Differential equations 1 paper 2019

1. Consider the following integral equation for $y(x)$ : $$ y(x)=f(x)+∫_a^x K(x, t) y(t) d t \text { for } x ∈[a, b] $$ where $f(x)$ and $K(x, t)$ are smooth and bounded functions on $[a, b]$.
  1. Show that the solution of (1) depends continuously on $f$. [Gronwall's inequality may be used without proof.]
  2. Hence or otherwise show that the solution is unique.
2. Consider now the differential equation system for $y(x)$ : \begin{gathered} y''=x^3 y, \\ y(0)=1, y'(0)=0 . \end{gathered}
  1. Show that if $y(x)$ is a solution of (2), then it is also a solution of the integral equation $$ y(x)=1+∫_0^x(x-t) t^3 y(t) d t $$
  2. Show that a solution to (2) will be unique for all $x$.
3. Consider the integral equation $$ y(x)=f(x)+λ ∫_a^b K(x, t) y(t) d t $$ where $λ$ is a constant, and $f$ and $K$ are given smooth functions satisfying $|f|<M$ on $[a, b]$ and $|K|<L$ on $[a, b] ×[a, b]$. Observe that the integral limits in (3) are independent of $x$.
  1. Formulate a sequence of Picard iterations $\left\{y_n(x)\right\}$ for (3).
  2. Show that $$ \left|y_n-y_{n-1}\right| ⩽ M_n $$ where $M_n$ is a bound depending on $λ$ which you should determine.
  3. Hence give a bound on $|λ|$ for which the sequence $\left\{y_n(x)\right\}$ converges.
  4. Let $f(x)=x^2, K(x, t)=x^2 t, a=0$ and $b=1$ in (3). Compute explicitly the first 2 Picard iterations, and show that in fact the sequence converges for a larger region of $λ$, which you should determine.
1. Consider the following PDE system for $u(x, y)$ : $$ \begin{aligned} & (1+u) u_x+y u_y=u, \\ & u(x, 1)=β x, 0 ⩽ x ⩽ 1, \end{aligned} $$ where $β$ is a constant.
  1. Show that the boundary data is Cauchy data for any value of $β$.
  2. Write down the differential equations satisfied along characteristics and integrate with appropriate boundary conditions to obtain a parametric solution.
  3. Find a sufficient and necessary condition on $β$ such that characteristic projections do not intersect.
  4. Sketch the region (in the x-y plane) where the solution is uniquely defined in the case $β=1$.
2. Suppose that $u=u(x, y, z)$ satisfies $$ u_x+u_y+u_z=u $$ with boundary data $u=x$ on the surface $x+y+2 z=1$ for $x ⩾ 1$.
  1. By adapting the method of characteristics to 3 independent variables, show that the characteristic projections are given by $$ x(r, s, t)=s+t, y(r, s, t)=t+1-s-2 r, z(r, s, t)=r+t . $$
  2. Obtain an explicit solution $u(x, y, z)$.
  3. What is the domain of definition for the solution?
1. Consider the following system for $(x(t), y(t))$ : $$ \begin{aligned} & \dot x=x\left(2-\sqrt{x^2+y^2}\right)-y \\ & \dot y=y\left(2-\sqrt{x^2+y^2}\right)+x \end{aligned} $$ where overdot denotes derivative with respect to $t$.
  1. Explain why this is an autonomous system.
  2. Show that $$ x^2+y^2=4 $$ is a solution trajectory.
  3. If $r^2=x^2+y^2$, show that $r(t)$ satisfies $$ \dot r=r(2-r) $$
  4. Hence show that $x^2+y^2=4$ is a limit cycle.
2. Consider the differential equation system for $u(t), v(t)$ : \begin{aligned} \dot u&=v \\ \dot v&=f(u, v) \end{aligned} where $f$ is a given smooth function satisfying $f(a, 0)=0$ for some constant $a$.
  1. Derive a condition on $f$ for which $(a, 0)$ is: (I) a stable spiral (II) a centre (III) a stable node.
  2. Suppose that $\frac{∂ f}{∂ u}(a, 0)>0$. Show that $(a, 0)$ is an unstable saddle.
  3. Consider now the second order differential equation for $y(t)$ : $$ \ddot y=\left(y^3-1\right) \cos \left(\dot y^2\right) $$ Find all equilibrium points and classify their stability.

1. 1. Parametrise boundary data as $x_0(s)=s,y_0(s)=1,0⩽s⩽1,u_0(s)=βs$.
    $(1+u_0)y_0'-y_0x_0'=-1≠0⇒$Cauchy data for any value of $β$
  2. characteristics satisfy \[\begin{aligned} \dot x&=1+u\\ \dot y&=y\\ \dot u&=u \end{aligned} \text{with} \begin{aligned} x(0)&=s\\ y(0)&=1\\ u(0)&=βs \end{aligned}⇒u=βse^t,y=e^t,x=t+βse^t+s(1-β)\]
  3. characteristics intersect if $J=\begin{vmatrix}x_t&x_s\\y_t&y_s\end{vmatrix}=0⇒(βe^t+1-β)e^t=0⇒e^t={β-1\overβ}$
    Thus we need ${β-1\overβ}≤0$ to ensure $J≠0$, which is true for $0≤β≤1$.
  4. $β=1$ characteristics do not intersect, and $u=se^t$ is well-defined for all $s,t$. Thus, the solution is defined in region spanned by the characteristics at $s=0$ and $s=1$.
    $s=0$: $x=t,y=e^t⇒x=\ln y$
    $s=1$: $x=t+e^t,y=e^t⇒x=y+\ln y$
2. 1. parametrised boundary data: $x_0=s,s⩾1,z_0=r,y_0=1-s-2r,u_0=s$
    characteristics $\begin{array}l\dot x=1\\\dot y=1\\\dot z=1\\\dot u=u\end{array}⇒\begin{array}lx=t+s\\y=t+1-s-2r\\z=t+r\\u=se^t\end{array}$
  2. Eliminate $(s,t,r):s=x-t,r=z-t⇒y=t+1-(x-t)-2(z-t)⇒t=\frac14(x+y+2z-1)⇒u(x,y,z)=\left[x-\frac14(x+y+2z-1)\right]\exp\left(\frac14(x+y+2z-1)\right)$
  3. $u$ has no singularities and the Jacobian$$J=\begin{vmatrix}x_t&x_s&x_r\\y_t&y_s&y_r\\z_t&z_s&z_r\end{vmatrix}=\begin{vmatrix}1&1&0\\1&-1&-2\\1&0&1\end{vmatrix}=-1-3≠0$$Thus the domain of definition is bounded only by the characteristic surface passing through the boundary of data surface $x=1$ on $x+y+2z≥1$.
    Set $s=1⇒x=t+1,y=t-2r,z=t+r$ gives $r=z-t⇒y=t-2(z-t)=3t-2z⇒3t=y+2z⇒3t=3x-3=y+2z$
    From $s≥1$, the domain of definition is $\{(x,y,z)|3x≥3+y+2z\}$
1. 1. Autonomous—no dependence on independent variable $t$ in $\dot x,\dot y$
  2. Take $\dot xx+\dot yy$, we get$$\frac12\frac d{dt}(x^2+y^2)=\dot xx+\dot yy=(x^2+y^2)\underbrace{(2-\sqrt{x^2+y^2})}_{=0\text{ if }x^2+y^2=4}$$Thus on $x^2+y^2=4$, $\frac d{dt}(x^2+y^2)=0$, so it is a solution trajectory.
  3. Define $r$ by $r^2=x^2+y^2$, so\[r\dot r=x\dot x+y\dot y=(x^2+y^2)(2-\sqrt{x^2+y^2})=r^2(2-r)⇒\dot r=r(2-r)\]
  4. Trajectories with $r>2$ has $\dot r<0$ so decrease to $r=2$.
    Trajectories with $r<2$ has $\dot r>0$ so increase to $r=2$.
    And $r=2$ is a solution trajectory, and is therefore a limit cycle.
2. At $\dot u(a, 0)=\dot v(a, 0)=0$, so $(a,0)$ is a critical point.
  1. Define $𝐮=(u,v)$ and let $𝐮=(a,0)+𝛏(x)$
    then $𝛏'(x)=M𝛏+O({|𝛏|}^2)$ with $M=\pmatrix{0&1\\f_u&f_v}\Bigg|_{(a,0)}$
    $𝛏=𝐜e^{λx}⇒(M-λI)𝐜=0⇒0=\det(M-λI)=λ^2-λf_v-f_u⇒λ_{1,2}=\frac{f_v±\sqrt{f_v^2+4f_u}}2$
    (I) a stable spiral if $f_v<0$ and $f_v^2+4f_u<0$
    (II) a centre if $f_v=0$ and $f_u<0$
    (III) a stable node if $λ_{1,2}<0,λ_{1,2}∈ℝ$, so $f_v<0,f_u<0$ and $f_v^2+4f_u>0$
  2. $λ_1λ_2=-f_u<0$, ∴point is a saddle.
  3. Let $u=y,v=\dot y$, we can use previous formulation.
    $f(a,0)=0$ only at $a=1$. We have $f_u(1,0)=3>0$. By part ii) saddle.

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