- #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.
MHB Integral equation by successive approximation
- Thread starter Suvadip
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The discussion centers on solving an integral equation using the method of successive approximation, also known as Picard Iterations. The user is exploring the convergence of the sequence defined by the iterations, specifically questioning the predictability of the leading terms in the polynomials generated. They express uncertainty about finding an explicit form for the approximations, denoted as $y_n(x)$. The integral equation relates to a differential equation with initial conditions, and the user provides specific values for $y_0$ and $y_1$. The conversation highlights the complexities involved in deriving solutions through this iterative method.
- #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
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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
$\chi$ $\sigma$