# Asymptote of solution of a differential equation without solving it

Consider the following differential equation (domain $\mathbb{R}$):

$$u(x) = 1 - u'(x)$$

and suppose $u(0) = 0$.

How can one prove that $u(x) \to 1$ for $x \to \infty$ without solving the equation explicitly?

• The substitution $u(x)=1-v(x)$ may aide your intuition. Jul 24, 2014 at 17:03
• @Semiclassical How does that help? It's just renaming $u'(x)$?
– Jam
Jul 24, 2014 at 17:05

Given the equation

$u(x) = 1 - u'(x), \tag{1}$

with the initial condition

$u(0) = 0, \tag{2}$

I begin by writing it in the more conventional form in which the derivative $u'(x)$ is isolated:

$u'(x) = 1 - u(x). \tag{3}$

Setting

$f(u) = 1 - u, \tag{4}$

we see that $f(u)$ is globally Lipschitz continuous with Lipschitz constant $K = 1$, since we have, for any $u_1, u_2 \in \Bbb R$,

$\vert f(u_1) - f(u_2) \vert = \vert (1 - u_1) - (1 - u_2) \vert = \vert u_2 - u_1 \vert; \tag{5}$

(5) in turn implies that (3) has a unique solulution for any initial condition $u_0 = u(x_0)$; furthermore any such solution extends to all $x \in \Bbb R$; these conclusions are simple consequences of the basic general theory of ODEs.

We next observe that, since solutions $u(x)$ of (1), (3) are continuously differentiable, $u'(x)$ is a continuous function of $x$, whence we may conclude from $u'(0) = 1 - u(0) = 1$ (see (2)) that for $\delta \in (0, 1)$ there exists $\epsilon > 0$ such that

$0 < 1 - \delta < u'(x) < 1 + \delta \tag{6}$

provided

$-\epsilon < x < \epsilon; \tag{7}$

thus for $x \in [0, \epsilon)$ we may write, using $u(0) = 0$,

$u(x) = u(x) - u(0) = \int_0^x u'(s) ds < \int_0^x (1 + \delta)ds = (1 + \delta) x < (1 + \delta)\epsilon < 1; \tag{8}$

if we take $\epsilon$ sufficiently small, i.e., $\epsilon < (1 + \delta)^{-1}$.

We next make note of the fact that the solution $u(x)$ of (1), (3) satisfying (2) cannot take the value $1$ for any finite $x_1 > 0$; for if $u(x_1) = 1$ then $u'(x_1) = 0$, and we conclude from this that $u(x) = 1$, $x \in \Bbb R$, is the unique solution such that $u(x_1) = 1$; but this contradicts the hypothesis that $u(0) = 0$; we conclude from this argument that $u(x) \ne 1$ for the solution $u(x)$ obeying (2). And since by (8) $u(x) < 1$ for positive $x$ sufficiently near $0$, we conclude, using the intermediate value theorem, that $u(x) < 1$ for all $x \ge 0$. Thus we have $u'(x) = 1 - u(x) > 0$; $u(x)$ is monotonically increasing for all $x \ge 0$.

A calculation similar to (6)-(8) shows that

$u(x) = u(x) -u(0) =\int_0^x u'(s) ds > \int_0^x (1- \delta) ds = (1 - \delta) x > 0, \tag{9}$

valid for sufficiently small $x > 0$; this, combined with the fact that $u(x)$ is monotonically increasing, implies that $u(x) \ge 0$ for $x \ge 0$, with strict inequality if $x > 0$. Though the non-negativity of $u(x)$ is not, strictly speaking, needed for the completion of our argument, it seems to me it is worth noting in passage, since it squares nicely with the overall spirit of qualitative analysis which motivates this question.

In any event, we have at this point, without ever explicitly solving (1)-(3), shown that:

A.) $0 \le u(x) < 1$ for $x \ge 0$, with $u(x) > 0$ for $x > 0$;

B.) $u(x)$ is a strictly increasing function of $x$ on $[0, \infty)$.

From points (A.) and (B.) we conclude that there is an $\alpha \in (0, 1]$ with

$\lim_{x \to \infty} u(x) = \alpha, \tag{10}$

indeed, we may take $\alpha = \sup \{u(x) \mid x \in [0, \infty)$}. If $\alpha = 1$, then we are done. If, on the other hand, $\alpha < 1$, then since $u(x) \le \alpha < 1$ we have $-1 < -\alpha \le - u(x)$ or

$0 < 1 - \alpha \le 1 - u(x) = u'(x); \tag{11}$

by (11), we have, for any $x_0 \in [0, \infty)$,

$u(x) - u(x_0) = \int_{x_0}^x u'(s) ds \ge \int_{x_0}^x (1 - \alpha) ds = (1 - \alpha)(x - x_0), \tag{12}$

or

$u(x) \ge u(x_0) + (1 - \alpha)(x - x_0); \tag{12}$

but now by taking $x$ sufficiently large we can force $u(x) > 1$; this contradiction implies we must have $\alpha = 1$. And we are done. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Assuming that $u(x)$ is at least $C^2(\mathbb{R})$ and that your domain is $\mathbb{R}$ (connected) you have:

$$u'(x)=1-u(x)$$

So $u'(x)=0 \iff u(x)=1$. Also $u'(0)=1$ so we just have to prove that $u'(x) \geq 0 \quad \forall x$

But $u'(x)$ is continuous so $u'(x) \geq 0 \quad \forall x$ (it can't go to a negative value because it should pass by zero and then the function is constant equal to 1 for all x because of the second derivative). So it's true that $u(x) \to 1$ when $x \to \infty$

You could be a bit more rigorous but at least that's the main idea.

EDIT: I am making use of the next fact:

$$u''(x)=-u'(x)$$

So $u''(x)=0 \iff u(x)=1$

• Would be nice if you could make it more rigerous Jul 24, 2014 at 17:29
• I'll do it when I get home because reading it again I've realized that what I said it's not completely true. Nevertheless, the reasoning can still be applied because if $u(x)>1$ then the derivative is negative. Jul 24, 2014 at 17:37
• Now it's better. Is it clearer now? Jul 24, 2014 at 18:23
• I don't really understand why it is sufficient to show that $u'(x) > 0$ for all $x$ and why this it the case. If $u'(x_0) = 0$ for some $x_0$, why is this automatically the case for all $x$? Oct 19, 2014 at 15:20