Triangular systems of differential equations

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 Triangular systems of differential equations.

We know how to solve linear systems with constant coefficients, and there are no methods for solving systems where the matrix $A$ is a matrix of functions unless we have a very special matrix, for example a triangular matrix.

Let's suppose that we have the following system: $(\begin{matrix} x^{'} \\ y^{'} \\ z^{'} \end{matrix}) = (\begin{matrix} a_{11} (t) & 0 & 0 \\ a_{21} (t) & a_{22} (t) & 0 \\ a_{31} (t) & a_{32} (t) & a 33 (t) \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) + (\begin{matrix} b_{1} (t) \\ b_{2} (t) \\ b_{3} (t) \end{matrix})$ i.e., a non homogeneous triangular linear system.

The idea is to solve the system step by step. Notice the first equation: $x^{'} (t) = a_{11} (t) \cdot x (t) + b_{1} (t)$

This is a linear equation like the ones we solved in the first unit of differential equations.

Therefore we can calculate its solution.

Now we concentrate on the second equation: $y^{'} (t) = a_{21} (t) \cdot x (t) + a_{2} 2 (t) \cdot y (t) + b_{2} (t)$ Now, $x (t)$ , is already known.

Replacing the function obtained in the previous step we have it $y^{'} (t) = a_{22} (t) \cdot y (t) + {\tilde{b}}_{2} (t), where {\tilde{b}}_{2} (t) = a_{12} (t) \cdot x (t) + b_{2} (t)$ is a well-known function.

Therefore we obtain another linear equation that we can already solve.

Finally, we take the third equation and replace the values of $x (t)$ and $y (t)$ , obtaining: $z^{'} (t) = a_{33} (t) \cdot z (t) + {\tilde{b}}_{3} where {\tilde{b}}_{3} (t) = a_{31} (t) \cdot x (t) + a_{32} (t) \cdot y (t) + b_{3} (t)$ which is a well-known function. This is another linear equation that we can solve.

Let's notice that we have considered a low triangular counterfoil, but we could use the same procedure if we had an upper triangular matrix but starting from below.

Thus we already solved the system.

Let's see it more clearly with an example.

Example

Let's consider the following system: $(\begin{matrix} x^{'} \\ y^{'} \\ z^{'} \end{matrix}) = (\begin{matrix} \frac{1}{t} & 0 & 0 \\ t & \frac{2}{t} & 0 \\ - t^{3} & t & \frac{3}{t} \end{matrix}) \cdot (\begin{matrix} x \\ y \\ z \end{matrix}) + (\begin{matrix} 2 t^{2} \\ - t^{4} \\ t^{6} \end{matrix})$ We are going to proceed as we described:

We take the first equation: $x^{'} = \frac{1}{t} \cdot x + 2 t^{2}$ Since this is a linear ODE, we solve it following the procedure described in other units:
- Resolution of the homogeneous part: $x_{h} (t) = k_{1} \cdot e^{\int \frac{1}{t} d t} = k_{1} \cdot e^{\ln t} = k_{1} \cdot t$
- Search for a particular solution: $x_{p} (t) = t \cdot u (t)$ , where $u (t)$ satisfies: $u^{'} = \frac{2 t^{2}}{t} = 2 t$ . Therefore, $u (t) = t^{2}$ and thus $x_{p} (t) = t^{3}$
- General solution $x (t) = k_{1} \cdot t + t^{3}$
We take the second equation, replacing the obtained value of x: $y^{'} = t \cdot x + \frac{2}{t} \cdot y - t^{4} = t \cdot (k_{1} \cdot t + t^{3}) + \frac{2}{t} \cdot y - t^{4} = \frac{2}{t} \cdot y + k_{1} \cdot t^{2}$
- Resolution of the homogeneous part: $y_{h} (t) = k_{2} \cdot e^{\int \frac{2}{t} d t} = k_{2} \cdot e^{\ln t^{2}} = k_{2} \cdot t^{2}$
- Search for a particular solution: $y_{p} (t) = t^{2} \cdot u (t)$ , where $u (t)$ satisfies: $u^{'} = \frac{k_{1} \cdot t^{2}}{t^{2}} = k_{1}$ . Therefore, $u (t) = k_{1} \cdot t$ and thus $y_{p} (t) = k_{1} \cdot t^{3}$
- General solution $y (t) = k_{2} \cdot t^{2} + k_{1} \cdot t^{3}$
We take the third equation, replacing the obtained values of $x$ and $y$ : $z^{'} = - t^{3} \cdot x + t \cdot y + \frac{3}{t} \cdot z + t^{6} =$ $= - t^{3} \cdot (k_{1} \cdot t + t^{3}) + t \cdot (k_{2} \cdot t^{2} + k_{1} \cdot t^{3}) + \frac{3}{t} \cdot z + t^{6} =$ $= \frac{3}{t} \cdot z + k_{2} \cdot t^{3}$
- Resolution of the homogeneous part: $z_{h} (t) = k_{3} \cdot e^{\frac{3}{t} d t} = k_{3} \cdot e^{\ln t^{3}} = k_{3} \cdot t^{3}$
- Search for a particular solution: $z_{p} (t) = t^{3} \cdot u (t)$ , where $u (t)$ satisfies: $u^{'} (t) = \frac{k_{2} \cdot t^{3}}{t^{3}} = k_{2}$ . Therefore, $u (t) = k_{2} \cdot t$ and thus $z_{p} (t) = k_{2} \cdot t^{4}$
- General solution $z (t) = k_{3} \cdot t^{3} + k_{2} \cdot t^{4}$

Therefore the solution of the system is: $(\begin{matrix} x (t) \\ y (t) \\ z (t) \end{matrix}) = (\begin{matrix} k_{1} \cdot t + t^{3} \\ k_{2} \cdot t^{2} + k_{1} \cdot t^{3} \\ k_{3} \cdot t^{3} + k_{2} \cdot t^{4} \end{matrix})$