I am looking to solve a non-autonomous differential inequality of the form $$\frac{dV}{d\tau}-\epsilon V-\frac{E^2}{2\epsilon}\leq\frac{A^2}{2\epsilon}\tau^2+\frac{2AE\tau}{\epsilon},$$where $V:\mathbb{R}_{\geq0}\to\mathbb{R}_{\geq0}$, and $\epsilon$, $E$, and $A$ are positive (non zero) scalar constants.
When I take the Laplace transform, and after some simple algebraic manipulation I obtain $$(s-\epsilon)\mathcal{V}(s)\leq\frac{A^2}{\epsilon s^3}+\frac{2AE}{\epsilon s^2}+\frac{E^2}{2\epsilon s}+V(0).$$ Now, the question is about how to properly dived both sides by $(s-\epsilon )$ since this, as far as I can tell, is sign indefinite. Now, if I assume that that there is no sign change when I do this operation, followed by partial fraction decomposition and then taking the inverse Laplace transform, I obtain $$V(\tau)\leq \left(\frac{A^{2}}{\epsilon^{4}}+\frac{E^{2}}{2\epsilon^{2}}+V(0)+\frac{2AE}{\epsilon^{3}}\right)\text{e}^{\epsilon\tau}-\left(\frac{A^{2}}{\epsilon^{3}}+\frac{2AE}{\epsilon^{2}}\right)\tau-\frac{E^{2}}{4\epsilon^{2}}\tau^{2}-\frac{E^{2}}{2\epsilon^{2}}-\frac{A^{2}}{\epsilon^{4}}-\frac{2AE}{\epsilon^{3}}.$$ This seems fine on initial inspection, since $V(\tau)=V(0)$ when $\tau=0$, however, things get odd when you change the value of $\epsilon$. For example, let $A=0.1$, $E=0.2$, $V(0)=0$, then when $\epsilon =0.1$ the value of $V(\tau)<0$ for $\tau\in(1,9)$ (rough bounds). This is a problem since $V$ is a positive definite function (time derivative is positive definite and initial condition here is zero). There is clearly an issue here, and my suspicion is in what I am doing while I am working in the Laplace domain. Now, I see that I can divide equation two by $(s-\epsilon)$ and take it in piecwise form such that $$\mathcal{V}_{u}(s)\leq\begin{cases} \frac{1}{s-\epsilon}\left(\frac{A^{2}}{\epsilon}\frac{1}{s^{3}}+\frac{2AE}{\epsilon}\frac{1}{s^{2}}+\frac{E^{2}}{\epsilon2}\frac{1}{s}+V_{u}(t_{0})\right), & s-\epsilon\geq0\\ -\frac{1}{s-\epsilon}\left(\frac{A^{2}}{\epsilon}\frac{1}{s^{3}}+\frac{2AE}{\epsilon}\frac{1}{s^{2}}+\frac{E^{2}}{\epsilon2}\frac{1}{s}+V_{u}(t_{0})\right), & s-\epsilon<0 \end{cases},$$ but how would one take the inverse Laplace transform of this while taking the conditions of the piecewise function into account? Would I do the following $$\mathcal{L}^{-1}\{s-\epsilon\}$$ and start to look at the time derivatives of the Dirac-delta function? That seems very bizarre to me, so any help on this issue would be greatly appreciated. perhaps I shouldn't use Laplace transforms and I should solve this differential inequality by a different method? Suggestions are welcome!
Thanks, Sage
$$\begin{align} &\iff \left(\frac{dV}{d\tau}-\epsilon V \right)\leq\frac{A^2}{2\epsilon}\tau^2+\frac{2AE\tau}{\epsilon} + \frac{E^2}{2\epsilon} \\ &\iff \underbrace{\left(\frac{dV}{d\tau}-\epsilon V \right)e^{-\epsilon\tau}}_{=\frac{d}{d\tau}\left(e^{-\epsilon\tau}V \right)}\leq \left(\frac{A^2}{2\epsilon}\tau^2+\frac{2AE\tau}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon\tau} \\ &\iff e^{-\epsilon\tau}V(\tau)-V(0) \leq\int_0^\tau \left(\frac{A^2}{2\epsilon}t^2+\frac{2AE t}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon t}dt \\ &\iff V(\tau)\leq V(0) e^{\epsilon\tau}+e^{\epsilon\tau}\int_0^\tau \left(\frac{A^2}{2\epsilon}t^2+\frac{2AE t}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon t}dt \end{align}$$
Remark: You can find the analytic expression of the integral easily.