# Solving linear differential inequality using linear differential equation.

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

• 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 $x_0\in X$. – Its_me Sep 26 '20 at 11:01

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 $$0 and $$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 $$x<0$$.