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Article

Regularity of a Parabolic Differential Equation on Graphs

Faculty of Organization and Informatics, University of Zagreb, 42000 Varaždin, Croatia
Submission received: 27 September 2023 / Revised: 22 October 2023 / Accepted: 25 October 2023 / Published: 27 October 2023

Abstract

:
In this paper, motivated by a biodegradable stent problem, we consider a parabolic differential equation on graphs. This kind of equation describes the stent material degradation in time. Since stents are mesh-like structures with thin struts, they can be modelled using a simple structure called the one-dimensional curved rod model. In this way, we obtain a graph-like domain for our parabolic equation. Here, we prove the regularity estimate for the unique solution of the equation, together with corresponding estimates.
MSC:
35B65; 35K10

1. Introduction

Our object of interest in this paper is parabolic equations defined on graphs. Let us first observe parabolic equations defined on a single segment. Let α 1 , and T be real numbers, with α 1 > 0 . For a given f L 2 ( 0 , ) , we have the boundary value problem
d t α 1 d x x = f , on ( 0 , T ) × ( 0 , ) ,
d x ( t , 0 ) = d x ( t , ) = 0 , t [ 0 , T ] ,
d ( 0 , x ) = 1 , x [ 0 , ] .
The unknown in this problem is
d : [ 0 , T ] × [ 0 , ] R ,
Please see references [1,2] for more details. Investigating partial differential equations on a segment is not rare and is carried out in many applications, like dealing with viscoelasticity and a class of partial integro-differential equations [3]. On the other hand, parabolic equations are used in many different contexts for modelling diffusion [4,5,6]. In the aforementioned articles, the researchers do not deal with the regularity of the problem solution like we do in this article. There are some researchers that investigate the properties of partial differential equations defined on graphs, like in reference [7], but for observing such boundary value problems here, we find motivation in modelling mesh-like structures, like biodegradable stents. A biodegradable stent is a mesh that is used to treat the narrow or closed part of the artery to open and restore normal blood flow, which is made of a biodegradable material [8,9,10] (see Figure 1, left).
However, since struts in the stent are thin, a simple structure model, called the one-dimensional curved rod model, can be used (see reference [11]). In the case of biodegradable elastic stents, more can be found in reference [12]. Thus, in the sequel, we consider the structure of the stent using this 1D description and as an object composed of N rods of length i ( i = 1 , , N ) whose ends are wedged together in M prominent joints. Therefore, we can idealize stents as graphs (with vertices and edges). Joints are represented by prominent points where the centre lines of the rods meet. Those highlighted points will be vertices and are marked with indices 1 , , M . The rods connecting the joints are modelled through their centre lines connecting the vertices. The rods (edges in the graph) are marked with indices 1 , , N . Also, due to the implementation of the conditions on contact force and contact moment, the edges of the graph will be directed, i.e., to each rod i, i { 1 , , N } ; an ordered pair of indices of the vertices that the rod connects is assigned. If we define a function o : [ 1 , , N ] [ 1 , , M ] × [ 1 , , M ] which is an injection, then with o ( i ) ( i = 1 , , N ) the orientation of each edge in the graph is defined. This gives us the mathematical structure of the graph, and an oriented graph G is assigned to each stent. Note, here, that the orientation is necessary for the formulation of the problem, but the solution will not depend on the orientation. In this kind of context, our function d will describe the amount of material left at time t at point x, so it is natural that d ( [ 0 , T ] × [ 0 , ] ) [ 0 , 1 ] . Note, also, that the initial condition in (3) is natural for the problem of degradation. On each graph edge i ( i = 1 , , N ) we have equation like in (1) with initial condition (3) defined. This gives us the system of N equations with their initial conditions and what there is left to do is to define the boundary conditions, i.e., the compatibility conditions in vertices. There, we demand the continuity of d in vertices and that the sum of all fluxes in each vertex equals zero [13,14,15], which results in the system
j = 1 , , M , α 1 > 0 d i ( t , 0 ) = d k ( t , k ) , i J j + , k J j , i J j + d x i ( t , 0 ) i J j d x i ( t , i ) = 0 , t [ 0 , T ] ,
where J j + and J j are sets of all edges that leave and enter the j-th vertex, respectively, for j = 1 , , M . Here, leaving and entering vertices means it is due to the parametrization of the edges. In this way, our problem is completely defined. Because of the nature of the compatibility conditions, we cannot observe the problem on each edge separately, but as a problem on the overall graph.
In this paper, we use, as much as we can, the result from reference [16] on the existence and uniqueness of the solution of an abstract parabolic problem. However, in order to get the regularity result and associated estimates, further analysis is necessary. These results will be used in the analysis of biodegradable stents with the diffusion of the material being more pronounced.
The structure of this paper is as follows: we start the analysis by defining the weak solution of the problem in the second section. In the third, we prove some auxiliary lemmas. In the fourth chapter, we first prove the existence by applying the analysis from reference [16] and then prove the regularity result and associated estimates. Finally, all the results are summarized in the conclusion and the major innovation point of the paper is emphasized.

2. Definition of the Weak Solution

Let G be a simple connected directed graph with M vertices and N edges of length i , i = 1 , , N . The problem that we have here is as follows: for given f i L 2 ( 0 , i ) , i = 1 , , N find d i , i = 1 , , N such that
i = 1 , , N d t i α 1 d x x i = f i , in ( 0 , T ) × ( 0 , i ) , d i ( 0 , x ) = 1 , x [ 0 , i ] ,
with conditions
j = 1 , , M , α 1 > 0 d i ( t , 0 ) = d k ( t , k ) , i J j + , k J j , i J j + d x i ( t , 0 ) i J j d x i ( t , i ) = 0 , t [ 0 , T ] .
Before moving on with the weak formulation of of the problem, we need to define the function spaces we will deal with. Let
V s = { c = ( c 1 , , c N ) H 1 ( 0 , 1 ) × × H 1 ( 0 , N ) :   c i ( 0 ) = c k ( k ) , i J j + , k J j , j = 1 , , M } , V s 2 = H 2 ( 0 , 1 ) × × H 2 ( 0 , N ) , H s = L 2 ( 0 , 1 ) × × L 2 ( 0 , N ) .
For the norms on V s 2 and H s , respectively, we choose
c = i = 1 N c i H 1 ( 0 , i ) 2 1 / 2 , | c | = i = 1 N c i L 2 ( 0 , i ) 2 1 / 2 ,
where both norms are induced by inner products. With ( · , · ) H we denote the inner product in H s , and with ( · , · ) V , the inner product in V s . We have that V s is densely and continously embedded in H s , i.e., V s H s . With a s : V s × V s R we define the bilinear form
a s ( d , c ) = α 1 i = 1 N 0 i d x i i c x i i d x i .
Trivially, we have
a s ( c , c ) + α 1 | c | 2 = α 1 c 2 .
Let us also define the space V s as a dual space of V s . Let d = ( d 1 , , d N ) , d 0 = ( d 0 1 , , d 0 N ) H s and ( f 1 , , f N ) L 2 ( 0 , T ; H s ) .
Definition 1.
We say that the function
d L 2 ( 0 , T ; V s ) with d t L 2 ( 0 , T ; V s )
is a weak solution of the problem (4) with conditions (5) if it holds that
i. 
d d t ( d , c ) H + a s ( d , c ) = ( f , c ) H , c V s , a . e . t [ 0 , T ] ,
ii. 
d ( 0 ) = d 0 .
Before proving the existence of the weak solution, we shall prove two lemmas which will be used in the proof of the theorem itself.

3. Auxiliary Results

In order to apply Theorems 1 and 2, on pages 512 and 513 in reference [16], we need coercivity of the form a s . Note that constant functions are in the kernel of a s . The following Poincaré-type lemma implies the coercivity of a s on a subspace of functions with a total average equal to zero.
Lemma 1
(Poincaré). There exists C p > 0 such that for each d V s holds
| d | 2 + i = 1 N 0 i d i 2 C p | d | 2 .
Proof. 
Let us assume the opposite, i.e., C p > 0 there exists d C p V s such that
| d C p | 2 + i = 1 N 0 i d C p i 2 < C p | d C p | 2
holds. Let us take C p = 1 k 2 and denote d k = d C p . Let
v k : = d k | d k | .
Now we have | v k | = 1 and
| v k | 2 + i = 1 N 0 i v k i 2 < 1 k 2 .
It follows that
| v k | = 1 v k is bounded in H s ,
| v k | < 1 k v k is bounded in H s , v k ( 0 , , 0 ) in H s ,
i = 1 N 0 i v k i < 1 k i = 1 N 0 i v k i 0 , in R .
From (6) and (7), it follows that ( v k ) is bounded in V s . The compactness of embedding V s H s implies that there exists a subsequence such that v k j v in H s strongly. From here, it follows that
| v | = 1 , i = 1 N 0 i v i = 0 .
Now, from v k j ( 0 , , 0 ) H s it follows that v = ( 0 , , 0 ) , from which we conclude that v i are constants for each i = 1 , , N . The continuity of the trace operator in edges implies that all these constants are equal, i.e., v = ( β , , β ) R N . From the integral in (9), it follows that v = ( 0 , , 0 ) which is in contradiction with | v | = 1 . □
The existence and uniqueness of the abstract parabolic problem in reference [16] is proven using the Galerkin method with a basis only in the space corresponding to V s . Since we are dealing with a regularity result, we need more smoothness for this basis. Thus, we now prove the edgewise smooth orthonormal basis.
Lemma 2
(Orthonormal basis). There exists an edgewise smooth orthonormal basis B V s for H s .
Proof. 
First, let us define spaces
V ¯ s = { c V s : i = 1 N 0 i c i = 0 } , H ¯ s = { c H s : i = 1 N 0 i c i = 0 } .
We shall find the orthonormal basis with the help of the eigenvalue problem solution: find a pair ( λ , d ) R × V ¯ s , λ 0 , d = ( d 1 , , d N ) such that
( d , c ) H = λ ( d , c ) H , c V ¯ s .
But before that, let us observe the next problem.
Problem 1.
For a given f H s , find d V ¯ s such that
( d , c ) H = ( f , c ) H , c V ¯ s .
Let us define the operator K : H s V ¯ s with K f = d , where d is a solution of (11). We shall prove that K is well defined, compact, symmetric, and positive, and then, by applying the spectral theorem for compact symmetric operators, the existence of edgewise smooth orthonormal basis follows.
1. Well defined. First of all, we need to show that this kind of operator K really exists, i.e., that problem (11) has a solution. We shall prove this by applying the Lax–Milgram lemma. The boundness and symmetry of the bilinear form are trivial to show, and the coercitivity follows from Lemma 1. Since d H ¯ s , it follows that
i = 1 N 0 i d i = 0 ,
so we have the estimate
1 C p | d | 2 | d | 2 = i = 1 N 0 i d x i d x i .
2. Compactness. From Lemma 1, we have the estimate
| d | c | d | , c = 1 C p R .
From (11), for c = d , we calculate
| d | 2 = ( f , d ) H | f | | d | c | f | | d | | d | c | f | .
Furthermore,
1 c 2 | d | 2 | d | 2 = ( f , d ) H | f | | d | | d | c 2 | f | .
Now we have
K f 2 = d 2 = | d | 2 + | d | 2 ( c 4 + c 2 ) | f | 2 K f c c 2 + 1 | f | .
From here, because of the compactness of the embedding V ¯ s H s , it follows that K : H s H s is a compact operator.
3. Symmetry. Let it hold that
K f = d and K g = v .
Then we have,
K f = d , for test function v ( K f , g ) H = ( d , v ) H K g = v , for test function d ( g , K f ) H = ( v , d ) H .
Since the inner product in H s is symmetric, we obtain that our operator is symmetric, i.e., ( f , K g ) H = ( K f , g ) H .
4. Positivity. We have
( K f , f ) H = ( d , f ) H = ( d , d ) H 0 .
Since K is a symmetric, positive, and compact linear operator, it follows that all its eigenvalues η i , i N are nonnegative and approach zero, and eigenvectors w i , i N form an orthonormal basis in H ¯ s . Let us notice now that, for η i 0 , we have that K w i = η i w i if and only if w i is a solution of our eigenvalue problem (10) for λ i = 1 / η i for i N . In this way, we describe the solutions of the eigenvalue problem (10). Notice that, in our wanted basis, we did not include constant vectors from V s (they are not in V ¯ s ). Hence, if we take d V ¯ s and r = ( a , , a ) R N , for a R we have
( d , r ) = i = 1 N 0 i d i a = a i = 1 N 0 i d i = 0 H s = H ¯ s ( a , , a ) R N , a R .
If we denote
w 0 = i = 1 N i 1 / 2 , , i = 1 N i 1 / 2 R N ,
then we have | w 0 | = 1 , with what we obtained the orthonormal basis B V s for H s .
5. Smoothness. On each edge of the graph and for each eigenpair ( λ k , d k ) , we have a differential equation
( d k i ) = λ k d k i .
Obviously d k i is smooth using the classical circular argument. □

4. Main Results

Now, we arrive to the main theorems in this paper. The first is the existence and uniqueness theorem, which now follows directly from Theorems 1 and 2 on pages 512 and 513 [16], after the applications of Lemmas 1 and 2. However, for the following theorem, we need to introduce some notation.
Theorem 1
(Existence and uniqueness on graph). There exists a unique weak solution from Definition (1).
Proof. 
The existence and uniqueness of the weak solution will be proved using the Galerkin method. Here, we refer to an important abstract result by Dautray and Lions in reference [16], and will sketch this proof only as much as we need for the Theorem 2 which follows. From Lemma 2, there exists
{ w k } k = 0 orthonormal basis for H s ,
with functions w k which are smooth on each edge. Let d 0 H s . We define d m : [ 0 , T ] V s
d m ( t ) = k = 0 m γ m k ( t ) w k ,
where ( γ m k ) k = 0 , , m satisfies the system of equations
( d m ) t ( t ) , w k H + a s ( d m ( t ) , w k ) = ( w k ) H , k = 0 , , m , t [ 0 , T ] , γ m k ( 0 ) = ( d 0 , w k ) H , k = 0 , , m .
Let
H s m = L { w 0 , , w m } .
System (12) sets the initial problem for a linear system of ordinary differential equations. The standard existence and uniqueness theorems of solutions of ordinary differential equations now imply that there exists the unique d m such that
d m C ( [ 0 , T ] , H s m ) , ( d m ) t L 2 ( 0 , T ; H s m ) .
In reference [16], we also have the corresponding a priori estimates for the problem (12):
max 0 t T | d m ( t ) | + 0 t d m ( τ ) 2 d τ C | d 0 | 2 + 0 T | f ( τ ) | 2 d τ ,
where C depends only on the domain, T and α 1 . This time dependence came from applying Gronwall’s inequality to finite-dimensional approximations d m . From (13), the convergence in (12) on a subsequence now follows, in corresponding topologies. The limit function d satisfies the equation and the initial condition from Definition 1, so it follows the existence of the weak solution such that
d L 2 ( 0 , T ; V s ) , and d t L 2 ( 0 , T ; V s ) .
The uniqueness follows from Theorem 1 on page 512 in reference [16]. □
Theorem 2
(Regularity on a graph). Let f L 2 ( 0 , T ; H s ) and d 0 V s . Let us also assume that d L 2 ( 0 , T , V s ) with d t L 2 ( 0 , T ; V s ) is a weak solution of the problem (4) with conditions (5). Then, in fact,
d L 2 ( 0 , T ; V s 2 ) L ( 0 , T ; V s ) and d t L 2 ( 0 , T ; H s ) ,
and we have the estimate
ess sup 0 t T d ( t ) + | d | L 2 ( 0 , T ; V s 2 ) + | d t | L 2 ( 0 , T ; H s ) C g r d 0 2 + | L 2 ( 0 , T ; H s ) ,
where C g r is a constant that depends only on the domain, T and α 1 .
Proof. 
Let us return to the eigenvalue problem (10). From its weak formulation. we see that our orthonormal basis actually satisfies
w j = λ j w j , j = 0 , , m .
Here, we have assigned the eigenvalue 0 to all constants. As it stands,
d m = j = 0 m γ m j w j = j = 0 m γ m j λ j w j H s m ,
in the Galerkin problem (12) we can take d m for the test function:
( ( d m ) t , d m ) H + α 1 ( d m , d m ) H = ( d m ) H .
We now apply integration by parts on the first term on the left hand side:
( ( d m ) t , d m ) H = i = 1 N 0 i ( d m i ) t ( d m i ) x i x i d x i , = i = 1 N ( d m i ) t ( i ) ( d m i ) x i ( i ) ( d m i ) t ( 0 ) ( d m i ) x i ( 0 ) + i = 1 N 0 i ( d m i ) t x i ( d m i ) x i d x i , = j = 1 M i J j + ( d m i ) t ( i ) ( d m i ) x i ( i ) i J j ( d m i ) t ( 0 ) ( d m i ) x i ( 0 ) ( * ) + ( ( d m ) t , d m ) H .
After applying integration over time and consider the boundary conditions in (5), expression ( ) equals zero. Next, let us notice that
d d t | d m ( t ) | 2 = 2 ( d m ( t ) , ( d m ) t ( t ) ) H .
Followed by the integration over time, this results in the next equality
0 t ( ( d m ) t , d m ) H d τ = 1 2 0 t d d t | d m ( τ ) | 2 d τ = 1 2 | d m ( t ) | 2 1 2 | d m ( 0 ) | 2 .
So, after applying integration by parts in the first term in (14) and integration over time, we obtain
1 2 | d m ( t ) | 2 + α 1 0 t ( d m , d m ) H d τ = 0 t ( f , d m ) H d τ + 1 2 | d m ( 0 ) | 2 .
From here, by the use of the Cauchy inequality with ϵ and positivity of the first term in (16), we obtain
α 1 0 t | d m | 2 d τ 0 t | f | | d m | d τ + 1 2 | d 0 | 2 , 0 t ε | f | 2 + 1 4 ε | d m | 2 d τ + 1 2 | d 0 | 2 .
For ε = 1 2 α 1 , we obtain
0 t | d m | 2 d τ 1 α 1 2 0 t | f | 2 d τ + 1 α 1 | d 0 | 2 .
Let us return now to (16), using positivity on the second term of the left hand side in (16), the Cauchy inequality, and the estimate in (17):
| d m ( t ) | 2 0 t | f | 2 + | d m | 2 d τ + | d 0 | 2 , 0 t | f | 2 d τ + 1 α 1 2 0 t | f | 2 d τ + 1 α 1 + 1 | d 0 | 2 ,
to obtain the estimate
| d m ( t ) | 2 1 α 1 2 + 1 0 t | f | 2 d τ + 1 α 1 + 1 | d 0 | 2 d τ , a . e . t [ 0 , T ] .
Now, we multiply the Equation (12) with ( γ m k ) t and sum from 0 to m:
( ( d m ) t , ( d m ) t ) H + a s ( d m , ( d m ) t ) H = ( f , ( d m ) t ) H , a . e . t [ 0 , T ] .
Furthermore,
a s ( d m , ( d m ) t ) = α 1 i = 1 N 0 i ( d m i ) x i ( d m i ) t x i d x i , = α 1 2 d d t d m , d m H .
We integrate (19) over time, like in (15), and take into account the previous equality
0 t | ( d m ) t | 2 d τ + α 1 2 | d m ( t ) | 2 = 0 t ( f , ( d m ) t ) H d τ + α 1 2 | d m ( 0 ) | 2 , a . e . t [ 0 , T ] .
Now, similarly to the estimations after (16), we have
0 t | ( d m ) t | 2 d τ 0 t | f | | ( d m ) t | d τ + α 1 2 | d 0 ( 0 ) | 2 , 0 t ε | f | 2 + 1 4 ε | ( d m ) t | 2 d τ + α 1 2 | d 0 ( 0 ) | 2 ,
so, for ε = 1 / 2 , we have
0 t | ( d m ) t | 2 d τ 0 t | f | 2 d τ + α 1 | d 0 ( 0 ) | 2 .
From (13), (17), (18), and (20), we have that, for sequences from the previous existence theorem, we can take their subsequences that also satisfy
( d m l ) t ( d ) t L 2 ( 0 , T ; H s ) , d m l d L 2 ( 0 , T ; V s 2 ) .
So, the solution of the parabolic partial differential equation on a graph also satisfies the additional regularity properties
d L 2 ( 0 , T ; V s 2 ) L ( 0 , T ; V s ) and ( d ) t L 2 ( 0 , T ; H s ) .
The estimate follows directly from (13), (17), (18), and (20). □

5. Conclusions

A parabolic partial differential equation on a graph was observed in this paper. The graph-like domain is obtained from the mathematical model of mesh-like structures. The general result from reference [16] was used regarding the existence and uniqueness of the solution of the abstract parabolic problem. The novelty is the additional regularity result and associated estimates on the boundness of the observed function which were proved here. This result is important for the analysis of the diffusion effect on the material degradation of biodegradable stents.

Funding

This research received no external funding.

Conflicts of Interest

The author declares no conflict of interest.

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Figure 1. 3D (left) and 1D (right) computer visualization of Palmaz-type stent.
Figure 1. 3D (left) and 1D (right) computer visualization of Palmaz-type stent.
Mathematics 11 04453 g001
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Žugec, B. Regularity of a Parabolic Differential Equation on Graphs. Mathematics 2023, 11, 4453. https://0-doi-org.brum.beds.ac.uk/10.3390/math11214453

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Žugec B. Regularity of a Parabolic Differential Equation on Graphs. Mathematics. 2023; 11(21):4453. https://0-doi-org.brum.beds.ac.uk/10.3390/math11214453

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Žugec, Bojan. 2023. "Regularity of a Parabolic Differential Equation on Graphs" Mathematics 11, no. 21: 4453. https://0-doi-org.brum.beds.ac.uk/10.3390/math11214453

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