Systems of non-linear ODEs.

    1. Let $\underline{g}:[a-h, a+h]→ℝ^2$ be any continuous function. Show that \[ \left‖\int_a^x \underline{g}(t) d t\right‖_1≤\left|\int_a^x\left‖\underline{g}(t)\right‖_1 d t\right| \] where we define the integral component-wise and where $‖⋅‖_1$ is the 1-norm of vectors in $ℝ^2$, i.e. $\left‖\left(y_1, y_2\right)\right‖_1=\left|y_1\right|+\left|y_2\right|$.
      [You may use that if $h:[a-h, a+h]→ℝ$ is continuous then $\left|\int_a^x h(t) d t\right|≤\left|\int_a^x{|h(t)|}d t\right|$.]
    2. Show that the functions
      i) $\underline{f}(x, \underline{y})=\left(\begin{array}{c}x \sin \left(y_2\right)+y_1 \\ y_2^2\end{array}\right)$ ii) $\underline{f}(x, \underline{y}):=\left(\begin{array}{c}y_2 \\ -\frac{y_1^2}{y_2}\end{array}\right)$
      satisfy the Lipschitz condition \[ \left‖\underline{f}(x, \underline{y})-\underline{f}(x, \underline{\tilde{y}})\right‖_1≤L\left‖\underline{y}-\underline{\tilde{y}}\right‖_1, \text { for all } x∈[-h, h], \quad \underline{y}, \underline{\tilde{y}}∈B_k\left(\left(1\atop1\right)\right) \] for suitable $L, h, k>0$ where $B_k(\underline{b}):=\left\{\underline{y}∈ℝ^2:\left‖\underline{y}-\underline{b}\right‖_1≤k\right\}$.
    1. Let $\underline g(t)=\big(g_1(t),g_2(t)\big)$, then\begin{align*} \left‖\int_a^x \underline{g}(t) d t\right‖_1&=\left|\int_a^x g_1(t) d t\right|+\left|\int_a^x g_2(t) d t\right|\\&≤\left|\int_a^x\left|g_1(t)\right|d t\right|+\left|\int_a^x\left|g_2(t)\right|d t\right|\\\small\text{non-negative}&=\left|\int_a^x\left|g_1(t)\right|+\left|g_2(t)\right| d t\right|\\&=\left|\int_a^x\left‖\underline{g}(t)\right‖_1 d t\right| \end{align*}
    2. Suppose there exists $L_1,L_2$ such that for $x∈[a-h,a+h]$ and $\underline u,\underline v∈B_k\left(\left(1\atop1\right)\right)$, \begin{aligned}\left|f_1\left(x, u_1, u_2\right)-f_1\left(x, v_1, v_2\right)\right| &≤L_1\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right)\quad\text{and} \\\left|f_2\left(x, u_1, u_2\right)-f_2\left(x, v_1, v_2\right)\right| &≤ L_2\left(\left|u_1-v_1\right|+\left|u_2-v_2\right|\right) \end{aligned}Then \begin{aligned}\left‖\underline{f}(x, \underline{y})-\underline{f}(x, \underline{\tilde{y}})\right‖_1&=\left|f_1\left(x,y_1,y_2\right)-f_1\left(x,\tilde y_1,\tilde y_2\right)\right|+\left|f_2\left(x,y_1,y_2\right)-f_2\left(x,\tilde y_1,\tilde y_2\right)\right|\\&≤(L_1+L_2)\left(\left|y_1-\tilde y_1\right|+\left|y_2-\tilde y_2\right|\right)\\&=(L_1+L_2)\left‖\underline y-\underline{\tilde y}\right‖_1\end{aligned} Hence it suffices to prove that the components $f_1$ and $f_2$ of $\underline{f}=\left(\begin{array}{l}f_1 \\ f_2\end{array}\right)$ satisfy a Lipschitz condition.
      i) By mean value theorem $\left|\sin u_2-\sin v_2\right|≤\left|u_2-v_2\right|$. For $L_1=\max\{1,h\},L_2=2(1+k)$,\begin{aligned}\left|f_1\left(x, u_1, u_2\right)-f_1\left(x, v_1, v_2\right)\right|&=\left|x\sin u_2+u_1-x\sin v_2-v_1\right|\\&≤\left|x\right|\left|\sin u_2-\sin v_2\right|+\left|u_1-v_1\right|\\&≤h\left|u_2-v_2\right|+\left|u_1-v_1\right|\\\left|f_2\left(x, u_1, u_2\right)-f_2\left(x, v_1, v_2\right)\right|&=\left|u_2^2-v_2^2\right|\\&=\left|u_2+v_2\right|\left|u_2-v_2\right|\\&≤2(1+k)\left|u_2-v_2\right|\\&≤2(1+k)(\left|u_1-v_1\right|+\left|u_2-v_2\right|)\end{aligned} ii) $L_1=1$, $L_2=\max\left\{\frac{2(1+k)}{1-k},\frac{(1+k)^2}{(1-k)^2}\right\}$,\begin{aligned}\left|f_1\left(x, u_1, u_2\right)-f_1\left(x, v_1, v_2\right)\right|&=\left|u_2-v_2\right|\\\left|f_2\left(x, u_1, u_2\right)-f_2\left(x, v_1, v_2\right)\right|&≤\left|f_2\left(x, u_1, u_2\right)-f_2\left(x, v_1, u_2\right)\right|+\left|f_2\left(x, v_1, u_2\right)-f_2\left(x, v_1, v_2\right)\right|\\&=\left|\frac{u_1^2}{u_2}-\frac{v_1^2}{u_2}\right|+\left|\frac{v_1^2}{u_2}-\frac{v_1^2}{v_2}\right|\\&=\frac{\left|u_1+v_1\right|}{u_2}\left|u_1-v_1\right|+\frac{v_1^2}{u_2v_2}\left|u_2-v_2\right|\\&≤\frac{2(1+k)}{1-k}\left|u_1-v_1\right|+\frac{(1+k)^2}{(1-k)^2}\left|u_2-v_2\right|\end{aligned}
  1. The aim of this question is to fill in the details of the proof of Theorem 1.6 in the lecture on Picard's theorem for a system of two first order ODEs via the CMT.
    Let $\underline{f}=\left(f_1, f_2\right): S:=[a-h, a+h]×B_k(\underline{b})→ℝ^2$ be a continuous function which satisfies the Lipschitz-condition \[\tag1 \left‖\underline{f}(x, \underline{y})-\underline{f}(x, \underline{\tilde{y}})\right‖_1≤L\left‖\underline{y}-\underline{\tilde{y}}\right‖_1 \text { for all } x∈[a-h, a+h], \underline{y}, \underline{\tilde{y}}∈B_k(\underline{b}) \] for some $L>0$ and set $M:=\max _{(x, y)\in S}\left‖\underline f(x, \underline{y})\right‖_1$.
    Given $0< η≤h$ we let $𝒞_η=C\left([a-η, a+η] ; B_k(\underline{b})\right)$ be the set of continuous functions $y:[a-η, a+η]→B_k(\underline{b}) \subset ℝ^2$, and equip this space with the sup norm defined by \[ {‖\underline{y}‖}_\text{sup}:=\sup _{x∈[a-η, a+η]}{‖\underline{y}(x)‖}_1, \text { for } \underline{y}∈𝒞_η . \] You can use that $\left(𝒞_η,{‖⋅‖}_\text{sup}\right)$ is a complete metric space.
    1. Show that if $Mη≤k$, then the map $T$ which assigns to each function $\underline{y}∈𝒞_η$ the function \[ (T \underline{y})(x)=\underline{b}+\int_a^x \underline{f}(s, \underline{y}(s)) d s \] has the property that $T \underline{y}∈𝒞_η$ for every $\underline{y}∈𝒞_η$. Show furthermore that if $ηL< 1$ then this map is a contraction on $\left(C_η,{‖⋅‖}_\text{sup}\right)$.
    2. Reformulate the initial value problem \[\tag2\begin{aligned} &y_1'(x)=f_1\left(x, y_1(x), y_2(x)\right) \\ &y_2'(x)=f_2\left(x, y_1(x), y_2(x)\right) \\ &y_1(a)=b_1, \quad y_2(a)=b_2 . \end{aligned}\] as an integral equation for $\underline{y}(x):=\left(\begin{array}{l}y_1(x) \\ y_2(x)\end{array}\right)$. Explain why the CMT and (a) imply that this initial value problem has a unique solution on $[a-η, a+η]$ for suitable $η>0$.
    3. Define what it means for the solution of (2) to depend continuously on the initial data and show that this holds true.
    4. Explain why, if the Lipschitz condition (1) holds for all $x∈[a-h, a+h]$ and $\underline{y}, \underline{\tilde{y}}∈ℝ^2$ (instead of only $\underline{y}, \underline{\tilde{y}}∈B_k(\underline{b})$ ), then there is a unique solution on the whole interval $[a-h, a+h]$
    1. For every $\underline{y}∈𝒞_η$,\[\left‖T\underline{y}-b\right‖_1\overset{\text{Q1}}≤\left|\int_a^x\left‖f(s,\underline{y}(s))\right‖_1ds\right|≤Mη≤k\]Taking sup over $x∈[a-η,a+η]$, we get $T\underline{y}∈B_k(\underline{b})$. By fundamental theorem of calculus, $T\underline{y}$ is continuous, so $T\underline{y}∈𝒞_η$.
      By Lipschitz condition,\[\left‖T\underline{y}-T\underline{\tilde y}\right‖_1\overset{\text{Q1}}≤\left|\int_a^x\left‖\underline{f}(s, \underline{y}(s))-\underline{f}(s, \underline{\tilde{y}}(s))\right‖_1ds\right|≤\left|\int_a^xL\left‖\underline{y}-\underline{\tilde{y}}\right‖_1ds\right|≤ηL\left‖\underline{y}-\underline{\tilde{y}}\right‖_1\]Taking sup over $x∈[a-η,a+η]$, we get $\left‖T\underline{y}-T\underline{\tilde y}\right‖_\text{sup}≤ηL\left‖\underline{y}-\underline{\tilde y}\right‖_\text{sup}$ for every $\underline{y},\underline{\tilde y}∈𝒞_η$. If $ηL< 1$, $T$ is a contraction on $\left(C_η,{‖⋅‖}_\text{sup}\right)$.
    2. The IVP can be reformulate as\[\underline y(x)=\underline b+\int_a^x\underline f(x,\underline y(s))ds\]that is, $T\underline y=\underline y$. By (a) and completeness of $\left(𝒞_η,{‖⋅‖}_\text{sup}\right)$ and CMT, $T$ has a unique fixed point, so there is a unique solution for $\underline y$.
    3. The solution depends continuously on the initial data if $∀ϵ>0$, for solutions $\underline y,\underline{\tilde y}$ with initial data $\underline b,\underline{\tilde b}$ respectively, $\left‖\underline b-\underline{\tilde b}\right‖_1< δ$ implies $\left‖\underline y-\underline {\tilde y}\right‖_\text{sup}< ϵ$.
      Proof 1: Take $δ=ϵ(1-Lη)$, \begin{align*}\left‖\underline y-\underline{\tilde y}\right‖_\text{sup}&=\left‖T\underline y-T\underline{\tilde y}\right‖_\text{sup}\\ \small\text{triangle inequality} &≤\left‖\underline b-\underline{\tilde b}\right‖_1+\left‖\int_a^x\underline f(x,\underline y(s))-\underline f(x,\underline{\tilde y}(s))ds\right‖_\text{sup}\\ \small\text{from 2.1.a} &≤δ+Lη\left‖\underline y-\underline{\tilde y}\right‖_\text{sup}\\\\ ⇒(1-Lη)\left‖\underline y-\underline{\tilde y}\right‖_\text{sup}&≤δ\\ ⇒\left‖\underline y-\underline{\tilde y}\right‖_\text{sup}&≤\fracδ{1-Lη}=ϵ \end{align*} Proof 2: As Q1.4b, Gronwall's inequality gives a sharper bound: \begin{align*}\left‖\underline y^{(1)}(x)-\underline y^{(2)}(x)\right‖_1&≤\left‖\underline b^{(1)}-\underline b^{(2)}\right‖_1+\left|\int_a^x\left‖f(s,\underline y^{(1)}(s))-f(s,\underline y^{(2)}(s))\right‖_1ds\right|\\ &≤\left‖\underline b^{(1)}-\underline b^{(2)}\right‖_1+\left|\int_a^xL\left‖\underline y^{(1)}(s)-\underline y^{(2)}(s)\right‖_1ds\right| \end{align*} By Gronwall's inequality, \[\left‖\underline y^{(1)}(s)-\underline y^{(2)}(s)\right‖_1≤\left‖\underline b^{(1)}-\underline b^{(2)}\right‖_1e^{L{|x-a|}}\] Taking $\sup_{s∈[a-η,a+η]}$, \[\left‖\underline y^{(1)}-\underline y^{(2)}\right‖_\text{sup}≤\left‖\underline b^{(1)}-\underline b^{(2)}\right‖_\text{sup}e^{Lη}\]
    4. [Corollary 1.5.]
      Proof: We look at $x≥a$ first. If $h< 1 / L$ we are done. (Take $η=h$.) Otherwise we choose $η_1< 1 / L$. Then, from Theorem 1.4, there exists a unique solution, $y_0$ say, on $\left[a, a+η_1\right]$.
      Now choose $η_2=\min \left\{2 η_1, h\right\}$, and look for a solution, $y_1$ say, on $\left[a+η_1, a+η_2\right]$, of the ODE with initial data $y_1\left(a+η_1\right)=y_0\left(a+η_1\right)$.
      Now define \begin{aligned} &y(x)=y_0(x), x \in\left[a, a+η_1\right] \\ &y(x)=y_1(x), x \in\left[a+η_1, a+η_2\right] \end{aligned} To construct $y_1$: As in Theorem 1.4, but we now work in the space $X_1:=𝒞\left(\left[a+η_1, a+η_2\right] ;[b-k, b+k]\right)$, and take (for $a+η_1≤x≤a+η_2$) \begin{align} \left(T_1 y\right)(x) &=y_0\left(a+η_1\right)+\int_{a+η_1}^x f(s, y(s)) d s\nonumber \\ &=b+\int_a^{a+η_1} f\left(s, y_0(s)\right) d s+\int_{a+η_1}^x f(s, y(s)) d s .\label1 \end{align} So $T_1: X_1 → X_1$ because from \eqref{1} \begin{aligned} \left\|T_1 y-b\right\|_\text{sup}&≤M η_1+M\left(x-\left(a+η_1\right)\right)=M(x-a)≤M η_2 \\ &≤Mh≤k . \end{aligned} Also $T_1$ is a contraction as the proof of claim 2 only requires that the length of the interval we work on, which for $T_1$ is $η_2-η_1$, is less than $1 / L$. Thus we obtain the existence of a unique solution on $\left[a, a+η_2\right]$. Repeating this argument, both in positive and negative direction, we continue to be able to extend the solution and after finitely many steps have reached the endpoint $a+h$ of the original interval, since we can carry out each step except the very last one (where we will be able to choose $η_j=h$ since we'll have $h-η_{j-1}<\frac{1}{L}$ ) with the same 'stepsize' $η_k-η_{k-1}=η_1$.
    1. Consider the second order differential equation \[\tag3 y''(x)=F\left(x, y(x), y'(x)\right), \text { with initial data } y(a)=c_1, y'(a)=c_2, \] where $F: S:=[a-h, a+h]×B_k(\underline{c})→ℝ$ is a continuous function. Suppose that $F$ satisfies a Lipschitz-condition, i.e. there exists $L$ such that \[ \left|F(x, \underline{y})-F(x, \underline{\tilde{y}})\right|≤L\left‖\underline{y}-\underline{\tilde{y}}\right‖\text { for all }(x, \underline{y}),(x, \underline{y})∈S . \] By writing (3) as a system of first order differential equations, and demonstrating that the conditions of Theorem 1.6 (see question 2.2 above) are satisfied, show that there exists $0< η≤h$ such that (3) has a unique solution on $[a-η, a+η]$.
    2. Now consider the second order linear differential equation for $y(x)$ \[\tag4p(x) y''+q(x) y'+r(x) y=s(x), \quad x∈[a, b]\] for continuous functions $p, q, r, s:[a, b]→ℝ$. Suppose that $p(x)≠0$ on all of $[a, b]$ and fix any $y_0, y_1∈ℝ$ and $x_0∈[a, b]$. Show that (4) with initial conditions $y\left(x_0\right)=y_0, y'\left(x_0\right)=y_1$ has a unique solution $y$ on $[a, b]$.
    3. (Taken from Collins) Consider the problem \[ y y''=-\left(y'\right)^2, \quad y(0)=y'(0)=1 . \] (i) Show that the problem has a unique solution on an interval containing 0.
      (ii) Find the solution and state where it exists.
    1. Let $z=y'$. Write (3) as \begin{cases} y'(x)=z\\ z'(x)=F\left(x, y(x), z(x)\right) \end{cases} Define $\underline f$ by $f_1(x,y,z)=z,f_2(x,y,z)=F(x,y,z)$. Then $\underline f$ is Lipschitz for $1+L$. For $η=\min\left(h,\frac k{\sup F}\right)$, (3) has a unique solution on $[a-η, a+η]$.
    2. Write the problem as $\underline y'=\underline f(x,\underline y)$ where $$ \underline y=\left(y(x)\atop y'(x)\right), \underline f(x,u,v)=\pmatrix{v\\\frac{s(x)-q(x)v-r(x)u}{p(x)}}. $$ Since $p(x)≠0$, $\underline f$ is continuous on $S$.$$\left|\frac{s(x)-q(x)v_1-r(x)u_1}{p(x)}-\frac{s(x)-q(x)v_2-r(x)u_2}{p(x)}\right|=\left|\frac{q(x)}{p(x)}\right|{|v_1-v_2|}+\left|\frac{r(x)}{p(x)}\right|{|u_1-u_2|}$$$\underline f$ satisfies Lipschitz condition on $S$, so $\underline y(x)$ has a unique solution.
    3. (i) Let $S=\left[-\frac13,\frac13\right]×\left[\frac13,\frac53\right]$. Write the problem as $\underline y'=\underline f(x,\underline y)$ where $$ \underline y=\left(y(x)\atop y'(x)\right), \underline f(x,u,v)=\pmatrix{v\\-\frac{v^2}u}. $$By 2.1.b(ii), $\underline f$ satisfies Lipschitz condition on $S$, so $\underline y(x)$ has a unique solution on $\left[-\frac13,\frac13\right]$.
      (ii) $(y^2)''=0⇒y^2$ is a linear function of $x$. Using $y(0)=y'(0)=1$, we have $y^2(0)=1,(y^2)'(0)=2$, so $y^2=2x+1⇒y=\sqrt{2x+1}$
  2. Autonomous systems of ODEs and the phase plane

  3. Consider the plane autonomous system \[ \frac{d x}{d t}=x(1-2 x-y), \quad \frac{d y}{d t}=y(1-x-2 y) . \]
    1. By showing that the axes of the phase plane and the line $x=y$ are solution trajectories explain why a solution starting in the octant $x>0,x< y$ must remain in this region for all time.
    2. Find all the critical points and analyse them to determine their local behaviour including the local direction of the trajectories and whether the points are stable.
    3. Sketch the phase plane.
      In an application, when suitably scaled, $x$ and $y$ represent species populations which are in competition for resources. Use the phase plane to interpret what happens to the populations in the long term.
    1. Substituting $x=0$, we get $\frac{d y}{d t}=y(1-2y)⇒t-\log C=\log{|y|}-\log{|1-2y|}⇒y=\frac{e^t}{C+2e^t}∈(0,\frac12)∪(\frac12,+∞)$. So the line $x=0$ is the union of two trajectories.
      Substituting $y=x$, we get $\frac{d x}{d t}=x(1-3x)⇒t-\log C=\log{|x|}-\log{|1-3x|}⇒x=\frac{e^t}{C+3e^t}∈(0,\frac13)∪(\frac13,+∞)$. So the line $y=x$ is the union of two trajectories.
      Since different trajectories never intersect, a solution starting in the octant $x>0, x< y$ must remain in this region for all time.
    2. To find all the critical points, we solve $x(1-2 x-y)=0,y(1-x-2 y)=0$. From the first equation, $x=0$ or $y=1-2x$.
      If $x=0$, the second equation is $y(1-2y)=0$, so $y=0$ or $y=1/2$.
      If $y=1-2x$, the second equation is $(1-2x)(-1+3x)=0$, so $x=1/2$ or $x=1/3$.
      So all the critical points are $(0,0)(0,1/2)(1/2,0)(1/3,1/3)$. \begin{array}l M=\begin{pmatrix}X_x&X_y\\Y_x&Y_y\end{pmatrix}=\begin{pmatrix}1-4x-y&-x\\-y&1-x-4y\end{pmatrix}\\ λ_1λ_2=\det M=4x^2+16xy+4y^2-5x-5y+1\\ λ_1+λ_2=\operatorname{tr}M=-5x-5y+2\\ (λ_1-λ_2)^2=(λ_1+λ_2)^2-4λ_1λ_2=9 x^2 - 14 x y + 9 y^2 \end{array} At $(0,0)$, $λ_1=λ_2=1>0,M=I⇒$unstable star
      At $(0,1/2)$ and $(1/2,0)$, $λ_1λ_2=-1/2< 0⇒$saddle
      At $(1/3,1/3)$, $λ_1λ_2=1/3>0,λ_1+λ_2=-4/3< 0,(λ_1-λ_2)^2=4/9>0⇒λ_1< λ_2< 0⇒$stable node
    3. Error! Click to view log.
      VectorPlot[{x(1-2x-y), y(1-x-2y)}, {x, -0.5, 1}, {y, -0.5, 1}]
      (1/3,1/3) is the only stable node, so two populations are equal in the long term. Competitive Lotka–Volterra equations
    1. Find and classify the types of all critical points of the system \[ \frac{d x}{d t}=\left(a-x^2\right) y, \quad \frac{d y}{d t}=x-y, \] in each of the cases
      (i) $a<-\frac14$
      (ii) $-\frac14< a< 0$
      (iii) $a>0$.
    2. Consider the case $a=-1/4$ and analyse in detail the behaviour at all the critical points. Hence sketch the phase plane in this case.
      Can you say anything about the trajectories in this case for $|x|$ large?
    3. Consider the case $a=1/2$ and analyse in detail the behaviour at all the critical points. Hence sketch the phase plane in this case.
  4. Solution.
    1. If $a< 0$ and $(a-x^2)y=x-y=0$, then $a-x^2< 0$, so $y=0$, so $(0,0)$ is the only critical point; If $a>0$, the critical points are $(0,0)(\sqrt a,\sqrt a)(-\sqrt a,-\sqrt a)$. \begin{array}l M=\begin{pmatrix}X_x&X_y\\Y_x&Y_y\end{pmatrix}=\begin{pmatrix}-2xy&a-x^2\\1&-1\end{pmatrix}\\ λ_1λ_2=\det M=x^2+2xy-a\\ λ_1+λ_2=\operatorname{tr}M=-2xy-1\\ (λ_1-λ_2)^2=(λ_1+λ_2)^2-4λ_1λ_2=4 x^2 y^2-4xy-4x^2+4a+1 \end{array} (i) At $(0,0)$, $λ_1λ_2=-a>0,λ_1+λ_2=-1< 0,(λ_1-λ_2)^2=4a+1< 0⇒$conjugate pair, $μ< 0⇒$stable spiral
      (ii) At $(0,0)$, $λ_1λ_2=-a>0,λ_1+λ_2=-1< 0,(λ_1-λ_2)^2=4a+1>0⇒0>λ_1>λ_2⇒$stable node
      (iii) At $(0,0)$, $λ_1λ_2=-a< 0⇒λ_1< 0< λ_2⇒$saddle
      At $(\sqrt a,\sqrt a)$ and $(-\sqrt a,-\sqrt a)$, $λ_1λ_2=2a>0,λ_1+λ_2=-2a-1< 0,(λ_1-λ_2)^2=(2a-1)^2≥0$.
      If $a≠1/2$, we have $0>λ_1>λ_2⇒$stable node; if $a=1/2$, we have $0>λ_1=λ_2$, but $Y_x=1$, so $M≠λI$, there is a stable inflected node.
    2. At (0,0), $λ_1=λ_2=-1/2,M=\pmatrix{0 & -\frac14 \\1 & -1}$, so $M≠λI$, there is a stable inflected node. eigenvector: $v=\pmatrix{1\\2}$. When $|x|$ large, $\pmatrix{x\\y}∼c_1te^{-t/2}\pmatrix{1\\2}$.
      phase plane: See the figure produced by Matlab
    3. At (0,0), $M=\pmatrix{0&\frac12\\1&-1},λ_1=\frac{-1-\sqrt3}2,λ_2=\frac{-1+\sqrt3}2⇒$saddle. eigenvectors: $v_1=\pmatrix{\frac{1-\sqrt3}2\\1},v_2=\pmatrix{\frac{1+\sqrt3}2\\1}$.
      At $\left(\frac1{\sqrt2},\frac1{\sqrt2}\right)$ and $\left(-\frac1{\sqrt2},-\frac1{\sqrt2}\right)$, $λ_1=λ_2=-1,M=\pmatrix{-1 & 0 \\1 & -1}$, so $M≠λI$, there is a stable inflected node. eigenvector: $\pmatrix{0\\1}$.
      phase plane: See the figure produced by Matlab