1
$\begingroup$

Suppose that the following differential inequality holds:

$$ \dot{y}(t) \leq -c\cdot y(t) + \lambda \ \ (1)$$

with $c,\lambda > 0 $ positive constants. Now let $\rho = \lambda /c > 0$, I'm wondering how the following is concluded from equation (1):

$$ y(t) \leq \rho + y(0)\cdot e^{-c\cdot t} - \rho\cdot e^{-c\cdot t} \ \ (2)$$

I know that the differential equation $\dot{y}(t) = -c\cdot y(t) + \lambda$ has the solution $y(t) = y(0)\cdot e^{-c\cdot t} + \rho$ but I can't figure out why (2) holds. Note that $y(t)$ is a positive definite function.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

$$\begin{align} (1)&\iff \dot{y}(t) + c\cdot y(t)\leq \lambda \\ &\iff e^{ct}\left(\dot{y}(t) + c\cdot y(t) \right) \le \lambda e^{ct} \\ &\iff (e^{ct}y(t))' \le \lambda e^{ct} \\ &\iff e^{ct}y(t) - y(0) \le \lambda \int_0^t e^{cs}ds \\ \end{align}$$ And you can deduce easily $(2)$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .