- #1
Suvadip
- 68
- 0
if
=(1+x)^2,%20y_2(x)=\frac{2}{3}x^3+3x^2+2x+1,%20y_3(x)=\frac{1}{3}x^4+2x^3+3x^2+2x+1)
, then what will be
)
. In fact I was solving the integral equation
=1+2\int_0^x(t+y(t))dt,%20y_0(x)=1)
by the method of successive approximation.
Integral equation by successive approximation
- Context: MHB
- Thread starter Suvadip
- Start date
Click For Summary
SUMMARY
The discussion focuses on solving integral equations using the method of successive approximation, also known as Picard Iterations. The integral equation in question is derived from the differential equation $\displaystyle y^{\ '} = f(x,y)$ with initial condition $y(0)= y_{0}$. The sequence defined by $y_{n} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{n-1}\}\ d t$ converges to the solution $y(x)$. In this specific case, with $y_{0}=1$ and $x_{0}=0$, the first iteration yields $y_{1} = (1 + x)^{2}$, indicating that $f(x,1) = 2(1+x)$.
PREREQUISITES- Understanding of integral equations and their formulations
- Familiarity with differential equations, specifically first-order equations
- Knowledge of Picard Iterations and their application in solving equations
- Basic calculus, including integration techniques
- Study the convergence criteria for Picard Iterations in integral equations
- Explore advanced techniques in solving integral equations, such as the Banach fixed-point theorem
- Learn about numerical methods for approximating solutions to differential equations
- Investigate the implications of initial conditions on the convergence of iterative methods
Mathematicians, students studying differential equations, and researchers interested in numerical analysis and approximation methods will benefit from this discussion.
- #2
Siron
- 148
- 0
I was not familiar with the name of the method until I noticed it's the same as Picard Iterations (after some googling). I tried some ways to see a pattern in the polynomials but I failed. Especially the leading terms are very unpredictable in my opinion. Perhaps it's not possible to find an explicit form for $y_n(x)$.
What's the background of this problem?
What's the background of this problem?
- #3
chisigma
Gold Member
MHB
- 1,627
- 0
suvadip said:if, then what will be. In fact I was solving the integral equationby the method of successive approximation.
Under appropriate conditions the solution of the differential equation ...
$\displaystyle y^{\ '} = f(x,y),\ y(0)= y_{0}\ (1)$
... must satisfy the following integral equation...
$\displaystyle y = y_{0} + \int_{x_{0}}^{x} f\{t, y(t)\}\ d t\ (2)$
If You define...
$\displaystyle y_{1} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{0}\}\ d t\ (3)$
... and...
$\displaystyle y_{n} = y_{0} + \int_{x_{0}}^{x} f\{t, y_{n-1}\}\ d t\ (4)$
... then the sequence of $y_{n}$ converges to the solution $y(x)$...
In Your case is $y_{0}=1$, $x_{0}=0$ and $y_{1} = (1 + x)^{2}$, so that is $f(x,1) = 2\ (1+x)$...
Kind regards
$\chi$ $\sigma$
Similar threads
Graduate
Help with Recurrence Equation
- · Replies 4 ·
- · Replies 3 ·
- · Replies 1 ·
High School
Uncomputable Numbers and Rational Approximations
- · Replies 4 ·
- · Replies 5 ·
- · Replies 4 ·
- · Replies 9 ·
- · Replies 6 ·
- · Replies 2 ·