Inverse Boundary Conditions Interface Problems for the Heat Equation with Cylindrical Symmetry

Abstract

In this paper, we study inverse interface problems with unknown boundary conditions, using point observations for parabolic equations with cylindrical symmetry. In the one-dimensional, two-layer interface problem, the left interval $0<r<l_1$, i.e., the zero degeneracy, causes serious solution difficulty. For this, we investigate the well-posedness of the direct (forward) problem. Next, we formulate and solve five inverse boundary condition problems for the interface heat equation with cylindrical symmetry from internal measurements. The finite volume difference method is developed to construct second-order schemes for direct and inverse problems. The correctness of the proposed numerical solution decomposition algorithms for the inverse problems is discussed. Several numerical examples are presented to illustrate the efficiency of the approach.

1. Introduction

Inverse boundary problems for parabolic problems determine the unknown boundary conditions of the equations from either final time measurements (backward problems) or local space or time measurements of the solutions. Such problems play a crucial role in applied mathematics and physics. For example, inverse and backward source and boundary problems are used for the identification of coefficient and boundary sources of pollution sources affecting water bodies such as rivers and lakes, heat-mass transfer, elasticity, heat conduction systems, medicine, etc. (see, e.g., [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15]).

It is challenging to solve inverse problems, even numerically, because, in general, they are ill-posed [2,16,17,18,19,20]. In [9], the authors provide theoretical reasons why the aquifer identification problem can be expected to be numerically unstable. Many papers have studied the inverse problem of identifying an unknown boundary in a parabolic equation. A class of such problems plays a key role in applied mathematics, physics, mechanics, and biology (see, e.g., [2,8,12,15,19,21,22,23]).

The uniqueness of the solution of inverse boundary problems in parabolic and time-fractional diffusion equations by observation of the temperature at a fixed point on the boundary is investigated in [15,24]. In [1], a meshless method is constructed for solving inverse boundary source problems for time-fractional diffusion equations. A semigroup approach is applied in [8,25] for studying inverse problems of recovering the right boundary condition in a linear and quasi-linear parabolic equation under the overspecified data of Dirichlet and Neumann type on the left boundary. A numerical method, based on a decomposition technique for solving the inverse problem for identifying the Dirichlet boundary condition in the heat conduction equation is proposed in [26].

Inverse and backward source and boundary problems are often used to model heat-mass transfer problems in composite materials. For example, in [23], the authors construct a numerical method, based on fundamental solutions and the Tikhonov regularization approach, for recovering a moving boundary in a one-dimensional heat equation on a multilayer domain.

In our previous works [27,28,29], we developed numerical methods, based on the decomposition technique or reduction method combined with the Saul’yev scheme, for recovering external boundary conditions in parabolic, time-fractional parabolic, and parabolic-hyperbolic problems, defined on disjoint domains. In order to solve inverse problems, we usually have to solve well-posed direct (forward) ones as well. There are many studies in this direction. We will mention some of them.

A global well-posedness result for 3D micropolar fluid equations is obtained in [30]. Lyapunov stability over a finite time interval provides well-posedness of the considered differential problem (see [31]). Stronger local and global finite-time stability for time-varying nonlinear systems are investigated. In [32], using comparison principles and the quasilinearization approach, the authors obtain monotone iterative sequences that uniformly and monotonically converge to the unique solution of integro-differential equations with nonlinear boundary conditions.

The work [33] provides an exact analysis of the stability of a multilayer diffusion-reaction problem, where the number of layers can be extensive. In [34], the authors construct a second-order numerical scheme for an elliptic interface system in the case of perfect and imperfect contact. The authors of [35] present a theoretical and computational investigation of an elliptic equation on a domain that consists of two sub-domains with substantially different thermal conductivities and fiber structures. A second-order accurate $L_2$ norm difference scheme for the axisymmetric Poisson equation is constructed and analyzed in [36]. Semi-analytical solutions of axisymmetric reaction–diffusion equations, describing heat conduction in composite media, are developed in [37].

In particular, inverse and backward source and boundary problems describe mass transfer processes in columns with cylindrical symmetry, taking into account the longitudinal mixing (see, e.g., [10]). Inverse source problems also have applications in multi-wave imaging, geophysics, [3,10,38] and identification of the rate of the outflow in oil reservoirs with walls [4]. Additionally, they have applications in numerous cylindrical constructs, such as pipelines for transporting substantial quantities of natural resources like gas and oil [39].

Recently, many models of biomechanics use cylindrical regions, such as bone and blood vessels. A scheme for determining the inhomogeneous elastic properties of an isotropic cylinder is developed in [5]. An inverse technique for estimating the inlet conditions of fluid in a thick pipe is adapted in [22].

The backward space-dependent source problem from noisy final data for the helium production-diffusion equation is studied in [38]. A coefficient inverse problem for the two-dimensional transient heat equation in a cylindrical coordinate system using the finite difference method is solved in [21]. An inverse problem for reconstructing a space-dependent source in the heat equation of a polar coordinate system from finite temperature measurements is studied in [40]. A modified quasi-boundary value method is developed in [7] to solve an inverse heat conduction problem with spherical symmetry for recovering the internal surface temperature distribution of a hollow sphere from observations at a fixed position inside it. A higher-order difference scheme is studied for the problem, formulated in cylindrical and spherical coordinate systems in [41].

In this article, we develop efficient numerical methods for solving inverse problems for identifying external Dirichlet boundary conditions in parabolic interface problems with cylindrical symmetry.

The main difficulties arise from the degeneration at $r=0$, the multi-layered structure of the domain, and the discontinuity of the coefficients (interface problem) across the different layers. They are overcome by constructing appropriate discretizations and decomposition techniques. Moreover, the formulated five inverse problems are reduced, as the method is the same regardless of whether the measurements are taken in the adjacent layer or further from the unknown boundary.

The rest of the paper is organized as follows: In Section 2, the direct (forward) and inverse problems are formulated. The well-posedness of the direct problem is proved in Section 3. Finite volume second-order difference schemes for the direct problem are derived in Section 4. A solution decomposition method for numerical solution of the inverse problems is proposed and studied for correctness in Section 5. The results from numerical experiments are presented and discussed in Section 6. The paper is finalized with some conclusions.

2. Direct and Inverse Problems

In this section, we formulate the basic interface direct (forward) problem for the heat equation with cylindrical symmetry.

2.1. Direct (Forward) Problem

We consider the interface initial-boundary value problem (IBVP):

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial u_1}{\partial t}}=d_1\left({\displaystyle \frac{{\partial }^2{u}_1}{\partial r^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u_1}{\partial r}}\right)+f_1(r,t),(r,t)\in Q_{1T}={\mathrm{\Omega }}_1\times (0,T),\hfill \end{array}$$

(1)

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial u_2}{\partial t}}=d_2\left({\displaystyle \frac{{\partial }^2{u}_2}{\partial r^2}}+{\displaystyle \frac{1}{r}}{\displaystyle \frac{\partial u_2}{\partial r}}\right)+f_2(r,t),(r,t)\in Q_{2T}={\mathrm{\Omega }}_2\times (0,T),\hfill \end{array}$$

(2)

where $u_i(r,t),i=1,2$ is the solution (temperature or concentration, etc.) at position $r\in {\mathrm{\Omega }}_i$, ${\mathrm{\Omega }}_1=(0,l_1)$, $r_2=(l_1,l_2)$, $l_2=L$ and time t, $d_i$ is the diffusion coefficient in the i-th layer, $i=1,2$.

The initial conditions are given by

$$u_i(r,0)=u_{i0}\left(r\right),i=1,2$$

(3)

and the external boundary conditions (BCs) are

$$\begin{array}{ccc}& & u_1(0,t)=g_0\left(t\right),0<t<T,\hfill \end{array}$$

(4)

$$\begin{array}{ccc}& & u_2(L,t)=g_L\left(t\right),0<t<T.\hfill \end{array}$$

(5)

Here, $g_0\left(t\right)$, $g_L\left(t\right)$ are specified time-dependent continuous functions.

The typical assumption is that the solution and the diffusive flux are continuous at the interface, meaning

$${\left[u\right]}_{r=l_1}=u_2(l_1,t)-u_1(l_1,t)=0,$$

(6)

$${\left[d{\displaystyle \frac{\partial u}{\partial r}}\right]}_{r=l_1}=d_2{\displaystyle \frac{\partial u_r}{\partial r}}(l_1,t)-d_1{\displaystyle \frac{\partial u_r}{\partial r}}(l_1,t)=0,$$

(7)

for $t\in (0,T)$.

Let us note that, in mechanics, physics, biology, etc., many real processes require different relations than those in (6) and (7) (see e.g., [42,43]).

The version in Cartesian coordinates of the problem (1)–(7) is the heat equation problem with an interface [42].

The direct (forward) problem is to find the solution $u_1(x,t)$, $u_2(x,t)$ of the IBVP (1)–(7) with given coefficients, right-hand sides, initial, boundary, and interface conditions.

2.2. Inverse Problems

We consider inverse problems where the functions $g_0\left(t\right)$ and $g_L\left(t\right)$ are unknown and must be identified using some overspecified data. Depending on the observation data, the inverse problems (IP) for determining the boundary functions in (4) and (5) can be formulated as follows:

IP1: Given observation $u_1(r_1^{*},t)={\mathrm{\Psi }}_1\left(t\right),r_1^{*}\in {\mathrm{\Omega }}_1$. Find $(u_1,u_2,g_0\left(t\right))$;

IP2: Given observation $u_2(r_2^{*},t)={\mathrm{\Psi }}_2\left(t\right),r_2^{*}\in {\mathrm{\Omega }}_2$. Find $(u_1,u_2,g_L\left(t\right))$;

IP3: Given observation $u_1(r_1^{*},t)={\mathrm{\Psi }}_1\left(t\right),r_1^{*}\in {\mathrm{\Omega }}_1$. Find $(u_1,u_2,g_L\left(t\right))$;

IP4: Given observation $u_2(r_2^{*},t)={\mathrm{\Psi }}_2\left(t\right),r_2^{*}\in {\mathrm{\Omega }}_2$. Find $(u_1,u_2,g_0\left(t\right))$;

IP5: Given observation $u_i(r_i^{*},t)={\mathrm{\Psi }}_i\left(t\right),r_i^{*}\in {\mathrm{\Omega }}_j,i=1,2$. Find $(u_1,u_2,g_0\left(t\right),g_L\left(t\right))$.

3. Well-Posedness of the Direct Problem

In this section, an analytical study of the direct problem is carried out to provide a rigorous proof of the well-posedness of the direct problem (1)–(7). The main difficulties in solving the direct problem come from the singularity at $r=0$ and the interface.

Theorem 1. 

Let $\underset{r\to 0}{lim}{u}_1(r,t)$ be bounded, $g_i\in C(0,T)$, $f_i\in Q_{iT}$, $u_{i0}\left(x\right)\in C\left({\mathrm{\Omega }}_i\right)$, $i=1,2$. Then, there exists a unique solution $u=\{u_1,u_2\}$, $u_1\in C^2\left({\mathrm{\Omega }}_1\right)\cap C^1\left(\left\{r\right|r=l_1\}\right)$, $u_2\in C^2\left({\mathrm{\Omega }}_2\right)\cap C^1\left(\left\{r\right|r=l_1\}\right)\cap C^1\left(\left\{r\right|r=R_2\}\right)$ of the problem (1)–(7) that continuously depends on the input data.

Proof. 

We consider the particular case of homogeneous Dirichlet BCs (4) and (5)

$$u_1(0,t)=g_0\left(t\right)=0,u_2(L,t)=g_L\left(t\right)=0.$$

First, we study the uniqueness of the solution, assuming existence. We multiply Equation (1) by $r{u}_1(r,t)$ and Equation (2) by $r{u}_2(r,t)$. Next, we integrate the new equations over the intervals ${\mathrm{\Omega }}_1$ and ${\mathrm{\Omega }}_2$, respectively, and add up the results to obtain:

$${\displaystyle \frac{\partial }{\partial t}}U\left(t\right)+d_1\underset{0}{\overset{l_1}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial r}}\right)}^{2}dr+d_2\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial r}}\right)}^{2}dr=\underset{0}{\overset{l_1}{\int }}r{u}_1{f}_{1}dr+\underset{l_1}{\overset{l_2}{\int }}r{u}_2{f}_{2}dr,$$

(8)

where

$$U\left(t\right)={\displaystyle \frac{1}{2}}\underset{0}{\overset{l_1}{\int }}r{u}_1^2(r,t)dr+{\displaystyle \frac{1}{2}}\underset{l_1}{\overset{l_2}{\int }}r{u}_2^2(r,t)dr.$$

(9)

Let us note that for the function $u_2(r,t)$, it is easy to derive a Friedrich-type inequality (see e.g., [22,44]),

$$\underset{l_1}{\overset{l_2}{\int }}r{u}_2^{2}dr\le C\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial r}}\right)}^{2}dr,C={\displaystyle \frac{l_1^2}{4}}\left({\displaystyle \frac{l_2^2}{l_1^2}}-2ln{\displaystyle \frac{l_2}{l_1}}-1\right)>0.$$

However, because of the singularity around $r=0$ in the Equation (1), it is not easy to obtain such an estimate for $u_1(r,t)$. Fortunately, in [45], the following inequality was derived:

$$\underset{0}{\overset{l_1}{\int }}r{u}_1^{2}dr+\underset{l_1}{\overset{l_2}{\int }}r{u}_2^{2}dr\le C\left(d_1\underset{0}{\overset{l_1}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial r}}\right)}^{2}dr+d_2\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial r}}\right)}^{2}dr\right),$$

(10)

where $C>0$ is a constant that is not dependent on the functions $u_1,u_2$.

Now, using (10), we deduce that the left-hand side of the equality (8) is greater than or equal to $t_0$:

$${\displaystyle \frac{\partial U}{\partial t}}\le {\displaystyle \frac{1}{C}}\left(\underset{0}{\overset{l_1}{\int }}r{u}_1^{2}dr+\underset{l_1}{\overset{l_2}{\int }}r{u}_2^{2}dr\right).$$

For the right-hand side of (8), using (10) again, we obtain:

$$\begin{array}{c}\underset{0}{\overset{l_1}{\int }}r{u}_1{f}_{1}dr+\underset{l_1}{\overset{l_2}{\int }}r{u}_2{f}_{2}dr\le {\left(\underset{0}{\overset{l_1}{\int }}r{f}_1^{2}dr\right)}^{1/2}{\left(\underset{0}{\overset{l_1}{\int }}r{u}_1^{2}dr\right)}^{1/2}\\ +{\left(\underset{l_1}{\overset{l_2}{\int }}r{f}_2^{2}dr\right)}^{1/2}{\left(\underset{l_1}{\overset{l_2}{\int }}r{u}_2^{2}dr\right)}^{1/2}.\hfill \\ \max (l_1.{\widehat{f}}_1,\sqrt{l_2^2-l_1^2}{\widehat{f}}_2)\left[{\left({\displaystyle \frac{1}{2}}\underset{0}{\overset{l_1}{\int }}r{u}_1^{2}dr\right)}^{1/2}+{\displaystyle \frac{1}{2}}{\left(\underset{l_1}{\overset{l_2}{\int }}r{u}_2^{2}dr\right)}^{1/2}\right)\\ \le \max (l_1{\widehat{f}}_1,\sqrt{l_2^2-l_1^2}{\widehat{f}}_2)\left(2\sqrt{2}\sqrt{U}\right)=F\left(t\right)\sqrt{U},\end{array}$$

(11)

where

$${\widehat{f}}_i\left(t\right)=\sqrt{U\underset{{\mathrm{\Omega }}_i}{max}\left|f_1(x,t)\right|},i=1,2.$$

Therefore, from (8) and (11), we obtain the inequalities

$${\displaystyle \frac{\partial }{\partial t}}U+cU\le F\left(t\right)U.c={\displaystyle \frac{1}{C}}min(d_1,d_2),$$

which implies

$$U\left(t\right)\le {\left(\sqrt{U_0}+{\displaystyle \frac{1}{2}}\underset{0}{\overset{t}{\int }}F\left(\tau \right)e^{c\tau }d\tau \right)}^2{e}^{-2ct},$$

(12)

where

$$U_0=U\left(0\right)={\displaystyle \frac{1}{2}}\underset{0}{\overset{l_1}{\int }}r{u}_{10}^2\left(r\right)dr+{\displaystyle \frac{1}{2}}\underset{l_1}{\overset{l_2}{\int }}r{u}_{20}^2\left(r\right)dr.$$

Therefore, from (9), we have:

$$\underset{0}{\overset{l_1}{\int }}r{u}_1^2(r,t)dr\le 2U\left(t\right),\underset{l_1}{\overset{l_2}{\int }}r{u}_2^2(r,t)dr\le 2U\left(t\right),$$

where $U\left(t\right)$ satisfies (12).

Next, we will prove the boundedness of the integrals

$$\underset{0}{\overset{l_1}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial r}}\right)}^{2}dr,C\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial r}}\right)}^{2}dr.$$

For this aim, we take the square of Equations (1) and (2), and then integrate the results:

$$\begin{array}{ccc}& & \underset{0}{\overset{t}{\int }}\underset{0}{\overset{l_1}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial t}}\right)}^{2}drds-2d_1\underset{0}{\overset{t}{\int }}\underset{0}{\overset{l_1}{\int }}{\displaystyle \frac{\partial u_1}{\partial t}}{\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_1}{\partial r}}\right)drds\hfill \\ & & +\underset{0}{\overset{t}{\int }}\underset{0}{\overset{l_1}{\int }}{d}_1^2.{\displaystyle \frac{1}{r}}{\left({\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_1}{\partial r}}\right)\right)}^{2}drds=\underset{0}{\overset{t}{\int }}\underset{0}{\overset{l_1}{\int }}r.f_1^{2}drds.\hfill \\ & & \underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial t}}\right)}^{2}drds-2d_2\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}{\displaystyle \frac{\partial u_2}{\partial t}}{\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_2}{\partial r}}\right)drds\hfill \\ & & +\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}{d}_2^2{\displaystyle \frac{1}{r}}{\left({\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_2}{\partial r}}\right)\right)}^{2}drds.\hfill \end{array}$$

Summing these equalities, and using integration by parts along with the initial, boundary, and interface conditions, we obtain

$$\begin{array}{c}\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial t}}\right)}^{2}drds+\underset{0}{\overset{t}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial t}}\right)}^{2}drds+\underset{0}{\overset{l_1}{\int }}r{\left({\displaystyle \frac{\partial u_1}{\partial r}}\right)}^{2}dr+\underset{l_1}{\overset{l_2}{\int }}r{\left({\displaystyle \frac{\partial u_2}{\partial r}}\right)}^{2}dr\\ +\underset{0}{\overset{t}{\int }}{d}_1^2{\displaystyle \frac{1}{r}}{\left({\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_1}{\partial r}}\right)\right)}^{2}drds+\underset{0}{\overset{t}{\int }}{d}_2^2{\displaystyle \frac{1}{r}}{\left({\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_2}{\partial r}}\right)\right)}^2\\ =\underset{0}{\overset{t}{\int }}\underset{0}{\overset{l_1}{\int }}r{f}_1^{2}drds+\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}r{f}_2^{2}drds.\end{array}$$

Therefore,

$$\underset{{\mathrm{\Omega }}_i}{\int }r{\left({\displaystyle \frac{\partial u_1}{\partial r}}\right)}^{2}dr\le {\displaystyle \frac{1}{d_1}}\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}r{f}_1^{2}drds+{\displaystyle \frac{1}{d_2}}\underset{0}{\overset{t}{\int }}\underset{l_1}{\overset{l_2}{\int }}r{f}_2^{2}drds.$$

Hence, the uniqueness and continuous dependence of the solution of the problem (1)–(7) on the input data follows.

The local existence of the solution of the problem (1)–(7) follows from the general theory of parabolic equations (see e.g., [22,44]). The global existence is a direct consequence of the a priory estimates above. □

4. Numerical Solution of the Direct Problem

First, we solve the direct problem (1)–(7). We rewrite the Equations (1) and (2) in divergent form

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial u_1}{\partial t}}={\displaystyle \frac{d_1}{r}}{\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_1}{\partial r}}\right)+f_1(r,t),(r,t)\in Q_{1T}={\mathrm{\Omega }}_1\times (0,T),\hfill \end{array}$$

(13)

$$\begin{array}{ccc}& & {\displaystyle \frac{\partial u_2}{\partial t}}={\displaystyle \frac{d_2}{r}}{\displaystyle \frac{\partial }{\partial r}}\left(r{\displaystyle \frac{\partial u_2}{\partial r}}\right)+f_2(r,t),(x,t)\in Q_{2T}={\mathrm{\Omega }}_2\times (0,T),\hfill \end{array}$$

(14)

At $r=0$, we impose the condition [42]

$$|u_1\left(0\right)|<\infty ,$$

(15)

or equivalently,

$$\underset{r\to 0}{lim}r{\displaystyle \frac{\partial u_1}{\partial r}}=0.$$

(16)

This condition can be replaced by

$${\displaystyle \frac{\partial u_1}{\partial r}}(0,t)=0.$$

(17)

Therefore, at $r=0$, in view of (15), we may consider either the Dirichlet boundary condition (4) or the natural boundary condition (17), based on (16).

Let us define the uniform spatial and temporal meshes

$${\overline{\omega }}_h=\{r_i=ih,j=0,1,\dots ,J,Jh=L\},{\overline{\omega }}_{\tau }=\{t_n=n\tau ,n=0,1,\dots ,N,N\tau =T\}.$$

Denote by $u_{i,j}^n$ the solution $u_i$ at the grid node $(r_j,t_n)\in {\overline{\omega }}_h\times {\overline{\omega }}_{\tau }$ and by $u_{i,j+1/2}^n$, the solution $u_i$ at the grid node $(r_{j+1/2},t_n)$, where $r_{j+1/2}=r_j+h/2$ is a node of the dual mesh $0=r_{-1/2}<r_{1/2}<r_{3/2}\cdots <r_{J-1/2}<r_{J+1/2}=L$.

In order to derive a conservative scheme and overcome the degeneracy at $r=0$, we follow [42] and apply the finite volume method to the problem (13), (14), (3), (17), and (5). Let $l_1$ be a grid node, where $l_1=r_{j^{*}}=j^{*}h$. At the inner spatial nodes, multiplying (13) and (14) by r, integrating over the volumes $(r_{j-1/2},r_{j+1/2})$, $j=1,2,\dots ,J-1$, and dividing by $h{r}_i$, we obtain

$$\begin{array}{c}{\displaystyle \frac{1}{r_{i}h}}\underset{r_{j-1/2}}{\overset{r_{j+1/2}}{\int }}r{\displaystyle \frac{\partial u_1}{\partial t}}dr-{\displaystyle \frac{d_1}{r_{j}h}}\left(r_{j+1/2}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{j+1/2}-r_{j-1/2}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{j-1/2}\right)\hfill \\ ={\displaystyle \frac{1}{r_{j}h}}\underset{r_{j-1/2}}{\overset{r_{j+1/2}}{\int }}{f}_1(r,t)rdr,j=1,2,\dots ,j^{*}-1,\hfill \end{array}$$

(18)

$$\begin{array}{c}{\displaystyle \frac{1}{r_{j}h}}\underset{r_{j-1/2}}{\overset{r_j}{\int }}r{\displaystyle \frac{\partial u_1}{\partial t}}dr+{\int }_{r_j}^{r_{j+1/2}}{\displaystyle \frac{\partial u_2}{\partial t}}dr-{\displaystyle \frac{1}{r_{j}h}}\left(d_2{r}_{j+1/2}{\displaystyle \frac{\partial u_2}{\partial r}}{|}_{j+1/2}-d_1{r}_{j-1/2}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{j-1/2}\right)\hfill \\ ={\displaystyle \frac{1}{r_{j}h}}\underset{r_{j-1/2}}{\overset{r_j}{\int }}{f}_1(r,t)rdr+{\displaystyle \frac{1}{r_{j}h}}{\int }_{r_j}^{r_{j+1/2}}{f}_2(r,t)rdr,j=j^{*},\hfill \end{array}$$

(19)

$$\begin{array}{c}{\displaystyle \frac{1}{r_{j}h}}\underset{r_{j-1/2}}{\overset{r_{j+1/2}}{\int }}r{\displaystyle \frac{\partial u_2}{\partial t}}dr-{\displaystyle \frac{d_2}{r_{j}h}}\left(r_{j+1/2}{\displaystyle \frac{\partial u_2}{\partial r}}{|}_{j+1/2}-r_{j-1/2}{\displaystyle \frac{\partial u_2}{\partial r}}{|}_{j-1/2}\right)\hfill \\ ={\displaystyle \frac{1}{r_{j}h}}\underset{r_{j-1/2}}{\overset{r_{j+1/2}}{\int }}{f}_2(r,t)rdr,j=j^{*}+1,\dots ,J-1.\hfill \end{array}$$

(20)

Approximating the integrals in (18)–(20) by the midpoint quadrature, the fluxes in (18)–(20) by finite differences are

$$r_{i-1/2}{\displaystyle \frac{\partial u_i}{\partial r}}=r_{i-1/2}{\displaystyle \frac{u_{i,j}-u_{i,j-1}}{h}},$$

and applying implicit time-stepping, we derive

$$\begin{array}{c}{\displaystyle \frac{u_{1,j}^{n+1}-u_{1,j}^n}{\tau }}-{\displaystyle \frac{d_1}{r_{j}h}}\left(r_{j+1/2}{\displaystyle \frac{u_{1,j+1}^{n+1}-u_{1,j}^{n+1}}{h}}-r_{j-1/2}{\displaystyle \frac{u_{1,j}^{n+1}-u_{1,j-1}^{n+1}}{h}}\right)\hfill \\ =f_{1,j}^{n+1},j=1,2,\cdots ,j^{*}-1\hfill \end{array}$$

(21)

$$\begin{array}{c}{\displaystyle \frac{u_{1,j}^{n+1}-u_{1,j}^n}{2\tau }}+{\displaystyle \frac{u_{2,j}^{n+1}-u_{2,j}^n}{2\tau }}-{\displaystyle \frac{1}{r_{j}h}}\left(d_2{r}_{j+1/2}{\displaystyle \frac{u_{2,j+1}^{n+1}-u_{2,j}^{n+1}}{h}}-d_1{r}_{j-1/2}{\displaystyle \frac{u_{2,j}^{n+1}-u_{2,j-1}^{n+1}}{h}}\right)\hfill \\ ={\displaystyle \frac{1}{2}}\left(f_{1,j}^{n+1}+f_{2,j}^{n+1}\right),j=j^{*},\hfill \end{array}$$

(22)

$$\begin{array}{c}{\displaystyle \frac{u_{2,j}^{n+1}-u_{2,j}^n}{2\tau }}-{\displaystyle \frac{d_2}{r_{j}h}}\left(r_{j+1/2}{\displaystyle \frac{u_{2,j+1}^{n+1}-u_{2,j}^{n+1}}{h}}-r_{j-1/2}{\displaystyle \frac{u_{2,j}^{n+1}-u_{2,j-1}^{n+1}}{h}}\right)\hfill \\ =f_{2,j}^{n+1},j=j^{*}+1,\dots ,J-1.\hfill \end{array}$$

(23)

To obtain the discretization at $r=0$, in the case of natural BCs (16) and (17), we consider the interval $I_0=(\epsilon ,r_{1/2}=h/2)$ and multiply (13) by r. Then, we integrate over the volume $I_0$ to obtain

$$\underset{\epsilon }{\overset{r_{1/2}}{\int }}r{\displaystyle \frac{\partial u_1}{\partial t}}dr-\left(r_{1/2}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{1/2}-\epsilon {\displaystyle \frac{\partial u_1}{\partial r}}{|}_{\epsilon }\right)=\underset{\epsilon }{\overset{r_{1/2}}{\int }}{f}_1(r,t)rdr.$$

Using the product approximation formula, we derive

$${\displaystyle \frac{h}{4}}\underset{\epsilon }{\overset{r_{1/2}}{\int }}{\displaystyle \frac{\partial u_1}{\partial t}}dr-\left(r_{1/2}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{1/2}-\epsilon {\displaystyle \frac{\partial u_1}{\partial r}}{|}_{\epsilon }\right)={\displaystyle \frac{h}{4}}\underset{\epsilon }{\overset{r_{1/2}}{\int }}{f}_1(r,t)dr,$$

Now, we apply the rectangle quadrature formula and let $\epsilon \to 0$. In view of (16) and (17), for the left external boundary condition, we obtain

$${\displaystyle \frac{h}{4}}{\displaystyle \frac{h}{2}}{\displaystyle \frac{\partial u_1}{\partial t}}(0,t)-{\displaystyle \frac{h}{2}}{\displaystyle \frac{\partial u_1}{\partial r}}{|}_{1/2}={\displaystyle \frac{h}{4}}{\displaystyle \frac{h}{2}}{f}_1(0,t).$$

Approximating the derivatives by finite differences and using an implicit time-stepping method, we obtain

$${\displaystyle \frac{u_{1,1}^{n+1}-u_{1,0}^{n+1}}{h}}={\displaystyle \frac{h}{4}}\left({\displaystyle \frac{u_{1,0}^{n+1}-u_{1,0}^n}{\tau }}-f_{1,0}^{n+1}\right).$$

(24)

In the case of Dirichlet BCs (4), the approximation is trivial

$$u_{1,0}^{n+1}=g_0\left(t_{n+1}\right).$$

(25)

So, we use either (24) or (25) for the computations.

The numerical scheme is completed by the left external boundary condition and initial condition, corresponding to (5) and (3), respectively

$$\begin{array}{ccc}& & u_{2,J}^{n+1}=u_2(L,t_{n+1}),\hfill \end{array}$$

(26)

$$\begin{array}{ccc}& & u_{i,j}^0=u_{i0}\left(r_j\right)=\left\{\begin{array}{cc}j=0,1,\dots ,j^{*},\hfill & i=1,\hfill \\ j=j^{*},j^{*}+1,\dots ,J,\hfill & i=2,\hfill \end{array}\right.u_{10}\left(l_1\right)=u_{20}\left(l_1\right).\hfill \end{array}$$

(27)

Denote

$$\begin{array}{ccc}& & u=[u_{1,0},u_{1,1},\dots ,u_{1,j^{*}}=u_{2,j^{*}},u_{2,j^{*}+1},\dots ,u_{2,J}],\hfill \\ & & u^0=[u_{10}\left(r_0\right),u_{10}\left(r_0\right),\dots ,u_{10}\left(r_{j^{*}}\right)=u_{20}\left(r_{j^{*}}\right),u_{10}\left(r_{j^{*}+1}\right),\dots ,u_{10}\left(r_J\right)]\hfill \end{array}$$

and rewrite the numerical scheme (21)–(23), (24) or (25), (26), (27) in more convenient form. For $n=0,1,\dots ,N$, we have

$$\begin{array}{c}\begin{array}{l}{P}_0{u}_0^{n+1}-D_0{u}_1^{n+1}=F_0^n,\\ -P_j{u}_{j-1}^{n+1}+Q_j{u}_j^{n+1}-D_i{u}_{j+1}^{n+1}=F_j^n,j=1,2,\dots ,J-1,\\ u_J^{n+1}=u_2(L,t_{n+1}).\end{array}\end{array}$$

(28)

where ${\displaystyle Q_j=\frac{1}{\tau }+P_j+D_j,}$

$$\begin{array}{ccc}& & P_j={\displaystyle \frac{r_{j-1/2}}{r_{j}h}}\left\{\begin{array}{cc}{d}_1,\hfill & j=1,2,\dots ,j^{*},\hfill \\ d_2,\hfill & j=j^{*}+1,\dots ,J-1,\hfill \end{array}\right.,D_j={\displaystyle \frac{r_{j+1/2}}{r_{j}h}}\left\{\begin{array}{cc}{d}_1,\hfill & j=1,2,\dots ,j^{*}-1,\hfill \\ d_2,\hfill & j=j^{*},\dots ,J-1,\hfill \end{array}\right.\hfill \\ & & F_j^n={\displaystyle \frac{u_j^n}{\tau }}+\left\{\begin{array}{cc}{f}_{1,j}^n,\hfill & j=1,2,\dots ,j^{*}-1,\hfill \\ \left(f_{1,j}^n+f_{2,j}^n\right)/2,\hfill & j=j^{*},\hfill \\ f_{2,j}^n,\hfill & j=j^{*}+1,\dots ,J-1,\hfill \end{array}\right.\hfill \\ & & P_0=\left\{\begin{array}{cc}{\displaystyle \frac{1}{\tau }+\frac{4}{h^2},}\hfill & \mathrm{BC}\left(24\right),\hfill \\ 1,\hfill & \mathrm{BC}\left(25\right),\hfill \end{array}\right.D_0=\left\{\begin{array}{cc}{\displaystyle \frac{4}{h^2},}\hfill & \mathrm{BC}\left(24\right),\hfill \\ 0,\hfill & \mathrm{BC}\left(25\right),\hfill \end{array}\right.\hfill \\ & & F_0^n=\left\{\begin{array}{cc}{\displaystyle \frac{u_j^n}{\tau }+\frac{h}{4}{f}_{1,0}^{n+1},}\hfill & \mathrm{BC}\left(24\right),\hfill \\ g_0\left(t_{n+1}\right),\hfill & \mathrm{BC}\left(25\right).\hfill \end{array}\right.\hfill \end{array}$$

Following the arguments in [42], we may deduce that the discretization (28) has an accuracy order $O(\tau +h^2)$.

5. Numerical Solution of the Inverse Problems

In this section, we propose a numerical method for solving inverse problems. To obtain numerical discretization for identifying external boundary conditions, we use a decomposition technique.

We begin with solving IP1-IP5. Note that in the case of natural BC (16) and (17), the problems IP1 and IP4 are omitted, as there is nothing to determine. Similarly, IP5 reduces to IP2 and IP3.

Consider IP2 and IP3. For the numerical method we construct, it does not matter whether the measurement is in the adjacent region of $g_L\left(t\right)$, or not. Therefore, we consider IP2 and IP3 together.

We seek the solution in the form

$$\begin{array}{ccc}\hfill & & u_j^{n+1}=y_j^{n+1}+g_L^{n+1}{v}_j\hfill \end{array}$$

(29)

and substitute it in (28). As a result, we obtain two problems for the unknown functions $y_j^{n+1}$ and $v_j^{n+1}$

$$\begin{array}{l}{P}_0{y}_0^{n+1}-D_0{y}_1^{n+1}=F_0^n,\\ -P_j{y}_{j-1}^{n+1}+Q_j{y}_j^{n+1}-D_i{y}_{j+1}^{n+1}=F_j^n,j=1,2,\dots ,J-1,\\ y_J^{n+1}=0,\\ u_j^0=u^0\left(j\right),\end{array}$$

(30)

$$\begin{array}{l}{P}_0{v}_0-D_0{v}_1=0,\\ -P_j{v}_{j-1}+Q_j{v}_j-D_j{v}_{j+1}=0,j=1,2,\dots ,J-1,\\ v_J=1.\end{array}$$

(31)

After solving (30) and (31), we find y and v in the whole computational domain. Then, from (29) and the measurement $u\left(r^{*}\right)=\mathrm{\Psi }\left(t\right)$, $r^{*}\in {\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2$, we find

$$g_L^{n+1}={\displaystyle \frac{{\mathrm{\Psi }}^{n+1}-y_{r^{*}}^{n+1}}{v_{r^{*}}}},$$

(32)

where $y_{r^{*}}^{n+1}$ and $v_{r^{*}}$ are numerical solutions $y^{n+1}$ and v at $r^{*}$. If $r^{*}$ is not a grid node, then interpolation is applied.

In order to establish the correctness of the numerical algorithm (29)–(32), we have to ensure that the recovered function $g_L^{n+1}$ and the solution $u_j^{n+1}$, $n=1,2,\dots ,N$, $j=1,2,\dots ,J-1$, exist and are unique. It is enough to prove that the systems (30) and (31) have a unique solution and $v_{r^{*}}\ne 0$.

Theorem 2. 

The systems (30) and (31) have a unique solution, $0<v_j\le 1$, $j=0,1,\dots ,J$, and, therefore, the algorithm (29)–(32) is correct.

Proof. 

For the proof, we follow the same line of considerations as in [27].

Since the systems (30) and (31) are non-homogeneous with a coefficient matrix that is strictly diagonally dominant and has positive elements on the main diagonal ($Q_j>0$, $P_j>0$, $j=0,1,\dots ,J$) and non-positive off diagonal elements, i.e., it is an M-matrix, we conclude that the solution is unique. Moreover, its inverse consists only of non-negative entries [46]. Taking into account that the right-hand side of (31) is a vector of non-negative elements, we deduce that $v_j\ge 0$, $j=0,1,\dots ,J$.

Further, we consider the $J-1$-th equation in (31)

$$Q_{J-1}{v}_{J-1}=P_{J-1}{v}_{J-2}+D_{J-1}\ge D_{J-1}\Rightarrow v_{J-1}>{\displaystyle \frac{D_{J-1}}{Q_{J-1}}}>0.$$

Then, from the $J-2$-th equation in (31), we obtain

$$v_{J-2}\ge {\displaystyle \frac{D_{J-2}{D}_{J-1}}{Q_{J-2}{Q}_{J-1}}}>0.$$

We proceed in the same manner up to the first equation and derive $v_j>0$, $j=0,1,\dots ,J$.

Finally, according to the discrete maximum principle [42], the following estimate holds

$$\underset{0\le j\le J}{max}\left|v_j\right|\le 1.$$

□

In the same manner, we deal with IP1 and IP4, if Dirichlet BC (4) at $r=0$ is imposed. We consider the decomposition

$$\begin{array}{ccc}\hfill & & u_j^{n+1}=y_j^{n+1}+g_0^{n+1}{v}_j\hfill \end{array}$$

(33)

and derive two problems: (30) and

$$\begin{array}{l}{P}_0{v}_0-D_0{v}_1=1,\\ -P_j{v}_{j-1}+Q_j{v}_j-D_j{v}_{j+1}=0,j=1,2,\dots ,J-1,\\ v_J=0.\end{array}$$

(34)

Computing (30) and (31) to determine y and v throughout the entire computational domain, and using (29) along with the measurement $u\left(r^{*}\right)=\mathrm{\Psi }\left(t\right)$, $r^{*}\in {\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2$, we recover the left boundary condition

$$g_0^{n+1}={\displaystyle \frac{{\mathrm{\Psi }}^{n+1}-y_{r^{*}}^{n+1}}{v_{r^{*}}^{n+1}}},$$

(35)

The correctness of (33)–(35) is proved as in Theorem 2.

Further, we proceed with numerically solving IP5 in the case of Dirichlet BC (4) at $r=0$ for a more general case of measurements

$$u\left(r_i^{*}\right)={\mathrm{\Psi }}_i\left(t\right),r_i^{*}\in {\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2,i=1,2.$$

(36)

Introducing the decomposition

$$\begin{array}{ccc}\hfill & & u_j^{n+1}=y_j^{n+1}+g_L^{n+1}{v}_j+g_0^{n+1}{w}_j\hfill \end{array}$$

(37)

and substituting it into (28) yields three problems for the unknown functions $y_j^{n+1}$, $v_j^{n+1}$, and $w_j^{n+1}$, respectively

$$\begin{array}{l}{y}_0^{n+1}=0,\\ -P_j{y}_{j-1}^{n+1}+Q_j{y}_j^{n+1}-D_j{y}_{j+1}^{n+1}=F_j^n,j=1,2,\dots ,J-1,\\ y_J^{n+1}=0,\\ u_j^0=u^0\left(j\right),\end{array}$$

(38)

$$\begin{array}{l}{v}_0=0,\\ -P_j{v}_{j-1}+Q_j{v}_j-D_j{v}_{j+1}=0,j=1,2,\dots ,J-1,\\ v_J=1.\end{array}$$

(39)

$$\begin{array}{l}{w}_0=1,\\ -P_j{w}_{j-1}+Q_j{w}_j-D_j{w}_{j+1}=0,j=1,2,\dots ,J-1,\\ w_J=0.\end{array}$$

(40)

Solving these problems, using (39) and taking into account (36), we obtain

$$\begin{array}{c}{\mathrm{\Psi }}_1^{n+1}=y_{r_1^{*}}^{n+1}+g_L^{n+1}{v}_{r_1^{*}}+g_0^{n+1}{w}_{r_1^{*}},\\ {\mathrm{\Psi }}_2^{n+1}=y_{r_2^{*}}^{n+1}+g_L^{n+1}{v}_{r_2^{*}}+g_0^{n+1}{w}_{r_2^{*}}.\end{array}$$

(41)

From (41), we obtain

$$\begin{array}{c}{g}_0^{n+1}={\displaystyle \frac{v_{r_1^{*}}({\mathrm{\Psi }}_2^{n+1}-y_{r_2^{*}}^{n+1})-v_{r_2^{*}}({\mathrm{\Psi }}_1^{n+1}-y_{r_1^{*}}^{n+1})}{v_{r_1^{*}}{w}_{r_2^{*}}-v_{r_2^{*}}{w}_{r_1^{*}}}},\\ g_L^{n+1}={\displaystyle \frac{({\mathrm{\Psi }}_1^{n+1}-y_{r_1^{*}}^{n+1})w_{r_2^{*}}-({\mathrm{\Psi }}_2^{n+1}-y_{r_2^{*}}^{n+1})w_{r_1^{*}}}{v_{r_1^{*}}{w}_{r_2^{*}}-v_{r_2^{*}}{w}_{r_1^{*}}}}.\end{array}$$

(42)

As before, the correctness (existence of a unique solution) of the inverse problem (37)–(42) can be established if the denominator in (42) is not zero and the systems (38)–(40) have unique solutions.

Theorem 3. 

The solution of the systems (38)–(40) is unique, $v_{r_1^{*}}{w}_{r_2^{*}}-v_{r_2^{*}}{w}_{r_1^{*}}\ne 0$, and, therefore, the algorithm (37)–(42) is correct.

Proof. 

Since the coefficient matrix of all systems above is one and the same, as in Theorem 2, we deduce that the solution of the systems (38)–(40) is unique and $v_j>0$, $w_j>0$, $j=0,1,\dots ,J$.

Further, we multiply each equation in the system (39) by $w_{r_2^{*}}$ and each equation in the system (40) by $v_{r_2^{*}}$ and subtract the resulting systems to obtain

$$\begin{array}{l}{w}_0{v}_{r_2^{*}}-v_0{w}_{r_2^{*}}=v_{r_2^{*}},\\ -P_j(v_{j-1}{w}_{r_2^{*}}-w_{j-1}{v}_{r_2^{*}})+Q_j(v_j{w}_{r_2^{*}}-w_j{v}_{r_2^{*}})\\ -D_j(v_{j+1}{w}_{r_2^{*}}-w_{j+1}{v}_{r_2^{*}})=0,j=1,2,\dots ,J-1,\\ v_J{w}_{r_2^{*}}-w_J{v}_{r_2^{*}}=w_{r_2^{*}}.\end{array}$$

(43)

Note that the coefficient matrix of (43) is the same M-matrix as for the systems (39) and (40), and the right-hand side is non-negative. Therefore, $w_0{v}_{r_2^{*}}-v_0{w}_{r_2^{*}}\ge 0$, $v_j{w}_{r_2^{*}}-w_j{v}_{r_2^{*}}\ge 0$, $j=1,2,\dots ,J$.

As in Theorem 2, we consequently estimate the diagonal elements of (43), starting from the last one and moving to the first equation, and conclude that $w_0{v}_{r_2^{*}}-v_0{w}_{r_2^{*}}>0$ and $v_j{w}_{r_2^{*}}-w_j{v}_{r_2^{*}}>0$, $j=1,2,\dots ,J$.

Therefore, since $r_1^{*}\in \{1,2,\dots ,J-1\}$, we complete the proof. □

6. Computational Examples

In this section, we verify the accuracy of the proposed numerical method. As a test example, we consider the problem (1)–(7) with parameters and functions

$$d_1=2,d_2=10,L=3,l_1=1,T=1,g_0\left(t\right)={\displaystyle \frac{\sqrt{2}}{2}}{e}^{t/2},g_L\left(t\right)=e^{t/2}{\displaystyle \frac{68+\sqrt{2}(1-\pi )}{2}}.$$

We take $f_1(r,t)$ and $f_2(r,t)$ such that the exact solution of the problem (1)–(7) is $u_1(r,t)=e^{t/2}cos(\pi x/4)$, $u_2(r,t)=e^{t/2}(a{x}^4-x^3+c)$, where $a=(120-\sqrt{2}\pi )/160$, $c=(40+\sqrt{2}(80+\pi ))/160$.

Example 1 

(Convergence: direct problem). We test the order of convergence of the direct problem. The numerical solution is computed using (28) for a fixed ratio $\tau =h^2$. Errors and the convergence rate of the numerical solution $u_i^n$, $i=1,2$ at the final time are estimated in the maximum and $L_2$ norms. Let $E_{i,j}=u_i(r_j,T)-u_{i,j}^M$ and

$$\begin{array}{ccc}& & {\mathcal{E}}_{1,\infty }={\mathcal{E}}_{1,\infty }\left(J\right)=\underset{0\le j\le j^{*}}{max}|E_{1,j}|,{\mathcal{E}}_{2,\infty }={\mathcal{E}}_{2,\infty }\left(J\right)=\underset{j^{*}\le J}{max}\left|E_{2,j}\right|,\hfill \\ & & {\mathcal{E}}_{1,L_2}={\mathcal{E}}_{1,L_2}\left(J\right)={\left[\sum _{j=0}^{j^{*}}h{\left(E_{1,j}\right)}^2\right]}^{1/2},{\mathcal{E}}_{2,L_2}={\mathcal{E}}_{2,L_2}\left(J\right)={\left[\sum _{j=j^{*}}^{J}h{\left(E_{2,j}\right)}^2\right]}^{1/2},\hfill \\ & & C{R}_{i,\infty }={log}_2{\displaystyle \frac{{\mathcal{E}}_{i,\infty }\left(J\right)}{{\mathcal{E}}_{i,\infty }\left(2J\right)}},C{R}_{i,L_2}={log}_2{\displaystyle \frac{{\mathcal{E}}_{i,L_2}\left(J\right)}{{\mathcal{E}}_{i,L_2}\left(2J\right)}},i=1,2,\hfill \end{array}$$

The results are listed in

Table 1. Errors and convergence rates of the recovered solution $u_i^n$, $i=1,2$, Dirichlet BC, Example 1.

J	${\mathcal{E}}_{1,\mathit{\infty }}$	${\mathit{CR}}_{1,\mathit{\infty }}$	${\mathcal{E}}_{2,\mathit{\infty }}$	${\mathit{CR}}_{2,\mathit{\infty }}$	${\mathcal{E}}_{1,{\mathit{L}}_2}$	${\mathit{CR}}_{1,{\mathit{L}}_2}$	${\mathcal{E}}_{2,{\mathit{L}}_2}$	${\mathit{CR}}_{2,{\mathit{L}}_2}$
30	5.1356 × 10−2		5.1356 × 10−2		4.3835 × 10−2		4.8458 × 10−2
60	1.2944 × 10−2	1.9882	1.2944 × 10−2	1.9882	1.1176 × 10−2	1.9717	1.2016 × 10−2	2.0118
120	3.2557 × 10−3	1.9913	3.2556 × 10−3	1.9913	2.8523 × 10−3	1.9702	2.9957 × 10−3	2.0040
240	8.1772 × 10−4	1.9933	8.1772 × 10−4	1.9933	7.2696 × 10−4	1.9722	7.4882 × 10−4	2.0002
480	2.0520 × 10−4	1.9946	2.0520 × 10−4	1.9946	1.8488 × 10−4	1.9753	1.8739 × 10−4	1.9986
960	5.1458 × 10−5	1.9956	5.1457 × 10−5	1.9956	4.6914 × 10−5	1.9785	4.6913 × 10−5	1.9980

and

Table 2. Errors and convergence rates of the numerical solution $u_i^n$, $i=1,2$, natural BC, Example 1.

J	${\mathcal{E}}_{1,\mathit{\infty }}$	${\mathit{CR}}_{1,\mathit{\infty }}$	${\mathcal{E}}_{2,\mathit{\infty }}$	${\mathit{CR}}_{2,\mathit{\infty }}$	${\mathcal{E}}_{1,{\mathit{L}}_2}$	${\mathit{CR}}_{1,{\mathit{L}}_2}$	${\mathcal{E}}_{2,{\mathit{L}}_2}$	${\mathit{CR}}_{2,{\mathit{L}}_2}$
30	6.2045 × 10−2		5.4563 × 10−2		6.0694 × 10−2		5.0790 × 10−2
60	1.5682 × 10−2	1.9842	1.3650 × 10−2	1.9990	1.4837 × 10−2	2.0324	1.2518 × 10−2	2.0205
120	3.9612 × 10−3	1.9851	3.4131 × 10−3	1.9998	3.6648 × 10−3	2.0174	3.1063 × 10−3	2.0107
240	1.0004 × 10−3	1.9853	8.5331 × 10−4	1.9999	9.1049 × 10−4	2.0090	7.7366 × 10−4	2.0054
480	2.5264 × 10−4	1.9855	2.1333 × 10−4	2.0000	2.2689 × 10−4	2.0046	1.9305 × 10−4	2.0027
960	6.3794 × 10−5	1.9856	5.3333 × 10−5	2.0000	5.6630 × 10−5	2.0024	4.8216 × 10−5	2.0014

. In Table 1, we provide errors and convergence rates obtained from computations imposing Dirichlet BC at $r=0$ (4), while in Table 2, we present results in the case of natural BC at $r=0$ (16) and (17). We verify that, in both cases, the accuracy is $O(\tau +h^2)$.

Example 2 

(Inverse problem: exact measurements). We numerically recover the functions $g_0$ and $g_L$ by using measurements equal to the numerical solution of the direct problem at $r_1^{*}$ and $r_2^{*}$ for exact external BCs, i.e., ${\mathrm{\Psi }}_1^n=u^n\left(r_1^{*}\right)$ and ${\mathrm{\Psi }}_2^n=u^n\left(r_2^{*}\right)$. The errors between the exact external BCs and the numerically restored ones ($g_0^n$, $g_L^n$) are denoted by $E_{g_0}=g_0\left(t_n\right)-g_0^n$, $E_{g_L}=g_L\left(t_n\right)-g_L^n$. We will also provide the error $E_u$ between the recovered solution $u^N=(u_1^N,u_2^N)$, obtained by numerically solving the inverse problems, and the numerical solution, obtained by the direct problem.

First, we examine the numerical method (29)–(32) in the case of natural BC at $r=0$ for recovering the function $g_L^n$ and the solution $(u_1,u_2)$ upon measurement at the point $r_1^{*}=0.5\in {\mathrm{\Omega }}_1$. The computations are performed for $\tau =h$ and $J=120$. In Figure 1 and Figure 2, we present the exact and recovered solution u and the function $g_L^n$, respectively, along with the corresponding errors.

Next, we present computational results obtained by solving (37)–(42), $\tau =h$, $J=120$, $r_1^{*}=0.5$, $r_2^{*}=2$ for recovering $g_0$, $g_L$, and the solution $(u_1,u_2)$. In Figure 3, Figure 4 and Figure 5, we present the exact and recovered solution u, the functions $g_0^n$, $g_L^n$, and the corresponding errors.

The numerical results show that the solution $(u_1,u_2)$ and the functions $(g_0,g_L)$, obtained by numerically solving the inverse problems, exactly recover the solution of the direct problem and exact external boundary conditions. Therefore, the order of convergence is the same as that for the forward problem.

Example 3 

(Inverse problems: noisy data). We perform the same experiments as in Example 2, but now we add perturbations to the measurements

$${\mathrm{\Psi }}_i^n=u^n\left(r_i^{*}\right)+2{\rho }_i({\sigma }_i^n-0.5)u^n\left(r_i^{*}\right),i=1,2,$$

where $u^n\left(r_i^{*}\right)$ are numerical solutions of the direct problem at time layer n and at spatial nodes $r_i^{*}$, ${\rho }_i$ are levels of noise, and ${\sigma }_i^n$ are random values, uniformly distributed on the interval $[0,1]$. The observations are at $r_1^{*}=0.5$ and $r_2^{*}=2$. We use polynomial curve fitting of degree 5 to smooth the data.

We estimate errors of the numerical solution $(u_1^n,u_2^n)$, computed by inverse problem, in the maximum and $L_2$ norms, using the same formulas as in Example 1, but $E_{i,j}$ is replaced by $E_{u_i}$, as in Example 2. For the errors of the recovered functions $g_0$, $g_L$, we use formulas

$${ϵ}_{s,\infty }=\underset{0\le n\le N}{max}\left|E_{g_s}\right|,{ϵ}_{s,L_2}={\left[\sum _{n=0}^N\tau E_{g_s}^2\right]}^{1/2},s=\{0,L\}.$$

The computations are performed for $\tau =h$ and $J=120$.

We consider the numerical method (29)–(32) in the case of natural BC at $r=0$, for determining the function $g_L^n$ and the solution $(u_1,u_2)$ under measurements at the point $r_1^{*}$. The computational results for different levels of noise $\rho ={\rho }_1$ are presented in

Table 3. Errors of the recovered solution $(u_1^N,u_2^N)$ and function $g_L$ for different levels of noise $\rho$, IP solved by (29)–(32), natural BC at $r=0$, Example 3.

$\mathit{\rho }$	${\mathcal{E}}_{1,\mathit{\infty }}$	${\mathcal{E}}_{2,\mathit{\infty }}$	${\mathcal{E}}_{1,{\mathit{L}}_2}$	${\mathcal{E}}_{2,{\mathit{L}}_2}$	${\mathit{ϵ}}_{\mathit{L},\mathit{\infty }}$	${\mathit{ϵ}}_{\mathit{L},2}$
0.01	2.2141 × 10−3	1.0083 × 10−2	1.1401 × 10−3	7.7277 × 10−3	1.8319 × 10−1	4.3869 × 10−2
0.03	6.1225 × 10−3	2.6653 × 10−2	3.2143 × 10−3	2.0666 × 10−2	6.6437 × 10−1	1.6369 × 10−1
0.05	1.0031 × 10−2	4.3224 × 10−2	5.2888 × 10−3	3.3605 × 10−2	1.1456	2.8357 × 10−1
0.1	1.9802 × 10−2	8.4650 × 10−2	1.0475 × 10−2	6.5954 × 10−2	2.3486	5.8330 × 10−1

.

In Figure 6, we depict the exact and recovered function $g_L$ and the corresponding error for $\rho =0.1$. The largest error appears at the initial time. Note that for $\rho =0.1$, the relative error in the maximum norm of the restored function $g_L$ is 7.3208 × 10⁻², while in the $L_2$ norm, it is 1.8183 × 10⁻², which is satisfactory for this level of noise.

Next, we perform experiments with the numerical method (37)–(42) for recovering the solution $(u_1,u_2)$ and the functions $g_0$ and $g_L$ with different levels of noise ${\rho }_1$ and ${\rho }_2$. The computational results are listed in

Table 4. Errors of the recovered solution $(u_1^N,u_2^N)$ and functions $g_0$ and $g_L$ for different levels of noise ${\rho }_1$, ${\rho }_2$, IP solved by (37)–(42), Example 3.

${\mathit{\rho }}_1$	${\mathit{\rho }}_2$	${\mathcal{E}}_{1,\mathit{\infty }}$	${\mathcal{E}}_{2,\mathit{\infty }}$	${\mathcal{E}}_{1,{\mathit{L}}_2}$	${\mathcal{E}}_{2,{\mathit{L}}_2}$	${\mathit{ϵ}}_{0,\mathit{\infty }}$	${\mathit{ϵ}}_{0,2}$	${\mathit{ϵ}}_{\mathit{L},\mathit{\infty }}$	${\mathit{ϵ}}_{\mathit{L},2}$
0.03	0.001	1.051 × 10−1	4.042 × 10−3	2.287 × 10−2	2.770 × 10−3	1.768 × 10−1	3.826 × 10−2	7.408 × 10−3	1.645 × 10−3
0.01	0.003	2.763 × 10−2	1.239 × 10−2	6.186 × 10−3	8.439 × 10−3	6.863 × 10−2	1.404 × 10−2	2.107 × 10−2	4.680 × 10−3
0.05	0.005	1.630 × 10−1	2.081 × 10−1	3.568 × 10−2	1.417 × 10−2	2.980 × 10−1	6.130 × 10−2	3.490 × 10−2	7.833 × 10−3

.

In Figure 7 and Figure 8, we plot the exact and numerically recovered functions $g_0$ and $g_L$, respectively, and the corresponding errors for ${\rho }_1=0.01$, ${\rho }_3=0.003$.

We observe that for the noisy measurements, the numerical approach recovers the numerical solution and external boundary conditions with satisfactory accuracy.

The computational simulations presented in this section to test the developed methods are implemented using MATLAB^® R2022a.

7. Conclusions

In the present work, we develop an efficient numerical method for solving five inverse problems for recovering the solution and the right and/or left boundary conditions in a two-layer interface parabolic equation with cylindrical symmetry. We prove the well-posedness of the direct differential problem and construct a second-order in space and first-order in time numerical approach for its solution. Due to the degeneracy of the problem at the left boundary $r=0$, we consider both the Dirichlet and natural boundary conditions. We construct a numerical method based on a decomposition technique for numerically solving the inverse problems. The correctness of the developed approaches is investigated.

Numerical experiments show the stability and the order of convergence—second in space and first in time—of the numerical method for solving the direct problem. Since for exact observations, numerical methods for solving inverse problems recover the numerical solution of the direct problem and external boundary conditions exactly, we conclude that the numerical algorithms for solving inverse problems are stable and achieve the same order of accuracy as the numerical method for the direct problem. Simulations with noisy observations illustrate that the recovered solution and external boundary conditions are of good accuracy.

Although in this study we consider continuous functions $g_0\left(t\right)$ and $g_L\left(t\right)$, the constructed numerical methods can be easily extended to a wide class of functions, for example, $L_2$. In future work, we plan to extend our investigation to inverse spherical geometry problems.