35  Problem set 5 solutions

Q1. Sensitivity of the climate

  1. Consider a perturbation of the solar radiation, say \(Q = Q_0 + \delta\) where \(\delta\) is small in comparison to \(Q_0\). Expand now the temperature into a series: \[ T = T_0 + \delta T_1 + \ldots \] Show that at \(O(\delta)\), the perturbation is governed by \[ C \frac{\mathrm{d}T_1}{\mathrm{d}t} = (1 - a(T_0)) - B(T_0) T_1 - A'(T_0) T_1 - T_0 B'(T_0) T_1 - Q_0 a'(T_0) T_1. \]
Solution

Each of the functions should be expanded in the usual way. For example, from Taylor series we have \[ A(T) = A(T_0) + A'(T_0)(T - T_0) + O(T - T_0)^2. \] Then using \(T = T_0 + \delta T_1 + \ldots\), we have \[ A(T) = A(T_0) + \delta A'(T_0) T_1 + O(\delta)^2. \] Do the same for both \(a(T)\) and \(B(T)\). Then substituting into the equation gives (remember to drop bars): \[ \text{LHS} = \delta C \frac{\mathrm{d}T_1}{\mathrm{d}t} + O(\delta^2) \] while for the right hand-side

\[\begin{align} \text{RHS} &= Q_0(1 - a(T_0)) - (A(T_0) + B(T_0) T_0) \\ & \quad + \delta \left[(1 - a(T_0)) - Q_0 a'(T_0)T_1 - A'(T_0)T_1 - B'(T_0)T_0 T_1 - B(T_0) T_1\right] + O(\delta)^2. \end{align}\]

Now the first, \(O(1)\), grouping of terms above is zero by assumption that \(T_0\) is steady-state. So considering only those terms that are multiplied by \(\delta\), we have \[ C \frac{\mathrm{d}T_1}{\mathrm{d}t} = (1 - a(T_0)) - B(T_0) T_1 - A'(T_0) T_1 - T_0 B'(T_0) T_1 - Q_0 a'(T_0) T_1. \]

  1. Consequently, show that the temperature perturbation can be written as \[ B(T_0) \tau \frac{\mathrm{\partial}T_1}{\mathrm{\partial}t} = [1 - a(T_0)] - \frac{B(T_0)}{g} T_1. \]
Solution

There is nothing tricky about this, but it just requires keeping track of bookwork. On the right hand-side, separate those terms proportional to \(T_1\) and write it in the desired form. Refer to the lecture videos where we completed this (modulo a possible sign!)

  1. Consider (Equation 28.3) at steady state, so therefore the perturbed equilibrium temperature is equal to \[ \delta T_1 = \frac{[1 - a(T_0)] \delta g}{B(T_0)}. \] If the CO2 level in the atmosphere doubles, then the radiative forcing might be adjusted as: \[ (1 - a(T_0)) \delta = 3.7 \, \mathrm{W} \cdot \mathrm{m}^{-2}. \] Assuming that the climate gain is \(g = 3\) and \(B(T_0) = 1.9 \mathrm{W} \cdot \mathrm{m}^{-2} \cdot ({}^\circ \mathrm{C})^{-1}\), what is the expected increase in temperature?
Solution

So we know that \((1 - a)\delta = 3.7 W/m^2\). So according to the equation, we have the fact that \[ \delta T_1 = 3.7 \frac{3}{1.9} \approx 5.8^\circ C. \] So this is the expected increase in temperature. Note that if a climate gain factor of \(g = 1\) is used then this gives an expected increase of 1.9 degrees instead.. The range of current model predictions, in fact, is about 1.5 to 4.5 degrees so this is not bad.

Q2. A van der Pol equation

\(\nextSection\)
  1. Determine \(f(x)\) so that this equation can be written as a Liénard phase plane system in the form

\[\begin{align} \epsilon \dot{x} &= f(x) + 4y, \\ \dot{y} &= -x. \end{align}\]

Solution

Start from the ODE and re-write \[ \frac{\mathrm{d}}{\mathrm{d}t} \left[\epsilon \dot{x} + (x^3 - 3x^2 - 9x)\right] = - 4x. \] Double check that the derivative of the LHS returns the original problem. Now set the quantity in brackets equal to one unknown: \[ \tilde{y} = \epsilon \dot{x} + (x^3 - 3x^2 - 9x), \] and solve for the derivative: \[ \epsilon \dot{x} = \tilde{y} - (x^3 - 3x^2 - 9x). \] The remaining equation is \[ \dot{\tilde{y}} = -4x. \]

Now in this question, we have chosen a slightly different choice to eliminate the factor of \(4\). Set \(\tilde{y} = 4y\) to get the system

\[\begin{align} \epsilon \dot{x} &= f(x) + 4y, \\ \dot{y} &= -x. \end{align}\]

It doesn’t really matter which choice you select. So here we see \(f(x) = -x^3 + 9x + 3 x^2\). What we often call the slow manifold in lectures is then \[ y = -f(x)/4 = \frac{x^3 - 3 x^2 - 9x}{4} \equiv S(x). \]

  1. For fixed \(\epsilon > 0\), find the equilibrium point(s) in the phase plane, find their eigenvalues, and classify their linear stability.
Solution

The only equilibrium point is \((x_*, y_*) = (0,0)\). Because the equilibrium point is at the origin, we can obtain the linear matrix by ignoring the higher-order powers in the differential equations. The matrix is \[ A = \begin{pmatrix} \frac{9}{\epsilon} & \frac{4}{\epsilon} \\ -1 & 0 \end{pmatrix}. \] Subtract \(\lambda\) from the diagonal elements and solve the resultant quadratic equation \(\lambda(\lambda - 9/\epsilon) + 4/\epsilon = 0\) to see that the eigenvalues are \(\lambda = (9 \pm \sqrt{81 - 16\epsilon}/(2\epsilon) > 0\), so the point is an unstable node.

  1. Use the expansions \(x(t) = x_0(t) + \epsilon x_1(t) + O(\epsilon^2)\) to determine the equations for the leading-order slow solution. Sketch the slow manifold, indicate the directions motion on each part, and identify the two attracting points on the curve.
Solution

As usual expand the solutions into a series in powers of \(\epsilon\). The slow manifold is \(y_0(x_0) = S(x_0) = (x_0^3 - 3 x_0^2 - 9 x_0)/4\). In image of the slow manifold is shown below in blue. Notice that for \(x < 0\), \(y\) is increasing and for \(x > 0\), \(y\) is decreasing (due to the second differential equation).

Figure 35.1: Illustration of slow manifold
  1. Use the expansions \(x(t) = X_0(T) + \epsilon X_1(T) + O(\epsilon^2)\) and \(y(t) = Y_0(T) + \epsilon Y_1(T) + O(\epsilon^2)\) with \(T = t/\epsilon\) to obtain the leading-order fast solution.
Solution

Rescale \(t = \epsilon T\) and re-write \(x = X(T)\) and \(y = Y(T)\). We get \[\begin{align} \frac{\mathrm{d}X}{\mathrm{d}T} &= f(X) + 4Y \\ \frac{\mathrm{d}Y}{\mathrm{d}T} &= -\epsilon X. \end{align}\] Now expand into powers of \(\epsilon\) and take the leading-order: \[\begin{align} \frac{\mathrm{d}X_0}{\mathrm{d}T} &= f(X_0) + 4Y_0 \\ \frac{\mathrm{d}Y_0}{\mathrm{d}T} &= 0. \end{align}\]

The second equation indicates that \(Y_0\) is constant. Its value, found in part (e) is either \(5/4\) or \(27/4\). The remaining equation for \(X_0\) is nonlinear and there is not much you can do with it. The important item of understanding is to examine its form: \[ \frac{\mathrm{d}X_0}{\mathrm{d}T} = f(X_0) + 4 Y_0. \] Now on the upper branch of the picture shown above \(f(X_0) = 4 Y_0 > 0\) and therefore the solution increases its value in \(X_0\) until it intersects with the blue slow manifold, at which point its rate of change is zero and it descends the slow manifold. A similar behaviour occurs on the lower branch of the fast evolution.

  1. Use the phase plane to determine the maximum and minimum values of \(x(t)\) during an oscillation. Sketch \(x(t)\) as a function of time.
Solution

By checking the derivatives, we can see that the slow manifold has local extrema at \((-1, 5/4)\) and \((3, 27/4)\). In the fast evolution stages, \(y\) is constant and \(x\) evolves from one extrema value to the other on the slow manifold, satisfying \(y = S(x)\).

Here is a plot of the key points.

  1. Using the time required for the slow motions in part c (neglecting the short times for the fast solutions in part d), determine the leading-order estimate for the period \(P\) of oscillation for the limit cycle.
Solution

In essence, we approximate the period only using the time during the slow phases. Begin from the slow manifold, \[ y = S(x) \] and remembering that both \(x\) and \(y\) are functions of time, we differentiate in \(t\). This gives (we drop the subscripts of zero): \[ 4\dot{y} = \dot{x}(3 x^2 - 6x - 9). \] From the second equation, we know \(\dot{y} \sim -x\). Inserting and solving now yields \[ \dot{x} = g(x) \equiv - \frac{4x}{4x^2 - 6x - 9}. \] This is effectively a single differential equation for \(x = x(t)\).

Now we can solve for time by separating and integrating. We get \[ t = \int^x \frac{dx'}{g(x')} \]

Therefore, to estimate the time spent during different phases, we only need to substitute in initial and final values of \(x\) in the above integral and integrate. The first slow phase starts at \(x = 5\) and ends at \(x = 3\). Thus \[ T_1 = \frac{1}{4} \int_{5}^3 \left(\frac{9}{x} + 6 - 3x\right) dx = 3 + \frac{9}{4}\log(3/5). \] The second slow phase starts at \(x = -3\) and ends at \(x = -1\). Thus \[ T_2 = \frac{1}{4} \int_{-3}^-1 \left(\frac{9}{x} + 6 - 3x\right) dx = 6 - \frac{9}{4}\log(3). \] Adding the two gives that the period is approximately \(P \sim 9 - (9/4) \log 5 \approx 5.38\).

Q3. Fast-slow dynamics with three variables

\(\nextSection\)
  1. For the system \[ \begin{aligned}[c] \dot{x} &= 2 - y, \\ \dot{y} &= x-z, \\ \epsilon\dot{z} &= y - y^2 + \frac{1}{3}y^3 - z, \end{aligned} \] with the initial conditions of \(x(0) = 1\), \(y(0) = 3\), and \(z(0) = 0\).

Identify the surface \(z = S(x, y)\) that defines the slow manifold. Find the equilibrium point of the leading=order slow phase plane system and show that it is asymptotically stable for \(t \to \infty\). Also determine the form of the initial layer that describes the transition from the initial conditions to the slow manifold.

Solution

Setting \(\epsilon = 0\) we see that the slwo manifold is \(z = S(y) = y - y^2 + y^3/3\). The leading-order slow system is

\[\begin{align} \frac{\mathrm{d}x_0}{\mathrm{d}t} &= 2 - y_0 \\ \frac{\mathrm{d}y_0}{\mathrm{d}t} &= x_0 - y_0 + y_0^2 - y_0^3/3. \end{align}\] The only equilibrium point is \((x^*, y^*) = (2/3, 2)\). Linearisation gives eigenvalues of \(\lambda = 1/2(-1 \pm 3i)\) so the point is a stable spiral. Consequently the \(t \to \infty\) solution will be \((x, y, z) = (2/3, 2, 2/3)\). Note also that \(2/3 = S(2)\).

Near the initial condition there is a boundary layer, found by setting \[ t = \epsilon T, \] and transforming \(x = X(T)\), \(y = Y(T)\), \(z = Z(T)\). We get at leading order,

\[\begin{align} \frac{\mathrm{d}X_0}{\mathrm{d}T} &= 0 \\ \frac{\mathrm{d}T_0}{\mathrm{d}T} &= 0 \\ \frac{\mathrm{d}Z_0}{\mathrm{d}T} &= Y_0 - Y_0^2 + Y_0^3/3 - Z_0. \end{align}\]

The initial condition sets \(X_0 = 1\) and \(Y_0 = 3\). We can finally solve the equation for \(Z_0\) to get \[ Z_0 = 3(1 - e^{-T}). \]

  1. For the system \[ \begin{aligned}[c] \dot{x} &= 2 - y, \\ \epsilon\dot{y} &= x-z, \\ \dot{z} &= y - y^2 + \frac{1}{3}y^3 - z, \end{aligned} \] with the initial conditions of \(x(0) = 0\), \(y(0) = 3\), and \(z(0) = 1\).

Show that the slow manifold reduces to a curve that could be written in parametric form as \(x = x(z), y = y(z), z = z\). Determine the asymptotic solution for \(t \to \infty\). Also determine the form of the initial layer that describes the transition from initial conditions to the slow manifold.

Solution

Setting \(\epsilon = 0\), we see that the second equation gives \(x = z\). Now if you use this result in the first equation you obtain \(\dot{z} = 2 - y\). This is now equated with the third equation which yields an algebraic equation: \[ 2 - y = y - y^2 + \frac{y^3}{3} - z \Longrightarrow z = \frac{y^3}{3} - y^2 + 2y - 2. \tag{35.1}\] Therefore we conclude that \(z = z(y)\). We can verify that this equation is monotonically increasing in \(y\), so we can invert without issue, yielding \(y = y(z)\). Similarly, we obtain \(x = z\). So the slow manifold is given by \[ (x, y, z) = (z, y(z), z), \] which is a curve in 3D.

In order to determine what happens as \(t\to\infty\), differentiate \[ \frac{\mathrm{d}y}{\mathrm{d}t} = \frac{\mathrm{d}y}{\mathrm{d}z} \frac{\mathrm{d}z}{\mathrm{d}t} = \frac{2 - y}{2 - 2y + y^2}, \] which follows from using the first equation of the system of ODEs and also the \(z\) derivative of (Equation 36.1). In particular, note that we do not have to invert this latter equation and we can take the \(y\)-derivative of (eq-PS5-Q3-tmp) and use \[ \frac{1}{\frac{\mathrm{d}z}{\mathrm{d}y}} = \frac{1}{2 - 2y + y^2}. \]

Take \(dy/dt = 0\) for the steady state gives \[ y^* = 2, \] as the only fixed point. At this value of \(y\), notice that \(z^* = 2/3\) from (Equation 36.1). So the fixed point is \[ (x^*, y^*, z^*) = (2/3, 2, 2/3). \]

The procedure for the initial layer is the same as in part (a). This time, the fast system satisfies at leading order,

\[\begin{align} \frac{\mathrm{d}X_0}{\mathrm{d}T} &= 0 \\ \frac{\mathrm{d}Y_0}{\mathrm{d}T} &= X - Z \\ \frac{\mathrm{d}Z_0}{\mathrm{d}T} &= 0. \end{align}\]

So we obtain \(X_0 = 0\) and \(Z_0 = 1\) using the initial condition. Solving for the final equation for \(Y_0\) with \(Y_0(0) = 3\) gives \[ Y_0 = -T + 3. \]