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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

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  • $\begingroup$ Are you aware of Gronwall’s inequality? $\endgroup$ Dec 4, 2023 at 17:43
  • $\begingroup$ ?? From $\mathcal{L}[f]\le\mathcal{L}[g]$ can we deduce $f \le g$ ?? $\endgroup$
    – GEdgar
    Dec 4, 2023 at 17:43
  • $\begingroup$ @GEdgar I don't know. I am confused about something, just not sure what. $\endgroup$ Dec 4, 2023 at 17:57
  • $\begingroup$ @Aruralreader I don't think I can write my differential inequality in the form required for the Gronwall inequality. Gronwall would require $\dot{V}\leq \beta (t) V(t)$, but I have something like $\dot{V}\leq \epsilon V(t)+f(t)$. $\endgroup$ Dec 4, 2023 at 18:02
  • $\begingroup$ @GEdgar My hunch is, "yes", based on the definition of a Laplace transform. But I don't have a proof in front me. $\endgroup$ Dec 4, 2023 at 18:15

2 Answers 2

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$$\begin{align} &\iff \left(\frac{dV}{d\tau}-\epsilon V \right)\leq\frac{A^2}{2\epsilon}\tau^2+\frac{2AE\tau}{\epsilon} + \frac{E^2}{2\epsilon} \\ &\iff \underbrace{\left(\frac{dV}{d\tau}-\epsilon V \right)e^{-\epsilon\tau}}_{=\frac{d}{d\tau}\left(e^{-\epsilon\tau}V \right)}\leq \left(\frac{A^2}{2\epsilon}\tau^2+\frac{2AE\tau}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon\tau} \\ &\iff e^{-\epsilon\tau}V(\tau)-V(0) \leq\int_0^\tau \left(\frac{A^2}{2\epsilon}t^2+\frac{2AE t}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon t}dt \\ &\iff V(\tau)\leq V(0) e^{\epsilon\tau}+e^{\epsilon\tau}\int_0^\tau \left(\frac{A^2}{2\epsilon}t^2+\frac{2AE t}{\epsilon} + \frac{E^2}{2\epsilon}\right)e^{-\epsilon t}dt \end{align}$$

Remark: You can find the analytic expression of the integral easily.

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  • $\begingroup$ Ok, so it seems I can solve this using the standard integration factor method. That sounds great (I just need to do the integration). That satisfies my problem on a practical level, but I'm still left with curiosity about Laplace transforms. Any insights there? $\endgroup$ Dec 4, 2023 at 18:43
  • $\begingroup$ @SageEdwards Laplace transform is just a (complex) way to find compute the two integrals (in the left and right hand sides of the third line). Normally,you might find the same result if you(could)take the inverse Laplace$$\begin{cases} \frac{1}{s-\epsilon}\left(\frac{A^{2}}{\epsilon}\frac{1}{s^{3}}+\frac{2AE}{\epsilon}\frac{1}{s^{2}}+\frac{E^{2}}{\epsilon2}\frac{1}{s}+V_{u}(t_{0})\right), & s-\epsilon\geq0\\ -\frac{1}{s-\epsilon}\left(\frac{A^{2}}{\epsilon}\frac{1}{s^{3}}+\frac{2AE}{\epsilon}\frac{1}{s^{2}}+\frac{E^{2}}{\epsilon2}\frac{1}{s}+V_{u}(t_{0})\right), & s-\epsilon<0 \end{cases}$$ $\endgroup$
    – NN2
    Dec 4, 2023 at 19:15
  • $\begingroup$ I see, after completing the computation I get: $V(t)\leq V(0)\text{e}^{\epsilon t}+\frac{A^{2}}{\epsilon^{4}}\text{e}^{\epsilon t}+\frac{2AE}{\epsilon^{3}}\text{e}^{\epsilon t}+\frac{E^{2}}{2\epsilon^{2}}\text{e}^{\epsilon t}-\frac{A^{2}(\epsilon^{2}t^{2}+2\epsilon t+2)}{2\epsilon^{4}}-\frac{2AE(\epsilon t+1)}{\epsilon^{3}}-\frac{E^{2}}{2\epsilon^{2}}$, which has a similar form, but is subtle enough to make the difference. I now see that I cant use Laplace transforms for differential inequalities, thanks to you and @GEdgar, but I am still left with the question of why. $\endgroup$ Dec 5, 2023 at 14:26
  • $\begingroup$ @SageEdwards I believe that if $A(t) \ge B(t)$ then $\mathcal{L}(A)(s) \ge \mathcal{L}(B)(s)$. So, it's correct to deduce that $\mathcal{V}(s) \le ....$. However, I'm not sure that we can do the same for the inverse Laplace, that means : if $\mathcal{A}(s) \ge \mathcal{B}(s) \implies \mathcal{L}^{-1}(\mathcal{A})(t) \ge \mathcal{L}^{-1}(\mathcal{B})(t)$ $\endgroup$
    – NN2
    Dec 5, 2023 at 14:36
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Solving inequalities using Laplace transform ... I do not think that works.

If we find $\mathcal{L}(f) \ge 0$, it does not follow that $f\ge 0$.
Example: $$ f(t) = \sin t,\qquad \mathcal{L}[f](s) = \frac{1}{1+s^2} $$ Here $\mathcal{L}[f](s) > 0$ for all $s$, but it is not true that $f(t) \ge 0$ for all $t$.

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  • $\begingroup$ Yeah, it doesn't seem to work, but I don't understand why I can't. There is something fundamental that I am missing here... $\endgroup$ Dec 5, 2023 at 14:28

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