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$(P)$ $$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \frac{dy}{dt} = f(t,y), y(t_o)=c_o $$ Remark: Our IVP, defined in $(P)$ carries the following assumptions: values of $f(t,y)$ are within the set of real numbers, and are continuous.

$$Lemma \, \, (0)$$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,$Our initial problem stated in $(P)$, can be solved by elementary means as follows: $$1.) \, \, \frac{dy}{dt}dt=f(t,y)dt$$ $$2.) \, \,\int dy=\int f(t,y)dt$$ $$3.) \, \,\int dy =\int f(t,\phi(t))dt$$ $$4.) \, \, y = \phi(t)$$

$$Lemma \, \, (1.1)$$ $\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $If $y=\phi(t)$ is a solution of problem $(P)$ lies withen the interval $$1.) \, \, \, \, \,|t-t_0| \, < a \leq a, \, \, then \, \,|\phi(t)-c_0| \, < b \, \, \, in \, \, |t-t_0| \,< \, a$$ Initially $1.)$ this can be stated as: $((t),\phi(t)) \in R((t_0,c_0),a,b)\, \, for \, \, |t-t_o| \, < a$ for $|t-t_o| < a$

Proof.: The intial move the author made was to assume $Lemma (1.1)$ was false and proceed by proof by contradiction. By the continuity of $\phi(t)$ there exists a positive number $\beta$ such that:

$$(I) \, \, \, \, \, \, \, \, \beta \, < \, a$$ $$(II) \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \begin{cases} | \vec{\phi(t)} - \vec{c_0} | < b \qquad &\text{for} \qquad |t-t_0| < \beta , \\ | \vec{\phi(t_0+\beta)} - \vec{c_0} | = b \qquad &\text{or} \qquad | \vec{\phi(t_0-\beta)} - \vec{c_0} | = b \end{cases}$$

From $(II)$ the initial assumption that is implied is that $(t,\phi(t)) \in R$ for the interval:$|t-t_0| \leq \beta$. Since our solution $\phi(t)$ is on our interval intuitively speaking our function is bounded withen our rectangular region: $R$. Rigsourly speaking: $|f(t,\phi(t)| \leq M$ for $|t-t_0| \leq \beta$, since: $\phi'(t)=f(t,\phi(t))$ and since $\phi(t_0)=c_0$ This leads to the following integral in $1.2)$

$(1.1.2)$ $$\phi(t) = \vec{c_0} + \int_{t}^{t_0}f(s,\phi(t))ds$$

This implies that we have the following operation in (1.1.3) $$(1.1.3)$$ $$| \vec{c_0} + \int_{t}^{t_0}f(s,\phi(t))ds-c_0|=|\int_{t}^{t_0}\vec{f}(s,\vec{\phi}(s))ds| \leq Max|\vec{f}(t,\vec{y})|t-t_o|$$

The initial crux of my question comes down to verifying the integral in (1.1.3), could it be possible to apply an integral inequality, what technique would be most useful in this particular scenario.

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  • $\begingroup$ From looking at previous examples of similar situations it seems like part of this defintion can be used:$|\int_a^b f|=\int_a^b|f|$ $\endgroup$ – Zophikel Mar 15 '17 at 17:40
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Yes, this follows from the triangle inequality as you've noted in your comments.

$$|\int_a^b g(x) dx| \le \int_a^b|g(x)|dx \le \max_{x \in [a,b]} |g(x)| \int_a^bdx = \max_{x \in [a,b]}|g(x)| (b-a) $$

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  • $\begingroup$ Interesting this should be useful in further proofs of this nature. $\endgroup$ – Zophikel Apr 5 '17 at 21:57
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    $\begingroup$ Note that while this is always try, some boundedness condition is required on $g$ to make it not trivial (if $g$ isn't bounded, the right hand side is just infinity which isn't interesting). This is a rather crude bound -- we're replacing $|g|$ everywhere by the biggest thing it can possibly be. Sometimes in analysis, a crude bound like this is enough. Other times, you will need to get more careful estimates. $\endgroup$ – anonymous Apr 5 '17 at 21:59
  • $\begingroup$ Interesting, I would like to pick up more on these tactic do you know a book or lectures notes with methods such as these. $\endgroup$ – Zophikel Apr 5 '17 at 22:04
  • $\begingroup$ Based on your other posts, you're reading Stein and Shakarchi, who will use techniques like these routinely. Rudin's Principles of Mathematical Analysis is another text you could look at. $\endgroup$ – anonymous Apr 5 '17 at 22:32

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