Solution. \(\dot V = 2x(-x + y^2) + 2y(-y + xy) = -2x^2 - 2y^2 + 4xy^2\). With \(r^2 = x^2 + y^2\), \(|x| \leq r\) and \(y^2 \leq r^2\), so \(|4xy^2| \leq 4r^3\) and \(\dot V \leq -2r^2 + 4r^3 < 0\) for \(0 < r < \frac 12\). So \(V\) is a strict Lyapunov function on the disc of radius \(\frac 12\), the origin is asymptotically stable, and every solution starting in \(x^2 + y^2 < \frac 14\) tends to \(0\). (A sharper bound, \(|x|y^2 \leq \frac {2}{3\sqrt 3}r^3\) on the circle of radius \(r\), enlarges the estimate.)

Solution. With \(v = x'\), \(E = \frac 12v^2 + \frac 12x^2 + \frac 14x^4\) satisfies \(\dot E = v(x'' + x + x^3) = 0\). \(E\) is positive definite, so the origin is stable (Theorem 11.1.2), and since every orbit stays on a closed level curve, nothing tends to \(0\). With \(x'' + bx' + x + x^3 = 0\), \(\dot E = -bv^2 \leq 0\); the largest invariant set in \(\{v = 0\}\) is the origin, so by LaSalle every solution tends to \(0\), and the origin is asymptotically stable with basin the whole plane (every sublevel set of \(E\) is compact).

Solution. With \(P = \begin {bmatrix} a & b\\ b & c\end {bmatrix}\), \(A^{\mathsf T}P + PA = \begin {bmatrix} -2a & 2a - 4b\\ 2a - 4b & 4b - 6c\end {bmatrix}\). Setting this equal to \(-I\): \(a = \frac 12\), \(2a - 4b = 0\) gives \(b = \frac 14\), and \(4b - 6c = -1\) gives \(c = \frac 13\). So \(P = \begin {bmatrix} 1/2 & 1/4\\ 1/4 & 1/3\end {bmatrix}\), with \(a > 0\) and \(\det P = \frac 16 - \frac 1{16} = \frac 5{48} > 0\): positive definite, as Theorem 11.3.1 requires since the eigenvalues \(-1\), \(-3\) are negative.

Solution. A gradient field has a symmetric Jacobian. Here \(\partial f_1/\partial y = -4xy\) while \(\partial f_2/\partial x = 0\), so the first system is not a gradient. With \(y' = -2y - 2x^2y\), \(\partial f_2/\partial x = -4xy\), and integrating \(-\Phi _x = f_1\) gives \(\Phi = -x^2 + \frac 12x^4 + x^2y^2 + h(y)\); then \(-\Phi _y = -2x^2y - h'(y) = -2y - 2x^2y\) gives \(h = y^2\). So \(\Phi = -x^2 + \frac 12x^4 + x^2y^2 + y^2\). The equilibria are \((0, 0)\), where \(D^2\Phi = \operatorname {diag}(-2, 2)\), a saddle of \(\Phi \) (unstable), and \((\pm 1, 0)\), where \(D^2\Phi = \operatorname {diag}(4, 4)\), strict minima (asymptotically stable).

Solution. \(q' = p\), \(p' = -q + q^2\), with equilibria \((0, 0)\) and \((1, 0)\). \(H\) has a strict minimum at the origin (\(H \approx \frac 12(p^2 + q^2)\)), which is therefore stable (Theorem 11.5.2(iv)). At \((1, 0)\) the Jacobian \(\begin {bmatrix} 0 & 1\\ 1 & 0\end {bmatrix}\) has eigenvalues \(\pm 1\), a saddle, with \(H(1, 0) = \frac 16\). The level set \(H = \frac 16\) contains the saddle and the two orbits that leave and return to it, a homoclinic loop enclosing the closed orbits around the origin; it is the boundary of the region of bounded oscillations.

Solution. Along a non-constant orbit \(\Phi \) strictly decreases (Theorem 11.4.2(i)), so it cannot return to its starting value, which a periodic orbit would require. For the competition model, \(\partial f_1/\partial y = -2x\) and \(\partial f_2/\partial x = -y\), which are not equal, so the Jacobian is not symmetric and \(f\) is not a gradient. The absence of cycles there has a different cause: the system is competitive, and Dulac’s criterion of the next section rules out cycles.