dreangannaa

2022-04-22

Showing that continuous and differentiable s(t), with $s\left(0\right)=0$ and $s}^{\prime}\left(t\right)\le 2ts\left(t\right)+\sqrt{s\left(t\right)$ for $t>0$, is identically zero for $t\ge 0$.

Let $s:[0,+\mathrm{\infty})\to [0,+\mathrm{\infty})$ be a continuous function that is differentiable on $(0,+\mathrm{\infty})$.

It also holds that ${s}^{\text{'}}\left(t\right)\le 2ts\left(t\right)+\sqrt{s\left(t\right)},\text{}\text{}t0s\left(0\right)=0$

Show that $s\left(t\right)=0,t\ge 0$.

Do we us some theorem, maybe Rolle or the mean value theorem?

Do we have to check maybe the sign of $2ts\left(t\right)+\sqrt{s\left(t\right)}$? This depends on the value of s(t), or not?

Brianna Sims

Beginner2022-04-23Added 19 answers

Step 1

Set$s\left(t\right)=\frac{1}{4}\mathrm{exp}\left({t}^{2}\right){\left({\int}_{0}^{t}\mathrm{exp}(-\frac{{\tau}^{2}}{2})d\tau \right)}^{2}(t\ge 0)$ . Then $s\left(t\right)\ge 0,s\left(0\right)=0$ and

${s}^{\prime}\left(t\right)=\frac{t}{2}\mathrm{exp}\left({t}^{2}\right){\left({\int}_{0}^{t}\mathrm{exp}(-\frac{{\tau}^{2}}{2})d\tau \right)}^{2}+\frac{1}{4}\mathrm{exp}\left({t}^{2}\right)2{\int}_{0}^{t}\mathrm{exp}(-\frac{{\tau}^{2}}{2})d\tau \mathrm{exp}(-\frac{{t}^{2}}{2})$

Step 2

$=2ts\left(t\right)+\frac{1}{2}\mathrm{exp}\left(\frac{{t}^{2}}{2}\right){\int}_{0}^{t}\mathrm{exp}(-\frac{{\tau}^{2}}{2})d\tau$

$=2ts\left(t\right)+\sqrt{s\left(t\right)}$

Edit: The idea to find this solution: Consider the ODE$s}^{\prime}=2ts+\sqrt{s$ . Then substitute $u=\mathrm{exp}(-{t}^{2})s$ to obtain $u}^{\prime}=\mathrm{exp}(-\frac{{t}^{2}}{2})\sqrt{u$ . Separation of variables then leads to the maximal solution of the ivp for u hence for s.

Set

Step 2

Edit: The idea to find this solution: Consider the ODE

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