Applied Mathematics E-Notes, 1(2001), 1-4 \begin{matrix}c \\ ◯ \end{matrix} Available free at mirror sites of http://math2.math.nthu.edu.tw/∼amen/

Closed Form Solutions of Iterative Functional Differential Equations ^{∗}

Wen-Rong Li^{†}, Sui Sun Cheng^{‡}, Tzon Tzer Lu^{§}

Received 20 July 2000

Abstract

Solutions of the form x(z) = λz^{μ} are found for the iterative functional differential equation x^{(n)}(z) = (x (x (...x (z))))^{k} .

Let h = h(z) be a function of a complex variable. The iterates of h are defined by h^{[0]} = I (the identity map), h^{[1]} = h, h^{[2]} = h ∘ h, ..., etc. A number of recent studies are concerned with iterative functional differential equations. In particular, in [1], the equation

x′(z) = x^{[m]}(z), m ≥ 2,

has been considered. Such an equation arises in problems related to motions of charged particles with retarded interactions [2]. Since it is quite different from the usual differential equations, the standard existence and uniqueness theorems cannot be applied. It is therefore of interest to find some or all of its solutions. It is shown in [1] that analytic solutions exist and series form can be given. However, we will show below that solutions of the form

x(z) = λz^{μ}  (1)

also exist. The idea of finding such solutions is new even though it reminds us of the usual idea of finding exponential solutions to linear ordinary differential equations with constant coefficients. The procedure for obtaining such solutions works equally well for more general equations. In this note, we will illustrate our procedure by considering a slightly more general equation of the form

x^{(n)}(z) = az^{j} (x^{[m]}(z))^{k} ,  (2)

where k, m, n are positive integers, j is a nonnegative integer, a is a complex number, and x^{(n)}(z) is the n-th derivative of x(z). We assume m ≥ 2 and a ≠ 0 to avoid trivial cases. It is also clear that the zero function is a trivial solution of (2), which is of no interest.

∗Mathematics Subject Classifications: 35C20, 35D10.

†Department of Mathematics, Binzhou Normal College, Binzhou, Shandong 256604, P. R. China

‡Department of Mathematics, Tsing Hua University, Hsinchu, Taiwan 30043, R. O. China

§Department of Applied Mathematics, Sun Yat-sen Univeristy, Kaohsiung, Taiwan 80424, R. O.

China

Substituting (1) into (2), we obtain

λμ(μ − 1) ⋯ (μ − n + 1)z^{μ−n} = aλ^{k(1+μ+...+μ^{m−1})}z^{kμ^{m}+j}.

This prompts us to consider the equations

λμ(μ − 1) ⋯ (μ − n + 1) = aλ^{k(1+μ+...+μ^{m−1})},  (3)

and

kμ^{m} + j = μ − n.  (4)

First of all, we assert that the polynomial

f(z) = kz^{m} − z + n + j

does not have any real roots if m is even, and has exactly one real root if m is odd. Indeed, for even m and real z, by solving

f′(z) = kmz^{m−1} − 1 = 0,

we see that the minimum of f occurs at the root ρ = 1/ ^{m−1} \sqrt{km} ∈ (0, 1). Hence

f(z) ≥ f(ρ) = ρ (\frac{1}{m} − 1) + n + j > (\frac{1}{m} − 1) + n + j > 0

for all real z. If m is odd, then f′ has two zeros ±ρ. Since

min_{z≥−ρ} f(z) = min{f(ρ), f(−ρ)} = f(ρ) > 0,

f does not have any real roots greater than or equal to −ρ. Furthermore, since f(−ρ) and f(−∞) have opposite signs, f has at least one real root in (−∞, −ρ). Finally, since f′(z) > 0 for all z < −ρ, f is increasing in (−∞, −ρ). So f has exactly one real root which is negative. As a consequence, the roots of f in either case cannot be 0, 1, ..., nor n − 1.

Next, we assert that f(z) has simple roots only. Suppose not, let r be a double root of f, then it is a root of f′ and

f(z) − \frac{z}{m}f′(z) = \frac{1 − m}{m}z + n + j.  (5)

Hence (5) implies that r = m(n + j)/(m − 1) is real and positive, which is impossible by our previous assertion.

Let μ_{1}, ..., μ_{m} be the roots of (4). In view of the above results, μ_{1}, ..., μ_{m} are pairwise distinct and each one of them is different from 0, 1, ..., or n − 1. Furthermore, in view of (3) and (4), we have

λμ_{i}(μ_{i} − 1) ⋯ (μ_{i} − n + 1) = aλ^{k(1−μ_{i}^{m})/(1−μ_{i})} = aλ^{(k+n+j−μ_{i})/(1−μ_{i})}

for i = 1, ..., m, from which we obtain

λ_{i} = [\frac{μ_{i}(μ_{i} − 1) ⋯ (μ_{i} − n + 1)}{a}]^{(1−μ_{i})/(k+n+j−1)} , i = 1, ..., m.  (6)

In other words, we have found m distinct solutions of the form:

x_{i}(z) = λ_{i}z^{μ_{i}}, i = 1, 2, ..., m,  (7)

where μ_{1}, ..., μ_{m} are roots of (4) and λ_{1}, ..., λ_{m} are defined by (6).

THEOREM 1. Let D be a domain of the complex plane C which does not include the negative real axis (nor the origin). Then there exist m distinct (single valued and analytic) power functions of the form (7) which are solutions of (2) defined on D.

We remark that each solution x_{i}(z) = λ_{i}z^{μ_{i}} has a nontrivial fixed point α_{i}. Indeed, from λ_{i}α_{i}^{μ_{i}} = α_{i}, we find

α_{i} = λ_{i}^{1/(1−μ_{i})} = [μ_{i}(μ_{i} − 1) ⋯ (μ_{i} − n + 1)]^{1/(k+n+j−1)} ≠ 0.  (8)

In terms of the fixed point α_{i}, we may therefore write x_{i}(z) in the form

x_{i}(z) = α_{i}^{1−μ_{i}}z^{μ_{i}}.  (9)

COROLLARY. Let μ_{1}, ..., μ_{m} be the roots of (4), and α_{1}, ..., α_{m} given by (8). Then in a neighborhood of each point α_{i}, i = 1, ..., m, equation (2) has an analytic solution of the form

x_{i}(z) = α_{i} + μ_{i}(z − α_{i}) + \frac{μ_{i}(μ_{i} − 1)}{2!α_{i}}(z − α_{i})^{2} + ...

+\frac{μ_{i}(μ_{i} − 1)...(μ_{i} − n + 1)}{n!α_{i}^{n−1}}(z − α_{i})^{n} + ... .

Indeed, in view of (9),

x_{i}(z) = α_{i}^{1−μ_{i}}z^{μ_{i}} = α_{i} (1 + \frac{z − α_{i}}{α_{i}} )^{μ_{i}}

= α_{i} [1 + \frac{μ_{i}}{1!} (\frac{z − α_{i}}{α_{i}} ) + \frac{μ_{i}(μ_{i} − 1)}{2!} (\frac{z − α_{i}}{α_{i}} )^{2} + ...]

as required.

As an example, consider the equation

x′(z) = x(x(z)).

Then (4) is reduced to

μ^{2} − μ + 1 = 0,

which has roots μ_{±} = (1 ± \sqrt{3}i)/2. And from (3), we find λ_{−} = μ_{−}^{1/μ_{−}} ≈ 2.145 − 1.238i, λ_{+} = μ_{+}^{1/μ_{+}} ≈ 2.145 + 1.238i. Since |μ_{±}| = 1 and μ_{±}^{6} = 1, α_{±} = μ_{±} are roots of unity. This shows that the requirements in the main Theorem in [1] does not hold. Therefore, we have found analytic solutions which cannot be guaranteed by the main Theorem in [1].

As our final remarks, note that the parameters j and k in (2) can be extended to real or even complex numbers. In this case, (4) might have more than one real roots, but the solutions (7) remain valid. It is possible that (4) has double roots, then the solutions in (7) corresponding to these roots are the same. Also some of its roots μ_{i} might equal to 0, 1, ..., or n − 1. But then λ_{i} = 0 by (3), and x_{i}(z) in (7) becomes the trivial solution.

References

[1] J. G. Si, W. R. Li and S. S. Cheng, Analytic solutions of an iterative functional differential equation, Computers Math. Applic., 33(1997)(6), 47-51.

[2] E. Eder, The functional differential equation x′(t) = x(x(t)), J. Diff. Eq., 54(1984), 390-400.