Two generalized Tricomi equations

In this note we give existence results for the generalized Tricomi equations Ru′′+Bu=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {R}}u'' + {\mathcal {B}}u = f$$\end{document} and (Ru′)′+Bu=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$({\mathcal {R}}u')' + {\mathcal {B}}u = f$$\end{document} with suitable boundary data where R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {R}}$$\end{document} may be an operator (or a function) depending also on time assuming positive, null and negative sign, while B\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {B}}$$\end{document} is an elliptic operator. To do that we also extend a result for equations like (Ru′)′+Au′+Bu=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$({\mathcal {R}}u')' + {\mathcal {A}}u' + {\mathcal {B}}u = f$$\end{document} to equations like Ru′′+Au′+Bu=f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\mathcal {R}}u'' + {\mathcal {A}}u' + {\mathcal {B}}u = f$$\end{document} and use these to derive the existence for the generalised Tricomi type equations mentioned above.


Introduction
The goal of this paper is to give some existence (and uniqueness) results for some elliptichyperbolic type equations ed in particular for some generalisation of the Tricomi equation which is of hyperbolic type where x > 0 and becomes elliptic where x < 0 and has some applications in the study of transonic flow. We refer to the review papers [2] and [6] (and the references therein) for an overview about Tricomi equation and more generally equations of elliptic-hyperbolic type, and their applications. Equations of elliptic-hyperbolic type like where A and B are "elliptic" operators and σ a function which may be non-negative or changing sign, have been considered by some authors, beside to [2] and [6] we mention here [3], where σ may be positive and negative and also first order equations with changing sign coefficient are considered, and a very general result due to Showalter and contained in [13]  where σ ≥ 0, but in return instead of a function σ an operator R is considered, i.e. equation like is considered, with R linerar non-negative operator, but not depending on time.
In [10] a general and abstract result for such type of equation is studied. The equation considered in [10] (with suitable boundary data) is where A and B are elliptic operators and R is a linear operator possibly depending on time.
This result extends the one contained in [13], both for the fact that R may depend also on time and for the fact that R may change sign.
In the present paper we consider another generalisation of the result contained in [13] and which is in some sense, complementary to the one contained in [10]. Indeed we consider an abstract equation like (2) with R linear operator possibly depending on time. This study is presented in Section 3. But this result is only functional to the study of a generalized Tricomi equation we present in Section 4 and which is the main subject of the present paper. Using also the result of [10] we study the two equations, with suitable boundary data, The idea is to approximate these equations with equations like where is a positive parameter and analyse the limit behaviour of the solutions when goes to zero. The result is not immediate, since a natural estimate of the solutions u of one of the two equations in (3) goes as −1/2 (see (34)). If R is the identity operator the approximation via the equations (3) is called parabolic regularisation and in that case, i.e. for the wave equation as done for instance in chapter 3 of [5], easily works. In our case this is more delicate due to the changing sign of the operator R, therefore some more accurate estimates are needed. As regards the boundary data, roughly speaking we assign a value for u at time 0 and for u t we give the initial datum at time 0 where R "is positive" and we prescribe a final datum at time T where R "is negative", no datum is prescribed where R = 0, both at time t = 0 and t = T . As regards the assumptions we stress that about the operators R and B and the datum f we have to require some regularity in time, but the assumptions about the operator R depend on the case: studying the equation Ru + Bu = f seems to require some more regularity (see subsection 4.1, and in particular point 5, for details) than studying the equation Anyway for the details we refer to Section 4 and Section 5 where some examples are shown.

Notations, hypotheses and preliminary results
Consider the following family of evolution triplets where H is a separable Hilbert space, V a reflexive Banach space which continuously and densely embeds in H and V the dual space of V , and we suppose there is a constant k which satisfies for every w ∈ H , v ∈ V and every t ∈ [0, T ]. Moreover we define the spaces endowed with the norms By V we denote the dual space of V endowed with the natural norm

Definition 2.1 Given a family of linear operators R(t) such that
R depends on a parameter t ∈ [0, T ] and R(t) ∈ L(H ), being L(H ) the set of linear and bounded operators from H in itself, instead of (7) we sometimes will write Now consider an abstract function R : [0, T ] −→ L(H ). We say that R belongs to the class Now, given two non-negative constants C 1 and C 2 , consider R ∈ E(C 1 , C 2 ). For every t ∈ [0, T ] we consider the spectral decomposition of R(t) (see, e.g., Section 8.4 in [7]) and define R + (t) and R − (t) as follows: since is a positive operator; then we can define the square root of R(t) 2 (see, e.g., Chapter 3 in [7]), which is a positive operator, 2 1/2 and then define the two positive operators By this decomposition we can also write and with respect to the norm w Clearly the operation depends on R. In this way = 0 (see, e.g., Theorem 10.37 in [7]) and R + (t) : Moreover we introduce the orthogonal projections Given an operator R ∈ E(C 1 , C 2 ) it is possible to define two other linear operators. First we can define the derivative of R which, unlike R, is valued in L(V , V ), i.e. the set of linear and bounded operators from V to V : since R ∈ E(C 1 , C 2 ) we can define a family of equibounded operators By the density of U in V we can extend R (t) to V . Then we can also define which turn out to be linear and bounded by the constant C 1 and, by density of U in V, an operator which turns out to be linear, self-adjoint and bounded by C 2 . As done before we can define, in a way analogous to that done for the spaces (9), with respect to the norm w H = |R| 1/2 w H , where |R| = R + + R − . Analogously, we define H + and H − and P + and P − the orthogonal projections fromH onto H + and H − respectively. H 0 is the kernel of R and P 0 the projection defined in H onto H 0 .

Now consider the two spaces
endowed respectively with the norms Since (Ru) = R u + Ru one gets that and then the two spaces W R and W R turn out to be the same and the two norms equivalent. We will simply denote by W R the two spaces, without specifying if not necesary. We now recall some results contained in [8] and [9].

Moreover the function t → (R(t)u(t), v(t)) H is continuous and there exists a constant c, which depends only on T , such that
Finally we recall a classical result (see, e.g., Section 32.4 in [15], in particular Corollary 32.26) for which we need some definitions, which we remind. We say that an operator S : X → X , X being a reflexive Banach space, is coercive if The same operator S is hemicontinuous if the map A monotone and hemicontinuous operator S is of type M if (see, for instance, Basic Ideas of the Theory of Monotone Operators in volume B of [15] or Lemma 2.1 in [14]), i.e. it satisfies what follows: for every sequence (u j ) j∈N ⊂ X such that

An existence result for a second order equation
In [10] an existence result for the abstract equation is presented, where R, A, B are operators defined in V or in one of its subspaces, where V is the reflexive Banach space defined in (6). We state the result at the end of this section (Theorem 3.5).
The aim of this section is to present an analogous result for an abstract equation like with suitable boundary data. Clearly when R is an operator independent of time the two results coincide. We need to consider three operators, R, A, B: about R we will require, even if R does not appear in the equation, one derivative in the weak sense given in Definition 2.1, i.e. we consider two non-negative constant C 1 and C 2 and (the class About the operator A we do not require, for the moment, any assumption, only that Assumptions about A will be clarify in Theorem 3.3. About B we make the following assumptions: that there is a family of operators (with the notations already used in the previous section), two non-negative constants In this way we define an abstract operator B as follows which turns out to be linear, monotone and symmetric. Thanks to assumptions (16) we moreover can define an operator B as follows: The assumption about the derivative of the operator B is needed because of the following result, which will be used below with Q = B (for the proof we refer to [10]).

Proposition 3.1 Consider Q : [0, T ] −→ L(V , V ) satisfying (16) and consider the two operators
We want to stress that the proof is based essentially in the following inequality which can be obtained following the proof of Proposition 2.3 in [10]:

The result
Now we want to study the problem and P − (T ) are the orthogonal projections defined in (10) and u will belong to the space The boundary conditions with respect to the variable t, i.e. the initial-final conditions, are given as follows: we give an initial condition for u at time zero where R is positive (i.e. the datum ϕ) while a final condition at time T where R is negative (i.e. the datum ψ). Where R is null, no conditions for u are given. About u we impose a datum at time t o ∈ [0, T ] (the datum η).

By Proposition 2.3 one has that
so the data ϕ, ψ and η makes perfectly sense. If R ≡ 0 the initial-final conditions about u make no sense and the problem simply becomes The initial/final conditions we require about u and u are easily understood by explaining how we prove the existence result: indeed the idea to solve problem (20) is to consider an operator J t o defined in Proposition 3.1 for some arbitrary t o ∈ [0, T ] and the change of variable v = u in (20) and then solve, once set g = f − Bη, the first order problem If R ≡ 0 the solution of (21) will be a function in the space Now to solve problem (22) we write the right hand term in the equation (22) as The idea is to use first, when ϕ = 0 and ψ = 0, Theorem 2.4 in the space Indeed one has (see Proposition 3.8 in [8]) that the operator bounded one can conclude. Then one can take general ϕ and ψ assuming that where (see (9) for the definition ofH − ,H 0 ,H + ). The following result follows from Theorem 3.13 in [8]. If there are two positive constants α, β such that A is continuous and Proof Point i) is an immediate consequence of Theorem 3.10 in [8], while points ii) and iii) follow by Theorem 3.13 in [8] since A is continuous and Then for every f ∈ V , ϕ ∈H + (0), ψ ∈H − (T ), η ∈ V problem (20) admits a unique solution u ∈ Z R and there is a positive constant c, depending only on α, β, C 3 , T and proportional to α −1/2 , such that Proof By Theorem 3.3 we get a function v solving (22) with g = f − Bη, then consider the function It is easy to verify that u is a solution of (20). The uniqueness is easily obtained since if u 1 , u 2 ∈ Z R are two solutions of (20) we have that both v 1 (t) := u 1 (t) and v 2 (t) := u 2 (t) are solutions of (22). By Theorem 3.3, point iii ), we get that Since u 1 , u 2 are two solutions of (20), we have that by which u 1 = u 2 . The estimate follows again by the estimate in point ii ) of Theorem 3.3.

The result for the equation (Ru ) + Au + Bu = f
Here we recall the result contained in [10], first because we need it in the next section and also because we slightly modify the assumptions. The proof in [10] is done assuming something stronger than (23), but this can be weakened as done in Theorem 3.3 and Theorem 3.4. Consider the following problem In fact the spaceẐ R coincides with the space Z R .
A is continuous and admits a unique solution u ∈Ẑ R and there is a positive constant c, depending only on α, β, C 3 , T and proportional to α −1/2 , such that

The existence result for two generalized Tricomi equations
In this section we want to give some existence results for some generalized Tricomi equation using the results of the previous section. We recall that Tricomi equation is , t), and then the equation is of hyperbolic type in the half-plane x > 0 and is of elliptic type in the half-plane x < 0.
and the two problems Notice that we do not simply require f ∈ V , but also that the datum f ∈ V admits a derivative and f ∈ V .
This is needed in the proof we present below. About the operators we will need that R satisfies (14), B satisfies (16) and: there are α, β, As usual R and R are defined as done in (11) and (12), B as done in (17) and B as done for A in (3).

The equation Ru + Bu = f
For problem (26) we suppose and, coherently with (16) and the fact that t o = 0, Moreover, since we lean on the results of the previous sections, we will also need (23). Finally we make the further assumption In this case the solution will belong to the space

Proofs
In this subsection we present the proofs of the two theorems just stated above. The computations are very similar, so we confine to prove Theorem 4.1, being the other proof very similar and actually equal in many parts. To prove the results we consider the family of second order problems Ru + Bu + Bu = f with suitable boundary data and take the limit when goes to zero. The main difference between the two problems we are going to consider is due to some difficulties when taking the limit to get the Tricomi equations: to get the existence of the solution to problem (26) we assume (32) which is not needed to get the existence of the solution to problem (27). This difficulty is well explained in point 5 below.
For the same reason the space Y R defined above and the space X R defined in Subsection 4.3 could be different, because, a priori, we do not know anything about R u .

1.
A family of approximating problems -The idea is to consider a second order problem, like those considered in the previous section, and choose A = B where is a positive parameter which will be sent to zero. Then for > 0 we consider the family of problems (remember that for f ∈ V , ϕ ∈H + (0), ψ ∈H − (T ), η ∈ V and denote by u the solution. Then we show that the family (u ) >0 , or some sequence selected from (u ) >0 , converges in some sense to a limit u which satisfies (26). Notice that the estimate in Theorem 3.4 does not help to have some boundedness of the solution, since the constant on the right hand side is proportional to −1/2 , so we have with c depending on T and max t∈[0,T ] B(t) L(V ,V ) . Another attempt can be done multiplying the equation by 2u , following what done in [5], chap. 3, section 8.5, but, as we will see, also this will be not sufficient. Anayway this procedure gives one of the two steps needed to get an estimate on u , (41) and (42).

2.
Modifying the initial datum η -Indeed we cannot choose every possible η. Since passing from problems (33) to (26) we loose some boundary conditions, and precisely P 0 (0)η does not appear in the limit problem (26), we can modify the information which will be lost without modifying the limit problem. This will allow to get uniqueness of the solution (see point 7). First of all we state the following lemma, needed only because we are concerned with moving spaces. Now, thanks to the previous lemma, we can consider the following problems (a family of problems depending on t). Once defined the space V 0 (t) := V ∩ Ker R(t), for every fixed t ∈ [0, T ] we solve the problem Denote by w the solution of problem (35) and byw the functioñ Finally consider the functionη defined bỹ Then we will consider, instead of (33), the following family of problems: 3. Boundedness for the solutions u -Denote by u the solution of equation (37), multiply (37) by 2u and integrate between 0 and t; we will derive (41). Notice that if R were positive (and so invertible) this would be sufficient to conclude. On the contrary, in our situation this estimate is not sufficient. We will couple this estimate with (42) and, since f is differentiable, get (47). Then we get (to lighten the notation we sometimes omit to write H as subscript in the scalar product of H and V × V as subscript in the duality product between V and V ) Since R ≤ 0 and B ≤ 0 we get and in particular It is clear that this procedure cannot lead to an estimate, since R is not necessarily positive and neither non-negative. Indeed if R were positive we would have a bound both on u 2 V and on u 2 H besides to a bound on u 2 V . But since this is not possible. Anyway, consider now t = T in (39) and divide by 2 to derive This last is the first one of the two inequalities we need to get the desired estimate. Notice that in the right hand side there is a term T 0 f (s), u (s) ds which cannot be controlled. So now we proceed and make a more detailed and refined estimate using (19) with t 1 = 0, t 2 = T , t o = 0 and with Q = B and v = u (we denote for simplicity by J the operator J t o ). Then we have We estimate − Bη, u V ×V as follows, using the assumptions (29) on B: for any ϑ,θ > 0. Then we estimate f , u V ×V as follows:

Now summing (41) and (42)
Using (43) and (44) in (46) we get Now using the estimates (43) and (44) with C 4 ϑ/2 = δ = α/16 and βθ/2 =δ = α/8 by (46) we finally get (47) Then we conclude, recalling that R is self-adjoint and R ≤ 0, that there is a positive constant c and a sequence (u j ) j∈N such that (we for simplicity write u instead of u j ) (48) 4. Taking the limit on the boundary conditions concerning u -In this step we want to show that where u il the limit in (48). Alongside we prove other estimates and convergence results regarding Ru , R u and Ru . First notice that since (48) we derive that also Ru is bounded in V and then it, or one of its subsequences, is weakly converging to some z ∈ V , that is and, by (48), there is w ∈ V such that (up to select, if necessary, a sequence) Then, since also Ru is bounded in V and Integrating Ru between t 1 and t 2 , we get

Ru (s)ds
and estimating we get Then we derive that the family >0 is equibounded and equicontinuous in [0, T ] with respect to the topology of H and then R(t)u (t) >0 is weakly relatively compact in H , i.e. there is a sequence ( j ) j and Now we see that y = Ru : for We denote byR the isomorphism and for each φ we denote by ψ the function Rφ ∈ H + ⊕ H − which has a derivative in H. Then, by (51), we have We conclude that Since R + (0)u (0) = R + (0)ϕ and R − (T )u (T ) = R − (T )ψ for every > 0 we in particular get that the conditions are obviously maintained also at the limit and, by (52), we get that where u is the limit of u defined in (48).

Taking the limit in the equation -By the previous point we know that
where u = lim →0 u in V-weak and moreover that by which we derive that Ru has a derivative in V and Now the goal is to show that Ru = z and R u = w.
Before going on we recall the following simple result, needed shortly (see, e.g., Proposition 21.27 in [15]).

Lemma 4.2
Consider X and Y Banach spaces and L : X → Y linear and continuous. Then L is weakly continuous, i.e. if (x n ) n∈N is a weakly converging sequence in X to x ∈ X then (Lx n ) n∈N is a weakly converging sequence in Y to L x ∈ X.
Observe that, by (54) and the lemma just stated, we in fact have that while about u we only know that Ru ∈ H. Now, since we are assuming (32), and since by Lemma 4.2 we also have that At this point we can conclude since we have that R u makes sense and w = R u and then, by (55), (Ru ) = R u + Ru and Ru = z.
Now, taking the limit in the equation Ru = − Bu − Bu + f (always for a special sequence j converging to 0) and using again Lemma 4.2, we get that u, the limit of u j , satisfies Ru + Bu = f .

6.
Passage to the limit on the boundary conditions concerning u -The last thing to be verified to have existence is that the condition about u(0) is maintained. We know that This does not allow to conclude that u (0) converges to u(0) (in some sense) because we have not enough information in Ker R. Anyway we have enough information in (Ker R) ⊥ . Indeed we have that there is a positive constant c such that

This because
As done in the step 4 we can get that R(t)u (t) >0 is weakly relatively compact in H , i.e. there is a sequence ( j ) j such that for every φ ∈ H This in particular holds for t = 0 and then we conclude that 7. Uniqueness -Point 2 is devoted to explain how to modify the initial datum η with a functioñ η defined in (36). This modification does not affect the limit problem (26) as got in (56), but, on the other side, forces this problem to have only one solution. Indeed suppose now problem (26) has two solutions u 1 and u 2 and suppose u j ( j = 1, 2) is obtained as limit of some sequences selected from the solutions (u (1) ) >0 and (u (2) ) >0 of the following problems with η 1 and η 2 , for the moment and a priori, different, but chosen in such a way to satisfy Now call u the function u 2 − u 1 ; then u satisfies Anyway u (2) − u (1) satisfies (33) with f = 0, ψ = 0, ϕ = 0; then in particular, by (34), we and this inequality has to be satisfied by each sequence selected from (u (1) ) >0 and from (u (2) ) >0 . It is clear that the choice (58) it is not sufficient, since P 0 (0)η 1 could differ from P 0 (0)η 2 . But if in (57) we replace η 1 withη 1 and η 2 withη 2 , the suitable modifications of η 1 and η 2 defined in point 2, we get that also P 0 (0)η 1 and P 0 (0)η 2 are the same. Indeed denote by w 1 and w 2 the solutions to (35) respectively with η 1 and η 2 . Then it is easy to check that that is,η 1 =η 2 (remember that f ≡ 0 and in particular f (0) = 0) by which we conclude.
8. Convergence of all the family {u } >0 -In all previous points we have shown the existence of a sequence { j } j∈N converging to 0 such that lim j→+∞ u j = u in V-weak and u solves (26). Now for every sequence selected from the family {u } >0 one can find a subsequence converging to some function v solving problem (26). By uniqueness saw in the previous step we get that v = u. Since this holds for every sequence selected from {u } >0 we conclude that all the family {u } >0 converges to u, solution of (26).

The equation (Ru ) + Bu = f
The differences between this case and the previous one are, first of all, that we will assume R ≥ 0 and B ≤ 0 ( 6 0 ) instead of (30) and than that we will not need R : H → H and therefore we drop assumption (32).
Also in this case we will also need (23). In this case the solution will belong to the space In this case we consider, the following family of approximating problems: These problems have a unique solution thanks to assumptions (60) (see Theorem 3.5).
One can go over the steps 1-8 of Subsection 4.2 and prove Theorem 4.3, the only difference, which we report here below, is in the third step. To do that we do not need to assume (32). All the other estimates and computations are exactly as in Subsection 4.2.

Examples
In this section we focus our attention only on the "Tricomi type" equations and present First we consider the following situation: for every ξ ∈ R n and for some positive α, β. By B we denote the operator defined as in (17). For simplicity, as done in the previous section, we consider t o = 0 and, in order that (16) and (29) are satisfied, we consider b such that and denoting by ∂b ∂t the matrix whose entries are for every u ∈ H 1 0 ( ). We fix now our attention on the operator R. Consider a function and Finally for every t ∈ [0, T ] we denote and (see also (9)) r + the positive part of r , r − the negative part of r ,

For a general r depending both on t and x the problem is
and, by (32), the problem has a solution if and, due to (30), if 5. Now consider the following R : [0, T ] → L(L 2 ( )). For a fixed function r such that and define If assumption (23) is satisfied we have the existence and uniqueness of the solution of the following problem In this case the difficulty is to characterise the spacesH + (0) andH − (0). 6. Unbounded coefficients. Another admissible situation is the following. Consider two functions μ, λ ∈ L 1 ( ).
Suppose λ > 0 a.e. while μ can change sign and also be zero. Denote by |μ| a suitable function (see [11] or [12] for this detail) such that |μ| > 0 a.e. (we choose |μ| = λ where μ ≡ 0) and μ(x) < 0 and the weighted Sobolev spaces (also for these details about this spaces we refer to [11] or to [12]) Then we consider verifying for every ξ ∈ R n and for some L ≥ 1.
Consider the spaces and the operator just introduced and once defined define the operators R : L 2 , |μ| → L 2 , |μ|(·, t) , R := P + − P − , P + : L 2 , |μ| → L 2 + , |μ| the orthogonal projection, In this way R turns out to be bounded even if μ is unbounded. Then for every ϕ ∈ has a unique solution. 7. One can adapt example 5 and consider an unbounded μ as in example 6. II -The equation (Ru ) + Bu = f 8. If r = r (t) we suppose r ∈ W 1,∞ (0, T ). Suppose, coherently with (60), that we have a hyperbolic equation and get that for every f ∈ H 1 (0, T ; H −1 ( )), ϕ ∈ L 2 ( ), η ∈ H 1 0 ( ) the following problem has a unique solution: the problem is an elliptic problem with a Dirichlet (at time 0) and a Nuemann (at time T ) boundary condition and has a unique solution for every ψ ∈ L 2 ( ), η ∈ H 1 0 ( ) and f ∈ H 1 (0, T ; H −1 ( )). An interesting situation is when In this case boundary data regarding u t are missing at all and the following problem has a unique solution for every f ∈ H 1 (0, T ; H −1 ( )) and η ∈ H 1 0 ( ) where the boundary datum in × {t o } is missing. This is coherent with the Fichera conditions given in his paper [1] for the well-posedness of a boundary value problem of elliptic-parabolic type (see also, e.g" [4] for a more recent paper discussing these conditions).
In But assumption (14) could be satisfied even if r does not admit a partial derivative with respect to time.
In this case we write the problem ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ The operator R (t) will be the operator defined as (w 1 , w 2 ∈ V = H 1 0 ( )) We show here a simple example, but for further details we refer to the analogous example in [9] and [8]. In dimension 1 consider r assuming only two values, 1 and −1.
In this case d dt R(t)w 1 , w 2 L 2 (a,b) = d dt
10. One can adapt examples 5 and 7 for the equation r (x, y, t)u t (y, t) t dy + Bu = f .
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