Mathematics
R. E. Vinograd
Submitted 1957-01-01 | RussiaRxiv: ru-195701.88877 | Translated from Russian

Abstract

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Mathematics

R. E. Vinograd

ON THE INSUFFICIENCY OF THE METHOD OF CHARACTERISTIC EXPONENTS IN APPLICATION TO NONLINEAR EQUATIONS

(Presented by Academician I. G. Petrovsky, December 10, 1956)

Consider a system of (n) differential equations in vector form

[
\frac{dx}{dt}=F(t,x),
\tag{1}
]

for which (F(t,0)\equiv 0) and the Lipschitz condition is satisfied

[
\left|F(t,x_1)-F(t,x_2)\right|\leq K\left|x_1-x_2\right|.
\tag{2}
]

From the estimate of the norms of solutions known under these conditions,

[
|x(t)|\leq |x(0)|e^{kt},
]

it follows that the characteristic exponents of all solutions are bounded:

[
\lambda=\varlimsup_{t\to\infty} t^{-1}\ln |x(t)|\leq K,
]

and therefore there exists (\sup \lambda=\Lambda\leq K). It is also known ((^1)) that in the case of linearity of system (1), i.e., when (F(t,x)=A(t)x), the following assertion is true:

For every (\varepsilon>0) there exists a constant (B_\varepsilon), the same for all solutions, such that

[
|x(t)|\leq |x(0)|B_\varepsilon e^{(\Lambda+\varepsilon)t}.
\tag{3}
]

In the nonlinear case, from the definitions of (\lambda) and (\Lambda) there likewise follows an inequality of the form (3), in which, however, (B_\varepsilon) depends on the choice of the solution (x(t)), or, what is the same, on its initial point (x(0)).

The question arises: is it possible also in the nonlinear case to choose (B_\varepsilon) independently of (x(0))?

A priori, inequality (3) may fail for two reasons:

a) (B_\varepsilon) exists for each sphere (|x(0)|=r<R), but grows without bound as (r\to 0) (or (r\to\infty));

b) for no sphere does (B_\varepsilon) exist (since from the existence of (B_\varepsilon) for (r=r_0) there follows its existence for (r<r_0)).

Case a) still leaves some possibility for using the number (\Lambda); thus, the following theorem holds:

Theorem 1. In case a), from the negativity of (\Lambda) there follows asymptotic stability of the trivial solution (x=0) of system (1).

In case b), however, the method of characteristic exponents proves unsuitable for studying solutions of system (1) in the sense in which this is done for linear systems; for example, from the condition (\Lambda<0) even ordinary stability of the solution (x=0) does not follow.

The following examples show that both cases are indeed encountered, and moreover with the right-hand side of (1) independent of (t).

Example 1. The system consists of one equation with one unknown (x) and is given in the strip (t \geqslant 0,\ |x|\leqslant 1):

[
\frac{dx}{dt}=\frac{2\ln^{2}x}{1+\ln^{2}x}\,x
\quad \text{for } x\neq 0;
]

[
\frac{dx}{dt}=0
\quad \text{for } x=0.
]

The Lipschitz condition is satisfied, since the derivative of the right-hand side with respect to (x) is bounded. The general solution has the form

[
x(t)=\pm e^{\,t-c-\sqrt{1+(t-c)^2}},
]

so that its characteristic exponent is (\lambda=0). The sphere (|x(0)|=r) consists of two points: (x(0)=\pm r), and for them (3) is satisfied; however, the constant (B_\varepsilon) cannot be uniform as (r\to 0).

Indeed, assuming (it is enough to consider (x(t)>0)) that

[
\frac{x(t)}{x(0)}

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Mathematics