Preface

This section presents derivation of simple pendulum equation.

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Glossary

Variational Iteration Method

The variational iteration method (VIM) was developed by Ji-Huan He in 2000--2007. This method is preferable over numerical methods as it is free from rounding off errors as it does not involve discretization and does not require large computer power or memory. The VIM gives rapidly convergent successive approximations of the exact solution if such a solution exists; otherwise, a few approximations can be used for numerical purposes. Many researches in variety of scientific fields applied this method and showed the VIM has many merits and to be reliable for a variety of scientific application.

To illustrate the basic concept of the He’s VIM, consider the following general nonlinear equation:

\[ L\left[ u(t) \right] = N\left[ u(t) \right] + g(t) , \]
where L is the linear differential operator of order m, N is the nonlinear operator, and g is the input (known) function. He has modified the general Lagrange multiplier method to an iteration method called as correction functional. The basic character of the method is to construct a correction functional for the aforementioned equation, which reads as
\[ u_{n+1} (t) = u_n (t) + \int_0^t \lambda (s) \left\{ L\left[ u_n (s) \right] - N\left[ \tilde{u}_n (s) \right] - g(s) \right\} {\text d}s , \]
where λ(s,t) is so called a general Lagrange multiplier, which can be identified optimally via variational theory. The subscript n denotes the n-th approximation, and \( \tilde{u}_n \) is a restricted variation., i.e., \( \delta\tilde{u}_n = 0 . \) Having λ obtained, an iteration formula should be used for the determinantion of the successive approximations un+1(t) of the solution u(t). The initial approximation u0(t) can be arbitrary function; however, it is usually selected based on the auxiliary conditions (such as the initial or boundary conditions). Consequently, the solution is given by
\[ u(t) = \lim_{n\to\infty}\,u_n (t) . \]

The main problem with VIM is the determination of the Lagrange multiplier, and there are known many forms of them. It could be achieved by making the correction functional stationary. Taking variation with respect to the variable un, noticing δun(0)=0, we get

\begin{eqnarray*} \delta u_{n+1} (t) &=& \delta u_n (t) + \delta \int_0^t \lambda \left\{ L\left[ u_n (s) \right] - N\left[ \tilde{u}_n (s) \right] - g(s) \right\} {\text d}s \\ &=& \delta u_n (t) + \left. \lambda\,\delta u_n^{(m-1)} \right\vert_{s=t} - \left. \lambda ' \,\delta u_n^{(m-2)} \right\vert_{s=t} + \cdots + \left. (-1)^{m-1} \lambda^{(m-1)} \delta u_n \right\vert_{s=t} + \int_0^t L^{\ast} \left[ \lambda \right] \delta u_n \,{\text d}s =0 , \end{eqnarray*}
for all variations δun. This yields the following equation
\[ L^{\ast} \left[ \lambda \right] =0 , \]
subject to some auxiliary conditions on the diagonal s = t. Here L* is adjoint differential operator for L. In other words, if L is a m-th order linear differential operator, we need to find an extremum form the following functional
\[ \int_0^t \lambda \,L\left[ u_n (s) \right] {\text d}s = \int_0^t \lambda \,f \left( s, u_n (s) , u'_n (s) , \ldots , u_n^{(m)} \right) {\text d}s . \]
So we consider the general functional
\[ I = \int_0^t F \left( s, y(s) , y' (s) , \ldots , y^{(m)} \right) {\text d}s \]
that attains an extremum on solutions of the generalized Euler--Lagrange equation (or simply the Euler equation):
\[ \frac{\partial F}{\partial y} - \frac{\text d}{{\text d}s} \left( \frac{\partial F}{\partial y'} \right) + \frac{{\text d}^2}{{\text d}s^2} \left( \frac{\partial F}{\partial y''} \right) - \cdots + (-1)^m \frac{{\text d}^m}{{\text d}s^m} \left( \frac{\partial F}{\partial y^{(m)}} \right) =0 . \]
Using the above equation, we see that extremum of the variational functional is given as follows
\[ \frac{\partial \left( \lambda \,f \right)}{\partial y} - \frac{\text d}{{\text d}s} \left( \frac{\partial \left( \lambda \,f \right)}{\partial y'} \right) + \frac{{\text d}^2}{{\text d}s^2} \left( \frac{\partial \left( \lambda \,f \right)}{\partial y''} \right) - \cdots + (-1)^m \frac{{\text d}^m}{{\text d}s^m} \left( \frac{\partial \left( \lambda \,f \right)}{\partial y^{(m)}} \right) =0 . \]
Solving these equations is sufficient to determine the Lagrange multipliers. We show how it works in a series on examples.

Example: Consider the IVP for the following Riccati equation

\[ y' = 3\,y - y^2 + 2\,x, \qquad y(0) =0. \]
First, we write the variational iteration sequence:
\[ y_{n+1} (x) = y_n (x) + \int_0^x \lambda (x,s) \left\{ y'_n (s) - 3\,y_n (s) + \tilde{y}_n^2 - 2\,s \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Taking variation, we obtain
\begin{eqnarray*} \delta y_{n+1} (x) &=& \delta y_n (x) + \delta \int_0^x \lambda \left\{ y'_n (s) - 3\,y_n (s) + \tilde{y}_n^2 - 2\,s \right\} {\text d}s \\ &=& \delta y_n (x) + \delta \int_0^x \lambda \left\{ y'_n (s) - 3\,y_n (s) \right\} {\text d}s \\ &=& \delta y_n (x) + \left. \lambda\,\delta y_n (s) \right\vert_{s=0}^{s=t} - \int_0^t \left[ \lambda ' + 3\,\lambda \right] \delta y_n (s) \, {\text d}s . \end{eqnarray*}
This gives us the equations for determination of the Lagrange multiplier:
\[ \begin{split} \lambda ' (s) + 3\,\lambda (s) &=0 , \\ 1 + \left. \lambda (s) \right\vert_{s=t} &= 0. \end{split} \]
So the multiplier can be identified as
\[ \lambda (x,s) = - e^{3(x-s)} . \]
Then the iteration formula becomes
\[ y_{n+1} = y_n - \int_0^x e^{3(x-s)} \left\{ y'_n (s) - 3\,y_n (s) + y_n^2 - 2\,s \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Using previously defined subroutine seriesDSolve, we find series solution of the given problem:
seriesDSolve[ y'[x] == 2*x + 3*y[x] - (y[x])^2, y, {x, 0, 10}, {y[0] -> 0}]
\[ y = x^2 + x^3 + \frac{3}{4}\,x^4 + \frac{x^5}{4} - \frac{5}{24}\, x^6 - \frac{25}{56}\,x^7 - \frac{187}{448}\, x^8 - \frac{2551}{12096}\, x^9 + \frac{1217}{40320}\, x^{10} + \cdots . \]
We start with an initial approximation y0 = x². This yields
\begin{eqnarray*} y_1 (x) &=& x^2 - \int_0^x e^{3(x-s)} \left\{ 2s - 3\,s^2 + s^4 - 2\,s \right\} {\text d}s = x^2 + \frac{1}{81} \left( 27\,x^4 -45\,x^2 -30\,x -10 \right) + \frac{10}{81}\, e^{3x} , \\ y_2 &=& y_1 - \int_0^x e^{3(x-s)} \left\{ y'_1 -3\,y_1 + y_1^2 -2\,s \right\} {\text d}s \approx x^2 - \frac{x^3}{3} - \frac{x^4}{4} - \frac{3}{20}\, x^5 - \frac{3}{40}\, x^6 + \frac{253}{840}\, x^7 + \cdots . \end{eqnarray*}
So we see that calculations become messy very shortly. We ask Mathematica for help.
y1[t_] = t^2 - Integrate[(s^4 - 3*s^2)*Exp[3*t - 3*s], {s, 0, t}];
y2[t_] = t^2 - Integrate[(D[y1[s], s] - 3*y1[s] + (y1[s])^2 - 2*s)* Exp[3*t - 3*s], {s, 0, t}];
y3[t_] = t^2 - Integrate[(D[y2[s], s] - 3*y2[s] + (y2[s])^2 - 2*s)* Exp[3*t - 3*s], {s, 0, t}];
Expand[y3[t]]
Series[%, {t, 0, 10}]
\[ y_3 (x) \approx x^2 + x^3 + \frac{3}{4}\,x^4 + \frac{x^5}{4} + \frac{31}{56}\, x^7 + \frac{205}{448}\, x^8 + \frac{4505}{12096} \, x^9 + \cdots , \]
which gives correct answer up to the power 6. Then we plot the third approximation y3 and the true solution.
sol = (y[x] /.
DSolve[{y'[x] == 2*x + 3*y[x] - (y[x])^2, y[0] == 0}, y[x], x])[[1]];
Plot[{Callout[sol, "exact"], Callout[y3[x], "three-term VIM approximation"]}, {x,0,1.2}, PlotStyle->Thick]
True solution and VIM approximations with N = 3 terms.

We demonstrate a modified variational iteration method that leads in our case to the recurrence:

\[ y_{n+1} (t) = y_n (t) + 3 \int_0^t \left( y_n (s) - y_{n-1} (s) \right) {\text d}s - \int_0^t \left( y_n^2 (s) - y_{n-1}^2 (s) \right) {\text d}s , \qquad n=1,2,\ldots ; \]
where \( y_{-1} = 0, \ y_0 = 0 \) (from the initial condition) and
\[ y_{1} (t) = y_0 + \int_0^t \left[ 3 \left( y_0 - y_{-1} \right) - \left( y_0^2 - y_{-1}^2 \right) + 2s \right] {\text d}s = t^2 . \]
Therefore, we can obtain the following successive approximations:
\begin{eqnarray*} y_2 (t) &=& t^2 + \int_0^t \left[ 3 \left( y_1 - y_{0} \right) - \left( y_1^2 - y_{0}^2 \right)\right] {\text d}s = t^2 + t^3 - \frac{t^5}{5} , \\ y_3 (t) &=& t^2 + t^3 + \int_0^t \left[ 3 \left( y_2 - y_{1} \right) - \left( y_2^2 - y_{1}^2 \right)\right] {\text d}s = t^2 + t^3 + \frac{3}{4}\,t^4 - \frac{t^5}{5} - \frac{13}{30}\, t^6 + \frac{t^8}{20} + \frac{2}{45}\,t^9 - \frac{t^{10}}{275} , \\ y_4 (t) &=& t^2 + t^3 + \int_0^t \left[ 3 \left( y_3 - y_{2} \right) - \left( y_3^2 - y_{2}^2 \right)\right] {\text d}s = t^2 + t^3 + \frac{3}{4}\,t^4 + \frac{t^5}{4} - \frac{13}{30}\, t^6 - \frac{19}{35}\, t^7 - \frac{107\,t^8}{560} + \frac{41}{432}\,t^9 + \frac{111}{700} \, t^{10} + \cdots . \end{eqnarray*}
y2[t_] = t^2 + t^3 - t^5 /5
y3[t_] = y2[t] + Integrate[3*y2[s] - 3*s^2 - (y2[s])^2 + s^4, {s, 0, t}]
y4[t_] = y3[t] + Integrate[3*y3[s] - 3*y2[s] - (y3[s])^2 + (y2[s])^2, {s, 0, t}]
■

Example: Consider the one dimensional steady-state heat conduction problem

\[ \frac{{\text d}^2 T}{{\text d} x^2} + T(x) + 2x =0, \qquad (0< x < \ell , \qquad T(0) = 4, \quad T(\ell ) =3 . \]
Here T is the temperature within the rod of length ℓ, and we set for simplicity ℓ=2. Its correction functional can we written as
\[ T_{n+1} (x) = T_n (x) + \int_0^x \lambda \left\{ T''_n (s) + T_n (s) + 2s \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Taking variation with respect to the dependent avriable Tn, and noticing that &deltaTn(0)=0, we have
\begin{eqnarray*} \delta T_{n+1} (x) &=& \delta T_n (x) + \delta \int_0^x \lambda \left\{ T''_n (s) + T_n (s) + 2s \right\} {\text d}s \\ &=& \delta T_n (x) + \left. \lambda \,\delta T'_n (s) \right\vert_{s=x} - \left. \lambda ' \,\delta T_n (s) \right\vert_{s=x} + \int_0^x \left[ \lambda '' + \lambda \right] \delta T_n (s) {\text d}s \end{eqnarray*}
for all variations δTn and δT'n. This yields the following equations:
\[ \begin{split} \lambda '' + \lambda &= 0, \\ \left. \lambda (s) \right\vert_{s=x} =0 , \\ \left. 1-\lambda ' (s) \right\vert_{s=x} =0. \end{split} \]
Therefore, the lagrange multiplier becomes
\[ \lambda (x,s) = \sin \left( s-x \right) , \]
which gives us the following iteration formula
\[ T_{n+1} (x) = T_n (x) + \int_0^x \sin \left( s-x \right) \left\{ T''_n (s) + T_n (s) + 2s \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
We choose the initial approximation as a function that satisfies the boundary conditions:
\[ T_{0} (x) = T(0) \,\frac{\ell -x}{\ell} + \frac{x}{\ell} \, T(\ell ) = 2 \left( 2-x \right) + \frac{3}{2}\, x = 4 - \frac{x}{2} . \]
Now we use iteration
\begin{eqnarray*} T_1 (x) &=& T_0 (x) + \int_0^x \sin \left( s-x \right) \left\{ T''_0 (s) + T_0 (s) + 2s \right\} {\text d}s = -2x + 4\,\cos x + \frac{3}{2}\, \sin x , \\ T_2 (x) &=& T_1 (x) + \int_0^x \sin \left( s-x \right) \left\{ T''_1 (s) + T_1 (s) + 2s \right\} {\text d}s = T_1 (x) . \end{eqnarray*}
So the iteration stops after first step. We compare it with the exact solution.
DSolve[{y''[t] + y[t] + 2*t == 0, y[0] == 4, y[2] == 3}, y[t], t] // Flatten
exact[t_] = y[t] /. %
\[ T(x) = -2x + 4\,\cos x - 4\,\cos 2\,\sin x + 7\,\sec 2\,\sin x . \]
T0[x_] = 4 - x/2;
T1[x_] = T0[x] + Integrate[Sin[s - x]*(D[T0[s], s, s] + T0[s] + 2*s), {s, 0, x}]
Plot[{Callout[exact[x], "exact"], Callout[T1[x], "VIM approximation"]}, {x,0,2}, PlotStyle->Thick]
True solution and VIM approximations with N = 3 terms.

So we see that the linear function as the initial approximation does not work. Instead, we use its complementary function \( T_0 (x) = A\,\cos x + B\,\sin x . \) Then

\[ T_{1} (x) = A\,\cos x + B\,\sin x + \int_0^x \sin \left( s-x \right) \left\{ T''_0 (s) + T_0 (s) + 2s \right\} {\text d}s = A\,\cos x + \left( B+2 \right) \sin x -2x . \]
By imposing the boundary conditions, we get A = 4 and \( B= 7\,\csc 2 - 4\,\cot 2 -2 . \) So
\[ T_{1} (x) = 4\,\cos x + \left( 7\,\csc 2 -4\,\cot 2 \right) -2\,x , \]
Next iteration provides a function that is far away from the true solution. ■

Example: Consider the IVP for harmonic oscillator

\[ \ddot{y} + \omega^2 y = f(t) \qquad\mbox{with}\quad f(t) = A\,\sin \omega t +B\,\cos t , \]
subject to the initial conditions
\[ y(0) =1, \qquad \dot{y} (0) =-1. \]
Its correction functional can be written as follows
\[ y_{n+1} (t) = y_n (t) + \int_0^t \lambda \left\{ y''_n (s) + \omega^2 \,y_n (s) - f(s) \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Making the above correction functional stationary, and noticing that δy(0) = 0, we get
\begin{eqnarray*} \delta y_{n+1} (t) &=& \delta y_n (t) + \delta \int_0^t \lambda \left\{ y''_n (s) + \omega^2 \,y_n (s) - f(s) \right\} {\text d}s \\ &=& \delta y_n (t) + \left. \lambda (s)\,\delta y'_n (s) \right\vert_{s=t} - \left. \lambda ' (s)\, \delta y_n (s) \right\vert_{s=t} + \int_0^t \left[ \lambda '' + \omega^2 \lambda \right] \delta y_n (s) \,{\text d}s . \end{eqnarray*}
From the right hand side, it follows
\[ \begin{split} \lambda '' + \omega^2 \lambda &= 0 , \\ \left. \lambda (s) \right\vert_{s=t} &= 0 , \\ \left. 1 - \lambda ' (s)\right\vert_{s=t} &= 0 . \end{split} \]
The Lagrange multiplier can be readily identified as
\[ \lambda (t, s) = \frac{1}{\omega}\,\sin \omega (s-t) . \]
With this in hands, we obtain the following iteration formula
\[ y_{n+1} (t) = y_n (t) + \frac{1}{\omega} \int_0^t \sin \omega (s-t) \left\{ y''_n (s) + \omega^2 \,y_n (s) - f(s) \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Now we consider the homogeneous equation corresponding to the given driven equation \( \ddot{y} + \omega^2 y =0 \) subject to the given initial conditions; its solution, called the complementary function, can be chosen as the initial approximation
\[ y_{0} (t) = C_1 \cos \omega t + C_2 \sin \omega t = \cos \omega t - \frac{\sin \omega t}{\omega} , \]
where C1 and C2 are such constants to satisfy the given initial conditions, y(0) = 1 and y'(0) = -1. Then the iteration formula gives
\begin{eqnarray*} y_1 (t) &=& y_0 (t) - \frac{1}{\omega} \int_0^t \sin \omega (s-t) \left[ A\,\sin \omega s + B\,\sin \omega s \right] {\text d}s \\ &=& C_1 \cos \omega t + C_2 \sin \omega t - \frac{A}{2} \left[ t\,\cos \omega t - \frac{\sin \omega t}{\omega} \right] + \frac{B\,\omega}{\omega^2 -1} \left[ \cos \omega t - \cos t \right] , \end{eqnarray*}
which is the general solution of the given differential equation (harmonic oscillator).

However, is we apply restricted variations to the correction function, then its exact solution can be arrived at only by successive iterations. Considering the corresponding homogeneous differential equation \( \ddot{y} + \omega^2 y =0 , \) we rewrite the correction functional as follows:

\[ y_{n+1} (t) = y_n (t) + \int_0^t \lambda \left\{ y''_n (s) + \omega^2 \,\tilde{y}_n (s) \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
Here \( \tilde{y}_n (s) \) is considered a restricted variation, then the stationary conditions (\( \delta\tilde{y}_n =0 \) ) of the above correction functional can be expressed as follows
\[ \begin{split} \lambda '' (s) &=0 , \\ \left. \lambda (s) \right\vert_{s=t} &= 0, \\ \left. 1 - \lambda ' (s) \right\vert_{s=t} &= 0 . \end{split} \]
Therefore, the Lagrange multiplier becomes \( \lambda = s-t . \) This leads to the following iteration formula
\[ y_{n+1} (t) = y_n (t) + \int_0^t (s-t) \left\{ y''_n (s) + \omega^2 \,y_n (s) - f(s) \right\} {\text d}s , \qquad n=0,1,2,\ldots . \]
It we choose the initial iteration as \( y_0 = 1 -t \) in order to satisfy the initial conditions, we have
\begin{eqnarray*} y_1 (t) &=& y_0 (t) + \omega^2 \int_0^t \left( s-t \right) y_0 (s)\, {\text d}s = 1-t + \omega^2 \left( -\frac{t^2}{2} + \frac{t^3}{6} \right) , \\ y_2 (t) &=& y_1 (t) + \int_0^t \left( s-t \right) \left( y''_1 (s) + \omega^2 y_1 (s) \right) {\text d}s = 1-t + \omega^2 \left( -\frac{t^2}{2} + \frac{t^3}{6} \right) + \omega^4 \left( \frac{t^4}{24} - \frac{t^5}{120} \right) , \\ y_3 (t) &=& y_2 (t) + \int_0^t \left( s-t \right) \left( y''_2 (s) + \omega^2 y_2 (s) \right) {\text d}s = 1-t + \omega^2 \left( - \frac{t^2}{2!} + \frac{t^3}{3!} \right) + \omega^4 \left( \frac{t^4}{4!} - \frac{t^5}{5!} \right) + \omega^6 \left( - \frac{t^6}{6!} + \frac{t^7}{7!} \right) , \\ y_n (t) &=& \sum_{k=0}^n \frac{(-1)^n}{(2k)!}\, \omega^{2k} t^{2k} - \frac{1}{\omega} \,\sum_{k=0}^n (-1)^k \frac{\omega^{2k+1} t^{2k+1}}{(2k+1)!} . \end{eqnarray*}
Thus, we obtaine
\[ \lim_{n\to\infty}\, y_n (t) = \cos\omega t - \frac{1}{\omega}\,\sin \omega t , \]
which is the exact solution of the corresponding homogeneous equation. ■

Example: Consider the IVP for the pendulum equation

\[ \ddot{\theta} + \omega^2 \sin \theta =0 , \qquad \theta (0) = \theta_0 , \quad \dot{\theta} (0) =0. \]
■

Example: Consider the IVP for harmonic oscillator ■

Example: Consider the third order differential equation

\[ u''' + 4\,u' = f(t) , \qquad u(0) = \]
■
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