Preface

This section demonstrates Picard's iteration scheme applicable to boundary value problems.

Picard Iterations

As early as 1893, Émile Picard constructively treated the second order differential equation subject to Dirichlet boundary conditions

$y'' (x) = f\left( x, y, y' \right) , \qquad y(a) = A , \quad y(b) = B ,$
where f(x,y,p) is a continuous functions of three independent variables.

A differential equation (as above) involves a derivative operator that is unbounded operator. This means that the derivative operator can map a bounded function into unbounded one. Therefore, iterations with noncontinuous operators may be problematic to deal with. This is avoided upon applying the (left) inverse operator $$\texttt{D}^{-2}$$ to obtain a fixed point equation with bounded operator:

$y(x) = \texttt{D}^{-2}f\left( x, y, y' \right) .$
The Picard iteration scheme is based on its equivalent integral reformulation (upon application of $$\texttt{D}^{-2}$$ ):
$y(x) = A + \left( x-a \right) \frac{B-A}{b-a} + \frac{1}{2} \int_0^x \left( x-t \right) f(t)\,{\text d}t - \frac{1}{2} \int_x^b \left( x-t \right) f(t)\,{\text d}t + \frac{1}{2} \,\frac{x-b}{a-b} \int_a^b \left( a-t \right) f(t) \,{\text d}t - \frac{1}{2} \,\frac{x-a}{b-a} \int_a^b \left( b-t \right) f(t) \,{\text d}t.$

Example: For our first example, we consider a standard "benchmark" problem:

$y'' + \pi^2 y =x, \qquad y(0) =1, \quad y \left( \frac{1}{2} \right) = -1.$
Application of Picard's iteration gives
$y_{m+1} (x) = 1-4x - \frac{\pi^2}{2} \int_0^x \left( x-t \right) y_m (t)\,{\text d}t + \frac{\pi^2}{2} \int_x^{1/2} \left( x-t \right) y_m (t)\,{\text d}t + \pi^2 x \int_0^{1/2} \left( \frac{1}{2} -t \right) y_m (t) \,{\text d}t - \left( x- \frac{1}{2} \right) \int_0^{1/2} t\, y_m (t) \,{\text d}t, \quad m=0,1,2,\ldots .$
As an initial approximation, we choose the linear function that satisfies the given bundary conditions:
$y_0 = -2x .$
The truue solution is
$y(x) = \cos (\pi x) - \frac{1 + 2\,\pi^2}{2\,\pi^2}\,\sin (\pi x) + \frac{x}{2\pi^2} .$
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Example: Consider the boundary value problem

$y'' = 6\, y^2 , \qquad y(0) = \frac{1}{4} , \quad y(1) = \frac{1}{9} .$
Its exact solution is
$y(x) = \left( x+2 \right)^{-2} .$
Using the initial approximation,
$y_0 (x) = \frac{1}{4} - \frac{5}{36}\, x ,$
we apply the formula for the inverse derivative operator $$\texttt{D}^{-2}$$ to obtain
$y_{m+1} (x) = \frac{1}{4} - \frac{5}{36}\,x + x \int_0^x \left( x-t \right) y_m^2 (t)\,{\text d}t - 3 \int_x^{1} \left( x-t \right) y_m^2 (t)\,{\text d}t + 3 \left( x-1 \right) \int_0^{1} t \, y_m^2 (t) \,{\text d}t - 3\,x \int_0^{1} t\, y_m^2 (t) \,{\text d}t, \quad m=0,1,2,\ldots .$
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Example:    ■

1. Agarwal, R.P. and Loi, S.L., On approximate Picard's iterates for multipoint boundary value problems, Nonlinear Analysis: Theory, Methods & Applications, 1984, Vol. 8, Issue 4, pp. 381--391.
2. Lai, M. and Moffatt, D., Picard;s successive approximation for non-linear two-point boundary value problems, Journal of Computational and Applied Mathematics, 1982, Vol. 8, No 4, pp. 233--236.
3. Robin, W.A., Solving differential equations using modified Picard iteration, International Journal of Mathematical Education in Science and Technology, 2010, Vol. 41, No. 5, pp.649--665; https://doi.org/10.1080/00207391003675182