long-time behaviour of nonhom. lin. ODE

**Debelix** · March 11th 2011, 01:56 AM

Hi

given the nonhomogeneous linear differential-equation
$x'(t) + a(t) x(t) = b(t)$
with
$a: [0, \infty[ \rightarrow [0,\infty[,$
$b: [0, \infty[ \rightarrow [0,\infty[$
and no explicit solution x(t) of the problem, i like to show that
$x(t) \rightarrow 0.$

in addition for $t \rightarrow \infty$ i'm able to show that:
$a(t) \rightarrow 0$ and $b(t) \rightarrow 0$

do you know any theory i'm allowed to use in this case? i just find some results in stabilitytheory for homogeneous case with a constant coefficient a(t).

thanks for any hints or suggestions in advance.
kind regards, debelix

**Ackbeet** · March 11th 2011, 02:03 AM

Ah, but you do know that, due to the integrating factor method, that

$\displaystyle x(t)=e^{-\int a(s)\,ds}\int e^{\int a(s)\,ds}\,b(s)\,ds+Ce^{-\int a(s)\,ds}.$

Does that get you started?

**Debelix** · March 11th 2011, 02:18 AM

thanks for your reply.

the problem is that for $t \rightarrow \infty$
$e^{- \int a(s) \ ds} \rightarrow 0$

but

$\int e^{\int a(s) \ ds} b(s) \ ds \rightarrow \infty$

is there any other way instead of analysing these general solution?

**Ackbeet** · March 11th 2011, 02:24 AM

You have a confusing statement in the OP that could use some clarification:

In addition for $t\to\infty$ I'm able to show that:
$a(t)\to 0$ and $b(t)\to 0.$

Does that mean you have already shown this? Because, given the conditions of the original problem, you definitely cannot conclude that. You need assumptions on the nature of those two functions.

**Debelix** · March 11th 2011, 02:42 AM

ok, to avoid a confusion (my english isn't the best):
i have
$x'(t) = \bigg( 1 / f(t)\bigg)' ( 1+ x(t)) - 2f(t) e^{\frac{-1}{f(t)}} x(t)$

my assumptitions on f are
1. $f:[0,\infty[ \rightarrow [0,\infty[$
2. $f(t) \downarrow 0$ and
3. $f(t) \geq \frac{c}{\ln{(2+t)}} \ \mbox{for every} \ t \in [0,\infty[ \ \mbox{and $c>0$ sufficiently large}$

the author of the article i'm working with says: "it's obvious that $x(t) \rightarrow0$ for $t \rightarrow \infty$ "
for me, it isn't that obvious

do you have any ideas?

EDIT (forgot the c in assumption 3)

**Debelix** · March 11th 2011, 04:50 AM

Originally Posted by Ackbeet

Does that mean you have already shown this?

yes, the coefficients go to zero. in addition to that i was able to show that x goes to zero for
$f:=\frac{c}{ln(2+t)} \ \mbox{with $c > 0$ sufficiently large}.$

perhaps by using this information it is easier to proof, that for a continuous differentiable function f with the 3 assumptions above, x(t) goes to zero as well?

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