1. 1. Let $μ ∈ ℝ$. Consider the initial value problem $$\tag† y'(x)=μ y(x)+x-μ,   y(0)=1 . $$
      1. Show that (†) has a unique solution on $ℝ$, stating carefully any results you use.
      2. Compute the first three Picard iterates: $y_0, y_1, y_2$.
      3. Hence, or otherwise, derive the general formula for $y_n(x)$, the $(n+1)$^th Picard iterate, and deduce the series representation for the solution of (†). Justify your answers.
      4. Consider now $y=y_μ(x)$ as a function of both $x$ and $μ$. Deduce that for any fixed $x ∈ ℝ, μ ↦ y_μ(x)$ is a differentiable function on $ℝ$ and let $z_μ(x)$ denote the partial derivative $z_μ(x)=\frac{∂}{∂ μ} y_μ(x)$. Show that for a fixed $μ ∈ ℝ$, the function $x ↦ z_μ(x)$ is the unique solution of an initial value problem which you should specify.
   2. Let $f: R → ℝ$ be a continuous function on $$ R:=\{(x,y,μ):{|x-a|}⩽h,{|y-b|}⩽k,{|μ|}⩽m\} $$ with ${|f(x,y,μ)|}⩽ M$ on $R$ and $Mh⩽k$. Assume further that on $R, f$ is Lipschitz continuous in the second variable with Lipschitz constant $L$ and Lipschitz continuous in the third variable with Lipschitz constant $Λ$. Consider the initial value problem $$\tag‡ y_μ'(x)=f\left(x, y_μ(x), μ\right),   y_μ(a)=b $$
      1. Citing a suitable result, explain briefly why, for any $μ ∈[-m, m]$, (‡) has a unique solution $y_μ:[a-h, a+h] →[b-k, b+k]$. Give an example of $f$ for which $μ ↦ y_μ$ is not differentiable at $μ=0$.
      2. State Gronwall's inequality. Hence, or otherwise, show that $$ \left|y_μ(x)-y_η(x)\right| ⩽ \fracΛL{|μ-η|}\left(\mathrm{e}^{L{|x-a|}}-1\right) $$ for any $x ∈[a-h, a+h]$ and any $μ, η ∈[-m, m]$. [You may find it helpful to consider the function $v(x)=Λ|μ-η|+L\left|y_μ(x)-y_η(x)\right|$.]
      3. Consider (†) as a special case of (‡) and, for given $h, k, m$, derive the smallest constants $M$, $L$ and $Λ$. Fix $h>0$. Allowing $k, m$ to vary, determine the maximal parameter domain $D$ so that (b)(i) implies that $y_μ(x)$ is well defined for all $(x, μ) ∈[-h, h] × D$.
2. Consider the PDE for $z(x, y)$ : $$\tag{*} P(x, y) z_x+Q(x, y) z_y=R(x, y, z) $$ along with data $x=x_0(s), y=y_0(s), z=z_0(s)$ for $s$ in a given range.
   1. 1. Derive the characteristic equations for the system above.
      2. State the condition satisfied by Cauchy data. Explain the geometric significance of this condition.
   2. Let $P(x, y)=-y, Q(x, y)=x$, and let the data curve be given by $x_0=s$, $y_0=0, z_0=0$ for $s ∈[p, q]$, with $0<p<q$.
      1. Show that the data is Cauchy.
      2. Let $R(x, y, z)=y \sqrt{x^2+y^2}$. Obtain an explicit solution to the PDE system, and define the domain of definition.
      3. Now let $R(x, y, z) ≡ 1$. Show that the parametric solution surface is not single-valued in $z$. Explain how to obtain a single-valued solution, and sketch the solution surface. [You do not need to explicitly compute $z(x, y)$.]
   3. Suppose that the characteristic projections of the PDE (*) are straight half-lines all passing through the point $(0,1)$.
      1. Let some initial data $z=z_0(s)$ be given along the curve $$ x_0(s)=3-2 \cos (s),   y_0(s)=1+2 \sin (s),   0 ⩽ s<α $$ where $α ∈(0, π)$ is a constant. Determine the range of $α$ for which the data is Cauchy. For $α$ in this range sketch the domain of definition. Does your answer depend on the function $R(x, y, z)$ in (*) ?
      2. Write down a PDE satisfying the given property of the characteristic projections.
3. 1. Consider a plane autonomous system of ODEs of the form: \begin{aligned} & \dot{x}=X(x, y), \\ & \dot{y}=Y(x, y), \end{aligned} where overdot denotes differentiation with respect to $t$ and $X, Y$ are Lipschitz continuous in $x$ and $y$.
      1. Define what is meant by a critical point of the system.
      2. Explain what is meant by a trajectory of the system in the phase plane.
      3. Show that trajectories never cross away from critical points.
   2. Let $a ∈ ℝ$ be a parameter and consider the system $$ X(x, y)=-y+a x\left(1-x^2-y^2\right),   Y(x, y)=x+y\left(1-x^2-y^2\right) . $$
      1. Show that the unit circle $\left\{(x, y): x^2+y^2=1\right\}$ is a closed solution trajectory of the system.
      2. Show that $(0,0)$ is a critical point of the system and classify it depending on the value of $a ∈ ℝ \backslash\{-1\}$.
      3. Let $a>0$. Show that $(0,0)$ is the only critical point. By considering $\dot{r}$, where $r^2=x^2+y^2$, or otherwise, show that every trajectory with $(x(0), y(0)) ≠(0,0)$ approaches the unit circle as $t → ∞$.
      4. Give a rough sketch of the phase plane for $a=1$.
   3. Consider the system in (b) with $a=-1$. Show that all trajectories within the unit circle are closed. [Hint: Derive the plane autonomous system of ODEs solved by $(u, v)=(x-y, x+y)$ and show that its trajectories are symmetric with respect to the axes.]

Solution

1. 1. 1. $f(x,y)=μy+x-μ$ is continuous for all $x,y∈ℝ$.
         ${|f(x,u)-f(x,v)|}={|μ|}{|u-v|}$, so $f$ satisfies Lipschitz condition in $y$ for all $x∈ℝ$.
         By Picard's theorem, (†) has a unique solution for all $x∈[a-h,a+h]$.
         By letting $h→∞$ we deduce that the solution exists on all of ℝ.
      2. \begin{split}y_0&=1\\y_1&=1+\int_0^xμy_0(t)+t-μ\mathrm{~d}t=1+\frac{x^2}2\\y_2&=1+\int_0^xμy_1(t)+t-μ\mathrm{~d}t=1+\frac{x^2}2+\frac{μx^3}6\end{split}
      3. $y_n(x)=1+\sum_{k=2}^{n+1}\frac{μ^{k-2}x^k}{k!}$
         Suppose this is true for $n$\begin{split}y_{n+1}(x)&=1+\int_0^xμy_n(t)+t-μ\mathrm{~d}t\\&=1+\frac{x^2}2+μ\int_0^x\sum_{k=2}^{n+1}\frac{μ^{k-2}t^k}{k!}\mathrm{~d}t \\&=1+\frac{x^2}2+μ\sum_{k=2}^{n+1}\frac{μ^{k-2}x^{k+1}}{(k+1)!} \\&=1+μ\sum_{k=2}^{n+2}\frac{μ^{k-2}x^k}{(k+1)!}\end{split}so it is true for all $n≥1$.
         The solution of (†) is $y(x)=1+\sum_{k=2}^∞\frac{μ^{k-2}x^k}{k!}=1+μ^{-2}(e^{μx}-1-μx)$ for all $x∈ℝ$.
      4. For any $x ∈ ℝ$, $y_μ(x)$ is a power series in $μ$, so it is differentiable wrt $μ$.
         $z_μ(x)$ is a power series in $x$, so it is differentiable wrt $x$.\begin{split}\frac∂{∂x}z_μ(x)&=\frac∂{∂x}\frac∂{∂μ}y_μ(x)\\&=\frac∂{∂μ}\frac∂{∂x}y_μ(x)\\&=\frac∂{∂μ}(μy_μ(x)+x-μ)\\&=y_μ(x)+μz_μ(x)-1\end{split}and $z(0)=y'(0)=0$, so $z(x)$ is the solution of \begin{cases}z'(x)=g(x,z)≔μ^{-2}(e^{μx}-1-μx)+μz(x)\\z(0)=0\end{cases}This has a unique solution by Picard's theorem since $g(x,z)$ is Lipschitz in $z$.
   2. 1. Fix ${|μ|}⩽m$, (‡) has a unique solution $y_μ:[a-h, a+h] →[b-k, b+k]$ by Picard.
         Consider $f(x,y,μ)={|μ|}y(x)+x-{|μ|},y(0)=1$. $f$ is Lipschitz in $μ$.
         By (a) $y_μ(x)=1+μ^{-2}(e^{|μ|x}-1-|μ|x)$ is not differentiable at $μ=0$.
      2. (Gronwall’s inequality)Suppose $A≥0$ and $b≥0$ are constants and $v$ is a non-negative continuous function satisfying\[v(x)⩽b+A\left|\int_a^xv(s)\mathrm{~d}s\right|\]Then $v(x)⩽b\exp(A{|x-a|})$.\begin{split}|y_μ(x)&-y_η(x)\mmlToken{mi}|=\left|\int_a^xf(s,y_μ(x),μ)-f(s,y_η(x),η)\mathrm{~d}s\right|\\&⩽\left|\int_a^x\left|f(s,y_μ(x),μ)-f(s,y_μ(x),η)\right|+\left|f(s,y_μ(x),η)-f(s,y_η(x),η)\right|\mathrm{~d}s\right|\\&⩽\left|\int_a^xΛ\left|μ-η\right|+L\left|y_μ(x)-y_η(x)\right|\mathrm{~d}s\right|\end{split}Let $v(x)=Λ{|μ-η|}+L\left|y_μ(x)-y_η(x)\right|$, we get\begin{split}\left|y_μ(x)-y_η(x)\right|&⩽\left|\int_a^xv(s)\mathrm{~d}s\right|\\⇒v(x)&⩽Λ{|μ-η|}+L\left|\int_a^xv(s)\mathrm{~d}s\right|\end{split}By Gronwall’s inequality\begin{split}v(x)&⩽{|μ-η|}\mathrm{e}^{L{|x-a|}}\\⇒\left|y_μ(x)-y_η(x)\right|&⩽\fracΛL{|μ-η|}\left(\mathrm{e}^{L{|x-a|}}-1\right)\end{split}
      3. In (†) $f(x,y,μ)=μ(y-1)+x,{|x|}⩽h,{|y-1|}⩽k,{|μ|}⩽m$.
         $M=\max_{(x,y,μ)∈R}{|f(x,y,μ)|}=mk+h$
         ${|f(x,u,μ)-f(x,v,μ)|}={|μ|}{|u-v|}⩽m{|u-v|}$, so $L=m$
         ${|f(x,y,μ)-f(x,y,η)|}={|y-1|}{|μ-η|}⩽k{|μ-η|}$, so $Λ=k$
         Fix $h>0$, we need $Mh⩽k⇒(mk+h)h⩽k⇒m⩽\frac 1h-\frac hk$
         $h,k>0⇒m<\frac1h$. So the maximal domain of μ is $\left(-\frac1h,\frac1h\right)$.
2. 1. 1. The solution surface Σ parametrized $(x,y,z(x,y))$ has normal vector $𝐧=(-z_x,-z_y,1)$ \begin{split} P(x, y) z_x+Q(x, y) z_y=R(x, y, z)&⇔(P(x,y),Q(x,y),R(x,y,z))⋅𝐧=0\\ &⇔(P(x,y),Q(x,y),R(x,y,z))\text{ is tangent to Σ} \end{split}Starting from a point $(x_0(s), y_0(s), z_0(s))$ we look for a curve $(x(t), y(t), z(t))$ whose tangent is $(P,Q,R)$\begin{cases}(\dot x(t),\dot y(t),\dot z(t))=(P,Q,R)\\(x(0),y(0),z(0))=(x_0(s), y_0(s), z_0(s))\end{cases}
      2. Cauchy data means $Py_0'(s)-Qx_0'(s)≠0$ for $(x,y)$ on the data curve $(x_0(s),y_0(s))$.
         Geometrically the characteristic projections are not tangent to the data curve,
         since their tangents $(P,Q),(x_0'(s),y_0'(s))$ are linearly independent.
   2. 1. $Py_0'(s)-Qx_0'(s)=-s≠0$ for $s∈[p,q]$, so the data is Cauchy.
      2. Solve the characteristic equations\begin{array}l(\dot x,\dot y)=(-y,x)&⇒(x,y,z)=(A\cos(t+B),A\sin (t+B)) \\(x(0),y(0))=(s,0)&⇒A=s,B=0 \end{array} $\begin{array}l\dot z(t)=y\sqrt{x^2+y^2}=s^2\sin(t)\\z(0)=0\end{array}\Bigg\}⇒z(t)=s^2-s^2\cos t=x^2+y^2-x\sqrt{x^2+y^2}$
         $(x(t),y(t))=(s\cos(t),s\sin(t))⇒$Solution is defined on the annulus $p^2<x^2+y^2<q^2$
      3. $\dot z(t)=1,z(0)=0⇒z=t⇒\tan z=y/x$
         Restrict $t$ to an interval of length 2π to obtain a single-valued solution eg. $t∈(-π,π]$
   3. 1. Data curve is circular arc of radius 2, center $(3,1)$
         Data is Cauchy when $α≤\arccos\frac23$ so that characteristic projections only intersect once
         Domain of definition is swept out by characteristic projections. This does depends on $R$: we require that $z$ has no singularities.
      2. A line through $(0,1)$ is $y=1+cx$. Let $y=\dot y$.
         Differentiating $y=1+cx$ we get $y=c\dot x⇒\dot x=\frac yc=\frac{xy}{y-1}$
         corresponding to the PDE $\frac{xy}{y-1}z_x+yz_y=R(x,y,z)$.
3. 1. 1. A point $(x_0,y_0)$ is a critical point if $X(x_0, y_0)=Y(x_0, y_0)=0$. So a critical point is a particular trajectory corresponding to solution $(x(t), y(t))$ that are constant in time.
      2. Given initial values, a solution defines a path in the phase plane called a trajectory.
      3. If $(x(t), y(t))$ and $(\tilde{x}(t), \tilde{y}(t))$ are trajectories through $(x_0,y_0)$
         By time-shift we set $(x(0), y(0))=(\tilde{x}(0), \tilde{y}(0))=(x_0,y_0)$. By Theorem 1.6. (Picard’s existence theorem for systems), the two solutions must be the same.
   2. 1. $x=\cos(t),y=\sin(t)⇒\dot x=-\sin(t)=-y,\dot y=\cos(t)=x$ satisfies the equations
      2. $X(0,0)=Y(0,0)=0⇒(0,0)$ is a critical point.\[M=\pmatrix{\frac{∂X}{∂x}&\frac{∂X}{∂y}\\\frac{∂Y}{∂x}&\frac{∂Y}{∂y}}=\pmatrix{a(1-y^2)-3ax^2&1-2xy\\-1-2axy&1-x^2-3y^2}\]At $(0,0)$, $M=\pmatrix{a&1\\-1&1}$ has eigenvalues $λ_1,λ_2$
         $λ_1+λ_2=λ_1λ_2=a+1,(λ_1-λ_2)^2=(a+1)(a-3)$
         For $a<-1$, $λ_1λ_2<0$, so $(0,0)$ is a saddle.
         For $a=3,λ_1=λ_2=2>0,M=\pmatrix{3&1\\-1&1}≠2I$, so $(0,0)$ is an unstable inflected node.
         For $a>3$, $λ_1>0,λ_2>0$, so $(0,0)$ is an unstable node.
         For $-1<a<3$, $λ_1=\overline{λ_2}∉ℝ$, $\operatorname{Re}λ_1=\frac{a+1}2>0$, so $(0,0)$ is an unstable spiral.
      3. $Y=0⇒y=a x\left(1-x^2-y^2\right)$……(1)
         $X=0⇒x+y\left(1-x^2-y^2\right)=0$, by (1) we get $x\left[1+a\left(1-x^2-y^2\right)^2\right]=0$
         For $a>0$, $[…]>0$, so $x=0$, by (1) $y=0$, so $(0,0)$ is the only critical point.$$r^2=x^2+y^2⇒r\dot r=x\dot x+y\dot y=(ax^2+y^2)\left(1-r^2\right)\tag2$$For initial point inside the circle, $\dot r>0$, but the trajectory can't intersect the trajectory $r=1$, so it tends towards $r=1$$$θ=\tan^{-1}\frac yx⇒\dotθ=\frac{x\dot y-y\dot x}{r^2}=1+(1-a)xy(r^{-2}-1)\tag3$$as $r→1$, $\dotθ→1$ it keeps rotating.
      4. when $a=1$ by (3) $\dotθ=1⇒θ=t+θ_0$
         by (2) $1={\dot r\over r^2(1-r^2)}=\dot r\left(\frac1{1-r^2}+\frac1{r^2}\right)$ Integrating from 0 to $t$\begin{split}t&=\left(\operatorname{tanh}^{-1}(r)-\frac1r\right)-\left(\operatorname{tanh}^{-1}(r_0)-\frac1{r_0}\right)\text{ for }r_0<1\\t&=\left(\operatorname{tanh}^{-1}\frac1r-\frac1r\right)-\left(\operatorname{tanh}^{-1}\frac1{r_0}-\frac1{r_0}\right)\text{ for }r_0>1\end{split}So the trajectory is\begin{split}θ-θ_0&=\left(\operatorname{tanh}^{-1}(r)-\frac1r\right)-\left(\operatorname{tanh}^{-1}(r_0)-\frac1{r_0}\right)\text{ for }r_0<1\\θ-θ_0&=\left(\operatorname{tanh}^{-1}\frac1r-\frac1r\right)-\left(\operatorname{tanh}^{-1}\frac1{r_0}-\frac1{r_0}\right)\text{ for }r_0>1\end{split}Sketch for $θ_0=0$ and $r_0=0.2,1,2$
   3. Dividing two equations to eliminate $t$\[[y+x(1-x^2-y^2)]dy+[x+y(1-x^2-y^2)]dx=0\]Integrating\[e^{-2 x y}(1-x^2-y^2) = \text{constant}\]Contour plot
      Poincare-Bendixson Theorem
      Consider new coordinates $(u,v)=(x+y,x-y)$\begin{cases}\dot u=\dot x+\dot y=v-v\left(1-x^2-y^2\right)=v(x^2+y^2)=v\frac{u^2+v^2}2≕U(u,v)\\\dot v=\dot x-\dot y=-u-u\left(1-\frac{u^2+v^2}2\right)≕V(u,v)\end{cases}and the unit disk $x^2+y^2≤1$ becomes $D:u^2+v^2≤2$
      Consider $\tilde u(t)=u(-t)$ and $\tilde v(t)=-v(-t)$. We have\begin{split}\dot{\tilde u}&=-\dot u(-t)=-v(-t)\frac{u(-t)^2+v(-t)^2}2=\tilde v(t)\frac{\tilde u(t)^2+\tilde v(t)^2}2=U(\tilde u,\tilde v)\\\dot{\tilde v}&=\dot v(-t)=-u(-t)-u(-t)\left(1-\frac{u(-t)^2+v(-t)^2}2\right)=-\tilde u-\tilde u\left(1-\frac{\tilde u^2+\tilde v^2}2\right)=V(\tilde u,\tilde v)\end{split}For any $(u_0,0)∈D,u(0)=u_0,v(0)=0$ by uniqueness of solution $(u(t),v(t))=(\tilde u(t),\tilde v(t))$ i.e. any solution passing through $(u_0,0)$ is symmetric about the $u$-axis. An analogous argument using $(-u(-t),v(-t))$ shows that a solution passing through $(0,v_0)$ is symmetric about the $v$-axis.
      Finally, a solution starting at $(u_0,0),u_0>0$, by the form of $U,V$, will enter the quadrant $u>0,v<0$ and both $u(t)$ and $v(t)$ will be decreasing. We know the solution stays in $D$.
      At some small $t_0>0$ we thus have $0<u(t_0)<u_0$ and $v(t_0)<0$.
      Thereafter $\dot u=v\frac{u^2+v^2}2<v⋅\frac{v^2}2<\frac{v(t_0)^3}2$ is bounded from above by a negative number, so the solution crosses the $v$-axis in finite time. This shows trajectories are closed and symmetric about both $v$- and $u$-axis.