Nonlinear stability

We consider a system of the form  

$$\dot x=Ax+f(x),$$ (69)

where it is assumed that f is a smooth function from ${\rm I\!R}^n$ to itself and that f is of quadratic order near the origin:  

$$\Vert f(x)\Vert\le C\Vert x\Vert^2$$ (70)

for small $\Vert x\Vert$. We replace (2.15) by the system  

$$\dot y=Ay+f({y+\bar y\over 2}),$$ (71)

where y is allowed to be complex: $y\in{\rm C \hskip -4pt\vrule height 6.1pt width 0.7pt depth -0.1pt\hskip 4pt}^n$. Of course, this system is equivalent to the original one if y is real, but by allowing a complex y, we can take advantage of the Jordan canonical form of A.

We assume that the linearized system is stable, i.e. that all eigenvalues of A are in the left half plane. Let $-\delta$ be an upper bound for the real parts of the eigenvalues. There exists a (complex) matrix Q such that B=Q^-1AQ has Jordan canonical form, i.e. on the diagonal of B we have the eigenvalues of A, and in the off-diagonal position of the Jordan blocks we can put any nonzero number $\epsilon$; we choose $\epsilon$ to be a positive number less than $\delta/2$. We can now set y=Qz and (2.17) becomes  

$$\dot z=Bz+\tilde f(z),$$ (72)

where  

$$\tilde f(z)=Q^{-1}f({Qz+\overline{Qz}\over 2}).$$ (73)

We take the inner product of (2.18) with z and consider the real part. The result is  

$${1\over 2}{d\over dt}(z,z)={\rm Re}\,(z,Bz)+{\rm Re}\,(z,\tilde f(z)).$$ (74)

We now have

$$\vert{\rm Re}\,(z,\tilde f(z))\vert\le \Vert z\Vert\Vert\tilde f(z)\Vert\le \tilde C\Vert z\Vert^3=\tilde C (z,z)^{3/2},$$ (75)

for some appropriately chosen constant $\tilde C$ and every z such that $\Vert z\Vert$ is small. Also, we decompose B in the form B=D+N, where D is the diagonal part, and N is the off-diagonal part. We have

   $$\begin{eqnarray} {\rm Re}\,(z,Dz)&\le& -\delta (z,z), \nonumber\ Vert{\rm Re}\,(z,Nz)\vert&\le&\epsilon (z,z)\le\delta (z,z)/2.\end{eqnarray}$$

Consequently, we find  

$${1\over 2}{d\over dt}(z,z)\le -{\delta\over 2}(z,z)+\tilde C(z,z)^{3/2}.$$ (76)

It is easy to show from this that $(z,z)\to 0$ as $t\to\infty$, as long as the initial value of (z,z) is less than $\delta^2/(4\tilde C)^2$ (and such that $\Vert z\Vert$ is small enough for (2.16) to hold). Hence linear stability of the zero solution also implies its stability as a solution of the nonlinear problem, as long as the initial disturbances remain sufficiently small.