# Preface

This section illustrates application of the modified decomposition method (MDM for short) in some systems of nonlinear ordinary differential equations.

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Introduction to Linear Algebra with Mathematica

# Euler Systems of ODEs

A linear system of differential equations
$t\,\dot{\bf y} = {\bf A} \, {\bf y} ,$
where $$\dot{\bf y} = {\text d}{\bf y}/{\text d}t$$ and A is a constant square matrix, is called an Euler system of equations or a Cauchy--Euler system of equation. It is sometimes referred to as an equidimensional equation because of its particularly simple equidimensional structure: the differential equation can be solved explicitly.
This vector equation with variable coefficients is an analog of a famous Euler equation for a single unknown function, discussed previously in section Euler equations. Every single Euler equation of the form
$a_n t^n y^{(n)} + a_{n-1} t^{n-1} y^{(n-1)} + \cdots + a_0 y =0 ,$
where a0, a1, ... , an ≠ 0 are given constants, can be reduced to an equivalent Euler system of equations.

Example: Using substitution

$y(t) = x_1 , \quad x_2 = t\,\dot{y} = t\,x'_1 , \quad x_3 = t\,\ddot{y} = t\,x'_2 , \quad \cdots \quad , x_n = t\, x'_{n-1} ,$
we calculate the derivatives in the Euler single equation
\begin{align*} x_1 &= y , \\ t\,y' &=x_2 , \\ t^2 y'' &= x_3 - x_2 , \\ t^3 y''' &= x_4 - 3\, x_3 + 2\, x_2 , \\ t^4 y^{(4)} &= x_5 - 6\, x_4 + 11\, x_3 - 6\, x_2 . \\ t^5 y^{(5)} &= x_6 -10\,x_5 + 35\, x_4 - 50\, x_3 + 24\,x_2 , \\ \vdots & \quad \vdots \\ t^{n-1} y^{(n-1)} &= x_n \end{align*}
■

Example: We consider the following Euler equation

$t^2 y'' + t\, y' - 4\, y =0 .$
If we seek its solution in the form y = tp, then upon substution into the Euler equation and canceling y, we obtain the indicial equation for p:
$p\left( p-1\right) +p - 4 =0 \qquad\mbox{or}\qquad p^2 - 4 =0.$
Since this algebraic equation has two real roots, the general solution of the given Euler equation is a linear combination of two power solutions:
$y (t) = c_1 t^2 + c_2 t^{-2} ,$
where c1 and c2 are arbitrary constants.

Now we convert this single differential equation of the second order to a system of first order equations by introducing new dependent variables:

$y (t) = x_1 , \qquad x_2 = t\, y' = t\, x'_1 .$
Then our single equation can be rewritten as
$t\, \frac{\text d}{{\text d}} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_2 \\ 4\,x_1 \end{bmatrix} = \begin{bmatrix} 0&1 \\ 4&0 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} .$
The characteristic polynomial for the corresponding matrix $${\bf A} = \begin{bmatrix} 0&1 \\ 4&0 \end{bmatrix}$$ is χ(λ) = det(λI - A) = λ² -4. Therefore, the eigenvalues of matrix A are ±2, and we obtain the same general solution as for single equation. ■
Using substitution
$t = e^x \qquad x = \ln t \qquad \Longrightarrow \qquad t\,\dot{\bf y} = t\,\frac{{\text d}{\bf y}}{{\text d}t} = \frac{{\text d}{\bf y}}{{\text d}x} ,$
we reduce the Euler system ty' = Ay to a constant coefficient system of differential equations:

$\frac{{\text d}{\bf y}}{{\text d}x} = {\bf A} \, {\bf y} .$
The general solution of the latter is known to be
${\bf y} (x) = e^{{\bf A} \,x} \, {\bf c} \qquad \Longrightarrow \qquad {\bf y} (t) = e^{{\bf A} \,\ln t} \, {\bf c} ,$
where c is a column vector of arbitrary constants. The latter formula is not friendly for applications, so we explain how to use it depending on eigenvalues of matrix A.

Suppose that n-by-n matrix A is diagonalizable and all its eigenvalues are real numbers. Then there exists a basis of eigenvectors:

$e^{{\bf A} \,x} = \left[ e^{\lambda_1 x} \, {\bf v}_1 , e^{\lambda_2 x} \, {\bf v}_2 , \ldots , e^{\lambda_n x} \, {\bf v}_n \right] = \left[ t^{\lambda_1} {\bf v}_1 , t^{\lambda_2} {\bf v}_2 , \ldots , t^{\lambda_n} {\bf v}_n \right] .$
In this case, the general solution of the Euler system of equations has the following form:
${\bf y} (t) = t^{\lambda_1} {\bf v}_1 c_1 + t^{\lambda_2} {\bf v}_2 c_2 + \cdots + t^{\lambda_n} {\bf v}_n c_n ,$
where $$c_1 , c_2, \ldots , c_n$$ are arbitrary constants.

Example: Solve the Euler vector equation $$t\,\dot{\bf y} = {\bf A}\, {\bf y} ,$$ subject to the initial condition $$\langle y_1 (1) , y_2 (1) \rangle^{\mathrm T} = \langle 3 , 2 \rangle^{\mathrm T} ,$$ where the matrix is

${\bf A} = \begin{bmatrix} 2 & 4 \\ 1 & -1 \end{bmatrix} .$

Substituting $${\bf y} (t) = t^{\lambda} {\bf \xi} ,$$ where λ is a scalar and ξ is unknown vector, into the given Euler equation, we obtain

$t\,\dot{\bf y} = t\,\lambda\, t^{\lambda -1} \,{\bf \xi} = t^{\lambda} \,{\bf A}\, {\bf \xi} .$
Canceling tλ from both sides, we get a vector equation
$\lambda\,{\bf \xi} = {\bf A}\, {\bf \xi} ,$
which indicates that λ is an eigenvalue and ξ is a corresponding eigenvector for the matrix A. Since the matrix A has two real eigenvalues λ = 3 and λ = -2, we obtain the general solution:
${\bf y} (t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = c_1 \,t^{-2} \begin{bmatrix} -1 \\ 1 \end{bmatrix} + c_2 \, t^3 \begin{bmatrix} 4 \\ 1 \end{bmatrix} ,$
where c1 and c2 are arbitrary constants. From the initial conditions, it follows that c1 = 1 and c2 = 1 and we get the solution:
${\bf y} (t) = \begin{bmatrix} x(t) \\ y(t) \end{bmatrix} = t^{-2} \begin{bmatrix} -1 \\ 1 \end{bmatrix} + t^3 \begin{bmatrix} 4 \\ 1 \end{bmatrix} .$

Example: We consider the following Euler equation

$t\,\dot{\bf y} = {\bf A}\,{\bf y}, \qquad {\bf A} = \begin{bmatrix} 1 & -1 \\ 1 & 3 \end{bmatrix} .$
Since matrix A has double eigenvalue λ = 2, we seek its general solution in the form:
${\bf y} = c_1 {\bf \xi} t^2 + c_2 t^2 \left( {\bf \xi}\,\ln |t| + {\bf \eta} \right) ,$
where c1, c2 are arbitrary constants and ξ and η are some vectors to be determined. ■

Example: We consider the following Euler equation

$t\,\dot{\bf y} = {\bf A}\,{\bf y}, \qquad {\bf A} = \begin{bmatrix} 1 & -1 \\ 1 & 1 \end{bmatrix} .$
A = {{1, -1}, {1, 1}}
Eigenvalues[A]
{1 + I, 1 - I}
Eigenvectors[A]
{{I, 1}, {-I, 1}}
The eigenvalues of matrix A are 1 ±j. We seek its solution in the form
${\bf y} = c_1 t\,\cos \left( \ln |t| \right) {\bf \xi} + t\,c_2 \sin \left( \ln |t| \right) {\bf \eta} ,$
for some vectors ξ and η. Since its derivative is
$\dot{\bf y} = c_1 \cos \left( \ln |t| \right) {\bf \xi} - c_1 \sin \left( \ln |t| \right) {\bf \xi} + c_2 \sin \left( \ln |t| \right) {\bf \eta} + c_2 \cos \left( \ln |t| \right) {\bf \eta} ,$
we get from the Euler equation that
$\left( c_1 {\bf \xi} + c_2 {\bf \eta} \right) \cos \left( \ln |t| \right) + \left( -c_1 {\bf \xi} + c_2 {\bf \eta} \right) \sin \left( \ln |t| \right) = {\bf A} \left[ c_1 \cos \left( \ln |t| \right) {\bf \xi} + c_2 \sin \left( \ln |t| \right) {\bf \eta} \right] .$
Since sine and cosine functions are linearly independent, we have the following equations for unknown vectors:
$\begin{split} c_1 {\bf \xi} + c_2 {\bf \eta} &= {\bf A}\, c_1 {\bf \xi} , \\ - c_1 {\bf \xi} + c_2 {\bf \eta} &= {\bf A}\, c_2 {\bf \eta} \end{split} \qquad\mbox{or} \qquad \begin{split} \left( {\bf A} - {\bf I} \right) c_1 {\bf \xi} &= c_2 {\bf \eta} \\ \left( {\bf A} - {\bf I} \right) c_2 {\bf \eta} &= - c_1 {\bf \xi} . \end{split}$
Multiplying the first equation by A - I, we obtain
$\left( {\bf A} - {\bf I} \right)^2 c_1 {\bf \xi} = \left( {\bf A} - {\bf I} \right) c_2 {\bf \eta} = - c_1 {\bf \xi} .$
B = MatrixPower[A - IdentityMatrix[2], 2]
{{-1, 0}, {0, -1}}
Eigenvalues[B]
{-1, -1}
Eigenvectors[B]
{{0, 1}, {1, 0}}
Therefore, the vector ξ = [0,1]T must be an eigenvector of matrix (A - I)² = -I corresponding to its eigenvalue (-1). Similar conclusion is true for another vector η = [1,0]T. ■

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