37 Problem set 5 solutions

Q1. Evolution

Consider again (Equation 29.1). Let \(T^*\) be a steady-state solution and set \(T = T^* + u(t)\) where \(u(t)\) is a small perturbation from the steady state.

Show that the perturbation satisfies \[ C \dot{u} = -D u + O(u^2). \] and hence solve for the general solution of the leading-order perturbation (ignoring quadratic terms). What are the conditions on \(T^*\) so that the steady state is linearly stable?

Solution Q1a

We linearise via \(T = T^* + u(t)\) where \(u \ll 1\). Then we have \[ C \frac{\mathrm{\partial}T}{\mathrm{\partial}t} = G(T^*) + G'(T^*)u(t) + O(u^2). \] The first term on the RHS is zero since \(T^*\) is a steady state. Ignoring quadratic terms, we see then that \[ C \dot{u} = -D u \] where \(D = -G'(T^*)\). The solution is thus \[ u(t) = A e^{-D t/C}, \] where \(A\) is a constant. In order to be linearly stable, we then need the above to tend to zero as \(t \to \infty\), which is guaranteed if \(D/C > 0\). Since \(C > 0\) then we only need \[ G(T^*) < 0. \]

Assuming \(T^*\) is linearly stable, find the typical response time to a perturbation. For instance, what is the time it takes for the perturbation to reach the value \(u(t) = 0\) if \(u(0) = 1\)? How does this response time change with \(C\)? What is the physical interpretation of this regarding the climate?

Solution Q1b

The question is somewhat ill-posed. By the behaviour of the system, we know that it would take infinite time for the perturbation, \(u\), to reach \(u = 0\). However, we can still provide a sensible answer.

If the system starts from \(u(0) = 1\), then to reach \(u = \delta\), it would need to take the time \[ e^{-D t/C} = \delta \Longrightarrow t = -\frac{C}{D} \log \delta = \frac{C}{D} \log(1/\delta). \] For instance, to reach \(\delta = 1/2\) (half its original perturbation) would take \(\log 2 \approx 0.69\) times (C/D).

This response is proportional to \(C\) so for instance, doubling \(C\) takes twice as long as before to reach the same value of \(\delta\). One can think of this as equivalent to decreasing the responsiveness of the planet to potential changes of stability.

Q2. Integral of energy over the planet

Solution Q2

This is a straightforward calculation without tricks. Since \(Q\) is constant, we have \[ I = 2\pi R_E^2 Q\int_{\varphi = -\pi/2}^{\pi/2} s(y = \sin\varphi) \cos\varphi \, \mathrm{d}\varphi. \] Converting now to integration in \(y\), we have \[ I = 2\pi R_E^2 Q\int_{y = -1}^{1} s(y) \, \mathrm{d}y. \] Examining the function \(s(y)\) in Equation 14.3 we note that it is symmetric in \(y\) and has been defined so its integral is one. Thus \[ I = 4\pi R_E^2 Q. \] This is the solar constant multiplied by the surface area of the planet (a sphere).

Q3. Mean temperature in the latitude-dependent EBM

Solution Q3a

We have from integration, \[ C \frac{\mathrm{\partial}}{\mathrm{\partial}t} \int_0^1 T \, \mathrm{d}y = Q\int_0^1 s(y) \, \mathrm{d}y - Q \int_0^1 s(y)a(y) \, \mathrm{d}y - (A + B\bar{T}) + k\bar{T} - k\bar{T}. \] Simplifying and using the property of \(s\) gives \[ C \frac{\mathrm{\partial}\bar{T}}{\mathrm{\partial}t} = Q(1 - \bar{a}) - (A + B\bar{T}), \] where we have defined \[ \bar{a} = \int_0^1 s(y)a(y) \, \mathrm{d}y = \int_0^{y_s} s(y) a_w \, \mathrm{d}y +\int_{y_s}^1 s(y) a_i \, \mathrm{d}y. \]

Solution Q3b

Again this question is free of tricks and only involves the integration using \[ s(y) = 1 - S_2 P_2(y) \] where \(S_2 = 0.482\) and \(P_2(y) = \frac{1}{2}(3y^2 - 1)\). It is useful to compute the two quantities

\[\begin{align} \int_0^{y_s} s(y) \, \mathrm{d}y &= y_s - \frac{S_2}{2} \left(y_s^3 - y_s\right), \\ \int_{y_s}^1 s(y) \, \mathrm{d}y &= 1 - y_s + \frac{S_2}{2} \left(y_s^3 - y_s\right). \end{align}\]

Multiply the first quantity by \(a_w\) and add to the second multiplied by \(a_i\) and simply to obtain \[ \bar{a} = a_i + (a_w - a_i)y_s \left[ 1 - \frac{S_2}{2}(y_s^2 - 1)\right]. \]

Q4. Sensitivity of the climate

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 Q4a

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 \[ C \frac{\mathrm{d}T_1}{\mathrm{d}t} = (1 - a(T_0)) - B(T_0) T_1 - A'(T_0) T_1 - T_0 B'(T_0) T_1 - Q_0 a'(T_0) T_1. \]

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 Q4b

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!)

Consider (Equation 30.3) 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 Q4c

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.