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My introductory differential equations textbook gives this theorem. I do not have enough time to learn the orginal Picard–Lindelöf theorem. Is this theorem true?

Let $I$ be an interval and $y:I\to\mathbb R$ and $y'(x)=y$ $\forall x\in I$.

$y=0$ is a solution. It seems like no other solution $=0$ for some $x\in I$, and I am trying to use this theorem to prove that.

I know $\forall x' \in I, \exists y: I'\mapsto \mathbb R:$ satisfies the given equation on $I'$ and $y(x')=0$. Also, $y(x)=0,x\in I\implies y(x)=0\forall x\in I'\subseteq I$.

I also attempted proof by contradiction. Suppose another $y$ satisfies this and such that for some $x \in I, y(x)=0$ and there does not exist an interval around $x$ such that $y(x)=0 \forall x$ in that interval. This will eventually lead to contradiction. But I do not know how to prove every other such $y$ is like this. And it seems unnecessarily complicated.

TFR
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  • Could you type out the initial value problem (2)? Also, please be more accurate with your symbols. It took a long time to work out what you meant by $\exists I' \land y$. The logical $\land$ symbol normally joins logical propositions, but for functions $f,g:I \to \mathbb{R}$ it sometimes means the pointwise maximum $(f \land g)(x) := \min{f(x),g(x)}$. I would say "$\forall x' \in I, \exists I' \subset I,y:I' \to \mathbb{R}$ (use $\to$ for stating codomains and ranges, and $\mapsto$ for stating function values, i.e. $y':I' \to \mathbb{R},x'' \mapsto 0$). – 1Rock Mar 28 '22 at 01:18
  • For this particular ODE $y'=y$, you could of course also just solve it using an integrating factor: https://math.stackexchange.com/a/58098/1242 – Hans Lundmark Mar 28 '22 at 06:43

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The theorem is indeed true. To use the theorem, first prove that $y_0(x):=0$ satisfies $y_0(0)=0$ and $y'_0(x)=y_0(x)$, $\forall x \in I$, and note that $(x,y) \mapsto y$ is continuous on $I \times [-1,1]$, then just say, "By Theorem 1.2.1, there is an interval $I_0 \subset I$ such that its restriction $y_0|_{I_0}$ is the only function $y :I_0 \to \mathbb{R}$ satisfying $y(0)=0$ and $y'(x)=y(x), \forall x \in I_0$. Thus, any other solution $y_1$ to $y_1(0)=0$ and $y_1'(x)=y_1(x)$ for all $x \in I$ must satisfy $y_1(x)=0$ for all $x \in I'$.

To say "there exists a unique function satisfying $y(0)=0$ and $y'(x)=y(x), \forall x \in I_0$" just means that if there are two solutions $y_0$ and $y_1$, then they have to be equal to each other on $I_0$. Thus, once you've found $y_0$, every solution has to be equal to $y_0$ on $I_0$.

1Rock
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  • I am trying to prove $y(x)=0 \forall x \in I$ is the unique solution on $I$, and $I$ is an arbitrary interval. Your proof just prooves some not determined open interval $I_0\subset I$. I assume you use $I'$ at the end of the first paragraph to mean $I_0$ – TFR Mar 28 '22 at 04:35
  • And $0$ may not be in $I$ – TFR Mar 28 '22 at 04:36