System of first order linear ODEs- complex case

Particular case of system of first order ODEs.

It is useful in order to study the real case.

Consider the equation

with $A:{\mathbb{C}}^n\mapsto {\mathbb{C}}^n$. If we want to study a real system, we deal with the complexificated one.

A solution for this equation with initial condition $\phi \left(t_0\right)=z_0$ is any

such that

for every $\tau \in I$, and $\phi \left(t_0\right)=z_0$.

Theorem

The solution of the previous equation with initial condition $\phi (0)=z_0\in {\mathbb{C}}^n$ is given by

Proof

Properties of exponential gives that $\phi$ verifies the equation.

Also, observe that there can not be solutions of other form. Consider $\psi (t)$ other solution. Then $\phi (t)=e^{At}\psi (0)$ has the same initial value, so they are the same. $◼$

So the only problem is to compute $e^{At}$, which is the matrix exponential. We compute the roots of the characteristic polynomial of $A$ and distinguish several cases:

1. All roots are distinct

In this case, $A$ is diagonalizable, i.e., we can write $A=M\cdot D\cdot M^{-1}$. By matrix exponential properties we have that

where $e^D=diag(e^{{\lambda }_1},e^{{\lambda }_2},\dots )$.

So solutions $\phi (t)$ has the form in every component of linear combinations of exponentials $e^{{\lambda }_1},e^{{\lambda }_2},\dots$

Can we "separate" these exponential maps? Yes. Observe that an operator $A$ with all eigenvalues different gives a decomposition of ${\mathbb{C}}^n=⨁{\mathbb{C}}_i$ in invariant complex lines.

Also,

and moreover,

From here we can conclude:

Theorem

[Arnold, page 185]

If the eigenvalues of $A$ are all distinct then

where ${\xi }_k$ are the eigenvectors, and $c_k$ are constants to be determined by the initial conditions.

2. $A$ is a Jordan block

If $A$ is a Jordan block, we write $A=\lambda Id+N$, as usual. Observe that $\lambda Id\cdot N=N\cdot \lambda Id$ so

By properties of matrix exponential we get that

Suppose that $A$ is a Jordan block but in other basis, that is, $A=M(\lambda Id+N)M^{-1}$. What we can conclude is that solutions of $\dot{z}=Az$ are

with $v\in {\mathbb{C}}^n$. If we define:

Definition

A quasi-polynomial with exponent $\lambda$ is an expression of the form $e^{\lambda t}p(t)$ where $p$ is a polynomial in $t$. $◼$

...we conclude that if $A$ could be carried to a Jordan block, $\phi (t)$ has the form, in every component, of a sum of quasi-polynomials with exponent $\lambda$ and degree less than $n$.

3. The rest of the cases

In general, matrix $A$ can be carried to a Jordan canonical form of a matrix, $$A=M J M^{-1}$$

Let's focus in the case $A=J$ and then follow with the general one. $J$ is made of blocks, and for each eigenvalue ${\lambda }_i$ we have two of such blocks: a diagonal block with size $d_i$ and a Jordan block of size $j_i$ (maybe $d_i=0$ or $j_i=0$!).

Then, observe that the algebraic multiplicity of ${\lambda }_i$ is $$n_i=d_i+j_i$$ and the geometric muliplicity of ${\lambda }_i$ is $$k_i=d_i+1$$. This is deduced by the form we proof the decomposition, view Jordan canonical form of a matrix.

Now, observe that matrix $e^{Jt}$, with $t\in \mathbb{R}$, is itself a block matrix, because of the following

Lemma

Let $B:V\mapsto V$ a linear map. Then, if $B(U)\subseteq U$ we have

$◼$

Proof

If $v\in V$, then $B(v)\in U$ and therefore

belongs to $U$ (I guess is a result of convergence...)
$◼$

It turns out then that if $B_i$ is a block of matrix $J$ then $e^{B_{i}t}$ is a block of $e^{Jt}$.

Moreover, for each ${\lambda }_i$ we see that $e^{Jt}$ has possibly two blocks:

• A scalar block, $e^{{\lambda }_{i}t}Id$, of size $d_i$.

• A quasipolynomial block (see the previous case) of exponent ${\lambda }_i$, degree $j_i-1$ and size $j_i$.

In conclusion, for a matrix made of Jordan blocks $J$ we get that $e^{Jt}$ is a block matrix with elements of the form $e^{\lambda }\cdot t^k$. If $A$ is \textbf{any matrix} then

And so,

But $M$ and $M^{-1}$ are made of (complex) constants and so only "mix" the elements of $e^{Jt}$. We can state that

Proposition

Let $\dot{z}=Az$ a system of first order linear equations, with $A:{\mathbb{C}}^n\mapsto {\mathbb{C}}^n$ a linear operator. Then, if $A$ has eigenvalues ${\lambda }_i$ with algebraic multiplicities $n_i$ and geometric multiplicities $k_i$ then every component of a solution $\phi (t)$ is \textbf{a sum of quasi-polynomials with exponent ${\lambda }_i$ and degree less or equal than $n_i-k_i$}. I. e., it has the form:

where $p_{j,i}(t)$ has degree less or equal than $n_i-k_i$ with constant complex coefficients. $◼$

Proof

It is all above, since

for $v\in {\mathbb{C}}^n$.

It only remains to show the degree of the quasipolynomials, but it is a simple computation. For every ${\lambda }_i$:

$◼$