Stability of equilibrium solution to inhomogeneous heat equation under a 3-point boundary condition

Abstract

We consider a one-dimensional heat equation with inhomogeneous term, satisfying three-point boundary conditions, such that the temperature at the end is controlled by a sensor at the point η. We show that the integral solution, in the space of continuous functions satisfying the boundary values, converges to the equilibrium solution. This answers a question posed for nonlinear Laplacians, but in the linear case only.

Keywords:

• three-point boundary value problems
• heat equation
• asymptotic stability

AMS Subject Classifications:

• 34B10
• 47D06

1. Introduction

In 1, the author considers the Cauchy problem on [0, ∞) × [0, 1],

where η ∈ (0, 1) and β > 1 are given, along with f and g. It is supposed that g : (a, b) → ℝ is an increasing homeomorphism and a < 0 < b.

It is shown that we have an integral solution to the Cauchy problem du/dt = Au − f with initial value u ₀, in the space of continuous functions, where A is the nonlinear Laplacian (g(u _x )) _x subject to the boundary conditions. The question is asked; does the solution converge to the equilibrium solution, A ⁻¹ f? In this note we show that this holds if g : ℝ → ℝ is linear, i.e. for some k ∈ ℝ, g(x) = kx, so that (1) becomes u _t (t, x) = ku _xx (t, x) − f(x), or after an adjustment, replacing t by τ = kt and f by f/k, we assume that we have

together with Equations (2)–(4). Note that the equilibrium solution has been investigated for the linear case in 2, as well as 1. Unfortunately, the Sobolev space setting of Guidotti and Merino 3 seems to be unavailable, and we rely on the space of continuous functions to describe our equations. The boundary conditions in 3 included u′(0) = 0. The paper 3 models the usage of a thermostat, and a nonlinear problem based on 3 was studied in the papers 4,5. It should be interesting to get stability for the situation in which f(x) is replaced by f(u(x)).

The convergence of the solution to the inhomogeneous heat equation

under other boundary conditions, such as Dirichlet, and Neumann, is well known, and the interested reader may consult and follow up [6, Ch 10.1] and [7, Ch 3.5] and the commentaries on these sections.

2. Preliminaries

Suppose β > 1 and η ∈ (0, 1) are given. Let X denote the Banach space of continuous functions u : [0, 1] → ℂ, satisfying u(0) = 0 and u(η) = βu(1), under the sup norm. We define a linear operator in X. Let D(L) consist of u ∈ X which have first and second continuous derivatives on [0, 1], i.e. one-sided derivatives at the endpoints. For u ∈ D(L) let Lu = u _xx .

LEMMA 1

Given β > 1 and η ∈ (0, 1), the equation

in the complex variable z has only real solutions.

Proof

(a) Suppose z = iy is a purely imaginary solution to (5). The identity

gives i sinh(ηy) = βi sinh(y) and y = 0.

(b) Now we suppose z = a + ib, a and b are real and ab ≠ 0. Define z(t) = sin(t(a + ib)) for t ∈ [0, 1]. We claim that if ab > 0 then arg(z(t)) is strictly decreasing on (0, 1), while if ab < 0 then it is strictly increasing. Suppose ab > 0. Write z(t) = x(t) + iy(t), x and y real; we claim that

We want the numerator to be negative, i.e.

Simplify this to give

which holds because the LHS is less than 1 and the RHS is greater than 1. This proves the claim for ab > 0. Suppose instead that ab < 0. Then (9) holds, so that (8) holds with the inequality reversed, and hence arg(z(t)) is strictly increasing.

Suppose a ≠ 0, b ≠ 0, a and b are real and

for t ∈ (0, 1). We claim that z(t) ≠ βz(1) for all t ∈ (0, 1). The curve t ↦ z(t) gives the solution to the initial value problem

where we choose , as we see by substituting (10) into (11). We have the RHS single valued on the cut plane given by a cut between −1 and 1, and we check that the solution does not cross the real axis between −1 and 1 for t > 0. Suppose y(t) = 0, then cos(ta) = 0, so sin(ta) = ±1 and cosh(tb) > 1, giving |z(t)| > 1. Thus we have the uniqueness of solutions of (11), and, in particular, the forward orbit does not intersect itself. Assume that ab > 0. Then we have a forward orbit spiralling clockwise out from the origin, so that if arg(z(t) decreases by 2π, then mod(z(t)) increases, so we cannot have z(t ₀) = βz(t ₀) for β ≥ 1 and t ₀ < 1.▪

LEMMA 2

Suppose β > 1. The eigenvalues of L consist of a sequence with and

with eigenvectors u _n = x ↦ sin(k _n x).

Proof

(a) We claim that for each n = 0, 1, … there is a unique k _n ∈ (π/2 + nπ, π/2 + (n + 1)π) with

Now k ↦ β sin(k) takes values β and −β at the two endpoints of (π/2 + nπ, π/2 + (n + 1)π), whereas k ↦ sin(ηk) has values in [0, 1], so there does exist k _n satisfying (13).

Suppose there are two or more solutions of (13), then the slope of k ↦ sin(ηk) at some point q with

is in absolute value at least as big as that of k ↦ β sin(k), i.e.

Thus

by (14), giving 1 ≥ β², contradicting β > 1.

(b) One checks that for each n = 0, 1, …, with , and u _n (x) = sin(k _n x), we have

(c) Suppose Lu = λu; λ ∈ ℂ, and u ≠ 0. We show that for some n, and u = u _n . Now λ ≠ 0, and we let λ = −k ². Since u _xx = −k ² u, we have

for some A, B. Since u(0) = 0, B = 0, and then u(η) = βu(1), which gives

By Lemma 1, k ∈ ℝ. Hence all eigenvectors of L are real, nonzero and (19) holds with eigenvector x ↦ sin(kx). Hence by (a), k = k _n for some n = 0, 1 …, and .▪

LEMMA 3

Let σ ∈ ℂ be not an eigenvalue of L. Then L − σI is surjective and has continuous inverse.

Proof

Note that L is surjective, with continuous single-valued inverse which is compact. Since L − σI is one to one, if f ∈ X is given, then Lu − σu = f iff u − σL ⁻¹ u = L ⁻¹ f, and I − σL ⁻¹ is one to one, so is open and surjective by the invariance of domain. Hence L − σI is surjective, and bounded by the closed graph theorem.▪

THEOREM 1 8

Let T be a positive C ₀ semigroup in a Banach lattice, with generator B. Then s(B) = ω₁(B).

In this result the only condition on B is that it is the generator of a positive C ₀ semigroup in a Banach lattice. We recall that T is called positive when for each t ≥ 0, T(t) maps the positive cone of the Banach lattice to itself. We recall [8, page 8] that s(B) ≔ sup{Re(λ) : λ ∈ σ(B)} in general, and hence in this article. Also,

Here ‖x‖ _D(B) ≔ ‖x‖ + ‖Bx‖. Note that by [1, Theorem 12], L is an m-dissipative operator in X, and hence is the generator of a C ₀ semigroup in a Banach lattice. We check that the semigroup is positive. The resolvent J _n = (I − nL)⁻¹, n a positive integer, is positive since if u − n ⁻¹ Lu = v, and v ≥ 0, then u ≥ 0, else u would be minimized at x ₀ with u(x ₀) < 0, and then Lu(x ₀) ≥ 0 because the three-point boundary condition implies that x ₀ < 1, and then v(x ₀) < 0, contradicting v ≥ 0. Hence the semigroup is positive, being given, for x ∈ X and t ≥ 0, by

this exponential being defined via the power series and hence mapping the positive cone of X to itself.

Corollary 1

The semigroup T generated by the operator L in X has the property that for all x ∈ X, T(t)x → 0 as t → ∞.

Proof

We check 1 that T is positive. By Theorem 1, there is M such that for all x ∈ D(L),

and T(t)x → 0. Hence for all x ∈ cl(D(L)) = X, since T is nonexpansive, T(t)x → 0 as t → ∞.▪

We consider the Cauchy problem: given u ₀ : [0, 1] → ℝ, find u(t, x) for x ∈ [0, 1] and t ≥ 0, satisfying

By 1, L is m-dissipative in X. By 9 there is a unique integral solution of the Cauchy problem (22) if u ₀ ∈ X.

THEOREM 2

Suppose β, η, X and L are as specified in Section 2. Let u ₀ and f be in X. Then the integral solution to (22) converges to L ⁻¹ f as t → ∞.

Proof

Let w ₀ = L ⁻¹ f. We know that L generates a nonexpansive semigroup T since L is m-dissipative. Let u(t) = T(t)(u ₀ − w ₀) + w ₀ for t ≥ 0. Suppose first that u ₀ ∈ D(L). Then u is C ¹ and u′(t) = Lu(t) − f, since u ₀ − w ₀ ∈ D(L); see [8, p. 3] on classical solutions. Then u(t) is an integral solution by [9, Theorem 5.5]. Then for general u ₀ in X we have u(t) the integral solution, by continuity. From the Corollary we have u(t) → w ₀.

Remark

(t, x) ↦ (T(t)(u ₀ − w ₀))(x) is a distributional solution of the heat equation and by hypoellipticity 10 it is C ^∞ on (0, ∞) × (0, 1). Hence the solution u(t) = T(t)(u ₀ − w ₀) + w ₀ is as smooth as L ⁻¹ f. From the boundary conditions, u(t) is smooth on the boundary x = 1 for t > 0.

Remark

The question arises as to whether the condition β > 1 is necessary for this article, or whether β > η suffices. In 11 it is shown that the condition β > 1 is necessary for their results. We note that a different case β < η has been discussed in 12, and the integral operator is then negative. Lemmas 1 and 2 use β > 1, but Corollary 1 may go through without their detailed conclusions, because we merely used s(L) < 0 when applying Theorem 1. However, for η < β < 1, we do not apply the theory of integral solutions, because we can show that we do not have L − ωI dissipative for any ω, and integral solutions concern such operators.

Proposition 1

Suppose β ∈ (0, 1), η ∈ (0, 1), ω > 0, a < 0 < b and g : (a, b) → ℝ is an increasing homeomorphism, and is C ¹. Then L − ωI is not dissipative in C([0, 1]).

Proof

Let

If ε > 0 is small, then u ∈ D(L), and we check that u − λ(L − ωI)u attains its maximum value at 1 for small λ > 0, but is less than u there.▪

Remark

On the other hand, we can still ask about other notions of solution of the Cauchy problem for η < β < 1, in case g(x) = x, and we can ask if the corresponding version of Theorem 2 will hold. But this study is not in the scope of this article. In fact, L is the generator of a positive C ₀ semigroup.

Acknowledgements

Thanks to Jeff Webb, Chaitan Gupta and the referees for their advice.