# 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