# Lower bound on ODE (Gronwall)

Consider a well-defined first-order ODE

$$\frac{du(t)}{dt} = F(t, u) + G(t),$$ which has a unique continuous solution on $$t\in[0,\infty)$$

Suppose that

1. The solution $$u(t)>0$$, and $$u(0)>0$$.
2. $$|F(t,u)| \leq f(t)\, u(t)$$, where $$f(t)>0$$ is continuous and uniformly bounded.
3. $$G(t)>C>0$$ s continuous and uniformly bounded.

Now I want to compose a positive lower bound of solution $$u(t)$$, such that

$$u(t)\geq \,?$$

on $$t\in[0,\infty)$$.

Upper bound can be directly obtained by Gronwall's lemma. Here are some papers could help with the proof on lower bound: 1 2

• Do you mean Grönwall's inequality? The name is sometimes spelled Gronwall in american publications, but i never heard of anyone named Grownwall. Commented May 28, 2020 at 14:56
• I edited your question changing "Gronwall's inequality" to "Gronwall's lemma". Anyways, you question has a simple answer: just define $v(t)=u(T-t)$. Commented May 28, 2020 at 15:42
• @LeanderTilstedKristensen Yes. Sorry I have difficulty memorizing names.
– anon
Commented May 28, 2020 at 16:47
• @JohnB Thanks, I fixed my question ^
– anon
Commented May 28, 2020 at 16:51

Here's an extremely simplistic lower bound using integrating factors.

$$u'(t) \geq G(t) - f(t) u(t)$$

If you write $$h(t) = \exp \int_0^t f(s) ds$$ then you have

$$(h(t) u(t))' \geq G(t) h(t)$$

and so you have

$$h(t) u(t) \geq u(0) + \int_0^t G(s) h(s) ~ds$$

or

$$u(t) \geq \frac{1}{h(t)} u(0) + \int_0^t \frac{h(s)}{h(t)} G(s) ~ds$$

• Is it possible to extend this to a non-linear case? That is, $|F(t,u)| \leq f(t)\, \beta(u(t))$, where $\beta$ is continuous positive increasing function.
– anon
Commented May 29, 2020 at 7:28
• e.g., let's say $\beta(u(t)) = u^p(t)$ for some $0<p<1$
– anon
Commented May 29, 2020 at 8:29
• @NathanExplosion: yes you can extend to to non-linear cases; but it really depends on exactly what you want. For example, notice that by concavity you have that if $f(t)$ is uniformly bounded in $t$, that $$f(t) u^p(t) \leq M u^p(t) \leq C_0 + M_0 u(t)$$ when $p \in (0,1)$. Then you can apply the same estimate as above. Whether this is useful for you depends on what you want to apply this to. Commented May 29, 2020 at 16:11
• But then we need to ensure that $G(t) - C_0\geq 0$, so that the lower bound is positive.
– anon
Commented May 29, 2020 at 17:13
• The mapping $u \mapsto M u^p$ is concave and passes through the origin. For any $\epsilon > 0$ you can find a tangent line to its graph that has $y$ intercept equal to $\epsilon$, and the graph lies below the tangent line. In particular, you can choose $0 < C_0 < C$ to your liking. Commented May 29, 2020 at 17:30