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Consider the IVP

$$y'(t)=f(y(t)), \ \ \ \ y(0)=a \in \mathbb{R}$$ $$f : \mathbb{R} \rightarrow \mathbb{R}$$

Which of the following is/are true

$(A)$ There exists a continuous function $f : \mathbb{R} \rightarrow \mathbb{R}$ and $a \in \mathbb{R}$ such that the above problem does not have a solution in any nbd of $0$.

$(B)$ The problem has unique solution for every $a \in \mathbb{R}$ when $f$ is Lipschitz continuous

$(C)$ When $f$ is twice continuously differentiable the maximal interval of existence for the above IVP is $\mathbb{R}$

$(D)$ The maximum interval of existence for the IVP is $\mathbb{R}$ when $f$ is bounded and continuously differentiable.

It is obvious to me that $A \ \& \ B$ are false from traditional existance and uniqueness theorems (viz Picards Theorem). I am not sure about $C \ \& \ D$ and this is really bugging me. Please could anyone shed light on this.

PS: Multiple correct options are allowed.

1 Answers 1

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A) should be answered with the existence theorem of Peano.

B) is an application of Picard-Lindelöf, it seems you only know about the theorem for differentiable $f$?

C) classical example $f(y)=y^2$.

D) if $|f(y)|\le M$, then $|y(t)-y(0)|\le M·|t|$ by the mean value theorem of your choice.

Details to D) By the assumption, local solutions exist everywhere. One has to show that one can continue thoes local solutions to the boundaries of their intervals, as one can restart an initial-value problem whose local solution extends the previous one. As $M$ bounds the derivative of any solution, any solution is Lipschitz-continuous which excludes any poles (and jumps and oscillations). Thus the limit at the boundary of a domain always exist.

To do this more systematically, consider the interval $[−T,T]$ for some (large) $T>0$. Set $b=M⋅T$, then there is a global Lipschitz constant $L$ on the compact set $[−T,T]×[−2b,2b]$. Set $a=\min(\frac1{2L},\frac T2)$. Then the Picard iteration is contractive over every interval $[t_0−a,t_0+a]⊂[−T,T]$ by the local quantified version of the Picard-Lindelöf theorem, which allows to find or piece together a unique solution over $[−T,T]$. As there are no restrictions on $T$, one can extend this to all of $R$.

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    So $C$ is false and $D$ is true. Is that correct? – 2017-01-07
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    Yes, that is correct. You should be able to justify all your choices, even if that is not asked in a multiple-choice test. It would help against mind games that are often part of those. – 2017-01-07
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    @LutzL how option D is correct? I mean by which result. – 2017-01-25
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    f may not Lipschitz even if it is bounded – 2017-01-25
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    @neelkanth: Continuously differentiable implies locally Lipschitz. As there can be no poles, any local solution can be infinitely extended. – 2017-01-25
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    But locally lip gives solution locally....please can you suggest how it has no poles? – 2017-01-25
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    In part C $y^{2}$ is also locally lip – 2017-01-25
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    I am confused how can we extant it to whole real line.... – 2017-01-25
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    Consider the interval $[-T,T]$. Set $b=M·T$, then there is a global Lipschitz constant $L$ on the compact set $[-T,T]×[-2b,2b]$. Set $a=\min(\frac1{2L}, \frac{T}2)$. Then the Picard iteration is contractive over every interval $[t_0-a,t_0+a]\subset[-T,T]$ by the local quantified version of the Picard-Lindelöf theorem, which allows to find or piece together a unique solution over $[-T,T]$. As there are no restrictions on $T$, one can extend this to all of $\Bbb R$. – 2017-01-25
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    if my d.e is $y'=\sin(y)$ ,$y(0)=1.$Then the the solution should be unique and on whole 4\mathbb{R}.? – 2017-08-29
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    @Upstart: As $f(t,y)=\sin y$ is globally Lipschitz in $y$ with constant $1$, you do not need any extra arguments for global existence. But for $y'=\sin(y^2)$ you would indeed need the boundedness argument. – 2017-08-29
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    still if $f(x,y)=\sin (y)$. The solution part would involve an intermediary step involving $\log(g(y))$. So for proceeding ahead of that step, wouldn't we have to make the assumptionn on $y$ and by default on $x$, which would thus shorten the maximum interval of existence – 2017-08-29
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    @Upstart : The roots of $0=\sin(y)$ are stationary points of the ODE, that is, the give constant solutions. By uniqueness, no other solution reaches a root of $sin(y)$. The separation calculation gives the global solutions without further problems. – 2017-08-29
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    Pardon me, but i didn't understand your statement – 2017-08-30
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    So B and D are correct answer. Can you please confirm. Thanks in advance. – 2018-12-08
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    @StammeringMathematician : Yes, these are the correct statements. – 2018-12-08