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Mixed Cauchy problem with lateral boundary condition for noncharacteristic degenerate hyperbolic equations


In this paper, in a cylindrical domain \(D=\Omega \times (0,T)\) with \(\Omega \subset {R}^{n}\), we consider a mixed Cauchy problem with a potential lateral boundary condition for the following noncharacteristic degenerated equation

$$\begin{aligned} Lu=u_{tt}-k(t)\Delta _{x}u(x,t)=f(x,t), \end{aligned}$$

where \(k(t)\geq 0\). As in the case for strictly hyperbolic equations, we first establish that \(u\in W_{2}^{1}(D)\) and \(u\in W_{2}^{2}(D)\) under the assumptions \(\Vert \frac{f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) and \(\Vert \frac{\mathrm{grad}_{x} f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) for every \(t\in [0,T]\), respectively.

1 Introduction

A number of studies have been devoted to the mixed Cauchy problem for noncharacteristically degenerate second-order hyperbolic equations, starting from the work of M.L. Krasnov [1]. Later these works were generalized for general degenerate higher-order equations by D.T. Dzhuraev [2], V.N. Vragov [3], and A.I. Kozhanov [4]. The study of boundary value problems for an equation of the mixed type, started by F.G. Tricomi [5], led to the study of new boundary value problems for hyperbolic equations in the characteristic cone, first investigated in the works of S. Gellerstedt [6], A.V. Bitsadze [7] A.M. Nakhushev [8], and T.S. Kal’menov [911]. In recent years, the well-posedness of the Cauchy problem for the wave equation with strongly singular coefficients has been investigated by M. Ruzhansky, N. Tokmagambetov [12]. More complete bibliography may be found in the monographs of M.M. Smirnov [13], I.E. Egorov, S.G. Pyatkov, S.V. Popov [14], E.V. Radkevich, O.A. Olejnik [15] and M. Ruzhansky, M. Sadybekov, D. Suragan [16].

In the study of the mixed Cauchy problem in a cylindrical domain, the lateral boundary conditions are usually local boundary conditions of the Dirichlet type or periodic boundary conditions.

In [17], the boundary condition for the Newton (volume) potential was found, which is a new integro-differential self-adjoint boundary condition for the Laplace equation. In this paper, we study the mixed Cauchy problem for one class of noncharacteristic degenerate hyperbolic equations using this boundary condition. Unlike other works devoted to this topic, where solutions of the mixed Cauchy problem with different lateral boundary conditions of the problems under consideration are obtained in weighted spaces; in this paper, all solutions of the mixed Cauchy problems under consideration are obtained in classical Sobolev spaces.

Note that in [17], the Newton potential (volume potential) is given by a self-adjoint integral operator

$$\begin{aligned} u(x)= \int _{\Omega}{\varepsilon (x,\xi )\rho (\xi )\,d\xi }, \end{aligned}$$

where \(\rho (\xi )\in L_{2}(\Omega )\) and \(\varepsilon (x,\xi )\) is a fundamental solution of the Laplace equation

$$\begin{aligned} -{\Delta }_{x}\varepsilon (x,\xi )=\delta (x-\xi ),\qquad \varepsilon (x, \xi )=\varepsilon (\xi,x), \end{aligned}$$

the function \(u\in W^{2}_{2}(\Omega )\) satisfies the equation

$$\begin{aligned} -{\Delta }_{x}u=\rho (x) \end{aligned}$$

and the lateral boundary condition

$$\begin{aligned} -\frac{u(x)}{2} + \int _{\partial \Omega} \biggl( \frac{\partial \varepsilon}{\partial n_{\xi }} (x,\xi )\cdot u(\xi )- \varepsilon (x,\xi )\cdot \frac{\partial u}{\partial n_{\xi }}(\xi ) \biggr)\,d\xi =0. \end{aligned}$$

Conversely, if \(u \in W^{2}_{2}(\Omega )\) satisfies equation (3) and boundary condition (4), then \(u(x)\) coincides with the Newton potential (1).

The aim of this paper is to study the mixed Cauchy problem with condition (4).

2 Preliminaries

Let \(\Omega \subset R^{n}\) be a finite domain with smooth boundary \(\partial \Omega \subset C^{2}\), \(D=\Omega \times [0, T]\) a cylindrical domain. In D, we consider the following mixed Cauchy problem.

Find a solution of the following equation

$$\begin{aligned} Lu=u_{tt}-k(t)\Delta _{x}u+b(t) \frac{\partial u}{\partial t}+a(t)u=f(x,t), \end{aligned}$$

which satisfies the initial conditions

$$\begin{aligned} u|_{t=0}=0, \qquad\frac{\partial u}{\partial t}\bigg|_{t=0}=0 \end{aligned}$$

and the lateral boundary condition

$$\begin{aligned} & N[u]\equiv -\frac{u(x,t)}{2}+ \int _{\partial \Omega} \biggl( \frac{\partial \varepsilon}{\partial n_{\xi }} (x,\xi )\cdot u(\xi,t)- \varepsilon (x,\xi )\cdot \frac{\partial u}{\partial n_{\xi }}(\xi,t) \biggr)\,d\xi =0, \\ & \quad 0< t< T, x\in \partial \Omega, \end{aligned}$$

where \(k\in C^{1+\alpha}[0,T]\), \(0<\alpha <1\), \(k(t)>0\), \(t>0\), \(k(0)=0\), \(k'(t)\geq 0\), and \(\varepsilon (x,\xi )\) is the fundamental solution of the Laplace equation (2).

The eigenfunctions of the Newton potential satisfies the following equation

$$\begin{aligned} -\Delta e_{m}(x)=\lambda _{m}e_{m}(x), \end{aligned}$$

and the boundary condition

$$\begin{aligned} N[e_{m}]\equiv -\frac{e_{m}(x)}{2}+ \int _{\partial \Omega} \biggl(\frac{\partial \varepsilon}{\partial n_{\xi }}(x,\xi )e_{m}( \xi )-\varepsilon (x,\xi )\frac{\partial e_{m}}{\partial n_{\xi }}( \xi ) \biggr)\,d\xi =0,\quad x\in \partial \Omega. \end{aligned}$$

According to relations (1)–(4), the set of eigenfunctions \(\{e_{m}(x)\}\) of self-adjoint boundary problem (8)–(9) forms a complete orthonormal system in \(L_{2}(\Omega )\).

For \(a(t)\equiv b(t)\equiv 0\), two-dimensional equation (5) is the Chaplygin equation, which is applied to model the supersonic flow of liquid and gas.

In what follows, the boundary condition (7) will be called a potential boundary condition. Although the boundary condition of problem (5)–(7) is cumbersome, the Green function of this problem coincides with the fundamental solution \(\varepsilon (x,\xi )\) of the Laplace equation, which means that the Green function is given explicitly in an arbitrary domain.

As in the case for strictly hyperbolic equations, we first show that \(u\in W_{2}^{1}(D)\) and \(u\in W_{2}^{2}(D)\) under the assumptions \(\Vert \frac{f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty, t\in [0,T]\) and \(\frac{a}{k} \in C^{1+\alpha}(\bar{D}), \frac{b}{k} \in C^{1+\alpha}( \bar{D})\), respectively.

3 Mixed Cauchy problem with the condition \(a(t)\equiv b(t)\equiv 0\)

Let us consider the problems (5)–(7) in the case \(a(t)\equiv b(t)\equiv 0\). Let

$$\begin{aligned} & Lu=u_{tt}-{k(t)\Delta}_{x}u=f(x,t),\quad k(t)>0, t>0, k(0)=0, k'(t) \geq 0, \end{aligned}$$
$$\begin{aligned} &u|_{t=0}=0, \qquad\frac{\partial u}{\partial t}\bigg|_{t=0}=0, \end{aligned}$$
$$\begin{aligned} &N[u]\equiv -\frac{u(x,t)}{2}+ \int _{\partial \Omega} \biggl( \frac{\partial \varepsilon}{\partial n_{\xi }} (x,\xi )\cdot u(\xi,t)- \varepsilon (x,\xi )\cdot \frac{\partial u}{\partial n_{\xi }}(\xi,t) \biggr)\,d\xi =0, \\ &\quad (x,t)\in \partial \Omega \times [0,T]. \end{aligned}$$

Due to the complexity of potential boundary condition (12), to establish an apriori estimates for problem (10)–(12), we will use the spectral decomposition method.

Let \(\{e_{m}(x) \}\) be a complete orthonormal system of eigenvectors of problem (8)–(9).

The solution of (10)–(12) can be written in the form

$$\begin{aligned} &u(x,t)=\sum^{\infty }_{|m|=1}{u_{m}(t)e_{m}(x)},\quad e_{m}=e_{m_{1}m_{2} \dots m_{n}}, \end{aligned}$$
$$\begin{aligned} & f(x,t)=\sum^{\infty }_{|m|=1}{f_{m}(t)e_{m}(x)}, \end{aligned}$$


$$\begin{aligned} u_{m}(t)= \int _{\Omega}u(x,t)e_{m}(x)\,dx, \qquad f_{m}(t)= \int _{\Omega}{f(x,t)}e_{m}(x)\,dx. \end{aligned}$$

Substituting (13) and (14) into equation (10), for \(u_{m}(t)\), we get the following one-dimensional Cauchy problem:

$$\begin{aligned} & \frac{d^{2}u_{m}}{dt^{2}}+\lambda _{m}k(t)u_{m}(t)=f_{m}(t), \end{aligned}$$
$$\begin{aligned} &u_{m}(0)=0,\qquad \frac{d}{dt}u_{m}(0) =0. \end{aligned}$$

Lemma 3.1

All solutions \(u_{m}\in W^{2}_{2}(0,T)\) of the Cauchy problem (15)(16) satisfy the inequality

$$\begin{aligned} \frac{1}{k(t)} \biggl\vert \frac{\partial u_{m}}{\partial t} \biggr\vert ^{2}(t)+ \lambda _{m}u^{2}_{m}(t)+ \int ^{t}_{0}\frac{k(\eta )}{k^{2}(\eta )} \biggl\vert \frac{\partial u_{m}}{\partial \eta} \biggr\vert ^{2}(\eta )\,d\eta \leq d_{1} \int ^{t}_{0} \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert ^{2}\,d \eta, \end{aligned}$$

where \(d_{1}\) is some positive constant independent of f.


Since \(u_{m}(0)=0\) and \(\frac{du_{m}}{dt}(0)=0\), we obtain

$$\begin{aligned} \frac{du_{m}}{dt}(t)= \int ^{t}_{0}{ \frac{d^{2}u_{m}}{d{\eta }^{2}}(\eta )\,d\eta } \quad\text{and}\quad u_{m}(t)= \int ^{t}_{0}{\frac{du_{m}}{d\eta }\,d\eta }. \end{aligned}$$

Integrating both sides of (15) from 0 to t and using equality (18), we get

$$\begin{aligned} \begin{aligned} \int ^{t}_{0}\frac{d^{2}u_{m}}{d{\eta }^{2}}( \eta )\,d\eta + \lambda _{m} \int ^{t}_{0}k(\eta )u_{m}(\eta )\,d \eta &=\frac{du_{m}}{dt} + \lambda _{m} \int ^{t}_{0}{k}(\eta ) (t- \eta )\frac{du_{m}}{d\eta} \,d\eta \\ &= \int ^{{t}}_{0}k(\eta )\frac{f_{m}(\eta )}{k(\eta )}\,d\eta. \end{aligned} \end{aligned}$$

Assuming that \(\frac{f_{m}}{k}\in L_{2}(0,T)\), integral equation (19) is a Volterra integral equation. Therefore, since \(k'\geq 0\), the inequalities

$$\begin{aligned} \int ^{t}_{0}k(\eta ) \biggl\vert \frac{f(\eta )}{k(\eta )} \biggr\vert \,d \eta \leq \sup_{0\leq \xi \leq t}k(\xi ) \int ^{t}_{0} \biggl\vert \frac{f(\eta )}{k(\eta )} \biggr\vert \,d\eta = k(t) \int ^{t}_{0} \biggl\vert \frac{f(\eta )}{k(\eta )} \biggr\vert \,d\eta \end{aligned}$$

hold, and we can see that

$$\begin{aligned} \biggl\vert \frac{\partial u_{m}}{\partial t} \biggr\vert (t)\leq C_{m}k(t) \int ^{t}_{0} \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert \,d \eta, \end{aligned}$$

where \(C_{m}\) depends on \({\lambda }_{m}\).

Now we obtain the necessary apriori estimates for problem (15)–(16) and rewrite this problem in the form

$$\begin{aligned} & \frac{1}{k(t)}\frac{d^{2}u_{m}}{dt^{2}}(t)+{\lambda }_{m}u_{m}(t)= \frac{f_{m}(t)}{k(t)}, \\ &u_{m}(0)=u'_{m}(0)=0. \end{aligned}$$

Let us calculate the inner product of (21) and \(\frac{du_{m}(t)}{dt}\) in \(L_{2}(0,T)\)

$$\begin{aligned} \begin{aligned} &\int ^{t}_{0}\frac{1}{2}k(\eta ) \frac{d^{2}u_{m}}{d\eta ^{2}}(\eta )\frac{du_{m}}{d\eta }\,d\eta + \lambda _{m} \int ^{t}_{0}u_{m}(\eta ) \frac{du_{m}}{d\eta}\,d \eta \\ &\quad=\frac{1}{2} \int ^{t}_{0}{\frac{1}{k(\eta )}\frac{d}{d\eta }{ \biggl(\frac{du_{m}}{d\eta } \biggr)}^{2}\,d\eta}+ \frac{\lambda _{m}}{2} \int ^{t}_{0}{\frac{du^{2}_{m}}{d\eta }}( \eta )\,d\eta \\ &\quad=\frac{1}{2}\frac{1}{k(t)} \biggl(\frac{du_{m}}{dt} \biggr)^{2}(t)- \frac{1}{2}\lim_{t\to 0} \frac{1}{k(t)}{ \biggl(\frac{du_{m}}{dt} \biggr)}^{2}(t)\\ &\qquad{}- \frac{1}{2} \int ^{t}_{0}{ \biggl( \frac{d}{d\eta } \frac{1}{k(\eta )} \biggr) \biggl( \frac{du_{m}}{d\eta } \biggr)}^{2}\,d \eta +\frac{\lambda _{m}}{2}u^{2}_{m}(t) \\ &\quad= \int ^{t}_{0}\frac{f_{m}(\eta )}{k(\eta )} \frac{du_{m}(\eta )}{d\eta } \,d\eta. \end{aligned} \end{aligned}$$

By inequality (20), we obtain

$$\begin{aligned} \lim_{t\to 0} \frac{1}{k(t)}{ \biggl( \frac{du_{m}}{dt} \biggr)}^{2}(t)=0. \end{aligned}$$

From the equality \(-\frac{d}{d\eta }\frac{1}{k(\eta )}=\frac{k'(\eta )}{k^{2}(\eta )}\) and

$$\begin{aligned} \int ^{t}_{0}\frac{f_{m}(\eta )}{k(\eta )} \frac{du_{m}(\eta )}{d\eta } \,d\eta \leq \frac{\varepsilon }{2} \int ^{t}_{0}{{ \biggl\vert \frac{du_{m}(\eta )}{d\eta } \biggr\vert }^{2}}\,d \eta +\frac{2}{\varepsilon } \int ^{t}_{0}{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert }^{2}\,d\eta, \end{aligned}$$

where ε is a positive real number, it is easy to verify that

$$\begin{aligned} \int ^{t}_{0} \biggl\vert \frac{\partial u}{\partial \eta} \biggr\vert ^{2}\,d \eta \leq t\cdot \sup_{0\leq \eta \leq t} \biggl\vert \frac{\partial u_{m}}{\partial \eta} \biggr\vert ^{2}. \end{aligned}$$

By (22)–(23), it is easy to check that

$$\begin{aligned} \begin{aligned} \frac{1}{2}\sup_{0\leq \eta \leq t} \frac{1}{k(\eta )} \biggl\vert \frac{du}{d\eta} \biggr\vert ^{2}&+ \frac{1}{2}{ \lambda }_{m}u^{2}_{m}(t)+ \frac{1}{2} \int ^{t}_{0} \frac{k'}{k^{2}} \biggl\vert \frac{du_{m}}{d\eta } \biggr\vert ^{2}\,d\eta \\ & \leq \frac{\varepsilon }{2} \int ^{t}_{0}{{ \biggl\vert \frac{du_{m}(\eta )}{d\eta } \biggr\vert }^{2}}\,d\eta + \frac{2}{\varepsilon } \int ^{t}_{0}{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert }^{2}\,d\eta \\ &\leq \frac{\varepsilon }{2}\cdot t\cdot \sup_{0\leq \eta \leq t} \biggl\vert \frac{\partial u_{m}}{\partial \eta} \biggr\vert ^{2}+ \frac{2}{\varepsilon } \int ^{t}_{0}{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert }^{2}\,d\eta. \end{aligned} \end{aligned}$$


$$\begin{aligned} \frac{1}{2}\sup_{0\leq \eta \leq t} \biggl\vert \frac{\partial u_{m}}{\partial \eta} \biggr\vert ^{2} \frac{1-\varepsilon k(\eta )t}{k(\eta )} + \frac{\lambda _{m}}{2}u^{2}_{m}(t)+ \frac{1}{2} \int ^{t}_{0}\frac{k'(\eta )}{k^{2}(\eta )} \biggl\vert \frac{\partial u_{m}}{\partial \eta } \biggr\vert ^{2}(\eta )\,d\eta \leq \frac{2}{\varepsilon } \int ^{t}_{0}{{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert }^{2}\,d\eta }. \end{aligned}$$

Since \(k(t)\) is bounded in \([0,T]\), for small ε, we have \(1-\varepsilon k(t)\cdot t>\delta \). Therefore, from the above inequality, we get

$$\begin{aligned} \frac{1}{2}\frac{1}{k(t)}{ \biggl(\frac{\partial u_{m}}{\partial t} \biggr)}^{2}(t)+\frac{{\lambda }_{m}}{2{\delta }_{1}}u^{2}_{m}(t)+ \frac{1}{2{\delta }_{1}} \int ^{t}_{0}{ \frac{k'(\eta )}{k^{2}(\eta )}{ \biggl\vert \frac{\partial u_{m}}{\partial \eta } \biggr\vert }^{2}(\eta )\,d\eta \le } \frac{2}{\varepsilon {\delta }_{1}} \int ^{t}_{0}{{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )} \biggr\vert }^{2}\,d\eta }, \end{aligned}$$

which is equivalent to

$$\begin{aligned} \frac{1}{k(t)} \biggl\vert \frac{\partial u_{m}}{\partial t} \biggr\vert ^{2}(t)+ \lambda _{m}u^{2}_{m}(t)+ \int ^{t}_{0}\frac{k(\eta )}{k^{2}(\eta )} \biggl\vert \frac{\partial u_{m}}{\partial \eta} \biggr\vert ^{2}(\eta )\,d\eta \leq \frac{4}{\varepsilon \cdot \delta _{1}}\cdot \int ^{t}_{0} \frac{f_{m}^{2}(\eta )}{k^{2}(\eta )}\,d\eta. \end{aligned}$$

This completes the proof. □

Remark 3.1

Note that to prove the main inequality (24), we have used inequality (23).

Lemma 3.2

Let \(k\in C^{1+\alpha }[0,T]\), \(1>\alpha >0\), \(k(t)>0\), \(t>0\), \(t>0\), \(k(0)=0\), \(k'(t)\geq 0\), \(\frac{f_{m}}{k}\in L_{2}(0,T)\). Then all solutions \(u_{m}\in W^{2}_{2}(0,T)\) of the mixed Cauchy problem (16)(19) satisfy the inequality

$$\begin{aligned} \begin{aligned} &\frac{1}{k(t)} \biggl(\sqrt{\lambda _{m}} \frac{du_{m}}{dt} \biggr)^{2}(t)+ \bigl(\lambda _{m}u_{m}(t) \bigr)^{2}+ \int ^{t}_{0} \biggl(\sqrt{\lambda _{m}} \frac{du_{m}}{dt}(t) \biggr)^{2}\,d\eta \\ &\quad\leq d_{2} \biggl[ \int ^{t}_{0} \biggl( \frac{\sqrt{\lambda _{m}}f_{m}(\eta )}{k(\eta )} \biggr)^{2}\,d\eta + \frac{f^{2}_{m}(t)}{k^{2}(t)} \biggr]. \end{aligned} \end{aligned}$$


Multiplying both sides of (17) in Lemma 3.1 by \(\lambda _{m}\), we get

$$\begin{aligned} \begin{aligned} &\frac{1}{k(t)}{ \biggl(\sqrt{\lambda _{m}} \biggl( \frac{d^{2}u_{m}}{dt^{2}} \biggr) \biggr)}^{2}+{ \lambda }^{2}_{m}u^{2}_{m}(t)+ \int ^{t}_{0}\frac{k'(\eta )}{k^{2}(\eta )}{ \biggl(\sqrt{ \lambda _{m}} \biggl(\frac{du_{m}}{d\eta } \biggr) \biggr)}^{2}( \eta )\,d \eta \\ &\quad\leq d_{3} \int ^{t}_{0} \biggl( \frac{\sqrt{{\lambda }_{m}}f_{m}(\eta )}{k(\eta )} \biggr)^{2}\,d\eta. \end{aligned} \end{aligned}$$

Hence, we obtain \({\lambda }^{2}_{m}u^{2}_{m}(t)\le d_{3}\int ^{t}_{0}{{ ( \frac{\sqrt{{\lambda }_{m}}f_{m}(\eta )}{k(\eta )} )}^{2}\,d\eta }\).

From (16) and the above inequalities for

$$\begin{aligned} \frac{1}{k(t)}\frac{\partial ^{2}u_{m}}{\partial t^{2}}(t)=-\lambda _{m}u_{m}(t)+ \frac{f_{m}(t)}{k(t)}, \end{aligned}$$

we have

$$\begin{aligned} \biggl\vert \frac{1}{k(t)}\frac{\partial ^{2}u_{m}}{\partial t^{2}} \biggr\vert ^{2} \leq 2 \biggl[\lambda ^{2}_{m}u^{2}_{m}(t)+ \frac{f^{2}_{m}(t)}{k^{2}(t)} \biggr]\leq 2 \biggl[d_{3} \int ^{t}_{0} \biggl(\frac{\sqrt{\lambda _{m}}f_{m}(\eta )}{k(\eta )} \biggr)^{2}\,d \eta +\frac{f^{2}_{m}(t)}{k^{2}(t)} \biggr]. \end{aligned}$$

By (26) and (27), it is easy to see

$$\begin{aligned} \begin{aligned} &\frac{1}{k(t)} \biggl(\sqrt{\lambda _{m}} \frac{d^{2}u_{m}}{dt^{2}} \biggr)^{2}+ \bigl(\lambda _{m}u_{m}(t) \bigr)^{2}+ \int ^{t}_{0} \biggl(\sqrt{\lambda _{m}} \frac{du_{m}}{dt} \biggr)^{2}\,d\eta \\ &\quad \leq d_{2} \biggl[ \int ^{t}_{0} \biggl( \frac{\sqrt{\lambda _{m}}f_{m}(\eta )}{k(\eta )} \biggr)^{2}\,d\eta + \frac{f^{2}_{m}(t)}{k^{2}(t)} \biggr], \end{aligned} \end{aligned}$$

which finishes the proof. □

4 General case

Now we will consider the mixed Cauchy problem.

Let \(\Omega \subset R^{n}\) be a finite domain with smooth boundary \(\partial \Omega \subset C^{2}\), \(D=\Omega \times [0, T]\) a cylindrical domain. Find a solution of the following equation in D

$$\begin{aligned} \frac{\partial ^{2}u}{\partial t^{2}}-k(t)\Delta _{x}u+b(t) \frac{\partial u}{\partial t}+a(t)u=f(x,t) \end{aligned}$$

that satisfies the initial conditions

$$\begin{aligned} u|_{t=0}=0 \quad\text{and}\quad \frac{\partial u}{\partial t}\bigg|_{t=0}=0 \end{aligned}$$

and the potential lateral boundary condition

$$\begin{aligned} N[u]\equiv -\frac{u(x,t)}{2}+ \int _{\partial \Omega} \biggl( \frac{\partial \varepsilon}{\partial \eta _{\xi }}(x,\xi )u(\xi,t)- \varepsilon (x,\xi )\frac{\partial u}{\partial \eta _{\xi }}(\xi,t) \biggr)\,d\xi =0. \end{aligned}$$

As in the case of Sect. 2, the solution of (28)–(30) has the form

$$\begin{aligned} &u(x,t)=\sum^{\infty }_{|m|=1}u_{m}(t)e_{m}(x), \end{aligned}$$
$$\begin{aligned} &f(x,t)=\sum^{\infty }_{|m|=1}f_{m}(t)e_{m}(x), \end{aligned}$$

where \(\{e_{m}(x)\}\) is the complete orthonormal system of functions of the following spectral problem

$$\begin{aligned} &{-}{\Delta }_{x}e_{m}(x)=\lambda _{m}e_{m}(x), \\ &N[e_{m}]|_{x\in \partial \Omega}\equiv 0, \\ &f_{m}(t)= \int _{\Omega}f(x,t)e_{m}(x)\,dx,\qquad u_{m}(t)= \int _{\Omega}u(x,t)e_{m}(x)\,dx. \end{aligned}$$

Substituting (31)–(32) into (28), we obtain the following Cauchy problem

$$\begin{aligned} & \frac{d^{2}u_{m}}{dt^{2}}(t)+k(t)\lambda _{m}u_{m}(t)+b(t) \frac{du_{m}}{dt}(t)+a(t)u_{m}(t)=f_{m}(t), \end{aligned}$$
$$\begin{aligned} & u_{m}(0)=0,\qquad u'_{m}(0)=0. \end{aligned}$$

Due to initial conditions (34), it is easy to verify

$$\begin{aligned} \frac{du_{m}}{dt}(t)= \int ^{t}_{0} \frac{\partial ^{2}u_{m}}{\partial \eta ^{2}}(\eta )\,d\eta,\qquad u_{m}(t)= \int ^{t}_{0}{\frac{\partial u_{m}}{\partial \eta }(\eta )\,d \eta }. \end{aligned}$$

Using these relations as in (20), we prove the following lemma.

Lemma 4.1

Let the following conditions be satisfied: \(k\in C^{1+\alpha }[0,T]\), \(1>\alpha >0\), \(k(t)>0\), \(t>0\), \(k(0)=0\), \(k'(t)\geq 0\), \(\frac{a}{k}\in C^{1+\alpha }[0,T]\), \(\frac{b}{k}\in C^{1+\alpha }[0,T]\) and \(\frac{f_{m}}{k}\in{L_{2}[0,T]}\). Then solution \(u_{m}\in W_{2}^{2}(0,T)\) to problem (33)(34) satisfy the following inequality

$$\begin{aligned} \biggl\vert \frac{du_{m}}{dt} \biggr\vert (t)\leq d_{4}\cdot \bigl\vert k(t) \bigr\vert \biggl\vert \int ^{t}_{0}\frac{f_{m}(\eta )}{k(\eta )}\,d\eta \biggr\vert . \end{aligned}$$

Let conditions \(b(t)\geq 0\), \(a(t)\geq 0\), \(\frac{\partial }{\partial t}\frac{a(t)}{k(t)}\leq 0\) and all the conditions of Lemma 4.1 be satisfied. Then, the regular solution \(u\in W^{2}_{2}(0,T)\) of the Cauchy problem (33)–(34) satisfies the following inequality

$$\begin{aligned} \begin{aligned} &\frac{1}{k(t)} \biggl( \frac{du_{m}}{dt} \biggr)^{2}(t)+\lambda _{m}u^{2}_{m}(t)+ \frac{a(t)}{k(t)}u^{2}_{m}(t) + \int ^{t}_{0} \frac{b(\eta )}{k(\eta )} \biggl( \frac{du_{m}}{d\eta} \biggr)^{2}\,d \eta \\ &\quad{}+\frac{1}{2} \int ^{t}_{0} \biggl[ \biggl( \frac{k'(\eta )}{k^{2}\eta } \frac{du_{m}}{d\eta} \biggr)^{2}- \frac{\partial}{\partial \eta }\frac{a(\eta )}{k(\eta )} \biggr]u^{2}_{m}( \eta )\,d\eta \leq d_{5} \int ^{t}_{0} \biggl\vert \frac{f(\eta )}{k(\eta )} \biggr\vert ^{2}\,d\eta. \end{aligned} \end{aligned}$$

Multiplying both sides of equation (33) by \(\lambda _{m}\), we have

$$\begin{aligned} \begin{aligned} &\frac{\lambda _{m}}{k(t)} \biggl( \frac{d^{2}u_{m}}{dt^{2}} \biggr)^{2}(t)+ \lambda ^{2}_{m}u^{2}_{m}(t)+ \lambda _{m}\frac{a(t)}{k(t)}u^{2}_{m}(t) \\ &\qquad{}+\lambda _{m} \int ^{t}_{0} \biggl( \frac{k'(\eta )}{k^{2}(\eta )}+ \frac{b(\eta )}{k(\eta )}- \frac{\partial }{\partial \eta} \biggl(\frac{a(\eta )}{k(\eta )} \biggr) \biggr)u^{2}_{m}(\eta )\,d\eta \\ &\quad\leq d_{6} \biggl[ \int ^{t}_{0} \biggl( \frac{\sqrt{\lambda _{m}} f_{m}(\eta )}{k(\eta )} \biggr)^{2}\,d\eta + \biggl(\frac{f_{m}(\eta )}{k_{m}(\eta )} \biggr)^{2} \biggr]. \end{aligned} \end{aligned}$$

By the Parseval equality, we rewrite (36) in terms of the space \(x, t\).

Let \(g\in L_{2}(D)\), then

$$\begin{aligned} &g(x,t)=\sum^{\infty }_{ \vert m \vert =1}g_{m}(t)e_{m}(x),\\ &\bigl\Vert g(x,t) \bigr\Vert ^{2}_{l_{2}(\Omega )}=\sum ^{\infty }_{ \vert m \vert =1} \bigl\vert g_{m}(t) \bigr\vert ^{2}< \infty,\quad t\in [0,T], \end{aligned}$$

where \(\{e_{m}(x) \}\) is the complete orthonormal system of eigenfunctions of the Newton (volume) potential corresponding to the eigenvalue \(\lambda _{m}\).

Let \(\alpha >0\), by \({ (-{\Delta }_{x} )}^{\alpha }\), we will denote the operator acting on \(g(x,t)\) by the formula

$$\begin{aligned} &(-\Delta _{x})^{\alpha}g=\sum ^{\infty }_{ \vert m \vert =1}g_{m}(t)\lambda ^{ \alpha}_{m}e_{m}(x), \end{aligned}$$
$$\begin{aligned} & \bigl\Vert (-\Delta _{x})^{\alpha }g \bigr\Vert ^{2}_{L_{2}(\Omega )}(t)= \sum^{\infty }_{ \vert m \vert =1} \bigl\vert g_{m}(t)\lambda ^{\alpha }_{m} \bigr\vert ^{2}< \infty,\quad t\in [0,T]. \end{aligned}$$

Since (37)–(38), inequality (35) can be rewritten as

$$\begin{aligned} \begin{aligned} & \biggl\Vert \frac{1}{\sqrt{k(t)}} \biggl( \frac{\partial u}{\partial t} \biggr) \biggr\Vert ^{2}_{L_{2}(\Omega )}(t)+ \bigl\Vert (-\Delta _{x})^{ \frac{1}{2}}u \bigr\Vert ^{2}_{L_{2}(\Omega )}(t) \\ &\quad{}+ \int ^{t}_{0} \biggl\Vert \biggl( \frac{k'(\eta )}{k^{2}(\eta )} \biggr)^{\frac{1}{2}} \biggl(\frac{\partial u}{\partial t} \biggr) \biggr\Vert ^{2}_{L_{2}(\Omega )}(\eta )\,d\eta \leq d_{7} \biggl\Vert \frac{f}{k} \biggr\Vert ^{2}_{L_{2}(\Omega )}(t). \end{aligned} \end{aligned}$$

Since (31) and (32), from (36), it follows

Theorem 4.1

Let \(k\in C^{1+\alpha }[0,T]\), \(1>\alpha >0\), \(k(t)> 0\), \(k(0)=0\), \(k'(t)\geq 0\). If \(\Vert \frac{f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) and \(\Vert \frac{{ (-{\Delta }_{x} )}^{\frac{1}{2}}f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) for all \(t\in [0,T]\), the solution \(u\in W^{2}_{2,k}(D)\) to the mixed Cauchy problem (10)(12) satisfies the inequality

$$\begin{aligned} \begin{aligned} { \Vert u \Vert ^{2}}_{W^{2}_{2,k}(D)}={}&{ \biggl\Vert \frac{1}{\sqrt{k(t)}}\frac{{\partial }^{2}u}{\partial t^{2}} \biggr\Vert ^{2}}_{L_{2} ( {\Omega } )}(t)+{{ \Vert { {\Delta }}_{x}u \Vert }^{2}}_{L_{2} ( {\Omega } )}(t) \\ &{}+ \int ^{t}_{0}{{{ \biggl\Vert { ({- {\Delta }}_{x} )}^{ \frac{1}{2}}\frac{\partial u}{\partial \eta } \biggr\Vert }^{2}}_{L_{2} ( {\Omega } )}(\eta )\,d\eta }+{{ \Vert u \Vert }^{2}}_{L_{2} ( {\Omega } )}(t) \\ \leq {}& d_{8} \biggl({ \biggl\Vert \frac{f}{k} \biggr\Vert }^{2}_{L_{2} ( { \Omega } )}(t)+{ \biggl\Vert \frac{{ ({- {\Delta }}_{x} )}^{\frac{1}{2}}f}{k} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(t) \biggr),\quad t\in [0,T]. \end{aligned} \end{aligned}$$

From (39), it follows that \(u\in W^{2}_{2,k}(D)\subset W^{2}_{2}(D)\).

Let us prove the existence of the solution of the mixed Cauchy problem (10)–(12). To do this, we will consider the regularized mixed Cauchy problem:

$$\begin{aligned} & L_{\varepsilon }u=\frac{1}{k(t)+\varepsilon } \frac{{\partial }^{2}u_{\varepsilon }}{\partial t^{2}}{- {\Delta }}_{x}u_{ \varepsilon }=\frac{f(x,t)}{k(t)+\varepsilon } \end{aligned}$$
$$\begin{aligned} & {u_{\varepsilon }|}_{t=0}= { \frac{\partial u_{\varepsilon }}{\partial t}\bigg|}_{t=0}=0,\qquad N [u_{\varepsilon } ]\equiv 0, \end{aligned}$$

where \(\varepsilon >0\) is an arbitrary positive number.

Since (31) and (32), using the spectral decomposition of \(u_{\varepsilon }(x,t)\) and \(f(x,t)\) by \(e_{m}(x)\), from (40)–(41), we obtain

$$\begin{aligned} &\frac{1}{k(t)+\varepsilon } \frac{{\partial }^{2}u_{\varepsilon m}}{\partial t^{2}}+{\lambda }_{m}u_{ \varepsilon m}= \frac{f_{m}(t)}{k(t)+\varepsilon } \\ &{u_{\varepsilon m}|}_{t=0}=0, \qquad{ \frac{\partial u_{\varepsilon m}}{\partial t}\bigg|}_{t=0}=0. \end{aligned}$$

Due to the properties of the solutions of the Cauchy problem, if \(f_{m}\in L_{2}(0,T)\), then its solution is \(u_{m}\in W^{2}_{2}[0,T]\). Similarly to inequalities (24) and (25), we verify the following inequalities

$$\begin{aligned} & \frac{1}{k(t)+\varepsilon }{ \biggl( \frac{\partial u_{\varepsilon m}}{\partial t} \biggr)}^{2}(t)+{ \lambda }_{m}u^{2}_{\varepsilon m}(t)+ \int ^{t}_{0}{ \frac{k'}{k(t)+\varepsilon }{ \biggl( \frac{\partial u_{\varepsilon m}}{\partial \eta } \biggr)}^{2}\,d\eta } \\ &\quad\le d_{9} \int ^{t}_{0}{{ \biggl\vert \frac{f_{m}(\eta )}{k(\eta )+\varepsilon } \biggr\vert }^{2}\,d\eta }, \end{aligned}$$
$$\begin{aligned} & { \biggl({ \biggl(\frac{1}{k(t)+\varepsilon } \biggr)}^{ \frac{1}{2}}\frac{{\partial }^{2}u_{\varepsilon m}}{\partial t^{2}}(t) \biggr)}^{2}+{ \bigl({ \lambda }_{m}u_{\varepsilon m}(t) \bigr)}^{2}+ \int ^{t}_{0}{\frac{k'}{k(t)+\varepsilon }{\lambda }^{ \frac{1}{2}}_{m}{ \biggl( \frac{\partial u_{\varepsilon m}}{\partial \eta } \biggr)}^{2} \,d\eta } \\ &\quad\le d_{10} \biggl[ \int ^{t}_{0}{{ \biggl( \frac{f_{m}(\eta )}{k(\eta )} \biggr)}^{2}\,d\eta +{ \biggl( \frac{{\sqrt{{\lambda }_{m}}f}_{m}(\eta )}{k(\eta )+\varepsilon } \biggr)}^{2} \,d\eta } \biggr]. \end{aligned}$$

Using the spectral decomposition of the functions \(u_{\varepsilon }(x,t)\) and \(f(x,t)\) in the terms of \(e_{m}(x)\) from (42) when \(\varepsilon \to 0\), we get \(u_{\varepsilon }\longrightarrow u\in W_{2}^{1}(D)\) and

$$\begin{aligned} \begin{aligned} { \Vert u \Vert }^{2}_{W^{1}_{2,k}(D)}={}&{ \biggl\Vert \frac{1}{\sqrt{k(t)}}\frac{\partial u}{\partial t} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(t)+{ \bigl\Vert { (-{\Delta }_{x} )}^{ \frac{1}{2}}u \bigr\Vert }^{2}_{L_{2} ( {\Omega } )}(t) \\ &{}+ \int ^{t}_{0}{{ \biggl\Vert \sqrt{ \frac{k'}{k}} \frac{\partial u}{\partial \eta } \biggr\Vert }^{2}_{L_{2} ( { \Omega } )}( \eta )\,d\eta } {+} { \Vert u \Vert }^{2}_{L_{2} ( {\Omega } )} {(} {t} {)} \\ \leq {}& d_{11} \int ^{t}_{0}{{ \biggl\Vert \frac{f}{k} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(\eta )\,d\eta }. \end{aligned} \end{aligned}$$

From (43), it also follows

Theorem 4.2

Let \(k\in C^{1+\alpha }[0,T]\), \(1>\alpha >0\), \(k(t)>0\), \(t>0\), \(k(0)=0\), \(k'(t)\ge 0\). If \(\Vert \frac{f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) and \(\Vert \frac{{ (-{\Delta }_{x} )}^{\frac{1}{2}}f}{k} \Vert _{L_{2}(\Omega )}(t)\), \(t\in [0,T]\), then there exists a unique solution \(u\in W^{2}_{2,k}(D)\) of the mixed Cauchy problem (10)(12) that satisfies the inequality

$$\begin{aligned} \begin{aligned} { \Vert u \Vert ^{2}}_{W^{2}_{2,k}(D)}={}& { \biggl\Vert \frac{1}{\sqrt{k(t)}}\frac{{\partial }^{2}u}{\partial t^{2}} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(t)+{ \Vert {\Delta }_{x}u \Vert }^{2}_{L_{2} ( {\Omega } )}(t) \\ &{}+ \int ^{t}_{0}{{ \biggl\Vert {\sqrt{ \frac{k'}{k}} (-{\Delta }_{x} )}^{\frac{1}{2}}\frac{\partial u}{\partial \eta } \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(\eta )\,d\eta }+ { \Vert u \Vert }^{2}_{L_{2} ( {\Omega } )}(t) \\ \leq{} & \int ^{t}_{0}{{ \biggl\Vert \frac{f}{k} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(\eta )\,d\eta } + \int ^{t}_{0}{{ \biggl\Vert \frac{{ (-{\Delta }_{x} )}^{\frac{1}{2}}f}{k} \biggr\Vert }^{2}_{L_{2} ( {\Omega } )}(\eta )\,d\eta }. \end{aligned} \end{aligned}$$

Corollary 4.1

Note that the weighted Sobolev space \(W_{2,k}^{2}(D)\) is a subspace of the classical space \(W_{2}^{2}(D)\). As in the case of strictly hyperbolic equations, we have first established that \(u\in W_{2}^{1}(D)\) and \(u\in W_{2}^{2}(D)\) under the condition \(\Vert \frac{f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) and \(\Vert \frac{\mathrm{grad}_{x} f}{k} \Vert _{L_{2}(\Omega )}(t)<\infty \) for all \(t\in [0,T]\), respectively.

Using the inequalities (35)–(36), the unique solvability of the mixed Cauchy problem (10)–(12) for the general equation \(\frac{{\partial }^{2}u}{\partial t^{2}}-k(t){\Delta }_{x}u+b(t) \frac{\partial u}{\partial t}+a(t)u=f(x,t)\) is established in exactly the same way.


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The authors thank Professor M. Sadybekov for a very careful reading of the manuscript and useful comments.


The work was supported by the grant (No. AP09260126) of the Science Committee of the Ministry of Education and Science of the Republic of Kazakhstan.

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Kakharman, N., Kal’menov, T. Mixed Cauchy problem with lateral boundary condition for noncharacteristic degenerate hyperbolic equations. Bound Value Probl 2022, 35 (2022).

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