{"id":17894,"date":"2023-07-19T17:48:44","date_gmt":"2023-07-19T15:48:44","guid":{"rendered":"https:\/\/djalil.chafai.net\/blog\/?p=17894"},"modified":"2024-01-14T16:44:28","modified_gmt":"2024-01-14T15:44:28","slug":"linear-odes-instabilities","status":"publish","type":"post","link":"https:\/\/djalil.chafai.net\/blog\/2023\/07\/19\/linear-odes-instabilities\/","title":{"rendered":"Linear ODEs instabilities"},"content":{"rendered":"
\"Photo<\/a>
Salvador Domingo Felipe Jacinto Dal\u00ed i Dom\u00e8nech, Marquess of Dal\u00ed of P\u00fabol (1904 - 1989) Probably nonlinear, unstable, and nonmonotonic<\/figcaption><\/figure>\n

This tiny post is devoted to a counter intuitive elementary phenomenon : for a linear ODE, with a time dependent matrix, stability is not always controlled by the spectrum of the matrix!<\/p>\n

Autonomous case.<\/strong> It is well known that the long time behavior of a linear differential equation $$x'(t)=Ax(t)$$ in $\\mathbb{R}^n$, when $A$ is diagonalizable, is dictated by the real part of the eigenvalues of $A$. Indeed, if $A=PDP^{-1}$ with $P$ complex invertible and $D$ complex diagonal, and $y(t):=P^{-1}x(t)$, then $$y'(t)=Dy(t)\\quad\\text{and}\\quad\\frac{\\mathrm{d}}{\\mathrm{d}t}\\|y(t)\\|^2=2\\Re\\langle Dy(t),y(t)\\rangle=2\\langle (\\Re D)y(t),y(t)\\rangle.$$ Hence the equation is (here exponentially) stable if all the eigenvalues of $A$ in $\\mathbb{C}$ have a negative real part, in the sense that $x(t)\\to0$ as $t\\to\\infty$ (exponentially fast) for all initial conditions $x(0)$. Another way to state it, using a matrix exponential, is $$x(t)=\\mathrm{e}^{tA}x(0)=P\\mathrm{e}^{tD}P^{-1}x(0)=P\\mathrm{diag}(\\mathrm{e}^{\\mathrm{i}t\\Im \\lambda_k})\\mathrm{diag}(\\mathrm{e}^{t\\Re\\lambda_k})P^{-1}x(0).$$<\/p>\n

If $A$ is not diagonalizable, we may use the Jordan factorization $A=PTP^{-1}$ where $T$ is upper triangular block diagonal with constant diagonal on each block. From the Jordan factorization we get the subtle decomposition $T=D+N$ where $D$ is diagonal and carries the eigenvalues of $T$, and $N$ is nilpotent and commutes with $D$, hence the exponential factorization $\\mathrm{e}^{tT}=\\mathrm{e}^{tD}\\mathrm{e}^{tN}$ and $\\mathrm{e}^{tN}$ is a polynomial in $t$. Thus the asymptotic behavior in long time of $$x(t)=\\mathrm{e}^{tA}x(0)=P\\mathrm{e}^{tT}P^{-1}x(0)=P\\mathrm{e}^{tD}\\mathrm{e}^{tN}P^{-1}x(0)$$ is dictated by the real part of the eigenvalues of $A$ : stability occurs when all the eigenvalues have a negative real part. Even if the eigenvalues of $A$ are all real and negative, then $$\\frac{\\mathrm{d}}{\\mathrm{d}t}\\|y(t)\\|^2=2\\Re\\langle Ty(t),y(t)\\rangle$$ can take positive values, see the two dimensional example below. But stability is about asymptotic analysis in long time<\/strong> which is not the monotonicity in time<\/strong>. The lack of monotonicity comes from the polynomial term $\\mathrm{e}^{tN}$, related to the geometry of the eigenspaces, a non-spectral information ($n\\geq2$). Of course $A$ is diagonalizable if and only if $N=0$.<\/p>\n

If we give up block sparsity for $T$, we can get $P$ unitary, and $A=P(D+N)P^*$ is then the Schur unitary factorization, however, in this case, the nilpotent part $N$ no longer commute with the diagonal part $D$ in general, making more complicated the computation of the exponential.<\/p>\n

The term ``stability'' comes from the fact that if $x(0)=0$ then $x(t)=0$ for all $t$, in other words $0$ is a stationary point, and we say that it is stable if there exists a neighborhood of $0$ such that the solution started from anywhere in this neighborhood converges in long time to $0$.<\/p>\n

The equation $x'(t)=Ax(t)$ is called autonomous because $A$ does not depend on $t$.<\/p>\n

Two dimensional case.<\/strong> The lack of monotonicity phenomenon is of geometric nature and requires at least two dimensions. Let us seek for a $2\\times 2$ real matrix $T$ with all eigenvalues real and negative and for which $\\langle Tx,x\\rangle>0$ for some $x\\in\\mathbb{R}^2$. This matrix cannot be symmetric. Let us denote $\\lambda_1<\\lambda_2\\leq0$ the eigenvalues, that we suppose distinct, and $\\theta\\in(0,\\pi)$ the angle between the eigenvectors. Let us take $T$ upper triangular, hence the name, of the form $$T:=\\begin{pmatrix} \\lambda_1 & (\\lambda_2-\\lambda_1)\\cot(\\theta)\\\\0&\\lambda_2\\end{pmatrix}.$$ For all $x\\in\\mathbb{R}^2$, the quantity $$\\langle Tx,x\\rangle=\\lambda_1x_1^2+\\lambda_2x_2^2+(\\lambda_2-\\lambda_1)x_1x_2\\cot(\\theta)$$ reaches its maximum $$\\frac{\\lambda_1+\\lambda_2+(\\lambda_2-\\lambda_1)\\cot(\\theta)}{2}$$ for $x=(\\sin(a),\\cos(a))$ with $a=(2\\theta+\\pi)\/4$. Also, with $\\rho:=\\lambda_2\/\\lambda_1\\in[0,1)$, we have $$\\langle Tx,x\\rangle>0\\text{ for some $x$}\\quad\\text{if and only if}\\quad\\sin(\\theta)<\\frac{1-\\rho}{1+\\rho}.$$ This is impossible when $\\theta=\\pi\/2$ ($T$ is symmetric). There exists a similar analysis for the case where $T$ has a double eigenvalue or two complex conjugate eigenvalues.<\/p>\n

Nonautonomous case.<\/strong> Let us consider now the nonautonomous linear differential equation $$x'(t)=A(t)x(t),\\quad\\text{for which}\\quad \\frac{\\mathrm{d}}{\\mathrm{d}t}\\|x(t)\\|^2=2\\langle A(t)x(t),x(t)\\rangle.$$ If $-A(t)$ is symmetric positive definite for all $t$ then the system is stable and monotonic. If $A(t)$ is diagonalizable for all $t$ and $P(t)=P$ does not depend on $t$, then the matrices $\\{A(t):t\\}$ commute<\/strong> and the solution can be expressed using a simple exponential as $$x(t)=\\mathrm{e}^{\\int_0^tA(s)\\mathrm{d}s}x(0)=P\\mathrm{e}^{\\int_0^t D(s)\\mathrm{d}s}P^{-1}x(0),$$ showing that the knowledge of the spectrum of $A(t)$ for all $t$ controls the stability.<\/p>\n

However, in general, when $P(t)$ depends on $t$, it turns out that we may have $\\|x(t)\\|\\to+\\infty$ as $t\\to\\infty$ while the eigenvalues of $A(t)$ are all real, negative, and constant in time! This means that the knowledge of the spectrum of $A(t)$ for all $t$ is not enough to describe the stability of the solution. In other words, the stability depends not only on the spectrum but also on the eigenspaces geometry dynamics through the matrix $P(t)$ in the Jordan or Schur factorization $$A(t)=P(t)T(t)P(t)^{-1}=P(t)(D(t)+N(t))P(t)^{-1}.$$ If we set $y(t):=P(t)^{-1}x(t)$ then $$y'(t)=P(t)^{-1}x'(t)+(P(t)^{-1})'x(t)=T(t)y(t)+(P(t)^{-1})'P(t)y(t).$$ The last term $(P(t)^{-1})'P(t)y(t)$, due to the dependency of $P(t)$ on $t$, is responsible of the instability phenomenon. This was apparently already known by Henri Poincar\u00e9 (1854 - 1912) and Aleksandr Mikhailovich Lyapunov (1857 - 1918), who are at the origin of the extensive study of the stability of ODEs. The phenomenon is of geometric nature and requires a dimension $n\\geq2$. Here is a concrete, simple, two dimensional counter example, due to Robert \u00c8. Vinograd : $$
\nA(t)=\\begin{pmatrix}
\n6 \\sin (12 t)-\\frac{1}{2} 9 \\cos (12 t)-\\frac{11}{2} & \\frac{9}{2} \\sin (12 t)+6 \\cos (12 t)+6 \\\\
\n\\frac{9}{2} \\sin (12 t)+6 \\cos (12 t)-6 & -6 \\sin (12 t)+\\frac{9}{2} \\cos (12 t)-\\frac{11}{2} \\\\
\n\\end{pmatrix},$$ for which the eigenvalues are $-1$ and $-10$ for all $t$, while $$x(t)=\\frac{\\mathrm{e}^{-13t}}{5}\\begin{pmatrix}
\n\\left(\\mathrm{e}^{15 t}+4\\right) \\cos (6 t)-2 \\left(\\mathrm{e}^{15 t}-1\\right) \\sin (6 t) & 2 \\left(\\mathrm{e}^{15 t}-1\\right) \\cos (6 t)-\\left(4 \\mathrm{e}^{15 t}+1\\right) \\sin (6 t) \\\\
\n\\left(\\mathrm{e}^{15 t}+4\\right) \\sin (6 t)+2 \\left(\\mathrm{e}^{15 t}-1\\right) \\cos (6 t) & 2 \\left(\\mathrm{e}^{15 t}-1\\right) \\sin (6 t)+\\left(4 \\mathrm{e}^{15 t}+1\\right) \\cos (6 t)\\end{pmatrix}x(0),$$ which may blow-up in long time for a well chosen initial condition $x(0)$ arbitrary close to $0$.<\/p>\n

In order to get a generic example of $A(t)$ in dimension $n=2$, we can conjugate say an upper triangular matrix $T$ as above by a time dependent orthogonal matrix, for instance the rotation $$P(t)=\\begin{pmatrix}\\cos(\\omega t) & -\\sin(\\omega t)\\\\\\sin(\\omega t) & \\cos(\\omega t)\\end{pmatrix},\\quad\\omega\\in\\mathbb{R},$$ which gives $A(t)=P(t)TP(t)^{-1}$, in particular $T=A(0)$. We have $$P(t)=\\mathrm{e}^{tQ}\\quad\\text{where}\\quad Q=\\begin{pmatrix}0 & -\\omega\\\\\\omega&0\\end{pmatrix}.$$ Also $x'(t)=A(t)x(t)=P(t)TP(t)^{-1}x(t)$, and with $y(t):=P(t)^{-1}x(t)$, we get $$y'(t)=Ty(t)+(P(t)^{-1})'x(t)=(T-Q)y(t),$$ hence $y(t)=\\mathrm{e}^{t(T-Q)}y(0)$, and $$x(t)=P(t)\\mathrm{e}^{t(T-Q)}P(t)^{-1}x(0).$$ The contribution of $-Q$ in $\\mathrm{e}^{t(T-Q)}$ is crucial for a possible blow-up. The problem is reduced to an autonomous ODE with matrix $T-Q$, which may have a positive eigenvalue even if $T$ does not. The stability is dictated by the real part of the eigenvalues of $T-Q$.<\/p>\n

The Vinograd counter example is obtained from the two dimensional example with $\\omega=-6$ in $Q$, and $\\lambda_1=-10$, $\\lambda_2=-1$, $\\cot(\\theta)=-4\/3$ in $T$. The eigenvalues of $T-Q$ are $-13$ and $2$.<\/p>\n

We could ask if the same phenomenon remains with a diagonal $T$, but the answer is no : indeed, if $T=D=\\begin{pmatrix}a&0\\\\0 &a+b\\end{pmatrix}$, then $T-Q$ has eigenvalues $\\frac{1}{2}(2 a\\pm\\sqrt{b^2-4 \\omega ^2}+b)$, the largest one $\\frac{1}{2}(2 a+\\sqrt{b^2-4 \\omega ^2}+b)$ cannot be positive if both $a$ and $a+b$ are negative.<\/p>\n

Conclusion.<\/strong> For the linear ODE $x'(t)=A(t)x(t)$, with $A(t)=P(t)(D(t)+N(t))P(t)^{-1}$, monotonicity as well as stability are not always dictated by the sole spectrum $D(t)$ : $N(t)$ acts as a perturbation of monotonicity while the dynamics of $P(t)$ acts as a perturbation of stability.<\/p>\n

In the same spirit.<\/strong> It can be shown that $x'(t)=A(t)x(t)$ can be stable while $A(t)$ has for all $t$ real positive eigenvalues! We refer to the review article by Kre\u0161imir Josi\u0107 and Robert Rosenbaum for more. The phenomenon says that the stability of the nonautonomous resolvent matrix equation $R'(t)=A(t)R(t)$ is not controlled solely by the spectrum of $A(t)$ for all $t$ in general. The same instability phenomenon takes place in discrete time, and corresponds to take long noncommutative words written with two matrix letters : the word may blow-up even if the letters are contractive, and this can be seen as a sort of geometric or directional resonance phenomenon. The phenomenas are related to the following topics between nonlinear aspects of linear algebra, analysis, ergodic theory, geometry, probability, and system control :<\/p>\n