1. Introduction and results

We consider the following initial-boundary value problem:

$\left\{ {\begin{array}{*{20}{l}} {{w_{tt}} = {c^2}\Delta w - {c^2}\Delta e + {m^2}w,}&{x \in \Omega ,\;t > 0,} \\ {{e_t} = \sigma \Delta e - (\gamma - 1){w_t},}&{x \in \Omega ,\;t > 0,} \\ {e = w = 0,}&{x \in \Gamma ,\;t ≥ 0,} \\ {w(x,0) = {w_0}(x),\;{w_t}(x,0) = {v_0}(x),\;e(x,0) = {e_0}(x),}&{x \in \Omega ,} \end{array}} \right.$

(1.1)

where $c>0$,$\sigma >0$,$m \in \mathbb{R}$ and $\gamma >1$ are constants. We assume that $\Omega$ is a fixed domain in ${\mathbb{R}^N}$ and that the boundary $\Gamma:=\partial \Omega$ is bounded and smooth. This problem originates from the following linearized equations of coupled sound and heat flow

$\frac{{\partial w}}{{\partial t}} = c\nabla \cdot {\mathbf{u}},$

(1.2)

$\frac{{\partial {\mathbf{u}}}}{{\partial t}} = c\nabla w - c\nabla e,$

(1.3)

$\frac{{\partial e}}{{\partial t}} = \sigma \Delta e - (\gamma - 1)c\nabla \cdot {\mathbf{u}}.$

(1.4)

As stated in [4,Section 1.4],these three equations appear in the flow of a compressible fluid. In such flow there are often considerable differences of temperature from one point to another,and the transfer of energy by thermal conduction may have a significant effect on the motion. The parabolic equation of heat flow is then coupled to the hyperbolic equations of fluid dynamics and the two phenomena must be calculated concurrently. This effect occurs also for infinitesimal or acoustic vibrations and is responsible for absorption of ultrasonic waves. Taking the divergence of both sides in (1.3) and eliminating $\nabla \cdot {\bf u}$ from the resulting system,we obtain two equations for the unknown scalar fields w(x,t) and e(x,t),namely,(1.2),(1.3),(1.4) are reduced to the two equations in (1.1) with m = 0.

Carasso ^[2] constructed and analyzed a least-squares procedure for approximately solving the problem (1.1) with m = 0. As a consequence existence and uniqueness of solutions were established.

The purpose of this paper is to give a simple proof of existence and uniqueness of solutions to (1.1) of Klein-CGordon type with $m \in \mathbb{R}$ by applying the Hille-CYosida theorem.

The first main result reads as follows.

Theorem 1.1 (existence and uniqueness). Assume that$w_0 \in H^2(\Omega) \cap H_0^1(\Omega)$,$v_0 \in H_0^1(\Omega)$ and $e_0 \in H^2(\Omega) \cap H_0^1(\Omega)$. Then there exists a unique solution (w,e) of (1.1) satisfying

$w \in C([0,\infty );{H^2}(\Omega ) \cap H_0^1(\Omega )) \cap {C^1}([0,\infty );H_0^1(\Omega )) \cap {C^2}([0,\infty );{L^2}(\Omega )),$

(1.5)

$e \in C([0,\infty );{H^2}(\Omega ) \cap H_0^1(\Omega )) \cap {C^1}([0,\infty );H_0^1(\Omega )) \cap {C^2}([0,\infty );{L^2}(\Omega )).$

(1.6)

Moreover,for some α > 0,the following estimates hold:

$\begin{gathered} \left\| {w(t)} \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| {v(t)} \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| {e(t)} \right\|_{{H^1}(\Omega )}^2 \hfill \\ ≤ {e^{2\alpha t}}\left( {\left\| {{w_0}} \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| {{v_0}} \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| {{e_0}} \right\|_{{H^1}(\Omega )}^2} \right)\quad \forall {\mkern 1mu} t ≥ 0,\hfill \\ \end{gathered}$

(1.7)

$\begin{gathered} \left\| {v(t)} \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| {{c^2}\Delta w(t) - {c^2}\Delta e(t) + {m^2}w(t)} \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| {\sigma \Delta e(t) - (\gamma - 1)v(t)} \right\|_{{H^1}(\Omega )}^2 \hfill \\ ≤ {e^{2\alpha t}}(\left\| {{v_0}} \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| {{c^2}\Delta {w_0} - {c^2}\Delta {e_0} + {m^2}{w_0}} \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| {\sigma \Delta {e_0} - (\gamma - 1){v_0}} \right\|_{{H^1}(\Omega )}^2)\quad \forall {\mkern 1mu} t ≥ 0. \hfill \\ \end{gathered}$

(1.8)

The second main result reads as follows.

Theorem 1.2 (regularity). Assume that the initial data w₀,v₀,e₀ satisfy

${w_0} \in {H^k}(\Omega ),\;{v_0} \in {H^k}(\Omega ),\;{e_0} \in {H^k}(\Omega )\qquad \forall {\mkern 1mu} k \in \mathbb{N},$

Then the solution (w,e) of (1.1) belongs to $C^\infty(\overline{\Omega} \times [0,\infty))\times C^\infty(\overline{\Omega} \times [0,\infty))$.

This paper is organized as follows. In the following section we will rewrite the initial-boundary value problem (1.1) as the Cauchy problem for a single abstract evolution equation dU/dt + AU = 0,where A is a matrix of operators. We will also collect theorems in the Hille-CYosida theory which will be used in this paper. Section 3 is devoted to the proof of Theorem 1.1. In Section 4 we will give the proof of Theorem 1.2.

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2. Abstract formulation toward the Hille-CYosida theory

Putting v := w_t,we rewrite equations in (1.1) as

$\left\{ {\begin{array}{*{20}{l}} {{w_t} = v,} \\ {{v_t} = {c^2}\Delta w - {c^2}\Delta e + {m^2}w,} \\ {{e_t} = \sigma \Delta e - (\gamma - 1)v,} \end{array}} \right.$

(2.1)

so that (2.1) becomes

$\begin{gathered} \left( {\begin{array}{*{20}{c}} {{w_t}} \\ {{v_t}} \\ {{e_t}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} v \\ {({c^2}\Delta + {m^2}I)w - {c^2}\Delta e} \\ { - (\gamma - 1)v + \sigma \Delta e} \end{array}} \right) \\ = \left( {\begin{array}{*{20}{c}} 0&I&0 \\ {{c^2}\Delta + {m^2}I}&0&{ - {c^2}\Delta } \\ 0&{ - (\gamma - 1)I}&{\sigma \Delta } \end{array}} \right)\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right) \\ = - \left( {\begin{array}{*{20}{c}} 0&{ - I}&0 \\ { - {c^2}\Delta - {m^2}I}&0&{{c^2}\Delta } \\ 0&{(\gamma - 1)I}&{ - \sigma \Delta } \end{array}} \right)\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right). \\ \end{gathered}$

Setting

$A: = \left( {\begin{array}{*{20}{c}} 0&{ - I}&0 \\ { - {c^2}\Delta - {m^2}I}&0&{{c^2}\Delta } \\ 0&{(\gamma - 1)I\;}&{ - \sigma \Delta } \end{array}} \right),\qquad U: = \left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right),$

(2.2)

we rewrite (1.1) as

$\left\{ {\begin{array}{*{20}{l}} {\frac{{dU}}{{dt}} + AU = 0,\quad t > 0,} \\ {U(0) = {U_0},} \end{array}} \right.$

(2.3)

where

${U_0}: = \left( {\begin{array}{*{20}{c}} {{w_0}} \\ {{v_0}} \\ {{e_0}} \end{array}} \right).$

We also note that

$AU = \left( {\begin{array}{*{20}{c}} { - v} \\ { - {c^2}\Delta w + {c^2}\Delta e - {m^2}w} \\ { - \sigma \Delta e + (\gamma - 1)v} \end{array}} \right).$

(2.4)

The following definition plays an important role in the Hille-CYosida theory.

Definition 2.1. An unbounded linear operator $A: D(A) \subset H \to H$ is said to be monotone if it satisfies

$(Av,v) \ge 0 \quad \forall\,v \in D(A).$

It is said to be maximal monotone if,in addition,R(I + A) = H,i.e.;

$\forall {\mkern 1mu} f \in H\;\exists {\mkern 1mu} u \in D(A)\;{\text{such}}\;{\text{that}}\;u + Au = f.$

We next introduce two useful theorems (for the proof see [1,Chapter 7]).

Theorem 2.1 (Hille-CYosida). Let A be a maximal monotone operator in a Hilbert space H. Then,given any $u_0 \in D(A)$ there exists a unique function

$u \in C^1 ([0,\infty);H) \cap C([0,\infty);D(A))$

satisfying

$\left\{ {\begin{array}{*{20}{l}} {\frac{{du}}{{dt}} + Au = 0}&{{\text{on}}\;[0,\infty ),} \\ {u(0) = {u_0}.}&{} \end{array}} \right.$

(2.5)

Moreover,

$\left\| {u(t)} \right\| ≤ \left\| {{u_0}} \right\|\quad {\text{and}}\quad \left\| {\frac{{du}}{{dt}}(t)} \right\| = \left\| {Au(t)} \right\| ≤ \left\| {A{u_0}} \right\|\quad \forall {\mkern 1mu} t ≥ 0.$

Theorem 2.2. Assume $u_0 \in D(A^k)$ for some integer k ≥ 2. Then the solution u of (2.5) obtained in Theorem 2.1 satisfies

$u \in C^{k-j} ([0,\infty);D(A^j))\quad \forall\,j = 0,1,\dots,k.$

In order to apply the Hille-CYosida theory to (2.3) derived from (1.1) we define the domain of A given by (2.2) as

$D(A) = ({H^2}(\Omega ) \cap H_0^1(\Omega )) \times H_0^1(\Omega ) \times ({H^2}(\Omega ) \cap H_0^1(\Omega )).$

Then A is an operator in the Hilbert space

$H:=H_{0}^1(\Omega)\times L^2(\Omega)\times H_{0}^1(\Omega)$

equipped with inner product

$({U_1},{U_2}): = \int_\Omega \nabla {w_1}\nabla {w_2} + \int_\Omega {{w_1}} {w_2} + \frac{1}{{{c^2}}}\int_\Omega {{v_1}} {v_2} + \frac{1}{{\gamma - 1}}\int_\Omega \nabla {e_1}\nabla {e_2} + \frac{1}{{\gamma - 1}}\int_\Omega {{e_1}} {e_2},$

where

${U_j}: = \left( {\begin{array}{*{20}{c}} {{w_j}} \\ {{v_j}} \\ {{e_j}} \end{array}} \right)\qquad (j = 1,2).$

Also,the norm in H is given by

${\left\| U \right\|^2} = (U,U) = \left\| w \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| v \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| e \right\|_{{H^1}(\Omega )}^2\quad {\text{for}}\;U = \left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right).$

(2.6)

In particular,we see from (2.4) that

${\left\| {AU} \right\|^2} = \left\| v \right\|_{{H^1}(\Omega )}^2 + \frac{1}{{{c^2}}}\left\| {{c^2}\Delta w - {c^2}\Delta e + {m^2}w} \right\|_{{L^2}(\Omega )}^2 + \frac{1}{{\gamma - 1}}\left\| {\sigma \Delta e - (\gamma - 1)v} \right\|_{{H^1}(\Omega )}^2.$

(2.7)

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3. Existence and uniqueness

In this section we prove Theorem 1.1 by using Theorem 2.1.

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3.1. Monotonicity

Let A and H be as in the end of Section 2. Let

$U = \left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right) \in D(A).$

Then it follows from the definition of the inner product and integration by part that

$\begin{gathered} (AU,U) = \int_\Omega \nabla ( - v)\nabla w - \int_\Omega v w + \frac{1}{{{c^2}}}\int_\Omega v ( - {c^2}\Delta w - {m^2}w + {c^2}\Delta e) \\ + \frac{1}{{\gamma - 1}}\int_\Omega \nabla [(\gamma - 1)v - \sigma \Delta e]\nabla e + \frac{1}{{\gamma - 1}}\int_\Omega e \left[{(\gamma - 1)v - \sigma \Delta e} \right] \\ = - \left( {1 + \frac{{{m^2}}}{{{c^2}}}} \right)\int_\Omega v w + \frac{\sigma }{{\gamma - 1}}\int_\Omega | \Delta e{|^2} + \frac{\sigma }{{\gamma - 1}}\int_\Omega | \nabla e{|^2} + \int_\Omega v e. \\ \end{gathered}$

Since the second and third terms on the right-hand side are nonnegative,we have

$\begin{gathered} (AU,U) ≥ - \left( {1 + \frac{{{m^2}}}{{{c^2}}}} \right)\int_\Omega v w + \int_\Omega v e \\ ≥ - \left( {1 + \frac{{{m^2}}}{{{c^2}}}} \right)\int_\Omega | v||w| - \int_\Omega | v||e| \\ ≥ - \frac{{1 + \frac{{{m^2}}}{{{c^2}}}}}{2}\int_\Omega {({v^2} + {w^2})} - \frac{1}{2}\int_\Omega {({v^2} + {e^2})} \\ = - \left( {1 + \frac{{{m^2}}}{{2{c^2}}}} \right)\int_\Omega {{v^2}} - \left( {\frac{1}{2} + \frac{{{m^2}}}{{2{c^2}}}} \right)\int_\Omega {{w^2}} - \frac{1}{2}\int_\Omega {{e^2}} . \\ \end{gathered}$

We define a positive constant α₀ as

${\alpha _0}: = {\text{max}}\left\{ {\frac{1}{2} + \frac{{{m^2}}}{{2{c^2}}},\quad {c^2}\left( {1 + \frac{{{m^2}}}{{2{c^2}}}} \right),\quad \frac{{\gamma - 1}}{2}} \right\}.$

Then we conclude that A + α₀ is monotone:

$\begin{gathered} ((A + {\alpha _0})U,U) ≥ - (1 + \frac{{{m^2}}}{{2{c^2}}})\int_\Omega {{v^2}} - \left( {\frac{1}{2} + \frac{{{m^2}}}{{2{c^2}}}} \right)\int_\Omega {{w^2}} - \frac{1}{2}\int_\Omega {{e^2}} \\ + {\alpha _0}\int_\Omega {{w^2}} + \frac{{{\alpha _0}}}{{{c^2}}}\int_\Omega {{v^2}} + \frac{{{\alpha _0}}}{{\gamma - 1}}\int_\Omega {{e^2}} \\ + {\alpha _0}\int_\Omega | \nabla w{|^2} + \frac{{{\alpha _0}}}{{\gamma - 1}}\int_\Omega | \nabla e{|^2} \\ = \left[{{\alpha _0} - \left( {\frac{1}{2} + \frac{{{m^2}}}{{2{c^2}}}} \right)} \right]\int_\Omega {{w^2}} + \left[{\frac{{{\alpha _0}}}{{{c^2}}} - \left( {1 + \frac{{{m^2}}}{{2{c^2}}}} \right)} \right]\int_\Omega {{v^2}} \\ + \left( {\frac{{{\alpha _0}}}{{\gamma - 1}} - \frac{1}{2}} \right)\int_\Omega {{e^2}} + {\alpha _0}\int_\Omega | \nabla w{|^2} + \frac{{{\alpha _0}}}{{\gamma - 1}}\int_\Omega | \nabla e{|^2} \\ ≥ 0. \\ \end{gathered}$

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3.2. Maximal monotonicity

We divide the proof into four steps.

Step 1. Writing down the aim. We will select α > 0 later. The aim is to show that A + αI is maximal monotone,i.e.,

$\forall \;F \in H\quad \exists \;U \in D(A)\quad {\text{s}}{\text{.t}}{\text{.}}\quad U + (A + \alpha I)U = F.$

To see this we take

$\left( {\begin{array}{*{20}{c}} f \\ g \\ h \end{array}} \right) \in H.$

Then we shall find

$\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right) \in D(A)\quad {\text{s}}{\text{.t}}{\text{.}}\quad \left\{ {\begin{array}{*{20}{l}} { - v + (\alpha + 1)w}&{ = f,} \\ { - {c^2}\Delta w - {m^2}w + {c^2}\Delta e + (\alpha + 1)v}&{ = g,} \\ {(\gamma - 1)v - \sigma \Delta e + (\alpha + 1)e}&{ = h.} \end{array}} \right.$

Step 2. Reducing the equations. We first delete v and then we have

$- {c^2}\Delta w + \left[{{{(\alpha + 1)}^2} - {m^2}} \right]w + {c^2}\Delta e = (\alpha + 1)f + g,$

(3.1)

$- \sigma \Delta e + (\alpha + 1)e + (\alpha + 1)(\gamma - 1)w = (\gamma - 1)f + h.$

(3.2)

Now let $\delta \not=0$ which will be fixed later. Making (3.1) × $\delta$ + (3.2),we have

$\begin{gathered} \hfill - \delta {c^2}\Delta w - (\sigma - \delta {c^2})\Delta e + \left[{\delta {{(\alpha + 1)}^2} - \delta {m^2} + (\alpha + 1)(\gamma - 1)} \right]w + (\alpha + 1)e \\ \hfill = [\delta (\alpha + 1) + (\gamma - 1)]f + \delta g + h. \\ \end{gathered}$

(3.3)

If there exists a constant k such that

$\frac{{\sigma - {c^2}\delta }}{{\delta {c^2}}} = \frac{{\alpha + 1}}{{\delta {{(\alpha + 1)}^2} - \delta {m^2} + (\alpha + 1)(\gamma - 1)}} = k,$

(3.4)

then (3.3) is reduced to

$- \delta {c^2}\Delta u + [\delta {(\alpha + 1)^2} - \delta {m^2} + (\alpha + 1)(\gamma - 1)]u = [\delta (\alpha + 1) + (\gamma - 1)]f + \delta g + h,$

(3.5)

where u := w + ke.

Step 3. Finding two kinds of ($\delta$,k) in (3.4). We rewrite (3.4) as

$\delta {c^2}(\alpha + 1) = (\sigma - \delta {c^2})[\delta {(\alpha + 1)^2} - \delta {m^2} + (\alpha + 1)(\gamma - 1)].$

Dividing the both sides,we have

$\begin{gathered} \delta {c^2} = (\sigma - \delta {c^2})\left[{\delta (\alpha + 1) - \frac{{\delta {m^2}}}{{\alpha + 1}} + (\gamma - 1)} \right] \\ = \sigma (\alpha + 1)\delta - \frac{{\sigma {m^2}}}{{\alpha + 1}}\delta + \sigma (\gamma - 1) - {c^2}(\gamma - 1) - {c^2}(\alpha + 1){\delta ^2} + \frac{{{c^2}{m^2}}}{{\alpha + 1}}{\delta ^2} - {c^2}(\gamma - 1)\delta . \\ \end{gathered}$

Therefore,

$\varphi (\delta ): = {c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]{\delta ^2} + \left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right]\delta - \gamma (\gamma - 1) = 0.$

(3.6)

In order to find two solutions $\delta= \delta_1,\delta_2$ of this equation we show that the discriminant D is positive for $\alpha>|m|-1$. Indeed,we observe that

$D = {\left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right]^2} + 4{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]\gamma (\gamma - 1),$

where if we take α as

$\alpha > |m| - 1,\quad {\text{i}}{\text{.e}}{\text{.}},\quad \alpha + 1 > |m|,$

then

$(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}} = \frac{{{{(\alpha + 1)}^2} - {m^2}}}{{\alpha + 1}} > 0.$

Thus we deduce that D > 0. Noting that $\varphi(0)=-\gamma(\gamma -1)<0$,we see that (3.6) has two solutions $\delta =\delta_1,\,\delta_2$ such that $\delta_1<0$ and $\delta_2>0$:

$\begin{gathered} {\delta _1} = \frac{{ - \left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] - \sqrt D }}{{2{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]}},\hfill \\ {\delta _2} = \frac{{ - \left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] + \sqrt D }}{{2{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]}}. \hfill \\ \end{gathered}$

So we find two kinds of ($\delta$,k) in (3.4):

$\begin{gathered} \frac{{\sigma - {c^2}\delta }}{{\delta {c^2}}} = \frac{{\alpha + 1}}{{\delta {{(\alpha + 1)}^2} - \delta {m^2} + (\alpha + 1)(\gamma - 1)}} \\ = \left\{ {\begin{array}{*{20}{l}} {{k_1},}&{{\text{if}}\quad \delta = {\delta _1},} \\ {[1mm]{k_2},}&{{\text{if}}\quad \delta = {\delta _2}.} \end{array}} \right. \\ \end{gathered}$

Step 4. Conclusion. Let us consider the 2 parameters of Step 3,$\delta =\delta_1 (<0)$ and $\delta=\delta_2(>0)$. Note that k₁ ≠ k₂. Hence we can find two solutions u = u₁,u₂ of (3.5) with $\delta=\delta_1,\delta_2$,respectively (see [1,Theorems 9.21 and 9.25]):

$\begin{gathered} - {\delta _1}{c^2}\Delta {u_1} + [{\delta _1}{(\alpha + 1)^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)]{u_1} \\ = [{\delta _1}(\alpha + 1) + (\gamma - 1)]f + {\delta _1}g + h,\\ \end{gathered}$

(3.7)

$\begin{gathered} - {\delta _2}{c^2}\Delta {u_2} + [{\delta _2}{(\alpha + 1)^2} - {\delta _2}{m^2} + (\alpha + 1)(\gamma - 1)]{u_2} \\ = [{\delta _2}(\alpha + 1) + (\gamma - 1)]f + {\delta _2}g + h,\\ \end{gathered}$

(3.8)

which are equivalent to

$- {c^2}\Delta {u_1} + [{(\alpha + 1)^2} - {m^2} + \frac{{(\alpha + 1)(\gamma - 1)}}{{{\delta _1}}}]{u_1} = [(\alpha + 1) + \frac{{(\gamma - 1)}}{{{\delta _1}}}]f + g + \frac{h}{{{\delta _1}}},$

(3.9)

$- {c^2}\Delta {u_2} + [{(\alpha + 1)^2} - {m^2} + \frac{{(\alpha + 1)(\gamma - 1)}}{{{\delta _2}}}]{u_2} = [(\alpha + 1) + \frac{{(\gamma - 1)}}{{{\delta _2}}}]f + g + \frac{h}{{{\delta _2}}},$

(3.10)

where the coefficients of the second terms on the left-hand sides are positive for some α > 0. Indeed,the coefficient of u₂ is positive when $\alpha>|m|-1$,because $\delta_2>0$. As to the coefficient of u₁,we see that

$\begin{gathered} \frac{1}{{{\delta _1}}} = - \frac{{2{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]}}{{\left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] + \sqrt D }} \\ = - \frac{{2{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]\left\{ {\left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] - \sqrt D } \right\}}}{{ - 4{c^2}\left[{(\alpha + 1) - \frac{{{m^2}}}{{\alpha + 1}}} \right]\gamma (\gamma - 1)}} \\ = \frac{{\left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] - \sqrt D }}{{2\gamma (\gamma - 1)}}. \\ \end{gathered}$

Hence it follows that

$\begin{gathered} {(\alpha + 1)^2} - {m^2} + \frac{{(\alpha + 1)(\gamma - 1)}}{{{\delta _1}}} \hfill \\ = {(\alpha + 1)^2} - {m^2} + (\alpha + 1)\frac{{\left[{{c^2}\gamma - \sigma (\alpha + 1) + \frac{{\sigma {m^2}}}{{\alpha + 1}}} \right] - \sqrt D }}{{2\gamma }} > 0 \hfill \\ \Leftrightarrow (2\gamma - \sigma ){(\alpha + 1)^2} + {c^2}\gamma (\alpha + 1) + (\sigma - 2\gamma ){m^2} > (\alpha + 1)\sqrt D . \hfill \\ \end{gathered}$

(3.11)

Let $\alpha>|m|-1$. First consider the case $2\gamma \ge \sigma$. In this case,since

$\begin{gathered} (2\gamma - \sigma ){(\alpha + 1)^2} + {c^2}\gamma (\alpha + 1) + (\sigma - 2\gamma ){m^2} \hfill \\ = (2\gamma - \sigma )\left[{{{(\alpha + 1)}^2} - {m^2}} \right] + {c^2}\gamma (\alpha + 1) \hfill \\ ≥ 0,\hfill \\ \end{gathered}$

we have

$\begin{gathered} \hfill (2\gamma - \sigma ){(\alpha + 1)^2} + {c^2}\gamma (\alpha + 1) + (\sigma - 2\gamma ){m^2} > (\alpha + 1)\sqrt D \\ \hfill \Leftrightarrow {\left[{(2\gamma - \sigma ){{(\alpha + 1)}^2} + {c^2}\gamma (\alpha + 1) + (\sigma - 2\gamma ){m^2}} \right]^2} > {(\alpha + 1)^2}D \\ \hfill \Leftrightarrow (\gamma - \sigma ){(\alpha + 1)^4} + {c^2}{(\alpha + 1)^3} - 2(\gamma - \sigma ){m^2}{(\alpha + 1)^2} \\ \hfill - {c^2}{m^2}(\alpha + 1) + (\gamma - \sigma ){m^4} > 0 \\ \end{gathered}$

$\Leftrightarrow (\gamma - \sigma ){\left[{{{(\alpha + 1)}^2} - {m^2}} \right]^2} + {c^2}(\alpha + 1)\left[{{{(\alpha + 1)}^2} - {m^2}} \right] > 0$

(3.12)

$\Leftrightarrow \left[{{{(\alpha + 1)}^2} - {m^2}} \right]\left[{(\gamma - \sigma ){{(\alpha + 1)}^2} + {c^2}(\alpha + 1) - (\gamma - \sigma ){m^2}} \right] > 0.$

(3.13)

Therefore,if $\gamma \ge \sigma$,then the coefficient of u₁ in (3.9) is positive in view of (3.12). If $2\gamma \ge \sigma > \gamma$,then from (3.13) it suffices to choose α such that

$(\gamma - \sigma ){(\alpha + 1)^2} + {c^2}(\alpha + 1) - (\gamma - \sigma ){m^2} > 0,$

that is,

$(\sigma - \gamma ){(\alpha + 1)^2} - {c^2}(\alpha + 1) - (\sigma - \gamma ){m^2} < 0.$

Solving this inequality and noting that $\alpha +1>|m|\ge 0$,we have

$(|m| < {\mkern 1mu} )\;\alpha + 1 < \frac{{{c^2} + \sqrt {{c^4} + 4{{(\sigma - \gamma )}^2}{m^2}} }}{{2(\sigma - \gamma )}}.$

(3.14)

Next consider the case $2\gamma<\sigma$. In this case,from (3.11) it suffices to take α such that

$(2\gamma - \sigma ){(\alpha + 1)^2} + {c^2}\gamma (\alpha + 1) + (\sigma - 2\gamma ){m^2} > 0,$

that is,

$(\sigma - 2\gamma ){(\alpha + 1)^2} - {c^2}\gamma (\alpha + 1) - (\sigma - 2\gamma ){m^2} < 0.$

Solving this inequality gives

$(|m| < {\mkern 1mu} )\;\alpha + 1 < \frac{{{c^2}\gamma + \sqrt {{c^4}{\gamma ^2} + 4{{(\sigma - 2\gamma )}^2}{m^2}} }}{{2(\sigma - 2\gamma )}}.$

(3.15)

Hence the same way as in the case $2\gamma \ge \sigma$ yields that the coefficient of u₁ in (3.9) is positive when α satisfies (3.14) and (3.15). Thus we can find two solutions u = u₁,u₂ of (3.5) with $\delta=\delta_1,\delta_2$,respectively. For k₁,k₂ and u₁,u₂ constructed above,we solve the following system with respect to w,e:

$\left\{ {\begin{array}{*{20}{l}} {w + {k_1}e = {u_1},} \\ {w + {k_2}e = {u_2}.} \end{array}} \right.$

Then we find

$w = \frac{1}{{{k_2} - {k_1}}}({k_2}{u_1} - {k_1}{u_2}),$

(3.16)

$e = \frac{1}{{{k_2} - {k_1}}}({u_2} - {u_1}).$

(3.17)

Moreover,setting

$v = \frac{{\alpha + 1}}{{{k_2} - {k_1}}}({k_2}{u_1} - {k_1}{u_2}) - f,$

(3.18)

we shall show that w,v,e are the desired functions in Step 1. Indeed,we see that (3.18) implies the required equation

$- v + (\alpha + 1)w = f.$

(3.19)

Making (k₂ - k₁)[(3.16) + k₁ × (3.17)] and (k₂ - k₁)[(3.16) + k₂ × (3.17)],we have

$\begin{gathered} ({k_1} - {k_2})(w + {k_1}e) = ({k_1} - {k_2}){u_1},\;{\text{i}}{\text{.e}}{\text{.,}}\;{u_1} = w + {k_1}e,\hfill \\ ({k_1} - {k_2})(w + {k_2}e) = ({k_1} - {k_2}){u_2},\;{\text{i}}{\text{.e}}{\text{.,}}\;{u_2}w + {k_2}e. \hfill \\ \end{gathered}$

Therefore,in view of (3.7) and (3.8),

$\begin{gathered} - {\delta _1}{c^2}\Delta (w + {k_1}e) + [{\delta _1}{(\alpha + 1)^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)](w + {k_1}e) = [{\delta _1}(\alpha + 1) + (\gamma - 1)]f + {\delta _1}g + h,\hfill \\ - {\delta _2}{c^2}\Delta (w + {k_2}e) + [{\delta _2}{(\alpha + 1)^2} - {\delta _2}{m^2} + (\alpha + 1)(\gamma - 1)](w + {k_2}e) = [{\delta _2}(\alpha + 1) + (\gamma - 1)]f + {\delta _2}g + h,\hfill \\ \end{gathered}$

of which the first equation is equivalent to

$\begin{gathered} - {\delta _1}{c^2}\Delta w + [{\delta _1}{(\alpha + 1)^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)]w \hfill \\ \quad + [- {\delta _1}{c^2}\Delta e + {\delta _1}{(\alpha + 1)^2}e - {\delta _1}{m^2}e + (\alpha + 1)(\gamma - 1)e]{k_1} \hfill \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad = [{\delta _1}(\alpha + 1) + (\gamma - 1)]f + {\delta _1}g + h. \hfill \\ \end{gathered}$

Recall the definition of k₁:

$\frac{{\sigma - {c^2}{\delta _1}}}{{{\delta _1}{c^2}}} = \frac{{\alpha + 1}}{{{\delta _1}{{(\alpha + 1)}^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)}} = {k_1}.$

Then it follows that

${k_1}{\delta _1}{c^2} = \sigma - {c^2}{\delta _1},\quad {k_1}\left[{{\delta _1}{{(\alpha + 1)}^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)} \right] = \alpha + 1,$

and hence

$\begin{gathered} \hfill - {\delta _1}{c^2}\Delta w - (\sigma - {\delta _1}{c^2})\Delta e + \left[{{\delta _1}{{(\alpha + 1)}^2} - {\delta _1}{m^2} + (\alpha + 1)(\gamma - 1)} \right]w + (\alpha + 1)e \\ \hfill = [{\delta _1}(\alpha + 1) + (\gamma - 1)]f + {\delta _1}g + h. \\ \end{gathered}$

(3.20)

In the same way as above we can deduce

$\begin{gathered} \hfill - {\delta _2}{c^2}\Delta w - (\sigma - {\delta _2}{c^2})\Delta e + \left[{{\delta _2}{{(\alpha + 1)}^2} - {\delta _2}{m^2} + (\alpha + 1)(\gamma - 1)} \right]w + (\alpha + 1)e \\ \hfill = [{\delta _2}(\alpha + 1) + (\gamma - 1)]f + {\delta _2}g + h. \\ \end{gathered}$

(3.21)

Making (3.20) - (3.21) and (3.20) × $\delta_2$ - (3.21) × $\delta_1$,we have

$\begin{gathered} \hfill - ({\delta _1} - {\delta _2}){c^2}\Delta w + ({\delta _1} - {\delta _2}){c^2}\Delta e + ({\delta _1} - {\delta _2}){(\alpha + 1)^2}w - ({\delta _1} - {\delta _2}){m^2}w \\ \hfill = ({\delta _1} - {\delta _2})(\alpha + 1)f + ({\delta _1} - {\delta _2})g,\\ \hfill - ({\delta _2} - {\delta _1})\sigma \Delta e({\delta _2} - {\delta _1})(\alpha + 1)(\gamma - 1)w + ({\delta _2} - {\delta _1})(\alpha + 1)e \\ \hfill = ({\delta _2} - {\delta _1})(\gamma - 1)f + ({\delta _2} - {\delta _1})h. \\ \end{gathered}$

Thus we arrive at (3.1) and (3.2). Making (3.1)-(3.19)×(α+1) and (3.2)-(3.19)×(γ-1),we obtain

$\begin{gathered} - {c^2}\Delta w - {m^2}w + {c^2}\Delta e + (\alpha + 1)v = g,\hfill \\ (\gamma - 1)v - \sigma \Delta e + (\alpha + 1)e = h. \hfill \\ \end{gathered}$

Consequently,we conclude that w,v,e are the desired functions which satisfy

$\left\{ {\begin{array}{*{20}{l}} { - v + (\alpha + 1)w}&{ = f,} \\ { - {c^2}\Delta w - {m^2}w + {c^2}\Delta e + (\alpha + 1)v}&{ = g,} \\ {(\gamma - 1)v - \sigma \Delta e + (\alpha + 1)e}&{ = h.} \end{array}} \right.$

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3.3. Proof of Theorem 1.1

Since A + α is maximal monotone as proved above,it follows from Theorem 2.1 that for $U_0 \in D(A)$ the problem

$\left\{ {\begin{array}{*{20}{l}} {\frac{{dV}}{{dt}} + AV + \alpha V = 0\qquad {\text{on}}\;[0,\infty ),} \\ {V(0) = {U_0}} \end{array}} \right.$

has a unique solution $V \in C^1 ([0,\infty);H) \cap C([0,\infty);D(A))$ such that

$\left\| {V(t)} \right\| ≤ \left\| {{U_0}} \right\|\quad {\text{and}}\quad \left\| {AV(t)} \right\| ≤ \left\| {A{U_0}} \right\|\quad \forall {\mkern 1mu} t ≥ 0.$

Setting

$U(t):=e^{\alpha t}V(t),$

we deduce that $U \in C^1 ([0,\infty);H) \cap C([0,\infty);D(A))$ satisfies

$\left\{ {\begin{array}{*{20}{l}} {\frac{{dU}}{{dt}} + AU = 0\qquad {\text{on}}\;[0,\infty ),} \\ {U(0) = {U_0},} \end{array}} \right.$

with the estimates

$\left\| {U(t)} \right\| ≤ {e^{\alpha t}}\left\| {{U_0}} \right\|\quad {\text{and}}\quad \left\| {AU(t)} \right\| ≤ {e^{\alpha t}}\left\| {A{U_0}} \right\|\quad \forall {\mkern 1mu} t ≥ 0.$

The properties (1.5),(1.6),(1.7),(1.8) follow from those for U. This completes the proof of Theorem 1.1.

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4. Regularity

We use the same notation as in the end of Section 2.

Proof of Theorem 1.2. We first recall the definition of D(A^k) which is given by induction as follows:

$D({A^1}): = D(A),\quad D({A^k}): = \{ U \in D({A^{k - 1}})|\;AU \in D({A^{k - 1}})\} ,\quad k ≥ 2.$

It is easy to see,by induction on k,that

$D({A^k}) = {D_k}: = \left\{ {\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right)\left| {\;\begin{array}{*{20}{l}} {w \in {H^{k + 1}}(\Omega ),\;}&{{\Delta ^j}w = 0\;{\text{on}}\;\Gamma }&{(0 ≤ \forall {\mkern 1mu} j ≤ [\frac{k}{2}])} \\ {v \in {H^k}(\Omega ),\;}&{{\Delta ^j}v = 0\;{\text{on}}\;\Gamma }&{(0 ≤ \forall {\mkern 1mu} j ≤ [\frac{{k + 1}}{2}] - 1)} \\ {e \in {H^{k + 1}}(\Omega ),\;}&{{\Delta ^j}e = 0\;{\text{on}}\;\Gamma }&{(0 ≤ \forall {\mkern 1mu} j ≤ [\frac{k}{2}])} \end{array}} \right.} \right\}.$

Indeed,when k = 1,we have

$\begin{gathered} D({A^1}) = ({H^2}(\Omega ) \cap H_0^1(\Omega )) \times H_0^1(\Omega ) \times ({H^2}(\Omega ) \cap H_0^1(\Omega )) \\ = \left\{ {\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right)\left| {\;\begin{array}{*{20}{l}} {w \in {H^2}(\Omega ),}&{w = 0\;}&{{\text{on}}\;\Gamma } \\ {v \in {H^1}(\Omega ),}&{v = 0\;}&{{\text{on}}\;\Gamma } \\ {e \in {H^2}(\Omega ),}&{e = 0\;}&{{\text{on}}\;\Gamma } \end{array}} \right.} \right\} = {D_1}. \\ \end{gathered}$

If D(A^k) = D_k holds for k,then the statement for k + 1 reads as follows:

$\begin{gathered} D({A^{k + 1}}) = \{ U \in D({A^k}) = {D_k}|\;AU \in D({A^k}) = {D_k}\} \\ = \left\{ {\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right)\left| {\;\begin{array}{*{20}{l}} {w \in {H^{k + 1}}(\Omega ),}&{{\Delta ^j}w = 0}&{{\text{on}}\;\Gamma }&{(0 ≤ \forall j ≤ [\frac{k}{2}])} \\ {[1mm]v \in {H^k}(\Omega ),}&{{\Delta ^j}v = 0}&{{\text{on}}\;\Gamma }&{(0 ≤ \forall j ≤ [\frac{{k + 1}}{2}] - 1)} \\ {[1mm]e \in {H^{k + 1}}(\Omega ),}&{{\Delta ^j}e = 0}&{{\text{on}}\;\Gamma }&{(0 ≤ \forall j ≤ [\frac{k}{2}])} \end{array}} \right.} \right\} \\ \cap \left\{ {\left( {\begin{array}{*{20}{c}} w \\ v \\ e \end{array}} \right)\left| {\;\begin{array}{*{20}{l}} { - v \in {H^{k + 1}}(\Omega ),\;{\Delta ^j}( - v) = 0\;{\text{on}}\;\Gamma \;(0 ≤ \forall j ≤ [\frac{k}{2}])} \\ {[1mm] - {c^2}\Delta w - {m^2}w + {c^2}\Delta e \in {H^k}(\Omega ),} \\ {{\Delta ^j}( - {c^2}\Delta w - {m^2}w + {c^2}\Delta e) = 0\;{\text{on}}\;\Gamma \;(0 ≤ \forall j ≤ [\frac{{k + 1}}{2}] - 1)} \\ {[1mm](\gamma - 1)v - \sigma \Delta e \in {H^{k + 1}}(\Omega ),} \\ {{\Delta ^j}((\gamma - 1)v - \sigma \Delta e) = 0\;{\text{on}}\;\Gamma \;(0 ≤ \forall j ≤ [\frac{k}{2}])} \end{array}} \right.} \right\} \\ = {D_{k + 1}}. \\ \end{gathered}$

In particular,$D(A^k)\subset H^{k+1}(\Omega)\times H^k(\Omega) \times H^{k+1}(\Omega)$ with continuous injection. Applying Theorem 2.2,we see that if U0 2 D(Ak),then the solution U of (2.3) satisfies

$\begin{gathered} U \in {C^{k - j}}([0,\infty];D({A^j})) \hfill \\ \subset {C^{k - j}}([0,\infty];{H^{j + 1}}(\Omega ) \times {H^j}(\Omega ) \times {H^{j + 1}}(\Omega ))\quad \forall \;j = 0,1,\ldots ,k. \hfill \\ \end{gathered}$

Therefore we conclude by [1,Corollary 9.15] that under the assumption of Theorem 1.2 (i.e.,$U_0\in D(A^k)\ \forall\ k \in \mathbb{N}$),$U \in {C^k}([0,\infty );{C^k}(\Omega ) \times {C^k}(\Omega ) \times {C^k}(\Omega ))\;\forall {\mkern 1mu} k \in \mathbb{N}$.

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Acknowledgments

The authors would like to the anonymous referees for helpful suggestions. T. Yokota is supported by Grant-in-Aid for Scientific Research (C),No. 16K05182.

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