Mem. Differential Equations Math. Phys. 19(2000), 133–141

M. Alves

ABOUT A PROBLEM ARISING IN CHEMICAL REACTOR THEORY

(Reported on December 23, 1998)

1. Notation

Throughout this paper C ≡ C[0, 1] denotes the Banach space of continuous functions x : [0, 1] ⟶ R^{1},

∥x∥_{C} =^{def} max_{0≤t≤1} |x(t)|;

L_{p} ≡ L_{p}[0, 1] (1 ≤ p < ∞) denotes the Banach space of summable in p-th degree functions x : [0, 1] ⟶ R^{1},

∥x∥_{L_{p}} =^{def} ( ∫_{0}^{1} |x(t)|^{p} dt)^{1/p};

L_{∞} ≡ L_{∞}[0, 1] denotes the Banach space of essentially bounded measurable functions x : [0, 1] ⟶ R^{1},

∥x∥_{L_{∞}} =^{def} vrai sup_{0≤t≤1} |x(t)|;

W_{p}^{2} ≡ W_{p}^{2}[0, 1] denotes the Banach space of continuous functions x : [0, 1] ⟶ R^{1} with the absolutely continuous derivative ẋ such that ẍ ∈ L_{p},

∥x∥_{W_{p}^{2}} =^{def} ∥ẍ∥_{L_{p}} + |x(0)| + |ẋ(0)|.

2. The Space of Solutions D_{p}

Consider the boundary-value problem

\{ \begin{matrix}(ℑ_{0}x)(t) =^{def} ẍ(t) + \frac{k}{t} ẋ(t) = f(t), t =∈ [0, 1], \\ ẋ(0) = 0, x(1) = α, \end{matrix}  (1)

where k > − \frac{1}{p′}, f ∈ L_{p}, 1 < p ≤ ∞, α ∈ R^{1}, \frac{1}{p} + \frac{1}{p′} = 1, p′ = 1 if p = ∞.

Considering this problem on the traditional space W_{p}^{2}, we see that ℑ_{0} is not defined as an operator acting from W_{p}^{2} into L_{p}. Following the scheme given in the monograph [1], we will investigate this problem on the space D_{p} ⊂ W_{p}^{2} of functions x : [0, 1] ⟶ R^{1}, such that ẋ(0) = 0 and defined by

x(t) = ∫_{0}^{t} (t − s)z(s) ds + β

1991 Mathematics Subject Classification. 34B15, 34K10.

Key words and phrases. Boundary value problem, functional differential equation, chemical reactor theory. for each pair {z, β} ∈ L_{p}×R^{1}. The space D_{p} is isomorphic to the direct product L_{p}×R^{1}. The isomorphisms J : L_{p} × R^{1} ⟶ D_{p} and J^{−1} : D_{p} ⟶ L_{p} × R^{1} we define by the equalities J = {Λ, Y }, J^{−1} = [δ, r], where

\{ \begin{matrix}(Λz)(t) = ∫_{0}^{t} (t − s)z(s) ds, (Y β)(t) = β, \\ δx = ẍ, rx = x(0). \end{matrix}

The space D_{p} becomes a Banach one under the norm

∥ x ∥_{D_{p}} =^{def}∥ ẍ ∥_{L_{p}} +|x(0)|.

The principal part of the operator ℑ_{0} : D_{p} ⟶ L_{p} is

(Qz)(t) =^{def} (ℑ_{0}Λz)(t) = z(t) + (Pz)(t),

where (Pz)(t) =^{def} \frac{k}{t} ∫_{0}^{t} z(s) ds is the Cesàro operator [2] on the space L_{p}. The functions

u(t) ≡ 1 and v(t) = t^{1−k} satisfy the equation ℑ_{0}x = 0. Nevertheless the fundamental system of ℑ_{0}x = 0 consists only of u(t) ≡ 1, such as the other element v(t) = t^{1−k} does not belong to the space D_{p}. By virtue of the results of [5, p. 102] it follows that, if

k > − \frac{1}{p′}, the operator Q : L_{p} ⟶ L_{p} has the bounded inverse

(Q^{−1}z)(t) = z(t) − kt^{−(1+k)} ∫_{0}^{t} s^{k}z(s) ds.

The solution of the problem (1) on the space D_{p} is given by the expression

x = Wf + α,

where the Green operator W : L_{p} ⟶ D_{p} is defined by

(Wf)(t) =^{def} ∫_{0}^{1} W (t, s)f(s) ds,

W (t, s) =^{def} \{ \begin{matrix}\frac{s^{k}(t^{1−k} − 1)}{1 − k} & if 0 ≤ s ≤ t ≤ 1, \\ \frac{s^{k}(s^{1−k} − 1)}{1 − k} & if 0 ≤ t < s ≤ 1, \end{matrix}

for k > − \frac{1}{p′} , k ≠ 1, or

W (t, s) =^{def} \{ \begin{matrix}s ln t & if 0 ≤ s ≤ t ≤ 1, \\ s ln s & if 0 ≤ t < s ≤ 1, \end{matrix}

for k = 1. Really, using the equality

ẋ(t) = ∫_{0}^{t} ẍ(s) ds

for x ∈ D_{p} we rewrite the problem (1) on the space D_{p} in the form

ẍ(t) = f(t) − kt^{−(1+k)} ∫_{0}^{t} s^{k}f(s) ds, t ∈ [0, 1], x(1) = α.

Immediate computations show that

\begin{aligned}x(t) & = ∫_{0}^{t} (t − s)[f(s) − ks^{−(1+k)} ∫_{0}^{s} τ^{k}f(τ) dτ] ds + x(0) = \\ & = ∫_{0}^{t} [t − s − ks^{k} ∫_{s}^{t} (t − τ)τ^{−(1+k)} dτ]f(s) ds + x(0). \end{aligned}

The condition x(1) = 0 gives

x(0) = − ∫_{0}^{1} [1 − s − ks^{k} ∫_{s}^{1}(1 − τ)τ^{−(1+k)} dτ]f(s) ds.

Consequently

x(t) = ∫_{0}^{1} W (t, s)f(s) ds + α, t ∈ [0, 1].

Bellow we will use results of [4] about estimation of the spectral radius ρ(H) of the isotonic operator H : C ⟶ C. We formulate this result in the form satisfying our aims:

Lemma 1. Suppose that the isotonic operator H enjoys the property (Hζ)(1) = 0 for each ζ ∈ C. The following statements are equivalent:

1) There exists y ∈ C such that

y(t) > 0, y(t) − (Hy)(t) > 0, t ∈ [0, 1);

2) ρ(H) < 1. □

Lemma 2. The integral operator W : L_{p} ⟶ C is completely continuous, for all 1 < p ≤ ∞.

Proof. We consider only the case k > − \frac{1}{p′}, k ≠ 1, the case k = 1 can be proved

analogously. To prove the compactness of the operator W it suffices to show [5, p. 102] that, for any t_{0} ∈ [0, 1] the equality

lim_{t→t_{0}} ∫_{0}^{1} |W (t, s) − W (t_{0}, s)|^{p′} ds = 0

holds. For 1 ≤ p′ < ∞, 0 < t_{0} < t ≤ 1 we have that

∫_{0}^{1} |W (t, s) − W (t_{0}, s)|^{p′} ds = ∫_{0}^{t_{0}} |\frac{s^{k}(t^{1−k} − 1)}{1 − k} − \frac{s^{k}(t_{0}^{1−k} − 1)}{1 − k} |^{p′} ds +

+ ∫_{t_{0}}^{t} |\frac{s^{k}(s^{1−k} − 1)}{1 − k} − \frac{s^{k}(t_{0}^{1−k} − 1)}{1 − k} |^{p′} ds ≤

≤ \frac{t_{0}^{p′k+1}}{|1 − k|^{p′}(p′k + 1)} (t^{1−k} − t_{0}^{1−k})^{p′} + O(t^{p′+1} − t_{0}^{p′+1}) → 0, t → t_{0}^{+}.

Analogously we prove the respective statement for 0 = t_{0} < t ≤ 1 and 0 ≤ t < t_{0} ≤ 1. □

3. The de la Vallée-Poussin Like Theorem

Consider the boundary-value problem

\{ \begin{matrix}(ℑx)(t) =^{def} (ℑ_{0}x)(t) − (T x)(t) = f(t), t ∈ [0, 1], \\ ẋ(0) = 0, x(1) = α, \end{matrix}  (2)

where k > − \frac{1}{p′}, T : C ⟶ L_{p} is a linear antitonic operator, f ∈ L_{p}. Denote A =^{def} WT :

C ⟶ C.

Lemma 3. The following statements are equivalent:

1) There exists an element y ∈ D_{p} such that

y(t) > 0, ϕ(t) =^{def} (ℑ_{0}y)(t) − (T y)(t) ≤ 0, t ∈ [0, 1), and

y(1) − ∫_{0}^{1} ϕ(s) ds > 0;

2) ρ(A) < 1;

3) The boundary value-problem (2) is uniquely solvable on D_{p} for each f ∈ L_{p}, α ∈ R^{1}, and its Green operator G is antitonic;

4) There exists a positive solution u ∈ D_{p} on [0, 1] of the homogeneous equation ℑx = 0.

Proof. Since y(·) satisfies

(ℑ_{0}x)(t) − (T x)(t) = ϕ(t), t ∈ [0, 1], x(1) = y(1)

on the space D_{p}, it follows that

y − Ay = Wϕ + y(1) > 0

on the space C. By virtue of Lemma 1 it follows that ρ(A) <= 1. The implication 1) ⟹ 2) is proved.

Supposing α ≥ 0 we consider the problem (2), which is equivalent to the equation

x = Ax + g

on the space C. Here

g(·) =^{def} ∫_{0}^{1} W (·, s)f(s) ds + α.

Since ρ(A) < 1, it follows that

G = (I + A + A^{2} + ⋯)W.

Consequently the implication 2) ⟹ 3) is proved.

The problem

ℑ_{0}x − T x = 0, x(1) = α

is equivalent to the equation

x = Ax + α.

Since ρ(A) < 1, we have x = α + Aα + A^{2}α + ⋯ ≥ 0 if α > 0. Thus the implication 3) ⟹ 4) is proved.

The implication 4) ⟹ 1), follows from Lemma 1 because the positive solution u(t) of the equation ℑx = 0 satisfies the inequalities

u(t) > 0, u(t) − (Au)(t) = α > 0, t ∈ [0, 1]. □

4. The Main Result

Consider the nonlinear boundary-value problem

ℑ_{0}x = f(·, Θx), ẋ(0) = 0, x(1) = α,  (3)

where Θ : C ⟶ L_{p} is a linear isotonic operator, 1 < p ≤ ∞, k > − \frac{1}{p′}, the function

f(·, ·) satisfies the Carathéodory conditions. By definition put v̄ = Θv, z̄ = Θz, [v̄, z̄] =^{def} {x ∈ L_{p} : v̄ ≤ x ≤ z̄}.

Following [5], we will say that the function f(·, ·) satisfies the condition L^{i}[v̄, z̄], i = 1, 2, if it is possible the decomposition

f[t, u(t)] = q_{i}(t)u(t) + M_{i}[t, u(t)], u ∈ [v̄, z̄],

where q_{i} ∈ L_{∞}, i = 1, 2, the operator M_{i} : [v̄, z̄]_{L_{p}} ⟶ L_{p} is defined by (M_{i}u)(·) =^{def} M_{i}[·, u(·)], M_{1} is isotonic and M_{2} is antitonic.

Theorem 1. Let v, z ∈ D_{p} be a pair of functions such that v(t) < z(t), t ∈ [0, 1], and

ℑ_{0}v ≥ f(·, Θv), ℑ_{0}z ≤ f(·, Θz), v(1) ≤ α ≤ z(1).  (4)

Suppose that the function f(·, ·) satisfies the condition L^{2}[v̄, z̄] with q_{2} ∈ L_{∞}, q_{2}(·) ≤ 0. Then the problem (3) has at least one solution x ∈ [v, z]_{D_{p}}.

If besides the L^{1}[v̄, z̄] condition is fulfilled with a coefficient q_{1} ∈ L_{∞}, and the Green operator of the auxiliary problem

ℑ_{1}x =^{def} ℑ_{0}x − q_{1}Θx = φ, x(1) = 0,  (5)

is antitonic, then the problem (3) has only one solution x ∈ [v, z].

Proof. Rewrite (3) in the form

(ℑ_{2}x)(·) =^{def} (ℑ_{0}x)(·) − q_{2}(·)(Θx)(·) = M_{2}[·, (Θx)(·)], x(1) = α

on the space D_{p}. This problem is equivalent to the equation

x = A_{2}x  (6)

with the completely continuous isotonic operator A_{2} : [v, z]_{C} ⟶ C, defined by

(A_{2}x)(·) =^{def} ∫_{0}^{1} G_{2}(·, s)M_{2}[s, (Θx)(s)] ds + u_{2}(·),

where u_{2}(·) is the solution of the semi-homogeneous problem

(ℑ_{2}x)(t) = 0, t ∈ [0, 1], x(1) = α,

G_{2}(·, ·) is the Green function of the problem

ℑ_{2}x = ξ, x(1) = 0.  (7)

We use here the fact that the Green operator G_{2} of the problem (7) has the representation G_{2} = WΓ [1,p.19], where Γ : L_{p} ⟶ L_{p} is a linear homeomorphism, consequently G_{2} is a completely continuous operator because of Lemma 2. Each continuous solution of the equation (6) belongs to the space D_{p}, because the operator A_{2} is defined on the order interval [v, z]_{C} of the space C and maps this interval into the space D_{p}. Obviously the isotonic operator Θ : C ⟶ L_{p} maps the order interval [v, z]_{C} into order interval [v̄, z̄]_{L_{p}}. The operator M_{2} : [v̄, z̄]_{L_{p}} ⟶ L_{p} is antitonic, therefore it maps the order

interval [v̄, z̄]_{L_{p}} into [M_{2}z̄, M_{2}v̄]_{L_{p}}. Let y =^{def} z − v. Then y(t) > 0, t ∈ [0, 1],

ℑ_{2}y ≤ M_{2}Θz − M_{2}Θv ≤ 0,

because of the antitonicity of M_{2} and

y(1) − ∫_{0}^{1}(ℑ_{2}y)(s) ds > 0.

Consequently, by Lemma 3 we have that the Green operator G_{2} : L_{p} ⟶ D_{p} ⊂ C of the problem (7) is antitonic. Thus

[G_{2}M_{2}v̄, G_{2}M_{2}z̄]_{D_{p}} ⊂ [G_{2}M_{2}v̄, G_{2}M_{2}z̄]_{C}.

Therefore the equation (6) may be considered in the order interval [v, z]_{C} of the space C. By virtue of the conditions (4) it follows that z(t) ≥ (A_{2}z)(t) and v(t) ≤ (A_{2}v)(t) for all t ∈ [0, 1]. Because of the isotonicity of the operator A_{2} : [v, z]_{C} ⟶ C this guarantees A_{2}[v, z]_{C} ⊂ [v, z]_{C}. For 1 < p ≤ ∞ the operator A_{2} : [v, z]_{C} ⟶ [v, z]_{C} is completely continuous as a product of the operators Θ : [v, z]_{C} ⟶ [v̄, z̄]_{L_{p}}, M_{2} : [v̄, z̄]_{L_{p}} ⟶ [M_{2}z̄, M_{2}v̄]_{L_{p}} and the completely continuous G_{2} : L_{p} ⟶ C.

Thus, the operator A_{2} maps the closed convex set [v, z]_{C} of the Banach space C into itself. In accordance with the Schauder’s fixed point theorem the equation (6) has at least one solution x ∈ [v, z]_{C}.

Let us show that the set of all solutions x ∈ [v, z]_{C} has a superior element x̄ ∈ [v, z]_{C} (the upper solution) and an inferior element \underline{x} ∈ [v, z]_{C} (the lower solution). Let x ∈ [v, z]_{C} be a solution of the equation (6). The sequence {z^{i}}, z^{i+1} = A_{2}z^{i}, z^{0} = z monotonically decreases and is bounded by x ∈ [v, z]_{C}, because the operator A_{2} maps the set [v, z]_{C} into itself. A compact monotone sequence {z^{i}} converges [2, p. 38] to x̄ = lim_{i→∞} z^{i}. Since this limit is a solution, the inequality x̄ ≥ x for any solution

x ∈ [v, z]_{C} is proved. Analogously we show the existence of the inferior solution \underline{x}.

Now we have to show that if the condition L^{1}[v̄, z̄] is fulfilled, the solution of the problem (3) is unique, i.e. x̄ = \underline{x}. Using the L^{1}[v̄, z̄], condition we rewrite the problem (3) in the form

(ℑ_{1}x)(·) = M_{1}[·, (Θx)(·)], x(1) = α.

This problem is equivalent to the equation

x = A_{1}x

on the order interval [v, z]_{C} of the space C with antitonic operator A_{1} : [v, z]_{C} ⟶ C, defined by

(A_{1}x)(·) =^{def} ∫_{0}^{1} G_{1}(·, s)M_{1}[s, (Θ = x)(s)] ds + u_{1}(·),

where G_{1}(·, ·) is the Green function of the problem (5), u_{1}(·) is the solution of semi-ho-mogeneous problem

(ℑ_{1}x)(t) = 0, t ∈ [0, 1], x(1) = α.

Consider the equality x̄ − \underline{x} = A_{1}x̄ − A_{1}\underline{x}. The left-hand side of the equality is non negative and the right-hand side is non positive, thus we get \underline{x} = x̄. □

5. Examples

Example 1. Consider the boundary-value problem

\{ \begin{matrix}ẍ(t) + \frac{1}{t} ẋ(t) = −β exp (− \frac{1}{|x(t)|}) , t ∈ [0, 1], \\ ẋ(0) = 0, x(1) = 0, \end{matrix}  (8)

where 0 ≤ β ≤ e^{2}. This problem describes processes arising in chemical reactor theory with cilindrical symmetry [7, p. 326], under the Arrhenius law. We consider this problem on the space D_{∞}.

As comparison functions we choose

v(t) ≡ 0, z(t) = \frac{β}{4} (1 − t^{2}) + \frac{1}{2} .

A trivial verification shows that the conditions (4) are fulfilled:

v̈(t) + \frac{v̇(t)}{t} + β exp (− \frac{1}{|v(t)|}) = 0,

z̈(t) + \frac{ż(t)}{t} + β exp (− \frac{1}{|z(t)|}) ≤ −β + β = 0, t ∈ [0, 1];

v(1) = 0 = x(1) < z(1) = \frac{1}{2} .

The function f(·, x) = −β exp (− \frac{1}{|x|}) satisfies the condition L^{2}[v, z] with the coeffi-

cient q_{2} ≡ 0. The boundary-value problem

ℑ_{0}x = ξ, x(1) = 0,

has for each ξ ∈ L_{∞} a unique solution x ∈ D_{∞}, and its Green function W (t, s) ≤ 0 on the square [0, 1] × [0, 1].

Besides, the function f(·, x) = −β exp (− \frac{1}{|x|}) satisfies the condition L^{1}[v, z] with the

coefficient q_{1} = −4βe^{−2}.

Taking the function y(t) = \frac{β}{4} (1 − t^{2}) we have:

(ℑ_{1}y)(t) = (ℑ_{0}y)(t) + 4βe^{−2}y(t) = −β + β^{2}e^{−2}(1 − t^{2}) < β(−1 + βe^{−2}) ≤ 0

and

y(1) − ∫_{0}^{1}(ℑ_{1}y)(s) ds = ∫_{0}^{1}[β − β^{2}e^{−2}(1 − s^{2})] ds = (β − \frac{2}{3} β^{2}e^{−2}) > 0,

since β ≤ e^{2}. Consequently, by Lemma 3, the Green operator G_{1} of the problem

ℑ_{0}x + 4βe^{−2}x = ξ, x(1) = 0

is antitonic. Then, because of Theorem 1 the problem (8) has a unique solution x ∈ D_{∞} such that

0 ≤ x(t) ≤ \frac{β}{4} (1 − t^{2}) + \frac{1}{2} , t ∈ [0, 1]. □

Example 2. Let

\{ \begin{matrix}ẍ(t) + \frac{2}{t} ẋ(t) = −β exp (− \frac{1}{|x(t)|}) , t ∈ [0, 1], \\ ẋ(0) = 0, x(1) = 0, \end{matrix}  (9)

be a nonlinear boundary-value problem, 37.28 ≤ β ≤ \frac{12}{17} e^{17/2}. This problem describes processes arising in chemical reactor with spherical symmetry [7, p. 326].

The problem (9) with such β has more than one solution on the space D_{p}, 1 < p ≤ ∞. Indeed, there are at least two pairs of functions

v_{1}(t) ≡ 0, z_{1}(t) = \frac{2(1 − t^{2})}{17} , v_{2}(t) = 4[erf(1) − erf(t^{2})], z_{2}(t) = \frac{β}{6} (1 − t^{2}).

The conditions (4) are fulfilled:

v̈_{1}(t) + \frac{2v̇_{1}(t)}{t} + β exp ( − \frac{1}{|v_{1}(t)|}) = 0,

z̈_{1}(t) + \frac{2ż_{1}(t)}{t} + β exp ( − \frac{1}{|z_{1}(t)|}) = β exp ( − \frac{17}{2(1 − t^{2})}) − \frac{12}{17} < 0,

v̈_{2}(t) + \frac{2v̇_{2}(t)}{t} + β exp ( − \frac{1}{|v_{2}(t)|}) = \frac{16 exp(−t^{4})}{\sqrt{π}} (4t^{4} − 3) +

+β exp (\frac{−0.25}{erf(1) − erf(t^{2})}) > 0,

z̈_{2}(t) + \frac{2ż_{2}(t)}{t} + β exp ( − \frac{1}{|z_{2}(t)|}) = β exp ( − \frac{6}{β(1 − t^{2})}) − β < 0,

since 37.28 ≤ β ≤ \frac{12}{17} e^{17/2}, t ∈ [0, 1]. The existence of solution of the problem (9) on each interval [v_{i}, z_{i}], i = 1, 2, follows from Theorem 1. Since the intervals [v_{1}, z_{1}], [v_{2}, z_{2}] are disjoint, the problem (9) has at least two solutions x_{1}, x_{2} ∈ D_{p}, 1 < p ≤ ∞, such that v_{1} ≤ x_{1} ≤ z_{1}, v_{2} ≤ x_{2} ≤ z_{2}. □

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Author’s address: Eduardo Mondlane University Department of Mathematics P.O. Box 257 - Maputo Mozambique