# ODE question: $y'+A(t) y =B(t)$, with $y(0)=0, B>0$ implies $y\ge 0$; another proof?

I am trying to prove that the solution for the different equation

$$y'+A(t) y =B(t)$$

with initial condition $y(0)=0$ and the assumption that $B\ge 0$, has non-negative solution for all $t\ge 0$, i.e. $y(t)\ge 0$ for all $t\ge 0$.

The one I know is constructive: Fisrt consider the homogeneous ODE $x'+\frac{1}{2} A(t) x=0$ and let $u'=\frac{B}{x^2}$ with $u(0)=0$. Then $y=\frac{ux^2}{2}$ will satisfy the origional ODE of $y$. Clearly, by the construction of $y$, $y\ge 0$ for $t\ge 0$.

But this proof is not natural in my opnion, there is no reason(at least I didn't see) to construct such $x(t)$ and $u(t)$. So is there any other way to prove this fact?

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Why not just solve it? In $\mathbb{R}$, $\exp(\cdot)$ is always positive, and you have $B$ nonnegative, and the integral of nonnegative functions will be nonnegative, and... – anon Oct 5 '11 at 1:38
All caps is considered "yelling" or "shouting". So, please don't shout. – Arturo Magidin Oct 5 '11 at 5:43

A natural approach is to start from the special case where $A(t)=0$ for every $t\geqslant0$. Then the ODE reads $y'(t)=B(t)$ hence $y'(t)\geqslant0$ for every $t\geqslant0$ hence $y$ is nondecreasing. Since $y(0)\geqslant0$, this proves that $y(t)\geqslant0$ for every $t\geqslant0$.

One can deduce the general case from the special one. This leads to consider $z(t)=C(t)y(t)$ and to hope that $z'(t)$ is a multiple of the LHS of the ODE. As you know, defining $C(t)=\exp\left(\int\limits_0^tA(s)\mathrm ds\right)$ fits the bill since $(C\cdot y)'=C\cdot (y'+A\cdot y)$ hence one is left with the ODE $z'=C\cdot B$.

Since $C(t)>0$, $C(t)\cdot B(t)\geqslant0$ hence the special case for the RHS $C\cdot B$ yields the inequality $z(t)\geqslant z(0)$ and since $z(0)=y(0)\geqslant0$, one gets $y(t)\geqslant C(t)^{-1}y(0)\geqslant0$ and the proof is finished.

Edit The same trick applies to more general functions $A$ and $B$. For example, replacing $A(t)$ by $A(t,y(t))$ and $B(t)$ by $B(t,y(t))$ and assuming that $B(s,x)\geqslant0$ for every $s\geqslant0$ and every $x\geqslant0$, the same reasoning yields $y(t)\geqslant C(t)^{-1}y(0)\geqslant0$ with $C(t)=\exp\left(\int\limits_0^tA(s,y(s))\mathrm ds\right)$.

The only modification is that one must now assume that $t\geqslant0$ belongs to the maximal interval $[0,t^*)$ where $y$ may be defined. Solutions of such differential equations when $A$ does depend on $y(t)$ may explode in finite time hence $t^*$ may be finite but $y(t)\geqslant0$ on $0\leqslant t<t^*$.

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Well since it is easy to actually explicitely solve such ODE's, the most natural solution to me would have been to analyse that solution. The solution is $$y(t) = \frac{ \int^t_0 \exp\left(\int^s_0 A(p) dp \right) B(s) ds }{ \exp \left(\int^t_0 A(p) dp \right)}$$

and since $\exp (x) > 0$ for all $x \in \mathbb{R}$ and it is given $B(s) \geq 0$, the numerator is non-negative while the denominator is strictly positive, and so $y(t) \geq 0.$

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Thanks, I forgot this explicit solution. Now what if $A$ depends on $y$? i.e. $A(y(t))$? – user17150 Oct 5 '11 at 1:59
Then the theorem is no longer true. For example, if $A(y) = -y$ then the solution of the ODE is $y=\tan x$ which is not always positive. – Ragib Zaman Oct 5 '11 at 2:28
@Ragib, this example is misleading since $\tan t$ becomes negative only after $t$ has crossed the point where the solution $y(t)$ of the differential equation stops to be defined. In other words $\tan$ is NOT a solution on the real line (nor on any interval containing both $0$ and some $t\geqslant0$ such that $\tan t<0$). – Did Oct 5 '11 at 7:39