1 Solving the Dirichlet problem constructively

The Dirichlet problem is of central importance in both applied and abstract potential theory. We prove the (perhaps surprising) result that the existence of solutions in the general case is an essentially nonconstructive proposition: there is no algorithm which will actually compute solutions for arbitrary domains and boundary conditions. A corollary of our results is the non-existence of constructive solutions to the Navier-Stokes equations of fluid flow. But not all the news is bad: we provide reasonable conditions, omitted in the classical theory but easily satisfied, which ensure the computability of solutions. 2010 Mathematics Subject Classification 03F60 (primary); 46S30 (secondary)


Introduction
It is a well-known theorem of classical mathematics that the following Dirichlet problem has a (unique) solution: Problem 1 (The Dirichlet boundary-value problem) Given an open, bounded integrable set Ω in R n and a uniformly continuous boundary condition f : ∂Ω → R, find a function u such that (1) ∆u = 0 on Ω, u(x) = f for all x ∈ ∂Ω.
Less well-known is the fact that, despite there being existence proofs of solutions which at first sight appear to be algorithmic, the theorem has no proof which is entirely constructive-in other words, there is no algorithm which will compute solutions to the Dirichlet problem for arbitrary open, bounded, integrable domains with arbitrary uniformly continuous boundary data.Accordingly, since Dirichlet boundary value problems are a special case of the Navier-Stokes equations of fluid flow, there is no • edge coherent if x ∈ Ω whenever x ∈ Ω and ρ (x, ∂Ω) > 0; • approximated internally by compact sets if for each ε > 0 there exists a compact-that is, complete and totally bounded-set K ⊂⊂ Ω such that if x ∈ Ω − K , then ρ (x, ∂Ω) < ε.
In keeping with the literature we adopt the usual conventions concerning the notation ∂Ω, C n (Ω), C n 0 (Ω), H 1 (Ω) and H 1 0 (Ω): C n (Ω) is the space of real-valued functions n times uniformly differentiable on compact subsets of Ω; C n 0 (Ω) is the space consisting of elements of C n (Ω) with compact support well contained in Ω; H 1 (Ω) is the space of square-summable functions whose weak derivatives are also in L 2 (Ω); and H 1 0 (Ω) is the completion of C 1 0 (Ω) in H 1 (Ω).In [6], Proposition 23 shows that if Ω is approximated internally by located sets, then it is edge coherent; and Proposition 24 says that if Ω is edge coherent, totally bounded, and has totally bounded boundary, then it is approximated internally by compact sets.
An important property applicable to an open subset of R N is the pointwise exterior cone condition: for each x ∈ ∂Ω, there exist r, θ > 0 and a right circular cone C with vertex x, vertex angle θ , and height r such that C ∩ Ω = {x}.To pass from this to the (uniform) exterior cone condition, we need the numbers r, θ to be constructed independent of the point x ∈ ∂Ω.Proposition 3.20 of [18] shows that if Ω ⊂ R N is edge coherent, totally bounded, and satisfies the uniform exterior cone condition, then Ω is approximated internally by compact sets.
The details of constructive integration theory are presented in [4] (Chapter 6), to which we refer the reader.However, it is convenient to present some aspects of that theory here.
Let X be a locally compact metric space, µ a positive measure on X , and L 1 (µ) the space of µ-integrable functions on X .By a complemented set in X we mean an ordered pair S = S 1 , S 0 of subsets of X such that ρ (x, y) > 0 for all x ∈ S 1 and y ∈ S 0 ; if the resulting characteristic function χ S : S 1 ∪ S 0 → {0, 1}, defined by is integrable, we call S an integrable set, with measure µ(S) ≡ χ S .We say that the complemented set S is compact (respectively, closed) if S 1 is compact (respectively, closed).When K is a compact subset of X , we write K for the complemented set (K, −K), where −K ≡ {x ∈ X : ρ (x, K) > 0} is the metric complement of K ;4 if, moreover, K is an integrable set, then we call it-or, loosely, K -an integrable compact set in X .
We say that a compact set K ⊂ X is strongly integrable if the following holds: In that case, K is integrable, and µ(K) = c.Here is the fundamental result about approximating integrable sets by strongly integrable ones.(Note that 'K < A' means that χ K ≤ χ A on a full set.) Theorem 2 Let µ be a positive measure on a locally compact metric space X, and let A be an integrable set with positive measure.Then for each ε > 0, there exists a strongly integrable compact set K such that K < A and µ(A − K) < ε.
The Dirichlet problem can be re-stated as: Under reasonable conditions on f , (2) is equivalent to (1), in the sense that existence of the solution to either version entails the existence of the solution in the other; see [14, p. 131].
By a weak solution to the Dirichlet problem (2) we mean a function u in H 1 0 (Ω) such that for all v ∈ H 1 0 (Ω).Before going into more detail, we remind the reader of some important but related issues.First is the (constructive) stability of solutions: Theorem 2 of [9] states that if u f is the weak solution of the Dirichlet problem (2), then for all functions g in L 2 (Ω), where γ is the constant in Poincaré's inequality.From this it follows that for a given function f in L 2 (Ω) the Dirichlet problem (2) has at most one weak solution.
Next, there is a group of problems which are constructively equivalent to the existence of weak solutions to (2) (for details, see [9, pp. 658-661] and [8, pp. 1159-1160]): • The total boundedness in the L 2 norm of the set • The uniform continuity in the double norm on H 1 0 (Ω) * of the mapping λ v → v(ξ) from S * to R.Here λ v denotes the bounded linear functional u → u, v on Here, by extending each u ∈ H 1 0 (Ω) by zero outside Ω, we regard H 1 0 (Ω) as a subset of H 1 0 (B R ) where B R = B(0, R) ⊂ R n .While these have been identified as equivalent problems, none of these have been solved.Thus in the present paper we establish that the constructive solution to any of these problems requires more than the usual classical hypotheses, and provide some quite reasonable constructively sufficient conditions.

The Dirichlet problem, classically
Associated with Problem 3 is the Dirichlet energy functional Dirichlet's principle states that the following two conditions are (classically) equivalent (see e.g.[17, pp. 178-179 and pp. 186-187]).
(i) The function u is a weak solution of the Dirichlet problem (2).That is, u satisfies (3) whenever v ∈ H 1 0 (Ω).(ii) The function u minimizes the Dirichlet energy functional (4) in H 1 0 (Ω).That is, J(u) ≤ J(v) whenever v ∈ H 1 0 (Ω).Minor modifications convert the classical proof of equivalence into a constructive proof.As with the equivalences identified at the end of the previous section, however, the proof remains silent on the question of how to actually compute the function u in H 1 0 (Ω) which solves (2).Typical classical approaches to finding a u which satisfies either condition proceed along these lines: and apply the Riesz Representation Theorem to find an element u of for all v in H 1 0 (Ω); then u solves the Dirichlet problem.(ii) Construct a minimizing sequence for the functional J , the infimum of which is guaranteed to exist by the least upper bound principle; then use (weak) sequential compactness to extract a (weakly) convergent subsequence, the limit of which solves the Dirichlet problem.
Constructively, (i) will not guarantee a solution to the Dirichlet problem, since the representability of ϕ is constructively equivalent to its normability.That is, ϕ can be represented only if its norm can be (explicitly) computed.The reason (ii) will not work constructively lies in the application of the least upper bound principle and the sequential compactness argument, as neither of these are provable using only constructive methods.See [8] for details.

Any general method must go constructively wrong
In this section we prove that the existence of a weak solution of the general Dirichlet problem for a domain Ω ⊂ R 2 cannot be proved constructively.To make the following examples easier to follow, we return to the version (1) from the introduction.
Still working within BISH, we remind the reader of two important principles: Markov's principle, MP: For each binary sequence (a n ) n≥1 , if it is impossible that a n = 0 for all n, then there exists n such that a n = 1, and the limited principle of omniscience, LPO: For each binary sequence (a n ) n≥1 , either a n = 0 for all n, or else there exists n such that a n = 1.
Markov's principle is consistent with (though not included in) BISH, is commonly employed in recursive constructive mathematics (essentially, BISH plus the Church-Markov-Turing thesis and Markov's principle-see [15]), but, with the aid of Brouwer's theory of the creating subject (see [11]) is provably false in intuitionistic mathematics.
On the other hand, LPO (which clearly implies MP) is recursively and intuitionistically false, and is regarded as essentially nonconstructive.
We are aiming for this result: The following are equivalent over BISH.
(ii) Markov's principle holds, and for every totally bounded, Lebesgue integrable, open subset Ω of R N , and every uniformly continuous function f : ∂Ω → R, the Dirichlet problem has a weak solution in H 1 0 (Ω).
Before proving this, though, we put on record a simple, but useful, preliminary proposition, and describe a key construction that we use in the proof: Proposition 5 LPO If S is a countable subset of R, then either S is bounded above or else it is unbounded above (that is, for each c > 0, there exists x ∈ S with x > c).Moreover, if S is bounded above, then sup S exists. 5roof Let f be a mapping of N + onto S. Given α, β ∈ R, construct a binary sequence λ ≡ (λ n ) n≥1 such that if λ n = 0, then f (n) < β , and if Applying LPO to λ, we see that either f (n) < β for all n or else there exists n such that f (n) > α.
Without loss of generality, we may assume that f (1) > 1. Setting µ 1 = 0, construct a binary sequence µ ≡ (µ n ) n≥1 such that for n ≥ 2, if µ n = 0, then there exists m such that f (m) > n, and if µ n = 1, then f (m) < n + 1 for all m.Applying LPO to µ, we see that either S is unbounded above or else it is bounded above.In the latter case, noting the first paragraph of this proof, we see from the constructive least-upper-bound principle ([5], Theorem 2.1.18)that sup S exists.
The following lemma embodies our key construction for the proof of Theorem 4.
Lemma 6 Let (a n ) n≥1 be an increasing binary sequence with a 1 = 0. Let D be the open unit disc in R 2 , and for each positive integer n define

D.S. Bridges and M. McKubre-Jordens
If Then Ω is totally bounded, located and integrable.
Proof First we prove that Ω is totally bounded and hence located.Observe that for each n, Ω n is totally bounded and Ω n+1 ⊂ Ω n , and that if a n = 0, then T n ⊂ Ω.
To show that Ω is integrable, first observe that for each n, Ω n is integrable.Construct the sequence of real numbers (I n ) n≥1 such that I n = χ Ωn for each n.Now for m ≥ n, and so (I n ) is a Cauchy sequence and converges to a limit I .A corollary of the constructive Lebesgue series theorem (a consequence of [4, Ch. 6 (2.17)]) now informs us that Ω is integrable and has measure lim n→∞ χ Ωn .
Figure 1 illustrates this construction.
Turning back to the Dirichlet problem, we give the proof of Theorem 4: Proof First assume LPO and consider the Dirichlet problem (1) with Ω ⊂ R N totally bounded, Lebesgue integrable, and open, and with f : ∂Ω → R uniformly continuous.Since LPO implies MP, it will suffice to prove that this Dirichlet problem has a weak solution; we base the argument on the Ritz-Galerkin method (see Chapter 9 of [2]).
Let (e n ) n≥1 be an orthonormal basis of H 1 0 (Ω), and for each n let H n be the finitedimensional subspace of H 1 0 (Ω) spanned by {e 1 , . . ., e n }.Define a linear functional φ on H 1 0 (Ω) by φ(v) ≡ − Ω vf .Standard estimates, using the Hölder and Poincaré inequalities [12] (Chapter 5, Theorem 3), yield v ∈ H 1 0 (Ω) , where γ is the constant in Poincaré's inequality.Thus the linear functional φ is bounded.For each n, since H n is finite-dimensional, the restriction of φ to H n is normed: 6  exists.We show that φ is normed, with φ = σ .Let P n be the projection of . On the other hand, given ε > 0 and choosing n such that 0 ≤ σ − φ| Hn < ε, we can find a unit vector v ∈ H n such that φ| Hn − ε < φ(v) ≤ φ| Hn .Since ε > 0 is arbitrary, we conclude that φ exists and equals σ .We are now able to apply the (constructive) Riesz representation theorem ([5], Theorem 4.3.6), to produce w ∈ H 1 0 (Ω) such that for each v ∈ H 1 0 (Ω), The function w is therefore the sought-after weak solution of the Dirichlet problem.Hence (i) implies (ii).
Suppose, conversely, that (ii) holds, and consider the Dirichlet problem where Ω is the domain from Lemma 6.Since (0, 0) / ∈ ∂Ω and we are assuming Markov's principle, for each x ∈ ∂Ω we have x > 0 and therefore log x welldefined.We show that the function f : x log x is uniformly continuous on ∂Ω.Let x, x ∈ ∂Ω be such that x − x < 1/2.Either min { x, x } > 1/2 or min { x, x } < 1.In the first case we must have x = x = 1, so log x = log x = 0.In the case min { x, x } < 1, we may assume that x < 1; since (as we observed above) x > 0, there exists N > 2 such that Ω = Ω N , x = 1/N , and therefore If a n = 0 for all n, then our Dirichlet problem has the unique (weak and strong) solution u(x) = 0 for all x ∈ Ω, and Ω u 2 = 0.If a n = 1 − a n−1 for some n, then (6) has the unique (weak and strong) solution u(x) = log x for all x ∈ Ω, and, with the usual polar coordinates r and θ , Moreover, the last expression is an increasing function of n that converges to π/2 as n → ∞, and in the case n = 2 is larger than 1/4.Assume that the Dirichlet problem (6) has a weak solution w ∈ L 2 Ω .Then either Ω w < 1/4 or Ω w > 0. In the first case, we must have a n = 0 for all n; in the second, it is impossible that a n = 0 for all n, so, by Markov's principle, there exists n such that a n = 1.This completes the proof that (ii) implies LPO.

Constructing a weak solution
In this section we examine the Dirichlet problem.Specifically, in this section we provide conditions that ensure the existence of a weak solution of (2): that is, an element u of the Hilbert space We call a subset Ω of R N a Wang domain if it has the following properties: (i) It is edge coherent, totally bounded, and open in R N .
(iv) There exists c 0 > 0 such that if Notice that conditions (i)-(iii) are classically trivial, in the sense that, using classical logic, any bounded domain is easily seen to satisfy them.Yet they yield important computational information: (i) and (ii) embody information about the boundary (classically taken for granted), and (iii) holds as a consequence of the (classical) divergence theorem.Condition (iv) is an inequality which provides information about the norm of candidates for the solution (elements of H 1 0 (Ω)); it asserts that the average value of u 2 in a band of width r about ∂Ω tends to zero with r and is motivated by the classical theory (see e.g.[17,p. 199,Theorem 6]).These conditions are easily seen to hold classically for domains with located boundary, so examples abound: the unit ball in R n is the canonical example.

Theorem 7
Let Ω be a Wang domain in R N , and let f ∈ L 2 (Ω) (relative to Lebesgue measure µ on Ω).Suppose that there exists a sequence (Ω n ) n≥1 of edge coherent, totally bounded, open subsets of Ω, each having compact boundary, such that for each n, We also get Since µ (∂Ω n ) = 0 and u = 0 on −Ω n , we see from (7) that Hence 4 and therefore u n − u N H 1 0 (Ω) < ε.We now see that u m − u n H 1 0 (Ω) < 2ε whenever m ≥ n ≥ N .Thus (u n ) n≥1 is a Cauchy sequence in the complete space H 1 0 (Ω).Let u be its limit.Given v ∈ C 1 0 (Ω), we compute Since v ∈ C 1 0 (Ω) is arbitrary, we conclude that u is the weak solution of the Dirichlet problem (1).

D.S. Bridges and M. McKubre-Jordens
Typical domains involving applications (such as the unit ball, unit cube, etc.) often satisfy the conditions of Theorem 7. The paper [10] shows that Green functions can be effectively constructed for a large class of domains, and hence on such domains the Dirichlet problem admits a constructive solution.
We now turn to some properties of the counterexample presented in Lemma 6.

Revisiting the Brouwerian example in Lemma 6
Recall that we already have established the locatedness and integrability of the set Ω constructed in Lemma 6.We now outline proofs of some properties of this domain.
(a) The domain Ω satisfies a pointwise exterior cone condition, but if it satisfies a uniform one, then ∀ n (a n = 0) ∨ ¬∀ n (a n = 0) .
Since for each x ∈ ∂Ω, either x > 1/2 or 0 < x < 1, we easily show that Ω satisfies the pointwise exterior cone condition.But if Ω satisfies the uniform exterior cone condition, then LPO holds.To see this, pick r, θ > 0 such that for each x ∈ ∂Ω, there exists a right circular cone with vertex x, vertex angle θ , and height r such that C ∩ Ω = {x}; then compute a positive integer N > 1/2r.Then it is impossible that a n = 1 − a n−1 for some n ≥ N , so if a N = 0, then a n = 0 for all n.
It should be noted that classically the domain Ω does satisfy the exterior cone condition.
(b) The domain Ω is edge coherent.
To see this, first note that Ω n is edge coherent for each n.Given x ∈ Ω with ρ (x, ∂Ω) > 0, pick r such that 0 < r < min 1  4 , ρ (x, ∂Ω) .Either x = 0 or x < r/2.In the first case, pick a positive integer N > 1/ x .If a N = 0, then x ∈ Ω N ⊂ Ω.If a N = 1, then there exists m ≤ N such that x ∈ Ω m = Ω.In the case x < r, for each y ∈ ∂Ω we have y ≥ x − y − x > r/2; whence ρ(0, ∂Ω) ≥ r/2.Computing an integer m > 2/r, we see that if a n = 1 − a n−1 for some n ≥ m, then Ω = Ω n and ρ (0, ∂Ω) = 1/n < r/2, a contradiction.It follows that if a m = 0, then a n = 0 for all n; whence Ω = D, and so x ∈ Ω.On the other hand, if a m = 1, then Ω = Ω n for some n ≤ m, again Ω is edge coherent, and so x ∈ Ω.
(c) If Ω is approximated internally by compact sets then (9) holds.
As K is compact, it is located.First, observe that the unit circle is in ∂Ω (and the measure of that component of (∂Ω) r is 4πr); call this component of ∂Ω the outer component.Call any other component of ∂Ω, if there is one, the inner component.Now either ρ(0, K) > r or ρ(0, K) < 2r.If ρ(0, K) > r, then there exists a positive integer N such that 1/r < N < 2/r.Suppose that there exists some n ≥ N such that a n = a n−1 − 1.Then there is an inner component of ∂Ω closer to 0 than 1/N .The measure of the corresponding inner component of (∂Ω) r is at least πr 2 , and so by (10) and our choice of ε there must be a corresponding inner component of K .Since N > 1/r and a n = 0 for all n < N , we have and so ρ(0, K) ≤ 1/N < r, a contradiction; hence if a n = 0 for each n ≤ N , then a n = 0 for all n.On the other hand, if ρ(0, K) < 2r ≤ 1/4, then it is impossible that a n = 0 for all n.
We see from comments (a) and (c)-(f) that in each case, if a certain property holds no matter what binary sequence (a n ) n≥1 is used to define Ω, then we can derive the following: WLPO: For each binary sequence (a n ) n≥1 , either a n = 0 for all n or it is impossible that a n = 0 for all n.
a lesser version of LPO.We can move from WLPO to the stronger omniscience property LPO in these situations if we allow ourselves the use of Markov's principle; since we made use of MP in the proof of Theorem 4, its further application here is not entirely without warrant.However, forgoing MP in (a) and (c)-(f) shows that each of the properties of Ω in question is genuinely nonconstructive.
(g) Assuming Markov's principle, we may compute a compact subset K ∞ of Ω such that Moreover, K ∞ − {0} is inhabited if and only if a n = 1 for some n.
To prove this, first observe that Proof We show that Ω is totally bounded, using an argument similar to that from Lemma 6.Given ε > 0, choose a positive integer N > 3/ε.Either there is a (least) m ≤ N such that a m = 1 or a m = 0 for each m ≤ N .In the former case, Ω = Ω m and an ε-approximation to Ω m is also an ε-approximation to Ω.In the latter case, note that Ω N ⊂ Ω and choose a finite ε/3-approximation {x 1 , x 2 , . . ., x n } to Ω N .Consider x ∈ D. Either | arg x| > 1/N or | arg x| < 2/N .In the first case, x ∈ Ω N , so there exists i with x − x i < ε.In the second case, so, choosing y ∈ Ω N so that x − y ≤ 2ε/3 and i such that y − x i < ε/3 we have x − x i < ε.Hence {x 1 , x 2 , . . ., x n } is a finite ε-approximation to D with points from Ω N ⊂ Ω; therefore, since Ω ⊂ D, it is an ε-approximation to Ω.
A proof analogous to that in Lemma 6 shows that, like the annular domain constructed there, this domain is integrable.
The construction is illustrated in Fig. 2. WLPO is sufficient to construct solutions to the Dirichlet problem.We conjecture that the existence of solutions to Dirichlet-type problems requires strong computability conditions on the boundary such as its locatedness (which condition is classically trivial).Since WLPO is not sufficient to locate sets in R n in general, it is unlikely that it would be sufficient for existence of solutions to the Dirichlet problem.
Finally, since solutions to the Dirichlet problem are solutions to a special case of the Navier-Stokes equations of fluid flow, we have the following corollary, indicating that any proof of existence of solutions to these equations will have to be non-constructive.
Corollary 11 There is no universal algorithm for computing solutions to the classical Navier-Stokes equations of fluid flow with arbitrary boundary conditions.

Problem 3 (
The Dirichlet problem) Given an open, bounded integrable set Ω in R n and f in L 2 (Ω), find a function u such that (2) ∆u = f on Ω, u(x) = 0 for all x ∈ ∂Ω,

Figure 1 :
Figure 1: The domain Ω n (left) when a n = 1 − a n−1 , used in the construction from Lemma 6; and the domain Ω (right), formed by intersecting the Ω n .