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.
1. Introduction
In Citation1, the author considers the Cauchy problem on [0, ∞) × [0, 1],
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 (Equation1) 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
The convergence of the solution to the inhomogeneous heat equation
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
Proof
(a) Suppose z = iy is a purely imaginary solution to (Equation5). The identity
(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
Suppose a ≠ 0, b ≠ 0, a and b are real and
LEMMA 2
Suppose β > 1. The eigenvalues of L consist of a sequence
with
and
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
Suppose there are two or more solutions of (Equation13), then the slope of k ↦ sin(ηk) at some point q with
(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
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 Citation8
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 [Citation8, page 8] that s(B) ≔ sup{Re(λ) : λ ∈ σ(B)} in general, and hence in this article. Also,
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 Citation1 that T is positive. By Theorem 1, there is M such that for all x ∈ D(L),
We consider the Cauchy problem: given u 0 : [0, 1] → ℝ, find u(t, x) for x ∈ [0, 1] and t ≥ 0, satisfying
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 (Equation22) 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 [Citation8, p. 3] on classical solutions. Then u(t) is an integral solution by [Citation9, 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 Citation10 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 Citation11 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 Citation12, 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
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.
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