Chapter 11

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-9 \sin x \\] If \((x, y)\) is a critical point, \(y=0\) and \(-9 \sin x=0 .\) Therefore \(x=\pm n \pi\) and so the critical points are \((\pm n \pi, 0)\) for \(n=0,1,2, \dots\)

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-2 x-y^{2} \\] If \((x, y)\) is a critical point, then \(y=0\) and so \(-2 x-y^{2}=-2 x=0 .\) Therefore (0,0) is the sole critical point.

The corresponding plane autonomous system is $$ x^{\prime}=y, \quad y^{\prime}=x^{2}-y\left(1-x^{3}\right) $$

Note that \(x=k\) is the only critical point since \(\ln (x / k)\) is not defined at \(x=0 .\) since \(g^{\prime}(x)=-k-k \ln (x / k)\) \(g^{\prime}(k)=-k<0 .\) Therefore \(x=k\) is an asymptotically stable critical point by Theorem 11.2

(a) If \(f(x)=x^{2} / 2, f^{\prime}(x)=x\) and so $$\frac{d y}{d x}=\frac{y^{\prime}}{x^{\prime}}=-g \frac{x}{1+x^{2}} \frac{1}{y}$$. We can separate variables to show that \(y^{2}=-g \ln \left(1+x^{2}\right)+c .\) But \(x(0)=x_{0}\) and \(y(0)=x^{\prime}(0)=v_{0}\) Therefore \(c=v_{0}^{2}+g \ln \left(1+x_{0}^{2}\right)\) and so $$y^{2}=v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right)$$. Now $$v_{0}^{2}-g \ln \left(\frac{1+x^{2}}{1+x_{0}^{2}}\right) \geq 0 \quad \text { if and only if } \quad x^{2} \leq e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1$$. Therefore, if \(|x| \leq\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2},\) there are two values of \(y\) for a given value of \(x\) and so the solution is periodic. (b) since \(z=x^{2} / 2,\) the maximum height occurs at the largest value of \(x\) on the cycle. From (a), \(x_{\max }=\) \(\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]^{1 / 2}\) and so $$z_{\max }=\frac{x_{\max }^{2}}{2}=\frac{1}{2}\left[e^{v_{0}^{2} / g}\left(1+x_{0}^{2}\right)-1\right]$$.

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x^{3} \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsilon x^{3}=0 .\) Hence \(x\left(-1+\epsilon x^{2}\right)=0\) and so \(x=0, \sqrt{1 / \epsilon},-\sqrt{1 / \epsilon} .\) The critical points are \((0,0),(\sqrt{1 / \epsilon}, 0)\) and \((-\sqrt{1 / \epsilon}, 0)\)

The corresponding plane autonomous system is \\[ x^{\prime}=y, \quad y^{\prime}=-x+\epsilon x|x| \\] If \((x, y)\) is a critical point, \(y=0\) and \(-x+\epsilon x|x|=x(-1+\epsilon|x|)=0 .\) Hence \(x=0,1 / \epsilon,-1 / \epsilon .\) The critical points \(\operatorname{are}(0,0),(1 / \epsilon, 0)\) and \((-1 / \epsilon, 0)\)

From \(x+x y=0\) we have \(x(1+y)=0 .\) Therefore \(x=0\) or \(y=-1 .\) If \(x=0,\) then, substituting into \(-y-x y=0\) we obtain \(y=0 .\) Likewise, if \(y=-1,1+x=0\) or \(x=-1 .\) We can conclude that (0,0) and (-1,-1) are critical points of the system.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-x+\left(\frac{1}{2}+3 y^{2}\right) y-x^{2}$$ and so \(\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}=0+\frac{1}{2}+9 y^{2} > 0 .\) Therefore there are no periodic solutions by Theorem 11.5.

From \(y^{2}-x=0\) we have \(x=y^{2} .\) Substituting into \(x^{2}-y=0,\) we obtain \(y^{4}-y=0\) or \(y\left(y^{3}-1\right)=0 .\) It follows that \(y=0,1\) and so (0,0) and (1,1) are the critical points of the system.

From \(x-y=0\) we have \(y=x .\) Substituting into \(3 x^{2}-4 y=0\) we obtain \(3 x^{2}-4 x=x(3 x-4)=0 .\) It follows that (0,0) and \((4 / 3,4 / 3)\) are the critical points of the system.

(a) Solving $$\begin{array}{l} x(-0.1+0.02 y)=0 \\ y(0.2-0.025 x)=0 \end{array}$$ in the first quadrant we obtain the critical point \((8,5) .\) The graphs are plotted using \(x(0)=7\) and \(y(0)=4\). (b) The graph in part (a) was obtained using NDSolve in Mathematica. We see that the period is around 40\. since \(x(0)=7,\) we use the FindRoot equation solver in Mathematica to approximate the solution of \(x(t)=7\) for \(t\) near \(40 .\) From this we see that the period is more closely approximated by \(t=44.65\).

Solving $$\begin{array}{l} x(20-0.4 x-0.3 y)=0 \\ y(10-0.1 y-0.3 x)=0 \end{array}$$ we see that critical points are \((0,0),(0,100),(50,0),\) and \((20,40) .\) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0.08(20-0.8 x-0.3 y) & -0.024 x \\ -0.018 y & 0.06(10-0.2 y-0.3 x) \end{array}\right)$$ and so $$\begin{array}{ll} \mathbf{A}_{1}=\mathbf{g}^{\prime}((0,0))=\left(\begin{array}{cc} 1.6 & 0 \\ 0 & 0.6 \end{array}\right) & \mathbf{A}_{2}=\mathbf{g}^{\prime}((0,100))=\left(\begin{array}{cc} -0.8 & 0 \\ -1.8 & -0.6 \end{array}\right) \\ \mathbf{A}_{3}=\mathbf{g}^{\prime}((50,0))=\left(\begin{array}{cc} -1.6 & -1.2 \\ 0 & -0.3 \end{array}\right) & \mathbf{A}_{4}=\mathbf{g}^{\prime}((20,40))=\left(\begin{array}{cc} -0.64 & -0.48 \\ -0.72 & -0.24 \end{array}\right) \end{array}$$ since \(\operatorname{det}\left(\mathbf{A}_{1}\right)=\Delta_{1}=0.96>0, \tau=2.2>0,\) and \(\tau_{1}^{2}-4 \Delta_{1}=1 >0,\) we see that (0,0) is an unstable node. since \(\operatorname{det}\left(\mathbf{A}_{2}\right)=\Delta_{2}=0.48>0, \tau=-1.4<0,\) and \(\tau_{2}^{2}-4 \Delta_{2}=0.04 >0,\) we see that (0,100) is a stable node. since \(\operatorname{det}\left(\mathbf{A}_{3}\right)=\Delta_{3}=0.48>0, \tau=-1.9<0,\) and \(\tau_{3}^{2}-4 \Delta_{3}=1.69 >0,\) we see that (50,0) is a stable node. since \(\operatorname{det}\left(\mathbf{A}_{4}\right)=-0.192< 0\) we see that (20,40) is a saddle point.

From \(x\left(10-x-\frac{1}{2} y\right)=0\) we obtain \(x=0\) or \(x+\frac{1}{2} y=10 .\) Likewise \(y(16-y-x)=0\) implies that \(y=0\) or \(x+y=16 .\) We therefore have four cases. If \(x=0, y=0\) or \(y=16 .\) If \(x+\frac{1}{2} y=10,\) we can conclude that \(y\left(-\frac{1}{2} y+6\right)=0\) and so \(y=0,12 .\) Therefore the critical points of the system are \((0,0),(0,16),(10,0),\) and (4,12)

Critical points are (1,0) and \((-1,0),\) and $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} 2 x & -2 y \\ 0 & 2 \end{array}\right)$$ At \(\mathbf{X}=(1,0), \tau=4, \Delta=4,\) and so \(\tau^{2}-4 \Delta=0 .\) We can conclude that (1,0) is unstable but we are unable to classify this critical point any further. At \(\mathbf{X}=(-1,0), \Delta=-4<0\) and so (-1,0) is a saddle point.

For \(\mathbf{X}=\left(K_{1}, 0\right), \tau=-r_{1}+r_{2}\left[1-\left(K_{1} \alpha_{21} / K_{2}\right)\right]\) and \(\Delta=-r_{1} r_{2}\left[1-\left(K_{1} \alpha_{21} / K_{2}\right)\right] .\) If we let \(c=1-K_{1} \alpha_{21} / K_{2}\) \(\tau^{2}-4 \Delta=\left(c r_{2}+r_{1}\right)^{2}>0 .\) Now if \(k_{1} >K_{2} / \alpha_{21}, c< 0\) and so \(\tau<0, \Delta>0 .\) Therefore \(\left(K_{1}, 0\right)\) is a stable node. If \(K_{1}< K_{2} / \alpha_{21}, c>0\) and so \(\Delta< 0 .\) In this case \(\left(K_{1}, 0\right)\) is a saddle point.

We have $$\mathbf{X}(0)=c_{1}\left(\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right)+c_{2}\left(\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right)+c_{3}\left(\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right)=\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{r} 1 \\ -4 \\ 6 \end{array}\right).$$ Thus, the solution of the initial-value problem is $$\mathbf{X}=\left(\begin{array}{c} t+1 \\ t \\ -2 t \end{array}\right)-4\left(\begin{array}{c} t \\ t+1 \\ -2 t \end{array}\right)+6\left(\begin{array}{c} t \\ t \\ -2 t+1 \end{array}\right).$$

\(y^{\prime}=2 x y-y=y(2 x-1) .\) Therefore if \((x, y)\) is a critical point, either \(x=1 / 2\) or \(y=0 .\) The case \(x=1 / 2\) and \(y-x^{2}+2=0\) implies that \((x, y)=(1 / 2,-7 / 4) .\) The case \(y=0\) leads to the critical points \((\sqrt{2}, 0)\) and \((-\sqrt{2}, 0) .\) We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rc} -2 x & 1 \\ 2 y & 2 x-1 \end{array}\right)$$ to classify these three critical points. For \(\mathbf{X}=(\sqrt{2}, 0)\) or \((-\sqrt{2}, 0), \tau=-1\) and \(\Delta<0 .\) Therefore both critical points are saddle points. For \(\mathbf{X}=(1 / 2,-7 / 4), \tau=-1, \Delta=7 / 2\) and so \(\tau^{2}-4 \Delta=-13<0 .\) Therefore \((1 / 2,-7 / 4)\) is a stable spiral point

From \(x^{2} e^{y}=0\) we have \(x=0 .\) since \(e^{x}-1=e^{0}-1=0,\) the second equation is satisfied for an arbitrary value of \(y .\) Therefore any point of the form \((0, y)\) is a critical point.

From \(\sin y=0\) we have \(y=\pm n \pi .\) From \(e^{x-y}=1,\) we can conclude that \(x-y=0\) or \(x=y .\) The critical points of the system are therefore \((\pm n \pi, \pm n \pi)\) for \(n=0,1,2, \ldots\)

since \(x^{2}-y^{2}=0, y^{2}=x^{2}\) and so \(x^{2}-3 x+2=(x-1)(x-2)=0 .\) It follows that the critical points are (1,1) \((1,-1),(2,2),\) and \((2,-2) .\) We next use the Jacobian $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{rr} -3 & 2 y \\ 2 x & -2 y \end{array}\right)$$ to classify these four critical points. For \(\mathbf{X}=(1,1), \tau=-5, \Delta=2,\) and so \(\tau^{2}-4 \Delta=17>0 .\) Therefore (1,1) is a stable node. For \(\mathbf{X}=(1,-1), \Delta=-2<0\) and so (1,-1) is a saddle point. For \(\mathbf{X}=(2,2), \Delta=-4<0\) and so we have another saddle point. Finally, if \(\mathbf{X}=(2,-2), \tau=1, \Delta=4,\) and so \(\tau^{2}-4 \Delta=-15<0 .\) Therefore (2,-2) is an unstable spiral point.

From \(x\left(1-x^{2}-3 y^{2}\right)=0\) we have \(x=0\) or \(x^{2}+3 y^{2}=1 .\) If \(x=0,\) then substituting into \(y\left(3-x^{2}-3 y^{2}\right)\) gives \(y\left(3-3 y^{2}\right)=0 .\) Therefore \(y=0,1,-1 .\) Likewise \(x^{2}=1-3 y^{2}\) yields \(2 y=0\) so that \(y=0\) and \(x^{2}=1-3(0)^{2}=1\) The critical points of the system are therefore \((0,0),(0,1),(0,-1),(1,0),\) and (-1,0)

From \(y^{2}-x^{2}=0, y=x\) or \(y=-x .\) The case \(y=x\) leads to (4,4) and (-1,1) but the case \(y=-x\) leads to \(x^{2}-3 x+4=0\) which has no real solutions. Therefore (4,4) and (-1,1) are the only critical points. We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} y & x-3 \\ -2 x & 2 y \end{array}\right)$$ to classify these two critical points. For \(\mathbf{X}=(4,4), \tau=12, \Delta=40,\) and so \(\tau^{2}-4 \Delta<0 .\) Therefore (4,4) is an unstable spiral point. For \(\mathbf{X}=(-1,1), \tau=-3, \Delta=10,\) and so \(x^{2}-4 \Delta<0 .\) It follows that (-1,-1) is a stable spiral point.

From \(-x\left(4-y^{2}\right)=0\) we obtain \(x=0, y=2,\) or \(y=-2 .\) If \(x=0,\) then substituting into \(4 y\left(1-x^{2}\right)\) yields \(y=0 .\) Likewise \(y=2\) gives \(8\left(1-x^{2}\right)=0\) or \(x=1,-1 .\) Finally \(y=-2\) yields \(-8\left(1-x^{2}\right)=0\) or \(x=1,-1\) The critical points of the system are therefore \((0,0),(1,2),(-1,2),(1,-2),\) and (-1,-2)

since \(x^{\prime}=-2 x y=0,\) either \(x=0\) or \(y=0 .\) If \(x=0, y\left(1-y^{2}\right)=0\) and \(s o(0,0),(0,1),\) and (0,-1) are critical points. The case \(y=0\) leads to \(x=0 .\) We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} -2 y & -2 x \\ -1+y & 1+x-3 y^{2} \end{array}\right)$$ to classify these three critical points. For \(\mathbf{X}=(0,0), \tau=1\) and \(\Delta=0\) and so the test is inconclusive. For \(\mathbf{X}=(0,1), \tau=-4, \Delta=4\) and so \(\tau^{2}-4 \Delta=0 .\) We can conclude that (0,1) is a stable critical point but we are unable to classify this critical point further in this borderline case. For \(\mathbf{X}=(0,-1), \Delta=-4<0\) and so (0,-1) is a saddle point.

We have \(d y / d x=y^{\prime} / x^{\prime}=-f(x) / y\) and so, using separation of variables, $$\frac{y^{2}}{2}=-\int_{0}^{x} f(\mu) d \mu+c \quad \text { or } \quad y^{2}+2 F(x)=c$$ . We can conclude that for a given value of \(x\) there are at most two corresponding values of \(y .\) If (0,0) were a stable spiral point there would exist an \(x\) with more than two corresponding values of \(y .\) Note that the condition \(f(0)=0\) is required for (0,0) to be a critical point of the corresponding plane autonomous system \(x^{\prime}=y, y^{\prime}=-f(x)\).

Note that \(\Delta=\mu+1\) and \(\tau=\mu+1\) and so \(\tau^{2}-4 \Delta=(\mu+1)^{2}-4(\mu+1)=(\mu+1)(\mu-3) .\) It follows that \(\tau^{2}-4 \Delta<0\) if and only if \(-1<\mu<3 .\) We can conclude that (0,0) will be a saddle point when \(\mu<-1\) Likewise (0,0) will be an unstable spiral point when \(\tau=\mu+1>0\) and \(\tau^{2}-4 \Delta<0 .\) This condition reduces to \(-1<\mu<3\)

(a) \(x^{\prime}=x(-a+b y)=0\) implies that \(x=0\) or \(y=a / b .\) If \(x=0,\) then, from $$-c x y+\frac{r}{K} y(K-y)=0$$ \(y=0\) or \(K .\) Therefore (0,0) and \((0, K)\) are critical points. If \(\hat{y}=a / b,\) then $$\hat{y}\left[-c x+\frac{r}{K}(K-\hat{y})\right]=0$$ The corresponding value of \(x, x=\hat{x},\) therefore satisfies the equation \(c \hat{x}=r(K-\hat{y}) / K\). (b) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} -a+b y & b x \\ -c y & -c x+\frac{r}{K}(K-2 y) \end{array}\right)$$ and so at \(\mathbf{X}_{1}=(0,0), \Delta=-a r<0 .\) For \(\mathbf{X}_{1}=(0, K), \Delta=n(K b-a)=-r b(K-a / b) .\) since we are given that \(K>a / b, \Delta<0\) in this case. Therefore (0,0) and \((0, K)\) are each saddle points. For \(\mathbf{X}_{1}=(\hat{x}, \hat{y})\) where \(\hat{y}=a / b\) and \(c \hat{x}=r(K-\hat{y}) / K,\) we can write the Jacobian matrix as $$\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\left(\begin{array}{cc} 0 & b \hat{x} \\ -c \hat{y} & -\frac{r}{K} \hat{y} \end{array}\right)$$ and so \(\tau=-r \hat{y} / K<0\) and \(\Delta=b c \hat{x} \hat{y}>0 .\) Therefore \((\hat{x}, \hat{y})\) is a stable critical point and so it is either a stable node (perhaps degenerate) or a stable spiral point. (c) Write $$\tau^{2}-4 \Delta=\frac{r^{2}}{K^{2}} \hat{y}^{2}-4 b c \hat{x} \hat{y}=\hat{y}\left[\frac{r^{2}}{K^{2}} \hat{y}-4 b c \hat{x}\right]=\hat{y}\left[\frac{r^{2}}{K^{2}} \hat{y}-4 b \frac{r}{K}(K-\hat{y})\right]$$ using $$c \hat{x}=\frac{r}{K}(K-\hat{y})=\frac{r}{K} \hat{y}\left[\left(\frac{r}{K}+4 b\right) \hat{y}-4 b K\right]$$. Therefore \(\tau^{2}-4 \Delta<0\) if and only if $$\hat{y}<\frac{4 b K}{\frac{r}{K}+4 b}=\frac{4 b K^{2}}{r+4 b K}$$ Note that $$\frac{4 b K^{2}}{r+4 b K}=\frac{4 b K}{r+4 b K} \cdot K \approx K$$ where \(K\) is large, and \(\hat{y}=a / b< K .\) Therefore \(\tau^{2}-4 \Delta< 0\) when \(K\) is large and a stable spiral point will result.

\(\tau=2 \alpha, \Delta=\alpha^{2}+\beta^{2}>0,\) and \(\tau^{2}-4 \Delta=-4 \beta<0 .\) If \(\alpha<0,(0,0)\) is a stable spiral point. If \(\alpha>0,(0,0)\) is an unstable spiral point. Therefore (0,0) cannot be a node or saddle point.

The equation $$x^{\prime}=\alpha \frac{y}{1+y} x-x=x\left(\frac{\alpha y}{1+y}-1\right)=0$$ implies that \(x=0\) or \(y=1 /(\alpha-1) .\) When \(\alpha>0, \hat{y}=1 /(\alpha-1)>0 .\) If \(x=0,\) then from the differential equation for \(y^{\prime}, y=\beta .\) On the other hand, if \(\hat{y}=1 /(\alpha-1), \hat{y} /(1+\hat{y})=1 / \alpha\) and so \(\hat{x} / \alpha-1 /(\alpha-1)+\beta=0\) It follows that $$\hat{x}=\alpha\left(\beta-\frac{1}{\alpha-1}\right)=\frac{\alpha}{\alpha-1}[(\alpha-1) \beta-1]$$ and if \(\beta(\alpha-1)>1, \hat{x}>0 .\) Therefore \((\hat{x}, \hat{y})\) is the unique critical point in the first quadrant. The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} \alpha \frac{y}{y+1}-1 & \frac{\alpha x}{(1+y)^{2}} \\ -\frac{y}{1+y} & \frac{-x}{(1+y)^{2}}-1 \end{array}\right)$$ and for \(\mathbf{X}=(\hat{x}, \hat{y}),\) the Jacobian can be written in the form $$\mathbf{g}^{\prime}((\hat{x}, \hat{y}))=\left(\begin{array}{cc} 0 & \frac{(\alpha-1)^{2}}{\alpha} \hat{x} \\ -\frac{1}{\alpha} & -\frac{(\alpha-1)^{2}}{\alpha^{2}}-1 \end{array}\right)$$ It follows that $$\tau=-\left[\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}+1\right]<0, \quad \Delta=\frac{(\alpha-1)^{2}}{\alpha^{2}} \hat{x}$$ and so \(\tau=-(\Delta+1) .\) Therefore \(\tau^{2}-4 \Delta=(\Delta+1)^{2}-4 \Delta=(\Delta-1)^{2}>0 .\) Therefore \((\hat{x}, \hat{y})\) is a stable node.

Letting \(y=x^{\prime}\) we obtain the plane autonomous system $$\begin{array}{l} x^{\prime}=y \\ y^{\prime}=-8 x+6 x^{3}-x^{5} \end{array}$$ Solving \(x^{5}-6 x^{3}+8 x=x\left(x^{2}-4\right)\left(x^{2}-2\right)=0\) we see that critical points \(\operatorname{are}(0,0),(0,-2),(0,2),(0,-\sqrt{2}),\) and \((0, \sqrt{2}) .\) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -8+18 x^{2}-5 x^{4} & 0 \end{array}\right)$$ and we see that \(\operatorname{det}\left(\mathbf{g}^{\prime}(\mathbf{X})\right)=5 x^{4}-18 x^{2}+8\) and the trace of \(\mathbf{g}^{\prime}(\mathbf{X})\) is 0. since det \(\left(g^{\prime}((\pm \sqrt{2}, 0))\right)=-8<0,(\pm \sqrt{2}, 0)\) are saddle points. For the other critical points the determinant is positive and linearization discloses no information. The graph of the phase plane suggests that (0,0) and (±2,0) are centers.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-x+\left(\frac{1}{2}-3 y^{2}\right) y-x^{2}$$ If \((x, y)\) is a critical point, \(y=0\) and \(s o-x-x^{2}=-x(1+x)=0 .\) Therefore (0,0) and (-1,0) are the only two critical points. We next use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -1-2 x & \frac{1}{2}-9 y^{2} \end{array}\right)$$ to classify these critical points. For \(\mathbf{X}=(0,0), \tau=1 / 2, \Delta=1,\) and \(\tau^{2}-4 \Delta<0 .\) Therefore (0,0) is an unstable spiral point. For \(\mathbf{X}=(-1,0), \tau=1 / 2, \Delta=-1\) and so (-1,0) is a saddle point.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=x^{2}-y\left(1-x^{3}\right)$$ and the only critical point is \((0,0) .\) since the Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ 2 x+3 x^{2} y & x^{3}-1 \end{array}\right)$$ \(\tau=-1\) and \(\Delta=0,\) and we are unable to classify the critical point in this borderline case.

Switching to polar coordinates, $$\begin{array}{l} \frac{d r}{d t}=\frac{1}{r}\left(x \frac{d x}{d t}+y \frac{d y}{d t}\right)=\frac{1}{r}\left(-x y-x^{2} r^{4}+x y-y^{2} r^{4}\right)=-r^{5} \\ \frac{d \theta}{d t}=\frac{1}{r^{2}}\left(-y \frac{d x}{d t}+x \frac{d y}{d t}\right)=\frac{1}{r^{2}}\left(y^{2}+x y r^{4}+x^{2}-x y r^{4}\right)=1 \end{array}$$ If we use separation of variables on \(\frac{d r}{d t}=-r^{5}\) we obtain $$r=\left(\frac{1}{4 t+c_{1}}\right)^{1 / 4} \quad \text { and } \quad \theta=t+c_{2}$$ since \(\mathbf{X}(0)=(4,0), r=4\) and \(\theta=0\) when \(t=0 .\) It follows that \(c_{2}=0\) and \(c_{1}=\frac{1}{256} .\) The final solution can be written as $$r=\frac{4}{\sqrt[4]{1024 t+1}}, \quad \theta=t$$ and so the solution spirals toward the origin as \(t\) increases.

In Problem \(5,\) Section \(11.1,\) we showed that \((0,0),(\sqrt{1 / \epsilon}, 0)\) and \((-\sqrt{1 / \epsilon}, 0)\) are the critical points. We will use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -1+3 \epsilon x^{2} & 0 \end{array}\right)$$ to classify these three critical points. For \(\mathbf{X}=(0,0), \tau=0\) and \(\Delta=1\) and we are unable to classify this critical point. For \((\pm \sqrt{1 / \epsilon}, 0), \tau=0\) and \(\Delta=-2\) and so both of these critical points are saddle points.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-\frac{\left(\beta+\alpha^{2} y^{2}\right) x}{1+\alpha^{2} x^{2}}$$ and the Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ \frac{\left(\beta+\alpha y^{2}\right)\left(\alpha^{2} x^{2}-1\right)}{\left(1+\alpha^{2} x^{2}\right)^{2}} & \frac{-2 \alpha^{2} y x}{1+\alpha^{2} x^{2}} \end{array}\right)$$ For \(\mathbf{X}=(0,0), \tau=0\) and \(\Delta=\beta .\) since \(\beta<0,\) we can conclude that (0,0) is a saddle point.

If \(\operatorname{det}(s \mathbf{I}-\mathbf{A})=0,\) then \(s\) is an eigenvalue of \(\mathbf{A} .\) Thus \(s \mathbf{I}-\mathbf{A}\) has an inverse if \(s\) is not an eigenvalue of \(\mathbf{A} .\) For the purposes of the discussion in this section, we take \(s\) to be larger than the largest eigenvalue of A. Under this condition \(s \mathbf{I}-\mathbf{A}\) has an inverse.

since \(\mathbf{A}^{3}=\mathbf{0}, \mathbf{A}\) is nilpotent. since $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{k} \frac{t^{k}}{k !}+\cdots.$$ if \(\mathbf{A}\) is nilpotent and \(\mathbf{A}^{m}=\mathbf{0},\) then \(\mathbf{A}^{k}=\mathbf{0}\) for \(k \geq m\) and $$e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2 !}+\cdots+\mathbf{A}^{m-1} \frac{t^{m-1}}{(m-1) !}.$$ In this problem \(\mathbf{A}^{3}=\mathbf{0},\) so $$\begin{aligned} e^{\mathbf{A} t}=\mathbf{I}+\mathbf{A} t+\mathbf{A}^{2} \frac{t^{2}}{2} &=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right)+\left(\begin{array}{rrr} -1 & 1 & 1 \\ -1 & 0 & 1 \\ -1 & 1 & 1 \end{array}\right) t+\left(\begin{array}{rrr} -1 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 1 \end{array}\right) \frac{t^{2}}{2} \\ &=\left(\begin{array}{ccc} 1-t-t^{2} / 2 & t & t+t^{2} / 2 \\ -t & 1 & t \\ -t-t^{2} / 2 & t & 1+t+t^{2} / 2 \end{array}\right) \end{aligned}$$ and the solution of \(\mathbf{X}^{\prime}=\mathbf{A X}\) is $$\mathbf{X}(t)=e^{\mathbf{A} t} \mathbf{C}=e^{\mathbf{A} t}\left(\begin{array}{l} c_{1} \\ c_{2} \\ c_{3} \end{array}\right)=\left(\begin{array}{c} c_{1}\left(1-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(t+t^{2} / 2\right) \\ -c_{1} t+c_{2}+c_{3} t \\ c_{1}\left(-t-t^{2} / 2\right)+c_{2} t+c_{3}\left(1+t+t^{2} / 2\right) \end{array}\right).$$

The differential equation \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{2}-2 x\right) / y\) can be solved by separating variables. It follows that \(y^{2} / 2=\left(x^{3} / 3\right)-x^{2}+c\) and \(\operatorname{since} \mathbf{X}(0)=\left(x(0), x^{\prime}(0)\right)=(1,0), c=\frac{2}{3} .\) Therefore The differential equation \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{2}-2 x\right) / y\) can be solved by separating variables. It follows that \(y^{2} / 2=\left(x^{3} / 3\right)-x^{2}+c\) and \(\operatorname{since} \mathbf{X}(0)=\left(x(0), x^{\prime}(0)\right)=(1,0), c=\frac{2}{3} .\) Therefore $$\frac{y^{2}}{2}=\frac{x^{3}-3 x^{2}+2}{3}=\frac{(x-1)\left(x^{2}-2 x-2\right)}{3}$$ But \((x-1)\left(x^{2}-2 x-2\right)>0\) for \(1-\sqrt{3} < x < 1\) and so each \(x\) in this interval has 2 corresponding values of \(y\) therefore \(\mathbf{X}(t)\) is a periodic solution.

The differential equation is \(d y / d x=y^{\prime} / x^{\prime}=\left(x^{3}-x\right) / y\) and so \(y^{2} / 2=x^{4} / 4-x^{2} / 2+c\) or \(y^{2}=x^{4} / 2-x^{2}+c_{1}\) since \(x(0)=0\) and \(y(0)=x^{\prime}(0)=v_{0},\) it follows that \(c_{1}=v_{0}^{2}\) and so $$y^{2}=\frac{1}{2} x^{4}-x^{2}+v_{0}^{2}=\frac{\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1}{2}$$The \(x\) -intercepts on this graph satisfy $$x^{2}=1 \pm \sqrt{1-2 v_{0}^{2}}$$ and so we must require that \(\left.1-2 v_{0}^{2} \geq 0 \text { (or }\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\right)\) for real solutions to exist. If \(x_{0}^{2}=1-\sqrt{1-2 v_{0}^{2}}\) and \(-x_{0} < x < x_{0},\) then \(\left(x^{2}-1\right)^{2}+2 v_{0}^{2}-1 > 0\) and so there are two corresponding values of \(y .\) Therefore \(\mathbf{X}(t)\) with \(\mathbf{X}(0)=\left(0, v_{0}\right)\) is periodic provided that \(\left|v_{0}\right| \leq \frac{1}{2} \sqrt{2}\)

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=\epsilon x^{2}-x+1$$ and so the critical points must satisfy \(y=0\) and $$x=\frac{1 \pm \sqrt{1-4 \epsilon}}{2 \epsilon}$$ Therefore we must require that \(\epsilon \leq \frac{1}{4}\) for real solutions to exist. We will use the Jacobian matrix $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ 2 \epsilon x-1 & 0 \end{array}\right)$$ to attempt to classify \(((1 \pm \sqrt{1-4 \epsilon}) / 2 \epsilon, 0)\) when \(\epsilon \leq 1 / 4\). Note that \(\tau=0\) and \(\Delta=\mp \sqrt{1-4 \epsilon}\) For \(\mathbf{X}=((1+\sqrt{1-4 \epsilon}) / 2 \epsilon, 0)\) and \(\epsilon<1 / 4, \Delta<0\) and so a saddle point occurs. For \(\mathbf{X}=((1-\sqrt{1-4 \epsilon}) / 2 \epsilon, 0)\) \(\Delta \geq 0\) and we are not able to classify this critical point using linearization.

The corresponding plane autonomous system is $$x^{\prime}=y, \quad y^{\prime}=-\frac{\alpha}{L} x-\frac{\beta}{L} x^{3}-\frac{R}{L} y$$ where \(x=q\) and \(y=q^{\prime} .\) If \(\mathbf{X}=(x, y)\) is a critical point, \(y=0\) and \(-\alpha x-\beta x^{3}=-x\left(\alpha+\beta x^{2}\right)=0 .\) If \(\beta>0\) \(\alpha+\beta x^{2}=0\) has no real solutions and so (0,0) is the only critical point. since $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cr} 0 & 1 \\ \frac{-\alpha-3 \beta x^{2}}{L} & -\frac{R}{L} \end{array}\right)$$ \(\tau=-R / L<0\) and \(\Delta=\alpha / L>0 .\) Therefore (0,0) is a stable critical point. If \(\beta<0,(0,0)\) and \((\pm \hat{x}, 0),\) where \(\hat{x}^{2}=-\alpha / \beta\) are critical points. At \(\mathbf{X}(\pm \hat{x}, 0), \tau=-R / L<0\) and \(\Delta=-2 \alpha / L<0 .\) Therefore both critical points are saddles.

If we let \(d x / d t=y,\) then \(d y / d t=-x^{3}-x .\) From this we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=-\frac{x^{3}+x}{y}$$ Separating variables and integrating we obtain $$\int y d y=-\int\left(x^{3}+x\right) d x$$ and $$\frac{1}{2} y^{2}=-\frac{1}{4} x^{4}-\frac{1}{2} x^{2}+c_{1}$$ Completing the square we can write the solution as \(y^{2}=-\frac{1}{2}\left(x^{2}+1\right)^{2}+c_{2} .\) If \(\mathbf{X}(0)=\left(x_{0}, 0\right),\) then \(c_{2}=\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}\) and so $$\begin{aligned} y^{2} &=-\frac{1}{2}\left(x^{2}+1\right)^{2}+\frac{1}{2}\left(x_{0}^{2}+1\right)^{2}=\frac{x_{0}^{4}+2 x_{0}^{2}+1-x^{4}-2 x^{2}-1}{2} \\ &=\frac{\left(x_{0}^{2}+x^{2}\right)\left(x_{0}^{2}-x^{2}\right)+2\left(x_{0}^{2}-x^{2}\right)}{2}=\frac{\left(x_{0}^{2}+x^{2}+2\right)\left(x_{0}^{2}-x^{2}\right)}{2} \end{aligned}$$ Note that \(y=0\) when \(x=-x_{0} .\) In addition, the right-hand side is positive for \(-x_{0}

(a) Letting \(x=\theta\) and \(y=x^{\prime}\) we obtain the system \(x^{\prime}=y\) and \(y^{\prime}=1 / 2-\sin x .\) since \(\sin \pi / 6=\sin 5 \pi / 6=1 / 2\) we see that \((\pi / 6,0)\) and \((5 \pi / 6,0)\) are critical points of the system. (b) The Jacobian matrix is $$\mathbf{g}^{\prime}(\mathbf{X})=\left(\begin{array}{cc} 0 & 1 \\ -\cos x & 0 \end{array}\right)$$ and so $$\mathbf{A}_{1}=\mathbf{g}^{\prime}=((\pi / 6,0))=\left(\begin{array}{cc} 0 & 1 \\ -\sqrt{3} / 2 & 0 \end{array}\right) \quad \text { and } \quad \mathbf{A}_{2}=\mathbf{g}^{\prime}=((5 \pi / 6,0))=\left(\begin{array}{cc} 0 & 1 \\ \sqrt{3} / 2 & 0 \end{array}\right)$$ since det \(\mathbf{A}_{1}>0\) and the trace of \(\mathbf{A}_{1}\) is \(0,\) no conclusion can be drawn regarding the critical point \((\pi / 6,0)\) since det \(\mathbf{A}_{2}<0,\) we see that \((5 \pi / 6,0)\) is a saddle point(c) From the system in part (a) we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{1 / 2-\sin x}{y}$$ Separating variables and integrating we obtain $$\int y d y=\int\left(\frac{1}{2}-\sin x\right) d x$$ and $$\frac{1}{2} y^{2}=\frac{1}{2} x+\cos x+c_{1}$$ or $$y^{2}=x+2 \cos x+c_{2}$$ For \(x_{0}\) near \(\pi / 6,\) if \(\mathbf{X}(0)=\left(x_{0}, 0\right)\) then \(c_{2}=-x_{0}-2 \cos x_{0}\) and \(y^{2}=x+2 \cos x-x_{0}-2 \cos x_{0} .\) Thus, there are two values of \(y\) for each \(x\) in a sufficiently small interval around \(\pi / 6 .\) Therefore \((\pi / 6,0)\) is a center.

(a) Writing the system as \(x^{\prime}=x\left(x^{3}-2 y^{3}\right)\) and \(y^{\prime}=y\left(2 x^{3}-y^{3}\right)\) we see that (0,0) is a critical point. Setting \(x^{3}-2 y^{3}=0\) we have \(x^{3}=2 y^{3}\) and \(2 x^{3}-y^{3}=4 y^{3}-y^{3}=3 y^{3} .\) Thus, (0,0) is the only critical point of the system. (b) From the system we obtain the first-order differential equation $$\frac{d y}{d x}=\frac{2 x^{3} y-y^{4}}{x^{4}-2 x y^{3}}$$ or $$\left(2 x^{3} y-y^{4}\right) d x+\left(2 x y^{3}-x^{4}\right) d y=0$$ which is homogeneous. If we let \(y=u x\) it follows that $$\begin{aligned} \left(2 x^{4} u-x^{4} u^{4}\right) d x+\left(2 x^{4} u^{3}-x^{4}\right)(u d x+x d u) &=0 \\ x^{4} u\left(1+u^{3}\right) d x+x^{5}\left(2 u^{3}-1\right) d u &=0 \\ \frac{1}{x} d x+\frac{2 u^{3}-1}{u\left(u^{3}+1\right)} d u &=0 \\ \frac{1}{x} d x+\left(\frac{1}{u+1}-\frac{1}{u}+\frac{2 u-1}{u^{2}-u+1}\right) d u &=0 \end{aligned}$$ Integrating gives $$\ln |x|+\ln |u+1|-\ln |u|+\ln \left|u^{2}-u+1\right|=c_{1}$$ or $$\begin{aligned} x\left(\frac{u+1}{u}\right)\left(u^{2}-u+1\right) &=c_{2} \\ x\left(\frac{y+x}{y}\right)\left(\frac{y^{2}}{x^{2}}-\frac{y}{x}+1\right) &=c_{2} \\ \left(x y+x^{2}\right)\left(y^{2}-x y+x^{2}\right) &=c_{2} x^{2} y \\ x y^{3}+x^{4} &=c_{2} x^{2} y \\ x^{3}+y^{2} &=3 c_{3} x y \end{aligned}$$ (c) We see from the graph that (0,0) is unstable. It is not possible to classify the critical point as a node, saddle, center, or spiral point.