Voofie | Solving system of first order linear differential equations with matrix exponential method

Content

1 The problem
2 Notation
3 The method

The problem

In this tutorial, the type of problem that we would like to solve is of the form:

$\f$\begin{cases} & \text{ } a_{1,1} x_1'(t) + a_{1,2} x_2'(t)+...+ a_{1,n} x_n'(t) + b_{1,1} x_1(t) +...+ b_{1,n} x_n(t) = g_1(t) \\ & \text{ } a_{2,1} x_1'(t) + a_{2,2} x_2'(t)+...+ a_{2,n} x_n'(t) + b_{2,1} x_1(t) +...+ b_{2,n} x_n(t) = g_2(t) \\ & \text{ ... } \\ & \text{ } a_{n,1} x_1'(t) + a_{n,2} x_2'(t)+...+ a_{n,n} x_n'(t) + b_{n,1} x_1(t) +...+ b_{n,n} x_n(t) = g_n(t) \\ \end{cases} \f$$ .................(1)

Notation

Let's denote:

$\f$\mathbf{x} (t) = \left\{ x_1(t), x_2(t), ... ,x_n(t) \right\} \f$$

$\f$\mathbf{g} (t) = \left\{ g_1(t), g_2(t), ... ,g_n(t) \right\} \f$$

$\f$A = \left\{ a_{i,j} \right\}\f$$

$\f$B = \left\{ b_{i,j} \right\}\f$$

And equation (1) can be written as:

$\f$A \mathbf{x}'(t) + B \mathbf{x} (t) = \mathbf{g} (t)\f$$

Notice that the matrix A need to be invertible, otherwise the system of equations cannot be solved.

Therefore we can multiple A inverse to both side, and we can consider the following equation instead.

$\f$\mathbf{x}'(t) + A \mathbf{x} (t) = \mathbf{g} (t)\f$$ .......................(2)

Of course here, A and g will be different from the original definition.

The method

If we treat the matrix A, vector x and g as scalar, (2) can be solved in a simple way using integrating factor. (I have written another tutorial about differential equation, and it teaches integrating factor as well if you need references.)

In short, the solution of equation (2) is given by:

$\f$\mathbf{x}(t) = e^{-A t} \left(\int e^{A t} \mathbf{g}(t) dt + \mathbf{C}\right)\f$$ ...................(3)

However, since we are dealing with matrices and vectors, we need to check if the solution is indeed valid. Now I will explain the meaning of exponential of matrix A in the next session.

Exponential of Matrix

Here we will use the following power series for exponential function:

$\f$e^x = \sum_{k=0}^{\infty }{ \frac{x^k}{k!}}\f$$

Therefore, for exponential of a matrix A,

$\f$e^A = \sum_{k=0}^{\infty }{ \frac{A^k}{k!}}\f$$

The infinity sum of matrix seems impossible to evaluate, but once we have diagonalize A, it's exponential can be found easily.

If A can be diagonalized, it can be written as:

$\f$A = P D P^{-1}\f$$

And D is a diagonal matrix. Powers of A can be found very easily since:

$\f$A^n = P D P^{-1} P D P^{-1} ... P D P^{-1}\f$$

$\f$A^n = P D (P^{-1} P) D (P^{-1} P) ... (P^{-1} P) D P^{-1}\f$$

$\f$A^n = P D^n P^{-1}\f$$

Therefore:

$\f$e^A = P \sum_{k=0}^{\infty }{ \frac{D^k}{k!}} P^{-1}\f$$

Differentiating exponential of matrix

What if we need to differentiate $\f$e^{ A t}\f$$ ?

We will get back to the power series expansion:

$\f$\frac{d}{dt} e^{ A t } = \frac{d} {dt} \sum_{k=0}^{\infty} { \frac{(A t)^k} {k!} }\f$$

$\f$= \sum_{k=0}^{\infty}{ \frac{d}{dt} \frac{A^k t^k}{k!} } \f$$

$\f$= \sum_{k=1}^{\infty}{ \frac{k A^k t^{k-1} }{k!} } \f$$

$\f$= A \sum_{k=0}^{\infty}{\frac{(A t)^k} {k!}} \f$$

$\f$= A e^{A t}\f$$

In general, if A is a function of t, it's derivative cannot be found easily in this way. Since:

$\f$\frac{d}{dt} A^2(t)= A'(t) A(t) + A(t) A'(t) \f$$

Since $\f$A'(t)\f$$ does not communte with $\f$A(t)\f$$ generally, therefore we cannot simplify it into $\f$2 A(t) A'(t)\f$$ . So the above derivation won't work in general.

Verifying the solution

Given (3), we would like to show that it indeed satisfies (2).

$\f$\frac{d \mathbf{x} (t) }{ d t } = \frac{d e^{-A t} }{ d t } (\int e^{A t} \mathbf{g}(t) dt + \mathbf{C}) + e^{-A t} \frac{d} {dt} \int e^{A t} \mathbf{g}(t) dt\f$$