Stability of nonlinear system given by $\dot{x} = f_1(x) + f_2(x)$

It is false in general that $x_0$ is stable equilibrium of the system

$\dot x = f_1(x) + f_2(x), \tag{1}$

even under the hypothesis that $x_0$ is an asymptotically stable equilibrium of

$\dot x = f_1(x) \tag{2}$

and

$\dot x = f_2(x) \tag{3}$

separately. Indeed, we may find counterexamples even amongst the class two-dimensional, real, linear, autonomous ordinary differential equations; these are relatively simple as far as ODEs are concerned.

As pointed out by user58533 in her/his answer (since deleted), we may without loss of generality, by a change of coordinates if necessary, assume that $x_0 = 0$; we do so since it makes the $\LaTeX$ go a tad on the easier side. Now using $y_1$, $y_2$ as a standard coordinate system in $\Bbb R^2$, and setting

$\mathbf r = \begin{pmatrix} y_1 \\ y_2 \end{pmatrix}, \tag{4}$

and defining the matrix $A_1$ as

$A_1 = \begin{bmatrix} \mu_1 & \nu \\ 0 & \mu 2 \end{bmatrix}, \tag{5}$

where $\mu_1, \mu_2, \nu \in \Bbb R$, we consider the system

$\dot {\mathbf r} = A_1 \mathbf r. \tag{6}$

The stability properties of (6) follow directly from an eigen-analysis of $A_1$; it is easy to see that $p_1(\lambda)$, the characteristic polynomial of $A_1$, is

$p_1(\lambda) = \lambda^2 - (\mu_1 + \mu_2) \lambda + \mu_1 \mu_2; \tag{7}$

the roots of $p_1(\lambda)$ are clearly $\mu_1$ and $\mu_2$; if $\mu_1, \mu_2 < 0$, the system (4) is as asymptotically stable as asymptotically stable can be. Likewise we may set

$A_2 = A_1^T = \begin{bmatrix} \mu_1 & 0 \\ \nu & \mu 2 \end{bmatrix}, \tag{8}$

and consider the system

$\dot {\mathbf r} = A_2 \mathbf r; \tag{9}$

it is easily seen that the characteristic polynomial of $A_2$, $p_2(\lambda) = p_1(\lambda)$; thus the eigenvalues of $A_2$ are also $\mu_1$ and $\mu_2$; (9) has the same stability properties as (6). But now set

$A = (A_1 + A_2) \tag{10}$

and consider the system

$\dot {\mathbf r} = A \mathbf r = (A_1 + A_2) \mathbf r = (A_1 + A_2)\mathbf r = (A_1 + A_1^T) \mathbf r = \begin{bmatrix} 2\mu_1 & \nu \\ \nu & 2\mu_2 \end{bmatrix} \mathbf r; \tag{11}$

the characteristic polynomial of $A$, $p_A(\lambda)$, is

$p_A(\lambda) = \lambda^2 - 2(\mu_1 + \mu_2) \lambda + (4 \mu_1 \mu_2 - \nu^2), \tag{12}$

and via the quadratic formula, the roots of $p_A(\lambda)$ are

$\lambda_\pm = \dfrac{1}{2}(2(\mu_1 + \mu_2) \pm \sqrt{4(\mu_1 + \mu_2)^2 - 4(4 \mu_1 \mu_2 - \nu^2)})$ $ = (\mu_1 + \mu_2) \pm \sqrt{(\mu_1 - \mu_2)^2 + \nu^2}. \tag{13}$

It is easy to see that if $\mu_1, \mu_2 < 0$ but $\vert \nu \vert$ is sufficiently large, $\lambda_+ > 0$ and thus (11) is unstable, even though (6) and (9) are asymptotically stable. Thus we have a linear counterexample to to the stability of $\dot x = f_1(x) + f_2(x)$.

A fully non-linear example may be constucted by adding to $A_1(x)$, $A_2(x)$ a nonlinear $C^\infty$ vector field $\phi$ which vanishes on some open set $V \subset U$ of the origin; i.e. by setting

$\dot x = f_1(x) = A_1(x) + \phi(x), \; \; \dot x = f_2(x) = A_2(x) + \phi(x) \tag{14}$

for such $\phi$; such a pair of systems will behave as we have demonstrated in $V$, though it will be nonlinear the portion of $U$ outside of $V$.

Addenda; Sunday 24 August 2014 11:00 PM PST: A few more remarks, as promised: The construction of a nonlinear system in the last paragraph, viz. (14), is in many ways somewhat artificial, though it does present a pair of ODEs $\dot x = f_i(x) = A_ix + \phi(x)$ which satisfies the OP's criterion of nonlinearity. But taking $\phi$ to be zero in some neighborhood $U$ is, in my opinion, overkill, and in fact doesn't fully illustrate what is going on. I was hurried when I wrote this example; as I explained previously, I needed to get to work and I wanted to leave all the essential points of the question at least addressed, even if not in the most elegant manner. A better way and indeed much more general way to see the existence of such nonlinear systems $\dot x = f_i(x)$ is to choose two nonlinear $C^1$ vector fields $\phi_1(x)$ and $\phi_2(x)$ on the domain $U$ such that:

$i.) \; \; \phi_1(0) = \phi_2(0) = 0 \tag{15}$

and

$ii.) \;\; \nabla \phi_1(0) = \nabla \phi_2(0) = 0, \tag{16}$

where $\nabla \phi_i(x)$ denotes the Jacobean matrix of $\phi_i(x)$ at $x$, and then take

$f_1(x) = A_1(x) + \phi_1(x), \;\; f_2(x) = A_2(x) + \phi_2(x) \tag{17}$

on $U$. If follows from (15)-(17) that

$f_1(0) = f_2(0) = f_1(0) + f_2(0) = 0 \tag{18}$

and that

$\nabla f_1(0) = A_1; \;\; \nabla f_2(0) = A_2; \;\; \nabla(f_1 + f_2)(0) = A_1 + A_2. \tag{19}$

If follows that $0$ is a hyperbolic equilibrium of each of the systems

$(a.)\;\dot x = f_1(x); \;\; (b.) \; \dot x = f_2(x); \;\; (c.) \; \dot x = f_1(x) + f_2(x), \tag{20}$

since the eigenvalues of the Jacobeans of each are real and nonvanishing there, as we have seen. We can thus invoke the stable manifold theorem to conclude that $0$ is an asymptotically stable fixed point of (a) and (b) but is a saddle of (c); this construction thus provides a vastly larger family of counterexamples than those given above.

Finally, there are certain conditions under which $\dot x = f_1(x) + f_2(x)$ will in fact be asymptotically stable, among them are

(A.) when $A_1 = \nabla f_1(0)$ and $A_2 = \nabla f_2(0)$ are negative definitite, that is, when $\mathbf r^T(\nabla f_i(0)) \mathbf r < 0$ for all $\mathbf r$ (and here we allow $\mathbf r \in \Bbb R^n$), $i = 1, 2$, then $\nabla(f_1 + f_2)(0)$ is negative definite as well, since $\mathbf r^T (\nabla f_1(0) + \nabla f_2(0)) \mathbf r = \mathbf r^T \nabla f_1(0) \mathbf r + \mathbf r^T \nabla f_2(0) \mathbf r < 0$ for all $\mathbf r$; but the negative definiteness of $(\nabla f_1 + \nabla f_2)(0)$ implies that all its eigenvalues are negative, so the system $\dot x = f_1(x) + f_2(x)$ is asymptotically stable in this case. It may be noted in this regard that the matrices $A_1, A_2 = A_1^T$ are only negative definite when $\vert \nu \vert$ is sufficiently small, since $\mathbf r^T A_1 \mathbf r = \mu_1 y_1^2 + \mu_2 y_2^2 + \nu y_1 y_2$ and so forth; recall from the above that $A_1 + A_2$ is not negative definite if $\nu$ is sufficiently large.

(B.) Suppose $\theta_1, \theta_2: U \to \Bbb R$ are functions such that $0$ is a non-degenerate local minimum for each. Then the gradient systems $\dot x = \nabla \theta_i(x)$, $i = 1, 2$, satisfy the criterion required in (A.); the system $\dot x = \nabla \theta_1(x) + \nabla \theta_2(x)$ is asymptotically stable at $0$. End of Addenda.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!