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I really would appreciate someone helping me solve this problem:

Consider the initial value problem $x'=f(t,x),x(0)=0$ where $f(t,x)=\begin{cases} 1 & t \geq 0 \text{ and }\forall x \\ -1 & t < 0 \text{ and }\forall x \end{cases} $

(a) Show the existence and uniqueness theorem does not apply to this IVP

(b) Show the initial value problem does not have a solution on any interval containing $t_0=0$

I have solved (a) already but am having an immense amount of trouble doing part(b). Could someone explain how I am supposed to $show$ that there is no solution to the IVP on any interval containing $t_0=0$. Thanks

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  • $\begingroup$ $|t|$ is an obvious 'solution', so it all depends on exactly what 'solution' means $\endgroup$ Oct 10, 2019 at 4:18
  • $\begingroup$ a differentiable function x(t) that when substituted into the differential equation makes an identity, so how do I implement this in the part(b) $\endgroup$
    – user707991
    Oct 10, 2019 at 4:21
  • $\begingroup$ Well you could prove that if $x$ is differentiable then $x'$ cannot have a jump singularity $\endgroup$ Oct 10, 2019 at 4:24
  • $\begingroup$ @CalvinKhor is there a way to prove it without assuming the |t| is a solution? $\endgroup$
    – user707991
    Oct 10, 2019 at 4:28
  • $\begingroup$ Would this be a sufficient proof?$x(t)=|t|\implies x'(t)=\frac{|t|}{t}$ which is identical to $f(t,x)$ for all $t \neq 0$ so the solution must be defined for all $t \neq 0$? $\endgroup$
    – user707991
    Oct 10, 2019 at 4:31

2 Answers 2

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This follows immediately from Darboux's theorem, which says that any function that is a derivative must satisfy the conclusion of the Intermediate Value Theorem. Its proof is not hard, but I don't think this result is part of standard syllabi. In any case, you can extract out a specific proof for your problem.

Here I adapt the first proof on Wikipedia. Suppose $a,b>0$ and $x$ is a solution on an interval $(-2a,2b)$. Then we must have $x'(-a)=-1$ and $x'(b)=1$. Then note that by the Extreme Value Theorem and continuity of $x$ on $[-a,b]$, $x$ achieves a minimum somewhere in $[-a,b]$. Note that $x$ can't be minimised at $b$, since this would mean that for $t<b$, $$ \frac{x(t)-x(b)}{t-b}\le 0 \implies x'(b) \le 0. $$ Similarly, the function cannot be minimised at $-a$. So its minimised at some $t_0\in(-a,b)$. But $x$ is differentiable at $t_0$, so $x'(t_0) = 0$, which is not allowed by the ODE; QED.

Alternative proof using ODE theory: Suppose $x$ is a solution on $(-a,a)$. Then it is a solution on $(\epsilon,a)$ as well; by uniqueness, which applies since $f$ is Lipschitz on this domain, it has to be $x(t) = c+t.$ Similarly for $(-a,-\epsilon)$, so we obtain

\begin{align} \epsilon < t < a \implies x(t) = c_1+|t|\\ -a < t < -\epsilon \implies x(t) = c_2+|t|\\ \end{align} since we can repeat this for any $\epsilon>0$, this holds on $(-a,a)\setminus \{0\}$. By continuity and the initial data $x(0)=0$ we deduce $c_1=c_2=0$, and $x=|t|$ on $(-a,a)$. But then $x$ is not differentiable at $0$, and therefore cannot be a solution to the IVP (as defined in the comments above).

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a) The existence theorems apply to functions $f$ that are continuous, which means simultaneous in both variables. Points with $t=0$ do not belong to the domain where this $f$ is continuous. For the existence theorems, also the initial point has to be inside the domain.

b) Any ODE solution can only extend to the boundaries of the domain of the ODE. The line $\{(t,x):t=0\}$ does not belong to this domain, no ODE solution can cross this line.

This does not preclude that different solutions of the different parts of the domain can be combined to get a continuous, piecewise differentiable function. But this function is not a solution to the ODE, only its restrictions to the parts of the domain of the ODE are solutions of the ODE.

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