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The Dirichlet problem for nonlocal operators

In this note we set up the elliptic and the parabolic Dirichlet problem for linear nonlocal operators. As opposed to the classical case of second order differential operators, here the"boundary data"are prescribed on the complement of a given bounded set. We formulate the problem in the classical framework of Hilbert spaces and prove unique solvability using standard techniques like the Fredholm alternative.


Introduction
Given an open and bounded set Ω ⊂ R d and functions f : Ω → R and g : ∂Ω → R, the classical Dirichlet problem is to find a function u : Ω → R such that (1.1b) A more general problem is to solve, instead of equation (1.1a), the equation where A(x) = (a ij (x)) is a d × d-matrix which is uniformly positive definite and uniformly bounded. When working in Hilbert spaces a standard assumption is f ∈ H −1 (Ω) = (H 1 0 (Ω)) * and g ∈ H 1 (Ω). H 1 (Ω) denotes the Hilbert space of all L 2 (Ω)-functions with a distributional derivate in L 2 (Ω) and H 1 0 (Ω) is the closure of C ∞ c (Ω) with respect to the norm of H 1 (Ω). The Riesz representation Theorem resp. the Lax-Milgram Lemma imply that there is a unique function u ∈ H 1 (Ω) such that u − g ∈ H 1 0 (Ω) and for every v ∈ H 1 0 (Ω) (A(·)∇u, ∇v) L 2 (Ω) = f, v .
(1. 3) This is one way to solve the Dirichlet problem. Note that the bilinear form on the left-handside of (1.3) is not necessarily symmetric. An important further extension is given when one considers additional terms of lower order in (1.2).
The aim of this article is to provide a similar set-up for the Dirichlet problem associated with nonlocal operators. Our main objects are nonlocal operators L of the form where k : R d × R d → [0, ∞] is measurable. Our focus is on kernels k with the following properties: (i) k is not necessarily symmetric, (ii) k might be singular on the diagonal and (iii) k is allowed to be discontinuous. These properties imply that Lu cannot be evaluated in the classical sense at a given point x ∈ R d even if u is smooth.
Assume Ω ⊂ R d is open, bounded and f ∈ L 2 (Ω). Given a function g : ∁ Ω → R, we want to find a solution u : R d → R to the problem Lu = f in Ω (1.5a) u = g on ∁Ω. (1.5b) In order to do so, we need to answer the following questions: Which is a natural space for the data g and for the solution u? In which sense is equation (1.5a) to be interpreted? Note that, opposed to the classical case of second order differential operators, here the data g are prescribed on the complement of a given bounded set. The term "complement value problem" would be more appropriate for (1.5) than "boundary value problem". (u(x) − u(y))k(x, y) dy .
In that case it would not make sense to describe data on ∁ Ω. For a large class of kernels one could study associated boundary value problems with boundary data g : ∂Ω → R. Note that such operators appear as generators of so-called censored processes. We do not study these cases in this article. ♣ In this work we establish a Hilbert space approach to the nonlocal Dirichlet problem (1.5). We prove results on existence and uniqueness of solutions under various assumptions on the kernels k : R d × R d → [0, ∞] which we assume to be measurable. The symmetric and anti-symmetric parts of k are defined by k s (x, y) = 1 2 (k(x, y) + k(y, x)) and k a (x, y) = 1 2 (k(x, y) − k(y, x)) .
Clearly, |k a (x, y)| ≤ k s (x, y) for almost all x, y ∈ R d . Throughout this article we assume that the symmetric part of the kernel satisfies the following integrability condition: x →R d 1 ∧ |x − y| 2 k s (x, y) dy ∈ L 1 loc (R d ). (L) We call a kernel k integrable if, for every x ∈ R d the quantity´R d k s (x, y) dy is finite and the mapping x →´R d k s (x, y) dy is locally integrable. We call a kernel non-integrable if it is not integrable in the sense above. In this work we deal with integrable and non-integrable kernels at the same time. In Section 6 we provide several characteristic examples. A simple integrable example is given by k(x, y) = 1 B 1 (x − y), a simple non-integrable example by k(x, y) = |x − y| −d−1 , both kernels being symmetric.
For a certain class of kernels k the Lax-Milgram Lemma turns out to be a good tool whereas for another class the Fredholm alternative is more appropriate. Rather than going through all assumptions in detail, for this introduction let us restrict ourselves to an example in the simple setting where Ω equals the unit ball B 1 ⊂ R d .
The part involving |x − y| −d−β can be seen as a lower order perturbation of the main part of the kernel resp. integro-differential operator produced by |x − y| −d−α 1 C 1 (x − y). We discuss this example in detail in Section 6. Define complement data g : R d → R by where γ is an arbitrary real number satisfying γ > α−1 2 . Note that g may be unbounded. Let f ∈ L 2 (Ω) be arbitrary. For such a kernel k and such data g and f our results imply that the Dirichlet problem (1.5) for Ω = B 1 ∈ R d has a unique variational solution u.
Another interesting example to which our theory applies, is given by under certain assumptions on the functions b : R d → (0, ∞) and α : R d → (0, 2), see Example 14 in Section 6.
In Corollary 5.5 we provide an existence result for parabolic equations (see (5.9)) where the operator contains a time-dependent kernel k t and where the complement data and the initial value are prescribed by a single function h(t, x). This result covers the case k t = k and h(t, x) = g(x) with g and k as above.
Let us comment on related results in the literature. It is remarkable that, although the questions and the framework of our studies are quite basic, the results have not been established yet. One reason might be that integro-differential operators not satisfying the transmission property have been studied only recently when considered in bounded domains. On the other hand, the proof of the weak maximum principle for nonsymmetric nonlocal operators, Theorem 4.1, turns out to be quite tricky.
To our best knowledge, the first profound study of a nonlocal Dirichlet problem for integrodifferential operators violating the transmission property is provided in [6, Ch. VII] using balayage spaces and the method of Perron, Wiener and Brelot. A detailed study of nonlocal Dirichlet problems including a Hilbert space approach can be found in [13]. Section 6 of [13] addresses the questions of this article but assumes the bilinear forms to be symmetric. As the proofs show, symmetry of k simplifies the situation greatly. Symmetry is not assumed in the previous sections of [13] but there, the symbol (i.e. the generator) is used which would lead to strong assumptions on the kernel k. Note that in our framework, typically the integrodifferential operator cannot be evaluated pointwise, even when applied to smooth functions.
The Dirichlet problem for nonlocal operators is studied for fully nonlinear problems in [4] using viscosity solutions. There, the complement data are chosen independently from the kernels which is very different from our approach where, for every k there is an appropriate function space for the data g.
Variational solutions to nonlocal problems have been already considered by groups working in the theory of peridynamics, e.g. in [1,8,5,20,21]. This theory describes a nonlocal continuum model for problems involving discontinuities or other singularities in the deformation. The corresponding integration kernels are often integrable but non-integrable kernels are considered too. What we call "complement data" in this article is called "volume constraints" in articles on peridynamics. An important tool in the articles above are appropriate Poincaré-Friedrichs inequalities. It occurs that Lemma 2.7, in the case of scalar functions, is more general than corresponding Poincaré-Friedrichs inequalities in the aforementioned articles. A major difference of our approach is that we study general (non-symmetric) kernels and non-zero complement data together with appropriate function spaces.
We do not at all discuss regularity of solutions. Regularity up to the boundary is studied carefully in [22] for the case of the fractional Laplacian.
Results on nonsymmetric nonlocal Dirichlet forms are related to our work, see our discussion of ( K) below. We refer the reader to [14], [17], [11] and [23]. We believe that our examples can be helpful when studying Hunt processes generated by nonsymmetric nonlocal (semi-)Dirichlet forms.
The article is organized as follows. In Section 2 we define the function spaces needed for our approach and explain their basic properties. We also define what a solution of (1.5) shall be. Section 2.3 is dedicated to nonlocal versions of the Poincaré-Friedrichs inequality. In Section 3 we prove a Gårding inequality, comment on the sector condition and apply the Lax-Milgram Lemma to the nonlocal Dirichlet problem. Section 4 is devoted to the weak maximum principle for integro-differential operators in bounded domains. This tool is applied when using the Fredholm alternative in Section 4.2. Our existence and uniqueness results are extended to the parabolic Dirichlet problem in Section 5. Finally, in Section 6 we provide many detailed examples of kernels k and discuss their properties. Although, formally, this section could be omitted, it presumably is one of the most important one for every reader. A short index with all conditions used in this work is given at the end of Section 6 on p. 28.

Function spaces and variational solutions
In this section we derive a weak resp. variational formulation of (1.5). We start with the definition of the relevant function spaces.

Function spaces.
Definition 2.1 (Function spaces). Let Ω ⊂ R d be open and assume that the kernel k satisfies (L). We define the following linear spaces: A seminorm on V (Ω; k) is given by [v, v] V (Ω;k) . Let us emphasize that Ω = R d is allowed in this definition. In this case we write V (R d ; k) = H(R d ; k) and a norm -which is obviously induced by a scalar product -on this space is defined by a) It is clear from this definition that and H(R d ; k) ⊂ V (Ω; k). Moreover, if g ∈ V (Ω; k) and g = 0 a.e. on ∁ Ω, then g ∈ In this case we recover the standard fractional Sobolev-Slobodeckij spaces and the norms are equivalent to the standard norms given on these spaces. More precisely, we obtain If additionally α = 1 then The following example illustrates that finiteness of the seminorm on V (Ω; k) requires some regularity of the function across ∂Ω: Let Ω = B 1 (0), α ∈ (0, 2) and define k : In this case, H(R d ; k) coincides with the fractional Sobolev space H α/2 (R d ) and we have We show that g ∈ V (Ω; k) if and only if β > α−1 2 . Since α is fixed we omit the factor α(2 − α). Then x. Then we may estimatê Then, by definition of · H(R d ;k) and the completeness of L 2 (R d ), (f n ) converges to some f in the norm of L 2 (R d ). We may chose a subsequence f n k that converges a.e. to f . Then v n k converges a.e. on Since (v n ) is a Cauchy sequence (and hence bounded) in Another application of Fatou's Lemma shows that This shows that f n k − f H(R d ;k) → 0 for k → ∞ for the subsequence (n k ) chosen above and thus f n − f H(R d ;k) → 0 as n → ∞, since (f n ) was assumed to be a Cauchy sequence. The completeness of H(R d ; k) is proved. The completeness of H Ω (R d ; k) follows immediately.
The mapping I 2) is isometric due to the definition of the norm in H(R d ; k). Having shown the completeness of H(R d ; k) we obtain that I(H(R d ; k)) is a closed subspace of the Cartesian product on the right-hand side of (2.2). This product is separable, which implies (cf. [ In order to prove well-posedness of this expression and that the bilinear form is associated to L, we need to impose an condition on how the symmetric part of k dominates the antisymmetric part of k. We assume that there exist a symmetric kernel k : for all u ∈ H(R d ; k), and at the same time A natural choice is k = k s because in this case ( K 1 ) trivially holds and k(x, y) = 0 implies k a (x, y) = 0. Assumption ( K) would then reduce to the condition extends to a regular lower bounded semi-Dirichlet form. Note that our assumption ( K) is weaker and thus we can extend [23, Thm 1.1], see Lemma 2.4. In Section 6 we provide an example illustrating the difference between ( K) and (K).
As mentioned above, our proof is an extension of the proof of [23, Theorem 1.1].
Splitting k in its symmetric and antisymmetric part yields The first integral is finite due to (L). In order to show the integrability of the second integrand we use ( K) with A = max(A 1 , A 2 ) and the Cauchy-Schwarz inequality: and that all expressions in this equality are well-defined. In particular, by dominated convergence lim

Now let us prove the continuity of
Again by the symmetry of k s and by ( K) we obtain This shows that E k is a continuous bilinear form on H(R d ; k) and on H Ω (R d ; k).
Finally, we are able to provide a variational formulation of the Dirichlet problem (1.5) with the help of the bilinear form E k : In the subsequent sections we will show how to solve this problem. There exists a constant C P > 0 such that for all u ∈ L 2 Ω (R d ) This inequality appears as an assumption, explicitly or implicitly, in all of our existence results. Below, we provide sufficient conditions on k s for (P) to hold. Note that (P) may hold for integrable kernels as well as for non-integrable kernels k, see the table in Section 6.
Assume that there is a symmetric, a.e. nonnegative function L ∈ L 1 (R d ) satisfying the following properties: |{L > 0}| > 0 and there is c 0 > 0 such that for all u ∈ L 2 (Ω) Then the following Poincaré-Friedrichs inequality holds: There is The example in [3,Remark 6.20] shows that Lemma 2.7 fails to hold if one replaces the domain of integration R d × R d by Ω × Ω in (2.5) and (2.6).
For the proof of the Poincaré-Friedrichs inequality, we need the following technical Lemma taken from [9, Lemma 10].
Lemma 2.8. Let q ∈ L 1 (R d ) be nonnegative almost everywhere and let supp q ⊂ B ρ (0) for some ρ > 0. Then for all R > 0 and all functions u: Proof of Lemma 2.7. Let L satisfy the assumptions of the lemma. Without loss of generality 0 ∈ Ω (otherwise shift Ω). Furthermore, we may assume that there is By the properties of L we have , which implies that we may find δ > 0 (depending on L) such that L 1 > 0 on B δ (0). By the property of the convolution there is m ∈ N depending on L and Ω such that L m > 0 on B R (0). Let u ∈ H Ω (R d ; k). Then we may estimate Iterated application of Lemma 2.8 (with ρ ′ = 2 m ρ and q = L j , j = m − 1, . . . , 0) yields , (2.8) and the assumption (2.5) imply This finishes the proof of Lemma 2.7.
For non-integrable kernels k we have the following Poincaré-Friedrichs inequality: Let Ω ⊂ R d open and bounded. Let k : R d × R d → R be measurable and nonnegative almost everywhere. We assume that for some α ∈ (0, 2), some λ > 0 and all Given α 0 ∈ (0, 2) and α ∈ [α 0 , 2), the constant C P can be chosen independently of α.
The proof of the main assertion is simple. The statement about the independence of C P on α is proved in [18, Theorem 1].

Gårding inequality and Lax-Milgram Lemma
In this section we discuss basic properties of the bilinear form E k which can be used in order to prove solvability of the Dirichlet problem. First, we establish a Gårding inequality under the conditions (L) and ( K). Then we comment on the sector condition. In a second subsection we show that if, in addition, the Poincaré-Friedrichs inequality and a certain cancellation property hold, the bilinear form E k is positive definite and coercive. This allows to establish a first existence result with the help of the well-known Lax-Milgram Lemma, Theorem 3.5.
if we choose ε sufficiently small such that Aε < 1 4 and then γ = γ(A) sufficiently large.
Remark 3.2. As the above proof shows, assumption ( K 2 ) is tailor-made for an estimate which shows that E ka is dominated by E ks . Thus, another consequence of (L) and ( K) is the estimate . This observation implies that the bilinear forms (E k , H(R d ; k)) and (E k , H Ω (R d ; k)) satisfy a sector condition under an additional assumption, see Proposition 3.6 below. ♣

3.2.
Application of the Lax-Milgram Lemma. To verify that the bilinear form E k is positive definite, we assume the following cancellation condition: As the proof below shows, assumption (C) can be relaxed. It is sufficient to assume with C P as in (P). The bilinear form E k would still be coercive. ♣ There are many cases for which condition (C) holds. If k a (x, y) depends only on x − y, then for every x ∈ R d and every ε > 0 one obtains´∁ Bε(x) k a (x, y) dy = 0 which trivially implies (C). But there are also many interesting cases for which condition (C) is not satisfied, see Section 6. Proof. In Lemma 2.4 it was shown that E k is a continuous bilinear form on H Ω (R d ; k). First, we show that (C) implies that E k is positive definite. Let u ∈ H Ω (R d ; k). Observe that k = k s + k a and for every ε > 0 k a (x, y) dy dx .

From (C) we obtain
Hence, i.e. E k (u, u) ≥ 0 for all u ∈ H Ω (R d ; k). By (P) and (3.2) which shows that E(u, u) is coercive.
Next, we show that the Dirichlet problem with suitable complement data g has also a unique solution.
Let Ω ⊂ R d be open and bounded and let k satisfy (L), (P) and (C). Assume further that there is k such thaẗ for all u ∈ V (Ω; k) and ( K 2 ) holds true. Then (D) has a unique solution u ∈ V (Ω; k). Moreover,

3)
where C = C(C P , A 1 , A 2 ) is a positive constant.
Proof. To prove the theorem we show that under the above assumptions on g the problem (D) can be transformed into a problem of the form (D 0 ). If u ∈ H Ω (R d ; k) is a solution to then u = u + g belongs to V (Ω; k) and solves (D). In order to apply Proposition 3.4 to (3.4) it remains to show that Lg = E k (g, ·) ∈ H * Ω (R d ; k). We have Since ϕ = 0 a.e. on ∁Ω an application of the Cauchy-Schwarz inequality yields The term I 2 can be estimated as follows: By ( K ′ 1 ) and ( K 2 ) This shows the continuity of E(g, ·) : H Ω (R d ; k) → R and hence (3.4) has a unique solution u ∈ H Ω (R d ; k).
In order to prove estimate (3.3) we apply u ∈ H Ω (R d ; k) as test function and obtain where the second term on the right-hand side is non-negative due to (C). Hence, The Young inequality and the fact that v = 0 a.e. on ∁Ω imply for ε > 0, to be specified later, Similarly, using ( K ′ 1 ) and ( K 2 ) we obtain Altogether we obtain Applying the Poincaré-Friedrichs inequality (P), and choosing ε = 1 16C P we deducë where c 1 ≥ 1 depends on C P and c 2 ≥ 1 depends on A and C P . Since H Ω (R d ; k) ⊂ V (Ω; k) and u = u + g the assertion (3.3) follows.
As mentioned in Remark 3.2, condition (C) ensures the sector condition in the sense of [16]: Proposition 3.6. Let k satisfy (L), ( K) and (C). Then the bilinear forms (E k , H(R d ; k)) and (E k , H Ω (R d ; k)) satisfy the weak sector condition. If, in addition, k satisfies (P), then the bilinear form (E k , H Ω (R d ; k)) satisfies the strong sector condition.

Weak maximum principle and Fredholm alternative
The goal in this section is to prove existence and uniqueness when the bilinear form E k is no longer positive definite. To this end we establish a weak maximum principle implying that the homogeneous equation has only the trivial solution. Then we apply Fredholm's alternative.

4.1.
Weak maximum principle. In this subsection we prove a weak maximum principle when the kernels k exhibit a non-integrable singularity at the diagonal. We need to impose two different assumptions on the class of admissible kernels.
We assume that for some α ∈ (0, 2), some λ > 0 and all u ∈ L 2 (R d ) the estimatë holds true. Further, we assume that there is D > 1 such that for almost every x, y ∈ R d |k a (x, y)| ≤ D −1 k s (x, y) (4.1) Condition (E α ) requires some minimal singularity of k s at the diagonal. It is not restrictive when considering non-integrable kernels. Concerning condition (4.1) note that, by definition, the inequality |k a (x, y)| ≤ k s (x, y) holds for almost every x, y ∈ R d . Condition (4.1) is satisfied by several examples, e.g for if 0 < β < α/2 and g ∞ ≤ 1 2 , cp. Example (11) in Section 6. But there are also examples which violate the condition, e.g.
Under the above conditions we can prove the following weak maximum principle: Then sup Ω u ≤ 0.

Remark 4.2.
As the proof reveals, it is possible to weaken assumption (4.1) significantly because the estimate under consideration is needed only in an integrated sense. However, it seems challenging to provide a simple appropriate alternative to (4.1). ♣ Proof. We apply a strategy which is often used in the proof of the weak maximum principle for second order differential operators (e.g. [12]). We first show that u attains its supremum on a set of positive measure. In a second step we show that this leads to a contradiction if the supremum is positive.
We choose as test function v = (u − k) + , where 0 ≤ k < sup Ω u. Then v ∈ H Ω (R d ; k) and and since the second term on the left hand side is positive

Now by Cauchy-Schwarz and the assumptions on k
By (E α ) and since v = 0 on ∁Ω, the Sobolev and the Hölder inequality imply that there is a (If d ≤ 2 the critical exponent may be replaced by any number greater than 2.) Thus This inequality is independent of k and therefore it holds for k ր sup Ω u. Therefore u must attain its supremum on a set of positive measure. This completes the first step of the proof.
We now derive a contradiction. Without loss of generality we may assume sup Ω u = 1.
We want to use v as a test function in (4.2) but it is not clear whether v belongs to H Ω (R d ; k).
Thus we look at approximations and define for small The function v ε is an admissible test function. However, in order to simplify the presentation, we use v instead of v ε and postpone this issue until the end of the proof. Plugging v into (4.3), we obtain This is equivalent tö An application of the Young inequality leads tö Using v = 0 on ∁Ω, v ≤ 1 and that k 2 a (x,y) ks(x,y) is symmetric, the right hand side can be estimated from above as follows: where we have applied Lemma 4.3 and (4.1). Now, we choose θ = D+1 2 such that θ 2 D < 1.

Combining the above estimate and (4.4) leads tö
for some positive constant c 1 = c 1 (D). Next, we want to estimate the left-hand side from below. We apply the inequality which holds for positive reals a, b, to a = 1 − v(y) and b = 1 − v(x). Thus we obtain Due to condition (E α ) we can apply the Sobolev inequality and obtain where c 2 ≥ 1 and w = log(1 − v). Recall that, in fact, we have proved w ε L 2d/(d−α) ≤ c 2 A|Ω| for w ε = log(1 − v ε ) and every ε ∈ (0, 1 2 ), where c 2 is independent of ε. By Fatou's lemma, this contradicts the fact that v = u + attains is supremum 1 on a set of positive measure. The proof is complete.  > 1 and a, b ∈ [0, 1). Then Proof. It is sufficient to establish assertion (4.5) since it implies (4.6). For the proof of (4.5) it is sufficient to assume a ≤ b. Assume θ > 1 and 0 ≤ a ≤ b < 1. Then, for t = b a 0 ≤ a < 1 ≤ t < 1 a and inequality (4.5) reads (4.7) which proves (4.7).

4.2.
Fredholm alternative. The aim of this subsection is to prove existence and uniqueness of solutions to (D) without assuming positive definiteness of the bilinear form E k , i.e. without assuming (C). As for the weak maximum principle we assume the kernel k to satisfy (E α ) and (4.1). Note that (E α ) implies (P) by Lemma 2.9.
We prove the following well-posedness result: Let Ω ⊂ R d be open and bounded. Let f ∈ H * Ω (R d ; k) and let k satisfy (L), (K), (4.1) and (E α ). Then the Dirichlet problem (D) has a unique solution u ∈ V (Ω; k). Moreover, there is a constant C = C(C P , A, D) > 0 such that Proof. We use the Fredholm alternative (see e.g. [10]).
Step 1: We will use (E α ) to show that the embedding H Ω (R d ; k) ֒→ L 2 Ω (R d ) is compact. Since the embedding L 2 (Ω) ֒→ H * Ω (R d ; k) is continuous we obtain then the compactness of the embedding where we used (E α ). Therefore A is bounded in H α/2 (B) and thus pre-compact in L 2 (B). By the definition of H Ω (R d ; k) we know u = 0 on ∁Ω and thus the set A is also pre-compact in L 2 Ω (R d ) and in H * Ω (R d ; k).
Step 2: Existence and uniqueness of (D 0 ). By Lemma 3.1 the bilinear form is coercive for some γ = γ(A) > 0 and therefore there is a unique solution u ∈ H Ω (R d ; k) to the problem Moreover, due to Lemma 3.1 the solution u satisfies This estimate together with Step 1 shows that the operator K : , which maps the inhomogeneity f to the solution u ∈ H Ω (R d ; k) ⋐ H * Ω (R d ; k) of (4.9), is a compact operator. Fredholm's theorem in combination with the weak maximum principle Theorem 4.1 shows that (D 0 ) has a unique solution u ∈ H Ω (R d ; k).
Step 3: The well-posedness of (D) follows in the same way as in the proof of Theorem 3.5. It remains to prove the estimate (4.8). Let u be the solution of (D). We apply v = u − g ∈ H Ω (R d ; k) as test function: As in the proof of Theorem 3.5 we may estimatë Due to v = 0 on ∁Ω, the last term can be estimated as follows: Hence, after choosing ε appropriately, where c 1 , c 2 > 0 depend on A. This implies (4.8).

Parabolic Problem
In this section we prove well-posedness of the initial boundary value problem   where Ω ⊂ R d is a bounded domain, Q T = (0, T ) × Ω and f ∈ L 2 (Q T ), u 0 ∈ L 2 (Ω). W (0, T ) is a Banach space endowed with the norm We consider the Gelfand triplet , where the two embeddings are continuous and each space is dense in the following one.
where we assume that k t (x, y) = a(t, x, y)k(x, y) with a measurable function a : R × R d × R d → 1 2 , 1 which is symmetric with respect to x and y, and a measurable kernel k : Definition 5.1 (Parabolic variational formulation). Let u 0 ∈ L 2 (Ω) and f ∈ L 2 (0, T ; H * Ω (R d ; k)). We say that u ∈ W (0, We refer to Definition 5.4 for the corresponding problem with nonzero complement data. b) The initial condition (5.5b) is well-defined due to the embedding W (0, T ) ֒→ C([0, T ]; L 2 (Ω)) , see e.g. [24]. c) Remark 2.6 holds analogously. ♣ A well-known parabolic analog (e.g. [25,Corollary 23.26]) of the Lax-Milgram Lemma asserts that there is a unique solution to (5.1) provided the bilinear form E(t; ·, ·) : B × B → R has the following properties: there is M > 0 such that for all t ∈ (0, T ) and u, v ∈ B we have there are m > 0 and L 0 ≥ 0 such that for all t ∈ (0, T ) and u ∈ B we have The measurability condition (5.6a) follows immediately from the measurability of k t . Due to the special structure (5.4) of k t we may refer to Lemma 2.4 for the proof of the continuity condition (5.6b) and to Lemma 3.1 for the Gårding inequality (5.6c). Therefore we obtain the following result: Theorem 5.3 (Well-posedness of the parabolic problem). Assume that k t is of the form (5.4), where k satisfies (L) and ( K). Then there is a unique solution u ∈ W (0, T ; H Ω (R d ; k)) of (5.1).
Moreover, for all u 0 ∈ L 2 (Ω) and f ∈ L 2 (0, T ; H * Ω (R d ; k)) there is a constant C > 0 such that The constant C depends on the constants M, m and L 0 in (5.6b),(5.6c).
In order to consider non-homogeneous complement data we assume that these values are prescribed by a function g : For the functional analytic treatment of this problem we need a modification of the space V (Ω; k) that turns this linear space into a Hilbert space.
One possibility is to define From the proof of Lemma 2.3 it can be seen that this space is a separable Hilbert space with the inner product defined by Definition 5.4 (Parabolic variational formulation). Let f ∈ L 2 (Q T ) and g ∈ W (0, T ; H(Ω; k)).
We say that u ∈ W (0, T ; H(Ω; k)) is a solution of   Corollary 5.5. Assume that k t is of the form (5.4), where k satisfies (L) and ( K). Then there is a unique solution u ∈ W (0, T ; H(Ω; k)) of (5.9).
Let us emphasize that the conditions on the complement data g are far from being optimal. On the one hand, it would be desirable to relax the condition on the spatial decay imposed by g(t, ·) ∈ L 2 (∁Ω) as in Section 3. On the other hand, one could try to follow the program as in the case of second order parabolic equations (e.g. [15]) in order to relax the regularity assumptions of g with respect to t and x.

Examples of kernels
We provide several examples of kernels k : R d × R d → [0, ∞] to which the theory above can be applied. Recall that all kernels studied in this work satisfy assumption (L). Further, we distinguish two cases. We call a kernel k integrable if, for every x ∈ R d the quantitý R d k s (x, y) dy is finite and the mapping x →´R d k s (x, y) dy is locally integrable. We call a kernel non-integrable if it is not integrable in the sense above. At the end of this section we list all examples together with their corresponding properties.
6.1. Integrable kernels. Let us start with a simple observation. Every kernel with the property that the antisymmetric part is of the form k a (x, y) = g(x − y) for some function g satisfies the assumption (C). This follows from the fact that for (1) k(x, y) := ½ B 1 (x − y). The kernel is obviously symmetric. Thus it satisfies (C). It also satisfies the Poincaré-Friedrichs inequality (P) as shown in Section 2.3.
Since k depends only on x − y, condition (C) holds. Concerning the Poincaré-Friedrichs inequality (P), k is not different from example (1). (4) This example is more general than Example 3. Set k(x, y) : where the set C is defined by C = {h ∈ R d | h |h| ∈ I} and I is an arbitrary nonempty open subset of S d−1 . If I is of the form I = B r (ξ) ∩ S d−1 for some ξ ∈ S d−1 and some r > 0, then C is a cone. In any case, we obtain In the examples above, k(x, y) depends only on x − y. As a result, one can choose L(z) = k(0, y − x) in the condition (2.5). Let us look at examples where this is not possible.
(5) k(x, y) := g(x, y)1 B 1 (x − y), where g is any measurable bounded function satisfying g ≥ c almost everywhere for some constant c > 0. Note that g does not need to be symmetric. Then . Condition (C) does not hold in general but (K) holds because k s (x, y) ≥ c1 B 1 (x− y) which allows us to apply the Poincaré-Friedrichs inequality (P) choosing L(z) = c1 B 1 (z) in (2.5).
By construction k a is antisymmetric and satisfies condition (C). Thus k is not a function of x − y but still satisfies (C). Conditions (K) and (2.5) hold true, too. where Similar results hold true for R d replaced by a bounded domain [7,18]. (9) Let I be an arbitrary nonempty open subset of S d−1 with the property I = −I. Set C = {h ∈ R d | h |h| ∈ I} and k(x, y) := |x − y| −d−α 1 C (x − y). Again, k is symmetric and satisfies (L). It turns out that k is comparable to example (8) in the sense of Lemma 2.9. The only difference is that the constant λ depends on I.
. This example is different from example (9) because k is not symmetric anymore. The symmetric and antisymmetric parts are given by Lemma 2.9 can still be applied but conditions (K) and ( K) do not hold. Condition (C) does hold, though.
(11) Assume 0 < β < α 2 and g : Additionally, we assume that k is nonnegative. This property does not follow in general under the assumptions above. However, for every choice of K there are many admissible cases with inf g = −K. We obtain k s (x, y) Since k s is nonnegative, we can apply Lemma 2.9 and (P) holds. Further ( K) is satisfied with k(x, y) = |x − y| −d−α ½ B 1 (x − y) and A 1 = 1. Conditions (C) and (K) hold for some but not for all choices of g.
The following example is an extension and, at the same time, a special case of Example (11). Example (12) shows that our condition ( K) is indeed a relaxation of (K) or [23, (1.1)].
Note that the kernel k depends only on x − y and is symmetric but condition (L) is not obvious. Using integration in polar coordinates one can show that there is C ≥ 1 such that for α = α ′ − (1/b − 1), h(z) = k(x, x + z) and every r ∈ (0, 1) Thus α is the effective order of differentiability of the corresponding integro-differential operator. In [9] the comparability of the quadratic forms needed for Lemma 2.9 is established. From the point of view of this article the kernel k is very similar to the kernel |x − y| −d−α 1 B 1 (x − y). Of course, one can now produce related nonsymmetric examples. (14) The following example is taken from [11], [23]. It provides a nonsymmetric kernel k with a singularity on the diagonal which is non-constant. Assume 0 < α 1 ≤ α 2 < 2 and let α : R d → [α 1 , α 2 ] be a measurable function. We assume that α is continuous and that the modulus of continuity ω of the function α satisfies (ω(r)| log r|) 2 r 1+α 2 dr < ∞ .
Let us slightly modify the example and look at k ′ (x, y) = ½ B R (x − y)k(x, y) for some R ≫ 1. Then conditions (L), (K) and (P) still hold true for k ′ .
Although we have provided several different examples, our class is still rather small. All examples of non-integrable kernels from above relate, in one way or another, to the standard kernel |x − y| −d−α for some α ∈ (0, 2) and the Sobolev-Slobodeckij space H α/2 (R d ). We could also study examples with kernels which relate to a generic standard kernel |x − y| −d φ(|x − y| 2 ) −1 where φ itself can be chosen from a rather general class of functions, e.g. the class of complete Bernstein functions. Condition (L) can be found on p. 2. Conditions ( K) and (K) on p. 7. Condition (P) on p. 9. Condition (C) on p. 12. Condition (E α ) on p. 16.