Gronwall's inequality with time-dependent terms

Question

Gronwall's inequality bounds the solution of a certain form of ODE or system of ODEs: citing from this question,

for a single differential inequality $$ u'(t) \leq a_1\,u(t) + a_0 $$ the Gronwall lemma guarantees that $$u(t)\leq\, u(0)\,e^{a_1 t} + \frac{a_0}{a_1}\, (e^{a_1t}-1) \,.$$

What would the inequality be if $a_0 = a_0(t)$?

Answer 1

You can still apply the solution formula, or at least the steps for it. With the integrating factor you get $$ (e^{-A_1(t)}u(t))'\le a_0(t), $$ where $A_1'(t)=a_1(t)$, $A_1(0)=0$. This can be integrated to $t>0$ giving $$ u(t)\le u(0)e^{A_1(t)}+\int_0^ta_0(s)e^{A_1(t)-A_1(s)}\,ds. $$ Not so coincidentally the right side is the solution formula for the solution of the equality version of the inequality.

Answer 2

For my own reference.

The proof of the inequality stated in the original question is: We find the solution of the ODE $$ u'(t) = a_1 u(t) + a_0 . $$

Define the integrating factor $I(t) = e^{\int_0^t -a_1 ds} = e^{-a_1t}$ so that $I'(t) = -I(t)a_1$.

Multiplying both sides of the equation by $I(t)$, we get

$$ u'(t) I(t) - a_1u(t)I(t) = a_0 I(t)$$ where the LHS is $\big(u(t)I(t)\big)' = u'(t)I(t) - u(t) I(t)a_1$.

Integrating both sides, $\int_0^t \big(u(s)I(s)\big)' ds = \int_0^t a_0 I(s)ds$ gives $$u(t)I(t) - u(0)I(0) = a_0 \int_0^t e^{-a_1s}ds.$$

Plugging in $I(t) = e^{-a_1 t}$ and $I(0)=1$, $$ u(t)e^{-a_1 t} - u(0) = a_0 \int_0^t e^{-a_1s}ds.$$ Hence, the solution is

$$ u(t) = u(0)e^{a_1t}+ a_0 \int_0^t e^{a_1t-a_1s } ds .$$

Then, it can be shown that $y(t) \leq u(t)$ for any $y(t)$ satisfying the inequality version of the ODE $$ y'(t) \leq a_1 y(t) + a_0$$ by using the product rule on $$ \frac{d}{dt} \frac{y(t)}{u(t)},$$ and showing that this derivative is negative. Hence, if $u(0)=y(0)$, then $y(t) \leq u(t)$. (I don't know how to do this part, though; while the proof in the wikipedia page is straightforward, the additional term $a_0$ changes things. )

Proving the time-dependent case:

Using the same method with integrating factors, now with $$ I(t) = e^{\int_0^t -a_1(s)ds}.$$ Also suppose that $A_1'(t) = a_1(t)$, so that $$ I(t) = e^{-A_1(t)}.$$

Then, the solution of the equation $u'(t) = a_1(t) u(t) + a_0(t)$ is

$$u(t) I(t) - u(0)= \int_0^t a_0(s) I(s) $$ is
$$u(t)e^{-A_1(t)} -u(0) = \int_0^t a_0(t) e^{-A_1(s)} $$ i.e. $$ u(t) = u(0)e^{A_1(t)} + \int_0^t a_0(t) e^{A_1(t) - A_1(s)} ds.$$

Showing the inequality, that any solution is less than or equal to the above solution, is similar to the above.