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I would like to solve the linear differential inequality: $$a_0y(x)+a_1y'(x)+...a_ny^{(n)}(x)\le f(x)$$

satisfied for all $x\in X$, where $X$ is some open subset of $\mathbb{R}$.

Here is my idea how to solve it. I would like to know whether this is a good idea.

Is this true that the solution are in the form $$y(x)\le g(x)$$ for all $x\in X$ where $g(x)$ is a solution to the differential equation: $$a_0y(x)+a_1y'(x)+...a_ny^{(n)}(x)=f(x)$$

If this is a bad idea, then how are the techniques of solving that kind of inequality.

Regards

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  • $\begingroup$ I think you need to include the initial conditions, i.e. $y(x)\le g(x)$ for all $x\in X$ if $y(x_0)\le g(x_0)$ for some $x_0\in X$. $\endgroup$ – Its_me Sep 26 '20 at 11:01
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No, the solutions are in general not in the form $y(x)\le g(x)$ for all $x\in X$.

Let $X = \mathbb{R}$, $f(x)=0$, $a_0 = 1$, $a_1 > 0$, and $a_n=0$ for $n\ge2$. Then, the solutions for

$$g(x)+a_1 g'(x) = 0$$

are exponentials $g(x)=g_0\exp\left(-\frac{1}{a_1}x\right)$ with $g(0)=g_0$. Let $g_0>0$, Now,

$$y(x) = g_0\exp\left(-\frac{1}{b_1}x\right)$$

with $0<b_1<a_1$ and $y(0) = g(0)$ fulfills the inequality

$$y(x)+a_1 y'(x) \le 0$$

for all $x\in\mathbb{R}$, because $1-\frac{a1}{b1}\le 0$. But, $(\frac{1}{b_1}-\frac{1}{a_1})x<0$ for $x<0$ and, thus,

\begin{eqnarray} g_0\exp\left(-\frac{1}{a_1}x\right)&<&g_0\exp\left(-\frac{1}{b_1}x\right)\\ g(x) &<& y(x) \end{eqnarray} for $x<0$.

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