# Preface

This section provides an illustration of application of the Adomian decomposition method (ADM for short) to second order singular differential equations.

# Modified Laplace Decomposition Method

The Laplace transform is totally incapable of handling nonlinear equations because of the difficulties that are caused by the nonlinear terms. However, the combination of Adomian (or its modification) decomposition method with the Laplace transform can be successfully applied to solve nonlinear differential equations in semi-infinite domain. For illustration, we consider nonhomogeneous logistic equation.

Example: Consider the initial value problem for the logistic equation

$\dot{y} = 2\, y(t) - y^2 (t) +1+t , \qquad y(0) =1.$
Here $$\dot{y} = \texttt{D}y = {\text d}y/{\text d}t$$ is the derivative with respect to independent time variable t. We ask Mathematica to find its explicit solution, which is expressed via Airy functions:
p[t_] = y[t] /. DSolve[{y'[t] == 2*y[t] - (y[t])^2 + 1 + t, y[0] == 1}, y[t], t] // Flatten
$y(t) = \frac{\mbox{Ai}'(2)\,\mbox{Bi}'(2+t) + \mbox{Ai}'(2)\,\mbox{Bi}(2+t) - \mbox{Ai}'(2+t)\,\mbox{Bi}'(2) - \mbox{Ai}(2+t)\,\mbox{Bi}'(2)}{ \mbox{Ai}'(2)\,\mbox{Bi}(2+t) - \mbox{Ai}(2+t)\,\mbox{Bi}'(2)} .$
Then we expand the above function in Maclaurin series
Series[p[t], {t, 0, 8}]
$y(t) = 1 + 2t + \frac{t^2}{2} - \frac{4}{3}\, t^3 - \frac{t^4}{2} + \frac{61}{60}\,t^5 + \frac{5}{9}\, t^6 - \frac{481}{630}\, t^7 - \frac{823}{1440}\, t^8 + \cdots .$
To apply the Adomian decomposition method, let
$y(t) = \sum_{n\ge 0} u_n (t) , \qquad y^2 (t) = \sum_{n\ge 0} A_n (t) ,$
where An are the Adomian polynomials depending on y0, y1, ... , yn. Upon substituting into the logistic equation, it can be rewritten as
$\texttt{D}y(t) = \sum_{n\ge 0} \texttt{D}u_n (t) = 2\,\sum_{n\ge 0} u_n (t) - \sum_{n\ge 0} A_n (t) + 1 +t, \qquad y(0) =1.$
Based on the recursion scheme of the ADM, we define
$u_0 (t) = 1 + t + \frac{t^2}{2} ,$
as the solution to the initial value problem $$\dot{y} = 1+t , \ y(0) =1 .$$ Then next terms un+1 are obtaind from the recursive initial value problems:
$\dot{u}_{n+1} = 2\,u_n - A_n (t), \qquad u_{n+1} (0) =0, \qquad n=0,1,2,\ldots .$
Note that the Adomian polynomials An(t) for the quadratic nonlinearty are written as
\begin{eqnarray*} A_0 (t) &=& u_0^2 (t) = \left( 1 + t + \frac{t^2}{2} \right)^2 , \\ A_1 (t) &=& 2\,u_0 (t)\,u_1 (t) = \left( 2 + 2t + t^2 \right) u_1 (t) , \\ A_2 (t) &=& 2\,u_0 (t)\,u_2 (t) + u_1^2 (t) = \left( 2 + 2t + t^2 \right) u_2 (t) + u_1^2 (t) , \\ A_3 (t) &=& 2\,u_0 (t)\,u_3 (t) + 2\,u_1 (t)\,u_2 (t) = \left( 2 + 2t + t^2 \right) u_3 (t) + 2\,u_1 (t)\,u_2 (t) , \\ A_4 (t) &=& 2\,u_0 (t)\,u_4 (t) + 2\,u_1 (t)\,u_3 (t) + u_2^2 = \left( 2 + 2t + t^2 \right) u_4 (t) + 2\,u_1 (t)\,u_3 (t) + u_2^2 , \\ \end{eqnarray*} and so on. Mathematica helps us with solving these recursive IVPs:
DSolve[{y'[t] == 2*(1 + t + t^2/2) - (1 + t + t^2/2)^2, y[0] == 0}, y[t], t]// Flatten
u[1][t_] = y[t] /. %
1/60 (60 t - 20 t^3 - 15 t^4 - 3 t^5)
Next iterations:
DSolve[{y'[t] == 2*u[1][t] - 2*u[1][t]*(1 + t + t^2/2), y[0] == 0}, y[t], t]// Flatten
u[2][t_] = y[t] /. %
(-960 t^3 - 360 t^4 + 192 t^5 + 200 t^6 + 72 t^7 + 9 t^8)/1440
and
DSolve[{y'[t] == 2*u[2][t] - 2*u[2][t]*(1 + t + t^2/2) -(u[1][t])^2, y[0] == 0}, y[t], t]// Flatten
u[3][t_] = y[t] /. %
(1/4989600)(-1663200 t^3 + 1995840 t^5 + 1386000 t^6 - 19800 t^7 - 360360 t^8 - 185570 t^9 - 43659 t^10 - 3969 t^11)
Other components can be calculated similarly.

Now we add all terms to obtain Adomian approximation and plot it along with the true solution.

phi3[t_]=Simplify[(1 + t + t^2/2) + u[1][t] + u[2][t]+u[3][t]]
1 + 2 t + t^2/2 - (4 t^3)/3 - t^4/2 + (29 t^5)/60 + (5 t^6)/12 + ( 29 t^7)/630 - (19 t^8)/288 - (241 t^9)/6480 - (7 t^10)/800 - ( 7 t^11)/8800
Plot[{Callout[p[t], "exact"], Callout[phi3[t], "ADM approximation"]}, {t,0,1.8}, PlotStyle->Thick]
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