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I'm trying to follow a proof for solving the ODE

$$\frac{df}{dx} = (f(x)+x)x$$

For $0 \leq x \leq 1$ with the initial condition $ f(0)=0$.

The proof I am following goes like this

Define $F:C[0,1]\rightarrow C[0,1]$ as $F(f)(x)=\int_0^x (f(u)+u)u du$ then the solution to the ODE is the fixed point of F. To apply contraction mapping theorem we need to be working in a complete space so we use the suprememum metric.

$d_\infty(F(f_1),F(f_2))=\sup_{t\in [0,1]}|F(f_1)(t)-F(f_2)(t)| $

$=\sup_{t\in [0,1]}|\int_0^t(f_1(u)+u)udu-\int_0^t(f_2(u)+u)udu |$

$=\sup_{t \in [0,1]}|\int_0^t(f_1(u)-f_2(u))udu |$

$\leq \sup_{t \in [0,1]}\int_0^t|f_1(u)-f_2(u)|udu$

$\leq \int_0^1|f_1(u)-f_2(u)|udu$ ($*$)

$\leq \sup_{t\in [0,1]} |f_1(t)-f_2(t)| \int_0^1 udu $ ($**$)

$= \sup_{t \in [0,1]}|f_1(t)-f_2(t)|/2$

$=d_\infty (f_1,f_2)/2$

Okay so firstly couldn't we use = instead of $\leq$ at ($*$)? I know it doesn't really matter in the proof but we have = everywhere so it would seem inconsistent making me doubt my belief we could use a =. If we can't use an = then why not?

My main issue with this proof is ($**$). It looks kind of like a theorem I know

$$|\int_a^bf(x)dx| \leq |b-a| ||f||_\infty$$

And although I believe the step, and it seems nearly intuitive I can't think of a justification for it.

Although I understand the rest of the proof I'll include it for completeness. Since F satisfies the contraction mapping theorem we can use initial condition with the zero function to get

$f_1(x)=F(f_0) = \frac{x^3}{3}$

$F(f_1)=\frac{x^5}{15} + \frac{x^3}{3}$

And in general $$f(x)=\sum_{i=1}^\infty \frac{x^{2i+1}}{(2i+1)!!}$$

Source for this proof is https://greenlees.staff.shef.ac.uk/mas331/MAS331notes7.pdf example 7.15

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For (*), I think you are correct. Once the integrand becomes positive, the $\sup$ is going to be the integral evaluated over the entire interval.

For (**), again consider Reimann sums with $C = \left|f_{1}(t)-f_{2}(t)\right|$ being the largest value. Now, summing a multiple of $C$ will be larger than summing over smaller positive elements.
So, the justification is: \begin{equation*} \int_{[0,1]} \left|f_{1}(u)-f_{2}(u)\right| u du \le \int_{[0,1]} C u du = C \int_{[0,1]} u du \end{equation*} which is similar to the justification that you cited.

A formal proof of (**) might go something like this:

Let $f:\mathbb{R}\to\mathbb{R}$, $g:\mathbb{R}\to\mathbb{R}$ with $g(x) \ge 0$ $\forall x\in\mathbb{R}$ and let $M = \sup_{x\in\mathbb{R}}\left|f(x)\right| < \infty$. Then, for some interval $[a,b]\in\mathbb{R}$ \begin{alignat*}{2} \int_{a}^{b} f(x)g(x)dx &\le \left|\int_{a}^{b} f(x)g(x)dx\right| \\ &\le \int_{a}^{b} \left|f(x)g(x)\right|dx \\ &= \int_{a}^{b} \left|f(x)\right|g(x)dx \\ &\le \int_{a}^{b} Mg(x) dx \\ &= M \int_{a}^{b}g(x) dx \end{alignat*}

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  • $\begingroup$ Thanks for your answer. For ($*$), I can't think of a case where we would have < rather than =. So although I understand why a $\leq$ works, I can't see why its necessary. For ($**$) you seem to be using that $\int_a^b f(x)g(x) dx \leq ||f||_\infty \int_a^b g(x)dx$? Do you have a formal proof for this? $\endgroup$ – FinalistBeel May 21 '14 at 17:51
  • $\begingroup$ I see what you mean now. I may have misread the proof before. I've changed my response to (*) and added a proof for (**) as requested. $\endgroup$ – jpb May 22 '14 at 0:16

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