Given that y'(x)+p(x)y(x)\geq 0 and $y(x_0)\geq 0$, how does one go about showing that $y(x)\geq0$ for all $x\in [x_0,\infty)$
Differential equation with inequalities
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1by the way, that expression is usually called a "differential inequality" or a "differential relation". :) – 2011-02-11
2 Answers
Define b=y'+py, so that you have, well, y'+py=b and $b\geq0$. Solve the inhomogeneous equation for $y$, and use now that $b$ is non-negative.
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0Thanks, I had tried that before and failed, but tried the same thing again, after your hint, and it worked. – 2011-02-11
Can't one reason in the following manner, whitout introducing the function $b$?
Multiplying both sides of the differential inequality $y^\prime (x)+p(x)y(x)\geq 0$ by the positive function $M(x):= \exp \left( \int_{x_0}^x p(t)\ \text{d} t\right)$ yields:
$y^\prime (x) M(x) +y(x)\ p(x)M(x) \geq 0$;
but $M^\prime(x)=p(x) M(x)$, hence the last inequality rewrites:
$\frac{\text{d}}{\text{d} x} \left[ y(x) M(x)\right]\geq 0$,
hence the function $y(x)M(x)$ (which is differentiable in $]x_0,+\infty[$, for it is product of differentiable functions) increases in $[x_0,+\infty[$; this fact implies:
$y(x)M(x)\geq y(x_0)M(x_0)=y(x_0)\geq 0$
and a fortiori $y(x)\geq 0$ for all $x\geq x_0$, which is the claim.
NOTE: The auxiliary funcion $M(x)$ is the reciprocal of the unique solution to the homogeneous ODE:
$\phi^\prime (x)+p(x)\phi(x)=0$
which satisfies $\phi(x_0)=1$. So $M(x)$ is not appeared out of nowhere; in fact the problem you're dealing with can be read as a comparison result between the solution and supersolutions of the problem:
- $\begin{cases} u^\prime (x)+p(x)u(x)=0 &\text{, in } ]x_0,+\infty[ \\ u(x_0)=y_0\end{cases}$
in the following way: each supersolution of problem 1 is greater than its unique solution in $[x_0,+\infty[$.
P.S.: I didn't check the date before answering. I don't know if this can be considered as necroposting or doesn't matter at all... I'm sorry anyway, next time I'll pay more attenction.
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0I don’t mind about necropostings. For me it suffices to check in the profile that OP is still with us. – 2017-08-27