Banach Fixed Point and Differential Equations

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SUMMARY

The discussion focuses on solving the integral equation \(\int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt=\frac{1}{2}\) using Banach's Fixed Point Theorem (BFPT) and Picard's Theorem. The participants confirm that these theorems guarantee a unique solution to the problem. The solution involves integrating the function and applying a substitution \(t=\tan(\theta)\) to simplify the integral. The final result is \(x=1.4743\), which satisfies the equation.

PREREQUISITES
  • Understanding of Banach's Fixed Point Theorem
  • Familiarity with Picard's Theorem
  • Knowledge of integral calculus, specifically integration techniques
  • Experience with trigonometric substitutions in integrals
NEXT STEPS
  • Study the application of Banach's Fixed Point Theorem in solving differential equations
  • Explore Picard's method for finding solutions to initial value problems
  • Learn advanced integration techniques, including trigonometric substitutions
  • Practice solving integral equations similar to \(\int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt\)
USEFUL FOR

Students and professionals in mathematics, particularly those studying differential equations and fixed point theorems, will benefit from this discussion.

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Homework Statement


Find the value of x, correct to three decimal places for which: [itex]\int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt=\frac{1}{2}[/itex].

Homework Equations


Banach's Fixed Point Theorem
Picard's Theorem?

The Attempt at a Solution


I'm not sure where to start with this type of problem.

From other BFPT problems, I will need to show a contraction mapping into itself.

Any pointers would be greatly appreciated.
 
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Banach's fixed point theorem and Picard's method show that this problem has a unique solution. Actually finding the solution doesn't require that.

You could, for example, actually integrate that:
[tex]\int_0^x \frac{t^2}{1+ t^2}dt= \int_0^x 1- \frac{1}{t^2+ 1}dt[/itex] <br /> is pretty easy to integrate.[/tex]
 
[itex]\int^{x}_{0}\frac{t^{2}}{1+t^{2}}dt=\int^{x}_{0}(1-\frac{1}{1+t^{2}})dt[/itex]

Let t=tanθ. Then dt=sec2θdθ.

[itex]\int^{x}_{0}(1-\frac{1}{1+t^{2}})dt=\int^{x}_{0}dt-\int^{x}_{0}\frac{1}{sec^{2}\vartheta}[/itex]sec2θdθ=[itex]\int^{x}_{0}dt-\int^{x}_{0}[/itex]dθ=t-arctan(t) from 0 to x=x-arctan(x).

x-arctan(x)=.5 at x=1.4743.

Are you familiar with any approaches that use BFPT?
 
Last edited:

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