ODE's homogeneous systems with constant coefficients

A linear system of differential equations is an ODE (ordinary differential equation) of the type: $x^{'} (t) = A (t) \cdot x + b (t)$ Where, $A (t)$ is a matrix, $n \times n$ , of functions of the variable $t$ , $b (t)$ is a dimension $n$ vector of functions of the variable $t$ , and $x$ is a vector of dimension $n$ that is the function that we want to find.

Example

An example of system of linear ODE would be: ${(\begin{matrix} x \\ y \end{matrix})}^{'} = (\begin{matrix} 0 & \ln t \\ - e^{t} & 3 \cos t \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix}) + (\begin{matrix} e^{t} \\ 3 e^{t} \end{matrix})$

A linear system of dimension $n$ has $n$ linearly independent solutions and solving the system means finding all of them. Any solution is a linear combination of these $n$ solutions. Therefore, given a vector of initial conditions ( $n$ ), we will determine the $n$ constants and will find the exact solution.

When we solve a linear system we will place the $n$ vectors that are the solution (linearly independent) in the columns of a matrix, the so-called fundamental matrix of the system ( $n \times n$ ). Therefore, solving the system is to find the fundamental matrix. By multiplying this matrix by a vector of arbitrary constants we will have the general solution.

An important property of the fundamental matrices is that if we multiply a fundamental matrix by a constant matrix with a determinant different from zero, the result is another fundamental matrix (it is important that the constant matrix multiplies from the right, if not this is not true).

There are no explicit methods to solve these types of equations, (only in dimension 1). Nevertheless, there are some particular cases that we will be able to solve: Homogeneous systems of ode's with constant coefficients, Non homogeneous systems of linear ode's with constant coefficients, and Triangular systems of differential equations.

In this unit we are going to explain the Homogeneous systems of ode's with constant coefficients.

Let's consider the problem: $x^{'} (t) = A \cdot x (t)$ . Then the fundamental matrix is: $ϕ (t) = e^{t A}$ . To do the calculation of raising $e$ to a matrix $A$ it is useful to have this matrix in Jordan's form.

Therefore, first of all let's calculate the Jordan form of the matrix $A$ . We have $A = S \cdot J \cdot S^{- 1}$ , where $J$ is the reduced Jordan form of the matrix $A$ and $S$ is the basis change.

Then it is satisfied that $e^{A} = e^{S J S^{-} 1} = S \cdot e^{J} \cdot S^{- 1}$ . Also, bearing in mind that $S^{- 1}$ is a constant matrix with determinant other than zero, if $ϕ (t) = e^{t A} = S \cdot e^{t J} \cdot S^{- 1}$ is a fundamental matrix, then $ϕ (t) = S \cdot e^{t J}$ is also a fundamental matrix.

Now we will tell how the exponential of a matrix is calculated in Jordan's form.

We will explain up to dimension three:

If $J$ is a diagonal matrix. Then $e^{t J} = (\begin{matrix} e^{λ_{1} t} & 0 \\ 0 & e^{λ_{2} t} \end{matrix}), e^{t J} = (\begin{matrix} e^{λ_{1} t} & 0 & 0 \\ 0 & e^{λ_{2} t} & 0 \\ 0 & 0 & e^{λ_{3} t} \end{matrix})$ for the cases of dimension 2 and 3, where $λ$ are the eigenvalues of the matrix. Namely the exponential of a diagonal matrix is the diagonal matrix with the exponential of the eigenvalues in the diagonal.
If $J$ is a matrix of the type: $J = (\begin{matrix} λ & 0 \\ 1 & λ \end{matrix})$ then $e^{t J} = (\begin{matrix} e^{λ t} & 0 \\ t \cdot e^{λ t} & e^{λ t} \end{matrix})$
If $J$ is a matrix of the type: $J = (\begin{matrix} λ_{1} & 0 & 0 \\ 0 & λ_{2} & 0 \\ 0 & 1 & λ_{3} \end{matrix})$ then $e^{t J} = (\begin{matrix} e^{λ_{1} t} & 0 & 0 \\ 0 & e^{λ_{2} t} & 0 \\ 0 & t \cdot e^{λ_{2} t} & e^{λ_{3} t} \end{matrix})$
If $J$ is a matrix of the type : $J = (\begin{matrix} λ & 0 & 0 \\ 1 & λ & 0 \\ 0 & 1 & λ \end{matrix})$ then $e^{t J} = (\begin{matrix} e^{λ t} & 0 & 0 \\ t \cdot e^{λ t} & e^{λ t} & 0 \\ \frac{t^{2}}{2} \cdot e^{λ t} & t \cdot e^{λ t} & e^{λ t} \end{matrix})$ Therefore, we see that the solution is: $x (t) = ϕ (t) \cdot C = S \cdot e^{t J} \cdot C$ , where $C$ is a vector of constants to be determined with the initial conditions. If we have $x (t_{0}) = x_{0}$ , vector of initial conditions, then $C = ϕ (t_{0})^{- 1} \cdot x_{0}$ and we have the complete solution (Problem of Initial Value).

Let's see more clearly with an example.

Example

Let's consider the system: ${(\begin{matrix} x \\ y \end{matrix})}^{'} = (\begin{matrix} 0 & 1 \\ - 1 & 2 \end{matrix}) \cdot (\begin{matrix} x \\ y \end{matrix})$ The matrixes have constant coefficients, therefore, as we already said, $ϕ (t) = e^{t A}$ is a fundamental matrix.

To calculate the exponential of a matrix we will calculate its Jordan matrix.

Let's then calculate the eigenvalues of the matrix: $| \begin{matrix} - λ & 1 \\ - 1 & 2 - λ \end{matrix} | = - λ (2 - λ) + 1 = 0 \Leftrightarrow λ^{2} - 2 λ + 1 = 0 \Rightarrow= {\begin{array}{rcl} λ_{1} & = & 1 \\ λ_{2} & = & 1 \end{array}$ So we have one eigenvalue of multiplicity 2.

To know if we can diagonalize the matrix we must calculate the rank of the matrix $(A - λ \cdot I d) = (\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix})$ which is clearly 1.

As the arithmetical and geometrical multiplicities do not coincide, the matrix cannot be diagonalized.

Therefore the Jordan form of the matrix will be: $J = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ Let's look now for a base of eigenvectors of the matrix.

We choose $v_{2} \in Ker (A - λ \cdot I d) and v_{1} \in Ker ((A - λ \cdot I d)^{2}), v_{1} \notin Ker (A - λ \cdot I d)$ .

Let's then write the matrixes: $A - λ \cdot I d = (\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}), (A - λ \cdot I d)^{2} = (\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix})$

Therefore choosing $v_{1} = (0, 1)$ , it is clear that $v_{1} \in Ker ((A - λ \cdot I d)^{2}), v_{1} \notin Ker (A - λ \cdot I d)$ .

As $v_{2}$ we take $v_{2} = (1, 1)$ because $(\begin{matrix} - 1 & 1 \\ - 1 & 1 \end{matrix}) \cdot (\begin{matrix} 1 & 1 \end{matrix}) = (\begin{matrix} 0 & 0 \end{matrix})$ and therefore $v_{2} \in Ker (A - λ \cdot I d)$ .

Thus, the change of basis matrix is: $S = (\begin{matrix} = 0 & 1 \\ 1 & 1 \end{matrix})$ As we said, the exponential in the Jordan form is: $e^{t J} = (\begin{matrix} e^{t} & 0 \\ t \cdot e^{t} & e^{t} \end{matrix})$

Therefore the fundamental matrix will be: $ϕ (t) = S \cdot e^{t J} = (\begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix}) \cdot (\begin{matrix} e^{t} & 0 \\ t \cdot e^{t} & e^{t} \end{matrix}) = (\begin{matrix} t \cdot e^{t} & e^{t} \\ e^{t} + t \cdot e^{t} & e^{t} \end{matrix}) = e^{t} (\begin{matrix} t & 0 \\ t + 1 & 1 \end{matrix})$

If, for example, we are asked for the solution that satisfies the initial condition: $(\begin{matrix} x (0) \\ y (0) \end{matrix}) = (\begin{matrix} 2 \\ 2 \end{matrix})$ the solution is: $(\begin{matrix} x (t) \\ y (t) \end{matrix}) = ϕ (t) \cdot ϕ (t_{0})^{- 1} \cdot (\begin{matrix} x (0) \\ y (0) \end{matrix}) = e^{t} (\begin{matrix} t & 0 \\ t + 1 & 1 \end{matrix}) (\begin{matrix} - 1 & 1 \\ 1 & 0 \end{matrix}) (\begin{matrix} 2 \\ 2 \end{matrix}) =$ $= e^{t} (\begin{matrix} t & 0 \\ t + 1 & 1 \end{matrix}) (\begin{matrix} 0 \\ 2 \end{matrix}) = e^{t} (\begin{matrix} 2 \\ 2 \end{matrix}) = (\begin{matrix} 2 e^{t} \\ 2 e^{t} \end{matrix})$