha (1-y)\big )}\textrm{d}z \end{aligned}$$

Taking $y=0$ or $y-1$, we get

$$\begin{aligned}&\partial _y\psi (0)=-\int _{0}^{1}w(z) \dfrac{\sinh \big (\alpha (1-z)\big )}{\sinh (\alpha )}\textrm{d}z,\qquad \partial _y\psi (1)=\int _{0}^{1}w(z)\dfrac{\sinh (\alpha z)}{\sinh (\alpha )}\textrm{d}z. \end{aligned}$$

$\square$

Appendix C: Elliptic Estimates

Lemma C.1

Let $\psi$ solve $(\partial _y^2-\alpha ^2)\psi =w,\ \psi |_{y=0,1}=0$ and $|\alpha |\ge 1$. There exists a decomposition $\psi =\psi _1+\psi _2$ so that

$$\begin{aligned}&\Vert (\partial _y,\alpha )\psi _1\Vert _{L^2}\le C|\alpha |\Vert \psi \Vert _{L^2},\quad \Vert \psi _2\Vert _{L^2}\le C\Vert y^2w\Vert _{L^2}. \end{aligned}$$

Remark C.1

We can also decompose $\psi$ as $\psi =\widetilde{\psi }_1+\widetilde{\psi }_2$ with

$$\begin{aligned} \Vert (\partial _y,\alpha )\widetilde{\psi }_1\Vert _{L^2}\le C|\alpha |\Vert \psi \Vert _{L^2},\quad \Vert \widetilde{\psi }_2\Vert _{L^2}\le C\Vert (1-y)^2w\Vert _{L^2}. \end{aligned}$$

Proof

We introduce $\gamma _0(y)=\dfrac{\sinh (\alpha (1- y))}{\sinh \alpha }$ and let $a=\langle \psi ,\gamma _0\rangle /\langle \gamma _0,\gamma _0\rangle$, $\psi _1=a\gamma _0$, $\psi _2=\psi -a\gamma _0$. Then we have

$$\begin{aligned} \psi =\psi _1+\psi _2,\quad \langle \psi _2,\gamma _0\rangle =0,\quad \langle \psi _2,\psi _1\rangle =0,\quad \Vert \psi \Vert _{L^2}^2=\Vert \psi _1\Vert _{L^2}^2+\Vert \psi _2\Vert _{L^2}^2. \end{aligned}$$

As $\Vert (\partial _y,\alpha )\gamma _0\Vert _{L^2}\le C|\alpha |^{\frac{1}{2}}$, $C^{-1}|\alpha |^{-\frac{1}{2}}\le \Vert \gamma _0\Vert _{L^2}\le C|\alpha |^{-\frac{1}{2}}$, we get

$$\begin{aligned} \Vert (\partial _y,\alpha )\psi _1\Vert _{L^2}\le C|\alpha |^{\frac{1}{2}}|a|\le C|\alpha |\Vert \psi _1\Vert _{L^2}\le C|\alpha |\Vert \psi \Vert _{L^2}. \end{aligned}$$

Let $\varphi$ solve $(\partial _y^2-\alpha ^2)\varphi =\psi _2,\ \varphi |_{y=0,1}=0$. Then $\partial _y\varphi (0)=-\langle \psi _2,\gamma _0\rangle =0$ and

$$\begin{aligned} \Vert \psi _2\Vert _{L^2}^2=\Vert \partial _y^2\varphi \Vert _{L^2}^2+2\alpha ^2\Vert \partial _y\varphi \Vert _{L^2}^2+\alpha ^4\Vert \varphi \Vert _{L^2}^2. \end{aligned}$$

By the Hardy’s inequality and $\varphi (0)=\partial _y\varphi (0)=0$, we have

$$\begin{aligned} \Vert \varphi /y^2\Vert _{L^2}\le C\Vert \partial _y\varphi /y\Vert _{L^2}\le C\Vert \partial _y^2\varphi \Vert _{L^2}\le C\Vert \psi _2\Vert _{L^2}. \end{aligned}$$

As $(\partial _y^2-\alpha ^2)\gamma _0=0,$ $\gamma _0(1)=0,$ $\psi _2=\psi -a\gamma _0$, $\psi (1)=0$, we have

$$\begin{aligned} (\partial _y^2-\alpha ^2)\psi _2=(\partial _y^2-\alpha ^2)\psi =w, \quad \psi _2(1)=0. \end{aligned}$$

Thus, due to $\varphi (0)=\partial _y\varphi (0)=\varphi (1)=\psi _2(1)=0$, we get

$$\begin{aligned} \Vert \psi _2\Vert _{L^2}^2&=\langle (\partial _y^2-\alpha ^2)\varphi ,\psi _2\rangle =\langle \varphi ,(\partial _y^2-\alpha ^2)\psi _2\rangle =\langle \varphi ,w\rangle \\ {}&\le \Vert \varphi /y^2\Vert _{L^2}\Vert y^2w\Vert _{L^2}\le C\Vert \psi _2\Vert _{L^2}\Vert y^2w\Vert _{L^2}, \end{aligned}$$

which gives $\Vert \psi _2\Vert _{L^2}\le C\Vert y^2w\Vert _{L^2}$. $\square$

Lemma C.2

If $\psi _1,\psi _2\in H^2(0,1)\cap H_0^1(0,1)$, $g\in H^2([0,1])$ satisfies $(\partial ^2_y-\alpha ^2)\psi _1=g(\partial ^2_y-\alpha ^2)\psi _2$ and $\alpha \ge 1$, then we have

$$\begin{aligned}&\alpha ^{\frac{1}{2}}\Vert (\partial _y,\alpha )\psi _1\Vert _{L^2}+|\partial _y\psi _1(0)|+|\partial _y\psi _1(1)|\\&\quad \le C\Vert g\Vert _{C^1}\big (\alpha ^{\frac{1}{2}}\Vert (\partial _y,\alpha )\psi _2\Vert _{L^2}+|\partial _y\psi _2(0)|+|\partial _y\psi _2(1)|\big ), \\&\qquad \Vert (\partial _y,\alpha )(\psi _1-g\psi _2)\Vert _{L^2}\le C\Vert \partial _yg\Vert _{H^1}\Vert \psi _2\Vert _{L^2}. \end{aligned}$$

Proof

We get by integration by parts that

$$\begin{aligned} \partial _y\psi _1(0)&=-\int _{0}^{1}\big ((\partial _y^2-\alpha ^2)\psi _1\big )(y)\dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\textrm{d}y\\&=-\int _{0}^{1}g(y)\big ((\partial _y^2-\alpha ^2){\psi _2}\big )(y)\dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\textrm{d}y\\&=g(0)\partial _y\psi _2(0)+\int _{0}^{1}\partial _y{\psi _2}\partial _y\bigg (g(y)\dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\bigg )\textrm{d}y\\&\quad +\alpha ^2\int _{0}^{1}{\psi _2}(y)g(y)\dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\textrm{d}y. \end{aligned}$$

Then we have

$$\begin{aligned} |\partial _y\psi _1(0)|\le&|g(0)||\partial _y\psi _2(0)|+\Vert g\Vert _{C^1}\Vert \partial _y{\psi _2}\Vert _{L^2}\left\| \dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\right\| _{H^1}\\&\quad +|\alpha |\Vert g\Vert _{L^\infty }\Vert \alpha \psi _2\Vert _{L^2}\left\| \dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\right\| _{L^2}\\&\le C\Vert g\Vert _{C^1}\big (\alpha ^{\frac{1}{2}}\Vert (\partial _y,\alpha )\psi _2\Vert _{L^2}+|\partial _y\psi _2(0)|\big ). \end{aligned}$$

Here we used the fact that

$$\begin{aligned} \Big \Vert \dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\Big \Vert _{H^1}\le C\alpha ^{\frac{1}{2}},\quad \Big \Vert \dfrac{\sinh (\alpha (1-y))}{ \sinh (\alpha )}\Big \Vert _{L^2}\le C\alpha ^{-\frac{1}{2}}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} |\partial _y\psi _1(1)|\le&C\Vert g\Vert _{C^1}\big (\alpha ^{\frac{1}{2}}\Vert (\partial _y,\alpha )\psi _2\Vert _{L^2}+|\partial _y\psi _2(1)|\big ). \end{aligned}$$

By integration by parts again, we get

$$\begin{aligned} \Vert (\partial _y,\alpha )\psi _1\Vert ^2_{L^2}&=\langle -(\partial _y^2-\alpha ^2)\psi _1,\psi _1\rangle = \langle -g(\partial _y^2-\alpha ^2)\psi _2,\psi _1\rangle \\&=\langle \partial _y\psi _2,\partial _y(\bar{g}\psi _1)\rangle +\alpha ^2\langle \psi _2,{\overline{g}}\psi _1\rangle \\&\le \Vert \partial _y\psi _2\Vert _{L^2}\Vert g\Vert _{C^1}\Vert \psi _1\Vert _{H^1}+\Vert g\Vert _{L^\infty }\Vert \alpha \psi _2\Vert _{L^2}\Vert \alpha \psi _1\Vert _{L^2}\\&\le C\Vert g\Vert _{C^1}\Vert (\partial _y,\alpha )\psi _2\Vert _{L^2}\Vert (\partial _y,\alpha )\psi _1\Vert _{L^2}, \end{aligned}$$

which gives

$$\begin{aligned} \Vert (\partial _y,\alpha )\psi _1\Vert _{L^2}\le C\Vert g\Vert _{C^1}\Vert (\partial _y,\alpha )\psi _2\Vert _{L^2}. \end{aligned}$$

Thanks to $(\partial _y^2-\alpha ^2)(g\psi _2-\psi _1)=(\partial _y^2g)\psi _2+2(\partial _yg)\partial _y\psi _2$, we get

$$\begin{aligned}&\Vert (\partial _y,\alpha )(g\psi _2-\psi _1)\Vert _{L^2}^2=-\langle (\partial _y^2-\alpha ^2)(g\psi _2-\psi _1),g\psi _2-\psi _1\rangle \\&\quad ={-}\langle (\partial _y^2g)\psi _2,g\psi _2-\psi _1\rangle {-}2\langle (\partial _yg)\partial _y\psi _2,g\psi _2-\psi _1\rangle \\&\quad \le \Vert \partial _y^2g\Vert _{L^2}\Vert \psi _2\Vert _{L^2}\Vert g\psi _2-\psi _1\Vert _{L^\infty } {+}2\langle \psi _2,\partial _y\big (\partial _y\bar{g}(g\psi _2-\psi _1)\big )\rangle \\&\quad \le \Vert \partial _y^2g\Vert _{L^2}\Vert \psi _2\Vert _{L^2}\Vert g\psi _2-\psi _1\Vert _{H^1} +2\Vert \psi _2\Vert _{L^2}\Vert (\partial _y\bar{g})(g\psi _2-\psi _1)\Vert _{H^1}\\&\quad \le C\Vert \partial _yg\Vert _{H^1}\Vert \psi _2\Vert _{L^2}\Vert (g\psi _2-\psi _1)\Vert _{H^1}\le C\Vert \partial _yg\Vert _{H^1}\Vert \psi _2\Vert _{L^2}\Vert (\partial _y,\alpha )(g\psi _2-\psi _1)\Vert _{L^2}. \end{aligned}$$

This shows that $\Vert (\partial _y,\alpha )(g\psi _2-\psi _1)\Vert _{L^2}\le C\Vert \partial _yg\Vert _{H^1}\Vert \psi _2\Vert _{L^2}$. $\square$