# Preface

This section gives an introduction to the homotopy pertubation method, first proposed by Ji-Huan He in 1998.

# Homotopy method

To find the solution x* of a nonlinear equation
$f(x) =0$
in the interval I = [a, b], we consider the homotopy
$H(x,p) = p\, f(x) + \left( 1- p \right) g(x) , \qquad x\in I, \quad p \in [0,1],$
which is a mapping I × [0,1] ↦ I. Here g(x) is an auxiliary function for which we know the solution g(x) = 0, and p is an embedding parameter. Usually, the function g is chosen in two forms:
• g(x) is a linear function, quadratic function, and so on;
• g(x) is the part function of f(x).
It is obvious that
$H(x,0) = g(x) = 0, \qquad H(x,1) = f(x) = 0, [0,1],$
and the changing process of p from 0 to 1 corresponds to moving H(x, p) from g(x) to f(x). In topology, this is called deformation.

Assuming that p is a small parameter from the interval [0,1], we can assume that the zero of the function H(x, p) can be expressed as a series in p:

$x= x_0 + p\,x_1 + p^2 x_2 + p^3 x_3 + \cdots .$
When p &maps; 1, we obtain (hopefully) the true solution of f(x) = 0, that is,
$x^{\ast} = \lim_{p\to 1} x = x_0 + x_1 + x_2 + x_3 + \cdots .$
To obtain its approximate solution of H(x, p) = 0, we first expand it into Taylor series
\begin{align*} f(x) &= f(x_0 ) + f' (x_0 ) \left( p\,x_1 + p^2 x_2 + p^3 x_3 + \cdots \right) + \frac{1}{2!}\,f'' \left( x_0 \right) \left( p\,x_1 + p^2 x_2 + p^3 x_3 + \cdots \right) + \cdots , \\ g(x) &= g\left( x_0 \right) + g' \left( x_0 \right) \left( p\,x_1 + p^2 x_2 + p^3 x_3 + \cdots \right) + \frac{1}{2!}\,g'' \left( x_0 \right) \left( p\,x_1 + p^2 x_2 + p^3 x_3 + \cdots \right) + \cdots . \end{align*}
Substituting these expansions into the equation H(x, p) = 0 and equating the coefficients of like powers of p, we get
\begin{align*} p^0 \,&: \ g\left( x_0 \right) =0; \\ p^1 \,&: \ g'\left( x_0 \right) x_1 + f\left( x_0 \right) - g\left( x_0 \right) =0; \\ p^2 \,&: \ g'\left( x_0 \right) x_2 + \left[ f'\left( x_0 \right) - g'\left( x_0 \right) \right] x_1 + \frac{1}{2}\, g'' \left( x_0 \right) x_1^2 =0; \\ p^3 \,&: \ g'\left( x_0 \right) x_3 + \left[ f'\left( x_0 \right) - g'\left( x_0 \right) \right] x_2 + \frac{1}{2}\left[ f'' \left( x_0 \right) - g'' \left( x_0 \right) \right] x_1 + \frac{1}{3!}\, g''' \left( x_0 \right) x_1^3 =0; \end{align*}
and so on. If $$g'\left( x_0 \right) \ne 0 ,$$ then the above equations can be solved to obtain
\begin{align*} x_0 & \ \mbox{ is exact null of the auxiliary function } \ g(x) = 0; \\ x_1 &= - \frac{f\left( x_0 \right)}{g'\left( x_0 \right)} ; \\ x_2 &= \frac{\left[ g'\left( x_0 \right) - f'\left( x_0 \right) \right] x_1 - \frac{1}{2}\, g'' \left( x_0 \right) x_1^2}{g'\left( x_0 \right)} ; \\ x_3 &= \frac{\frac{1}{2} \left[ g'' \left( x_0 \right) - f'' \left( x_0 \right) \right] x_1 - \left[ f'\left( x_0 \right) - g'\left( x_0 \right) \right] x_2 - g'' \left( x_0 \right) x_1 x_2 - \frac{1}{3!}\, g''' \left( x_0 \right) x_1^3}{g'\left( x_0 \right)} . \end{align*}
Particularly, if we choose the function $$g(x) = f(x) - f\left( x_0 \right) ,$$ where x0 is an initial approximation, then
\begin{align*} x_1 &= - \frac{f\left( x_0 \right)}{g'\left( x_0 \right)} ; \\ x_2 &= - \frac{1}{2!}\,\frac{f'' \left( x_0 \right)}{f'\left( x_0 \right)} \, x_1^2 = - \frac{1}{2!}\,\frac{f'' \left( x_0 \right)}{f'\left( x_0 \right)} \left\{ \frac{f\left( x_0 \right)}{g'\left( x_0 \right)} \right\}^2 ; \\ x_3 &= - \frac{1}{2!}\,\frac{f'' \left( x_0 \right)}{f'\left( x_0 \right)} \, x_1 x_2 - \frac{1}{3!} \, \frac{f''' \left( x_0 \right)}{f'\left( x_0 \right)} \, x_1^3 . \end{align*}
Therefore, we can obtain
$x= x_0 + x_1 \quad\mbox{with first order approximation},$
and
$x= x_0 + x_1 + x_2 \quad\mbox{with second order approximation},$
and
$x= x_0 + x_1 + x_2 + x_3 \quad\mbox{with third order approximation},$
From above, we can write down the iteration forms:
\begin{align*} x_{n+1} &= x_n - \frac{f\left( x_n \right)}{f'\left( x_n \right)} ; \\ x_{n+1} &= x_n - \frac{f\left( x_n \right)}{f'\left( x_n \right)} - \frac{f'' \left( x_n \right)}{2\,f'\left( x_n \right)} \left\{ \frac{f\left( x_n \right)}{f'\left( x_n \right)} \right\}^2 ; \\ x_{n+1} &= x_n - \frac{f\left( x_n \right)}{f'\left( x_n \right)} - \frac{f'' \left( x_n \right)}{2\,f'\left( x_n \right)} \left\{ \frac{f\left( x_n \right)}{f'\left( x_n \right)} \right\}^2 + \left\{ \frac{f''' \left( x_n \right)}{6\, f'\left( x_n \right)} - \frac{1}{2} \left[ \frac{f'' \left( x_n \right)}{f' \left( x_n \right)} \right]^2 \right\} \left\{ \frac{f\left( x_n \right)}{f'\left( x_n \right)} \right\}^3 . \end{align*}
Obviously, the first iteration formula is the well-known Newton's iteration formula. In order to avoid computing derivatives, we replace $$f'\left( x_n \right)$$ and $$f'' \left( x_n \right)$$ by the centered diﬀerences
$\frac{f \left( x_n +h \right) -f \left( x_n -h \right) }{2h} \qquad\mbox{and} \qquad \frac{f \left( x_n +h \right) + f \left( x_n -h \right) - 2\,f \left( x_n \right)}{h^2} ,$
respectively, where $$h = \alpha\, f\left( x_n \right)$$ is a step size and α ≠ 0 is a parameter. So we get the following second-order iterative method:
$x_{n+1} = x_n - \frac{2h\,f\left( x_n \right) }{f\left( x_n +h \right) - f\left( x_n -h \right)}$
and the third order iterative method
$x_{n+1} = x_n - \frac{2h\,f\left( x_n \right) }{f\left( x_n +h \right) - f\left( x_n -h \right)} - \frac{4h\,f^2 \left( x_n \right) \left[ f\left( x_n +h \right) + f\left( x_n -h \right) -2\,f\left( x_n \right) \right]}{\left[ f\left( x_n +h\right) - f\left( x_n -h \right) \right]^3} ,$
respectively.

Example: Suppose that we need to solve the rtanscendent equation

$e^{\cos x} -1-x =0 , \qquad x\in [0,1].$
Of course, Mathematica can find its root:
FindRoot[Exp[Cos[x]] == 1 + x, {x, 1}]
{x -> 0.884511}
which is an approximation to 0.8845106161658526.    ■

Example: Solve the polynomial equation

$x^9 = 1+x \quad\mbox{in the interval} \quad [1,2].$
FindRoot[x^9 == 1 + x, {x, 1}]
{x -> 1.08507}
which is an approximation to 1.085070245491451.    ■

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