- #1
Suvadip
- 74
- 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
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- Thread starter Suvadip
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In summary, the conversation discusses the method of successive approximation for solving an integral equation and its relation to Picard Iterations. The method involves defining a sequence of functions that converges to the solution of the differential equation, and the starting value and initial conditions are also mentioned.
- #2
Siron
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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
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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
$\chi$ $\sigma$
Related to Integral equation by successive approximation
1. What is an integral equation by successive approximation?
An integral equation by successive approximation is a method used to approximate the solution to an integral equation by breaking it down into a series of simpler equations that can be solved iteratively. This method is commonly used in mathematical physics and engineering.
2. How does the method of successive approximation work?
The method of successive approximation works by starting with an initial guess for the solution of the integral equation and then iteratively improving the approximation until a desired level of accuracy is achieved. This is done by substituting the initial guess into the equation and using the resulting value to improve the guess in the next iteration.
3. What are the advantages of using the integral equation by successive approximation?
One advantage of using this method is that it can be applied to a wide range of integral equations, making it a versatile tool for solving complex problems in various fields. Additionally, it is a relatively straightforward and efficient method, making it a popular choice for many scientists and engineers.
4. Are there any limitations to using the integral equation by successive approximation?
While the method of successive approximation can be effective in solving many integral equations, it does have some limitations. It may not always converge to the exact solution, and the convergence rate can be slow for certain types of equations. Additionally, the method may be computationally intensive for larger and more complex equations.
5. How is the convergence of the successive approximation method determined?
The convergence of the successive approximation method is determined by monitoring the difference between successive approximations and comparing it to a predetermined tolerance level. If the difference falls below the tolerance level, the method is considered to have converged to a satisfactory solution. However, if the difference does not decrease sufficiently, the method may need to be repeated with a different initial guess or other modifications.
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