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

$$\dot{\boldsymbol{z}}=A \boldsymbol{z}, \quad \boldsymbol{z} \in \mathbb{C}^{n}$$

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 $\varphi\left(t_{0}\right)=z_{0}$ is any

$$\varphi : I \rightarrow \mathbb{C}^{n}$$

such that

$$\left.\frac{d \varphi}{d t}\right|_{t=\tau}=A \varphi(\tau)$$

for every $\tau\in I$, and $\varphi\left(t_{0}\right)=z_{0}$.

Theorem

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

$$\varphi(t)=e^{A t} z_{0}$$

Proof

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

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

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 $e^{D}=diag(e^{\lambda_1}, e^{\lambda_2}, \ldots)$.

So solutions $\varphi(t)$ has the form in every component of linear combinations of exponentials $e^{\lambda_1}, e^{\lambda_2}, \ldots$

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

$$\varphi(t)=\sum_{k=1}^{n} c_{k} e^{\lambda_{k} t} \xi_{k}$$

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 $t$. $\blacksquare$

• ...we conclude that if $A$ could be carried to a Jordan block, $\varphi(t)$ 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

$\blacksquare$

Proof

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

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

$\blacksquare$

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

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 $\varphi(t)$ 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. $\blacksquare$

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 $\lambda_i$:

$\blacksquare$

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Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es