The Uniqueness of the Periodic Solution for A Class of Differential Equations^{1}

Zhaosheng Feng^{2,3}

2Department of Mathematics Texas A&M University, College Station, TX, 77843-3368 USA 3zsfeng@math.tamu.edu

Abstract. In this paper we are concerned with a class of nonlinear differential equations and obtaining the sufficient conditions for the uniqueness of the periodic solution by using Brouwer’s fixed point theory and the Sturm Theorem. Keywords. uniqueness, fixed point, existence, control theory method, Sturm comparison theorem. AMS (MOS) subject classification: 34C25

1 Introduction

This paper is concerned with the uniqueness of periodic solutions for the following differential equations

μ(t, x, x′)x′ = f(t, x)  (1)

where (t, x) ∈ [0, 2π]× ℝ and μ(t, x, z) ∈ C′([0, 2π]×ℝ×ℝ), f ∈ C^{2}([0, 2π]× ℝ), and μ(t, x, x′), f(t, x) are 2π-periodic functions with respect to t.

It is easy to see that equation (1) is more general than the classical ordinary differential equation

x′ = f(t, x) for all (t, x) ∈ [0, 2π] × ℝ  (2)

During the past three decades, with the use of topological degree theory, general critical point theory, fixed point theory, boundary value condition theory and cross-ratio method, some profound results on the existence and the number of periodic solutions for equation (2) have been presented ( see references [1-15] and the reference therein ). But none of these papers are concerned with the uniqueness of the periodic solutions for equation (1). However, when does the equations (1) or (2) have a unique 2π-periodic solution ?

In the present paper, using the Brouwer’s fixed pointed theorem, the Sturm Theorem, and some results of the optimal control theory method, we are trying to obtain two theorems for the sufficient conditions which guarantee that equation (1) has a unique 2π-periodic solution.

Consider the following conditions

(H1): Suppose that f(t, x) = [x − x_{0}(t)] · G(t, x). Here f(t, x) and x_{0}(t) are 2π-periodic continuous functions, and x_{0}(t) separates the domain 0 ≤ t ≤ 2π into two parts, denoted by Ω_{1} and Ω_{2} ( we assume that Ω_{1} is above x = x_{0}(t) and Ω_{2} is below x = x_{0}(t) ). Suppose that there exist two sets S_{1}: {(t, x)|x_{1} ≤ x ≤ x_{2}, 0 ≤ t ≤ 2π} and S_{2}: {(t, x)|x_{3} ≤ x ≤ x_{4}, 0 ≤ t ≤ 2π} in the domains Ω_{1} and Ω_{2}, respectively, such that G(t, x) has the same sign for all (t, x) in S_{1} and S_{2}. Here we assume that either x = x_{1} or x = x_{4} doesn’t intersect with x = x_{0}(t).

1Supported by the NSF.

(H2): Suppose that f_{tx}, f_{xx} : ℝ^{2} → ℝ and [\frac{f(t,x)}{μ(t,x,z)}]_{x} : ℝ^{3} → ℝ exist, and are 2π-periodic continuous with respect to t, where f_{tx} denotes the partial derivative with respect to t and x, and [\frac{f(t,x)}{μ(t,x,z)}]_{x} denotes the derivative of

the quotient \frac{f(t,x)}{μ(t,x,z)} with respect to x. Suppose that there exist two positive real numbers L and M, one non-negative integer N, and two non-negative continuous functions u_{1}(t) and u_{2}(t), such that

L ≤ μ(t, x, x′) ≤ M

−u_{1}(t) ≤ f_{tx} + f_{xx} \frac{f(t, x)}{μ(t, x, z)} + f_{x}[\frac{f(t, x)}{μ(t, x, z)}]_{x} ≤ −u_{2}(t)

and

(N + 1)^{2} ≩ \frac{u_{1}(t)}{L} ≥ \frac{u_{2}(t)}{M} ≩ N^{2}

where ≩ indicates “greater than or equal to” but not identically equal.

(H3): Suppose that f_{x}, f_{tx}, and f_{xx} are continuous and 2π-periodic with respect to t. Let (k−1)^{2} < A < k^{2} < B < ∞, where k is the minimal positive integer suiting the inequality. Assume that there exists a β(x) ∈ C[0, 2π] such that

−A ≥ f_{tx} + f_{xx} · f + (f_{x})^{2} ≥ −β(x) ≥ −B, ∫_{0}^{2π} β(t)dt < 2πA + 2(B − A)α_{k}

where α_{k} is the minimal positive root of

\sqrt{A} cot (\sqrt{A}\frac{π − x}{2k}) = \sqrt{B} tan (\sqrt{B} \frac{x}{2k} )

Our main results are the following theorems:

Theorem 1: If H(1) and H(2) hold, then equation (1) has a unique 2π-periodic solution.

Theorem 2: If H(1) and H(3) hold, then equation (2) has a unique 2π-periodic solution.

2 The Proof of Main Results

Proof of Theorem 1. Consider the case G(t, x) < 0 for all (t, x) in S_{1} and S_{2}. Obviously, in the domain Ω_{1}, we have f(t, x(t)) < 0 for any x(t) > x_{0}(t).

In the domain Ω_{2}, we have f(t, x(t)) > 0 for any x(t) < x_{0}(t). In the compact set [0, 2π]×[x_{1}, x_{2}], f(t, x) has the maximal value, denoted by m_{1}, (m_{1} < 0). Hence, we can choose some negative number k_{1} such that k_{1} > \frac{m_{1}}{L} and the whole segment l_{1}: x_{1}(t) = k_{1}t + x_{2} is inside the set S_{1} whenever 0 ≤ t ≤ 2π. Similarly, in the compact set [0, 2π] × [x_{3}, x_{4}], f(t, x) has the minimal value, denoted by m_{2}, (m_{2} > 0). Thus, we also can choose some positive number k_{2} such that k_{2} < \frac{m_{2}}{M} and the whole segment l_{2}: x_{2}(t) = k_{2}t + x_{3} is inside the set S_{2} whenever 0 ≤ t ≤ 2π.

\{ \begin{matrix}x_{2} > x_{1} \\ \frac{dx_{1}(t)}{dt} & = & k_{1} \\ & ≥ & _{\frac{1}{μ((t,x_{1}(t),x′_{1}(t))}}f[t, x_{1}(t)]  t ∈ [0, 2π] \end{matrix}

and

\{ \begin{matrix}x_{4} > x_{3} \\ \frac{dx_{4}(t)}{dt} & = & k_{2} \\ & ≤ & _{\frac{1}{μ((t,x_{2}(t),x′_{2}(t))}}f[t, x_{2}(t)]  t ∈ [0, 2π]. \end{matrix}

Letting L_{1} = x_{3}, L_{2} = x_{2}, L_{3} = x_{4}, and L_{4} = x_{1}, clearly, we can see that L_{1} < L_{2} and L_{3} < L_{4}.

Define an operator T_{1}: C[L_{1}, L_{2}]→ C[L_{3}, L_{4}]:

∀x(0) ∈ C[L_{1}, L_{2}], T_{1}x(0) = x(2π).

Since C[L_{3}, L_{4}]⊆C[L_{1}, L_{2}], we get that T_{1} is continuous and maps x(t) from C[L_{1}, L_{2}] to C[L_{3}, L_{4}]. Consequently,

T_{1}(C[L_{1}, L_{2}]) ⊆ C[L_{1}, L_{2}].

By the Brouwer’s fixed point theorem, we obtain that T_{1} has at least one fixed point in C[L_{1}, L_{2}]. That is, equation (1) has at least one 2π-periodic solution.

In the case G(t, x) > 0 for all (t, x) in S_{1} and S_{2}, the proof is similar to the above. In the domain Ω_{1}, we have f(t, x(t)) > 0 for any x(t) > x_{0}(t). In the domain Ω_{2}, we have f(t, x(t)) < 0 for any x(t) < x_{0}(t). In the compact set [0, 2π] × [x_{1}, x_{2}], f(t, x) has the minimal value, denoted by n_{1}, (n_{1} > 0). Hence, we can choose some positive number such that k_{3} < \frac{n_{1}}{M} and the whole segment l_{3}: x_{3}(t) = k_{3}t + x_{1} is inside the set S_{1} whenever 0 ≤ t ≤ 2π. Similarly, in the compact set [0, 2π] × [x_{3}, x_{4}], f(t, x) has the maximal value, denoted by n_{2}, (n_{2} < 0). Thus, we also can choose some negative number k_{4} such that k_{4} > \frac{n_{2}}{L} and the whole segment l_{4}: x_{4}(t) = k_{4}t + x_{4} is inside the set S_{2} whenever 0 ≤ t ≤ 2π.

\{ \begin{matrix}x_{2} > x_{1} \\ \frac{dx_{3}(t)}{dt} & = & k_{3} \\ & ≤ & _{\frac{1}{μ((t,x_{1}(t),x′_{1}(t))}}f[t, x_{1}(t)]  t ∈ [0, 2π] \end{matrix}

and

\{ \begin{matrix}x_{4} > x_{3} \\ \frac{dx_{4}(t)}{dt} & = & k_{4} \\ & ≥ & _{\frac{1}{μ((t,x_{2}(t),x′_{2}(t))}}f[t, x_{2}(t)]  t ∈ [0, 2π]. \end{matrix}

Letting L_{1} = x_{3}, L_{2} = x_{2}, L_{3} = x_{4}, and L_{4} = x_{1}, clearly, we can see that L_{1} < L_{2} and L_{3} < L_{4}.

Define an operator T_{2}: C[L_{1}, L_{2}]→ C[L_{3}, L_{4}]:

∀x(2π) ∈ C[L_{1}, L_{2}], T_{2}x(2π) = x(0).

Since C[L_{3}, L_{4}] ⊆C[L_{1}, L_{2}], we have that T_{2} is continuous and maps x(t) from C[L_{1}, L_{2}] to C[L_{3}, L_{4}]. Consequently,

T_{2}(C[L_{1}, L_{2}]) ⊆ C[L_{1}, L_{2}].

By the Brouwer’s fixed point theorem, we obtain that T_{2} has at least one fixed point in C[L_{1}, L_{2}]. That is, equation (1) has at least one 2π-periodic solution.

Differentiating both sides of equation (1) with respect to t, we have

(μ(t, x, x′)x′)′ = F (t, x)

where F (t, x) = f_{t}(t, x) + f_{x}(t, x) · \frac{f(t,x)}{μ(t,x,x′)}. Here f_{t}(t, x) denotes the partial derivative with respect to t, and F_{x}(t, x) denotes the partial derivative with respect to x. By recalling the assumption (H2), we know that −u_{1}(t) ≤ F_{x}(t, x) ≤ −u_{2}(t).

Define an operator T: C_{2π}^{1}[0, 2π] → C_{2π}^{1}[0, 2π], where C_{2π}^{1}[0, 2π] denotes the set of all the 2π-periodic continuous differentiable functions in ([0, 2π] × ℝ). For any ω ∈ C_{2π}^{1}[0, 2π], assume that T_{ω} = T_{ω}(t) is a 2π-periodic solution of the equation

(μ(t, ω, ω′)x′)′ = ∫_{0}^{1} F_{x}(t, θω)dθx + f(t, 0).  (3)

Next, we prove that equation (3) has at most one 2π-periodic solution under the conditions (H2). Suppose that T_{y} = T_{y}(t) is another 2π-periodic solution of equation (3). Then b(t) = T_{ω}(t) − T_{y}(t) must be a 2π-periodic solution of

(μ(t, ω, ω′)x′)′ = ∫_{0}^{1} F_{x}(t, θω)dθx.  (4)

Compare equation (4) with

(Mx′)′ = −B(t)x.  (5)

Letting dt = μ(t, ω, ω′)dτ, t(τ) = ∫_{0}^{τ} μ[t(s), ω(t(s)), ω′(t(s))]ds, then equations (5) and (4) are equivalent to the following systems, respectively,

\{ \begin{matrix}\frac{dx}{dτ} = \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]}{M}x_{1} \\ \frac{dx_{1}}{dτ} = −μ[t(τ), ω(t(τ)), ω′(t(τ))] + B(t)x \end{matrix}  (6)

and

\{ \begin{matrix}\frac{dx}{dτ} = x_{1} \\ \frac{dx_{1}}{dτ} = μ[t(τ), ω(t(τ)), ω′(t(τ))] · ∫_{0}^{1} F_{x}(t, θω)dθx. \end{matrix}  (7)

Assume that (6) has the solution of the form

( \begin{matrix}x \\ x_{1} \end{matrix} ) = ( \begin{matrix}cos X(τ) & sin X(τ) \\ − sin X(τ) & cos X(τ) \end{matrix} ) · ( \begin{matrix}C_{1}(τ) \\ C_{2}(τ) \end{matrix} )  (8)

where X(τ) is to be determined, and Y (τ), C(τ) are as follows

Y (τ) = ( \begin{matrix}cos X(τ) & sin X(τ) \\ − sin X(τ) & cos X(τ) \end{matrix} ) C(τ) = ( \begin{matrix}C_{1}(τ) \\ C_{2}(τ) \end{matrix} )

X′(τ) · ( \begin{matrix}− sin X(τ) & cos X(τ) \\ − cos X(τ) & − sin X(τ) \end{matrix} ) · C(τ) + Y (τ)C′_{1}(τ)

= ( \begin{matrix}0 & \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]}{M} \\ −μ[t(τ), ω(t(τ)), ω′(t(τ))]B(t) & 0 \end{matrix} ) · Y (τ)C(τ).

Simplifying the above, we have

C′(τ) = ( ( \begin{matrix}E_{11} & \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]}{M} − μ[t(τ), ω(t(τ)), ω′(t(τ))] · ∫_{0}^{1} F_{x}(t, θω)dθ sin^{2} X \\ E_{21} & E_{22} \end{matrix} )

+X′(τ) · ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ) · C(τ)

where

\begin{matrix}E_{11} = μ[t(τ), ω(t(τ)), ω′(t(τ))]B(t(τ)) sin X cos X − \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]) sin X cos X}{M} \\ E_{21} = −μ[t(τ), ω(t(τ)), ω′(t(τ))]B(t(τ)) cos^{2} X − \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]) sin^{2} X}{M} + X′ \\ E_{22} = −μ[t(τ), ω(t(τ)), ω′(t(τ))]B(t(τ)) cos X sin X + \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]) sin X cos X}{M} \end{matrix} \} .

Let X_{1}(τ) denote the solution of the following initial value problem

\{ \begin{matrix}X′_{1} = \frac{μ[t(τ),ω(t(τ)),ω′(t(τ))]}{M} cos^{2} X_{1} + μ[t(τ), ω(t(τ)), ω′(t(τ))]B(t(τ)) sin^{2} X \\ X_{1}(0) = 0. \end{matrix} (9)

Similarly, assume that (7) has the solution of the form (8), then

C′(τ) = ( ( \begin{matrix}∗ & cos^{2} X − μ[t(τ), ω(t(τ)), ω′(t(τ))] sin^{2} X ∫_{0}^{1} F_{x}(t, θω)dθ \\ ∗ & ∗ \end{matrix} )

+X′(τ) · ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} ) ) · C(τ).

Letting X_{2}(τ) denote the solution of the following initial value problem

\{ \begin{matrix}X′_{2} = cos^{2} X_{2} − μ[t(τ), ω(t(τ)), ω′(t(τ))] sin^{2} X_{2} ∫_{0}^{1} F_{x}(t, θω)dθ \\ X_{2}(0) = 0. \end{matrix}  (10)

Note that for any τ_{0} ∈ ℝ^{+} \ {0}, by the Sturm comparison theorem for the first order ordinary differential equations, we know that X_{1}(τ_{0}) < X_{2}(τ_{0}). From (6) and (7), if τ_{1}, τ_{2} ∈ ℝ^{+} \ {0} are zeros of the nontrivial solutions x = x(t), x_{1} = x_{1}(t) of equations (6) and (7), respectively, and satisfy the initial value problems

x(0) = 0, x_{1}(0) = 0

then

X_{1}(τ_{1}), X_{2}(τ_{2}) ∈ {nπ, n = 0, 1, 2, ⋯ }.  (11)

On the other hand, if τ_{1}, τ_{2} ∈ ℝ^{+}\{0} satisfy (11), then (6),(7) must have the nontrivial solutions x = x(t) and x_{1} = x_{1}(t), such that x(0) = x(t(τ_{1})) = 0 and x_{1}(0) = x_{1}(t(τ_{2})) = 0, respectively.

(I). In the case N ≥ 1, notice that X_{2}[τ(2π)] ≥ X_{1}[τ(2π)]. By using the Sturm comparison theorem, we obtain that X_{1}[τ(2π)] > 2Nπ. Thus, X_{2}[τ(2π)] > 2Nπ. Again by the Sturm comparison theorem, we obtain that every nontrivial solution of equation (4) must have at least 2N zeros in the open interval (0, 2π). Without the loss of generality, we assume that b(0) = b(2π) = 0.

Compare equation (4) with

(Lx′)′ = −u_{2}(t)x.

By using the similar arguments as the above, and letting X_{3}(τ)andX_{4}(τ) denote the solutions of the following initial value problems, respectively, we have

\{ \begin{matrix}X′_{3} = \frac{L}{μ[t(τ),ω(t(τ)),ω′(t(τ))]} cos^{2} X_{3} − L sin^{2} X_{3} ∫_{0}^{1} F_{x}[t(τ), θω]dθ \\ X_{3}(0) = 0 \end{matrix}

and

\{ \begin{matrix}X′_{4} = cos^{2} X_{4} + Lu_{1}(t(τ)) sin^{2} X_{4} \\ X_{4}(0) = 0. \end{matrix}

Notice that X_{4}[τ(2π)] ≥ X_{3}[τ(2π)]. Also by using the Sturm comparison theorem, we obtain that X_{4}[τ(2π)] < 2(N + 1)π. Thus,

X_{3}[τ(2π)] < 2(N + 1)π.  (12)

Since b(t) is 2π-periodic, it is impossible for b(t) to have an odd number of zeros in [0, 2π]. By our previous hypothesis b(0) = b(2π) = 0 and the above conclusion that b(t) has at least 2N zeros in (0, 2π), we can conclude that b(t) has at least 2(N + 1) zeros in [0, 2π]. Hence, X_{3}[τ(2π)] ≥ 2(N + 1)π. This yields a contradiction with (12).

(II). In the case N = 0, we can conclude that b(t) has zeros in ℝ by the Sturm comparison theorem. Without the loss of generality, we assume that b(0) = 0. Due to the periodicity, b(t) has at least two zeros in (0, 2π). The rest of the proof is similar to that of the case (I), so it is omitted.

¿From the above proof, we know that equation (1) has at least a 2πperiodic solution. However, we know that every 2π-periodic solution of equation (1) must be 2π-periodic solution of equation (μ(t, x, x′)x′)′ = F (t, x) . Therefore, we can conclude that under the assumptions (H1) and (H2) equation (1) has a unique 2π-periodic solution. The proof is complete.

Proof of Theorem 2. It is well-known that the following result of the optimal control theory can be widely applied. For the detailed proof, we may go back to [18]. Some interesting applications of the optimal control theory method to several boundary value problems for ordinary differential equations can be found in [16-19]. Let (k − 1)^{2} < A < k^{2} < B, where k is the minimal positive integer suiting the inequality. Suppose that u ∈ L[0, 2π] satisfying

A ≤ u(t) ≤ B and ∫_{0}^{2π} u(t)dt < 2Aπ + 2(B − A)α_{k}

where α_{k} = α(\frac{1}{2k}), the minimal positive root of

\sqrt{A} cot (\sqrt{A}λ(π − x)) = \sqrt{B} tan(\sqrt{B}λx)

for λ = \frac{1}{2k}. Then the periodic boundary value problem

\{ \begin{matrix}y′′ + u(t)y = f(t) \\ u(t) = u(t + 2π) \\ y(0) = y(2π), y′(0) = y′(2π) \end{matrix}

has a unique 2π-periodic solution for each 2π-periodic function f ∈ L[0, 2π].

To prove that equation (2) has at least one 2π-periodic solution, we can use the same arguments as that of theorem 1. To prove the uniqueness, by differentiating both sides of equation (2) with respect to t, we have

x′′ = F (t, x)  (13)

where F (t, x) = f_{t} + f · f_{x} and F_{x}(t, x) = f_{tx} + f_{xx} · f + (f_{x})^{2}. Let X_{1}(t) and X_{2}(t) be any two 2π-periodic solutions of equation (13). Then b(t) = X_{1}(t) − X_{2}(t) is a 2π-periodic solution of

x′′ = ∫_{0}^{1} F_{x}[t, x_{2}(t) + θx_{1}(t)]dθx.  (14)

¿From the assumption, we see

A ≤ − ∫_{0}^{1} F_{x}[t, x_{2}(t) + θx_{1}(t)]dθx ≤ β(x).

By using the above result of optimal control theory, b(t) ≡ 0 for all t ∈ ℝ. Therefore, equation (2) has a unique 2π-periodic solution. The proof is complete.

3 Conclusion

¿From Section 2, we can see that using the Sturm Theorem as well as the Brouwer’s fixed pointed theorem is really an effective approach for equations (1) and (2). It is easily noted that even when a(t, x, x′) = 1, our conditions are different from all those in the previous references [1-8]. We also can use this method to study the sublinear Duffing equations investigated in [20].

4 Acknowledgements

The author thanks Prof. Goong Chen and the referee for their helpful suggestions.

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