Non-convex Control Problems`Convex' Control ProblemsThe Shallow Lake ProblemOptimal Control of the Shallow LakeA Close Up of the Steady StatesThe Optimal Consumption PathsNon-convex Control Problems ARE 263 - Fall 0810 Non-convex Control ProblemsThis section is a hybrid of three papers with some minor changes. Thesepapers are all published in a special issue of the Journal Environmental andResource Economics (vol 26) on the economics non-convex ecosystems. Thearticles are Dasgupta & M¨aler (2003), Brock & Starrett (2003), and M¨aler,Xepapadeas & de Zeeuw (2003). Before we discuss a non-convex controlproblem, let me briefly review what constitutes a convex problem.10.1 ‘Convex’ Control ProblemsSo far⋄ max∞Z0u(c, x)e−ρtdt (1)subject to ˙x = f(c, x), x(0) = x0⋄ H(c, x, λ) = u(c, x) + λf(c, x) (current value)⋄ Necessary conditions for an optimum:∂Hct= 0,∂Hxt= ρλt−˙λt,∂Hλt= ˙xt,and the transversality condition limt→∞e−ρtλt= 0→ candidate(s) for optimum (c∗t, x∗t)t∈[0,∞)In a convex problem we have in addition⋄ Sufficient conditions for an optimum (here Mangasarian):Let λ ≥ 0 and and H(c, x, λ) be concave in (c, x) for all λ.Then the candidate(s) (c∗t, x∗t)t∈[0,∞)are globally optimal paths.Moreover, if H(c, x, λ) is strictly concave in (c, x) for all λ,then (c∗t, x∗t)t∈[0,∞)is unique.1Non-convex Control Problems ARE 263 - Fall 08Note that H(c, x, λ) concave for λ ≥ 0 is equivalent to u(c, x) and f(c, x)being concave in (c, x). Such a control problem is generally referred to as“convex”. Why?!1. Historically people were analyzing min-problems rather than max prob-lems, so that problem (1) turns intominZ∞0−u(c, x)e−ρtdtwith −u convex rather than concave.2. For a typical “capital stock investment ( ˙x) versus consumption (c)”model with free disposal (‘≤’) where the constraint of motion is˙x ≤ F (x) − cwe find that F (x) concave implies that the set of feasible programs isconvex:Let (c◦t, x◦t)t∈[0,∞)and (c⋄t, x⋄t)t∈[0,∞)be two feasible programs. Then{˜ct, ˜xt}t∈[0,∞)with˜ct= γ c◦t+ (1 − γ) c⋄t˜xt= γ x◦t+ (1 − γ) x⋄tis feasible as well for all λ ∈ [0, 1].Proof: By feasibility of {c◦t, x◦t}t∈[0,∞)and {c⋄t, x⋄t}t∈[0,∞)we haveγ ˙x◦t≤ γ F (x◦t) − γ c◦t(1 − γ) ˙x⋄t≤ (1 − γ) F (x⋄t) − (1 − γ) c⋄tand adding these up we find˙˜xt= γ ˙x◦t+ (1 − γ) ˙x⋄t≤ γ F (x◦t) + (1 − γ) F (x⋄t) − (γ c◦t+ (1 − γ) c⋄t)≤ F (γ x◦t+ (1 − γ) x⋄t) − (c◦t+ (1 − γ) c⋄t)≤ F (˜xt) − ˜ct.2Non-convex Control Problems ARE 263 - Fall 08The second inequality is due to concavity of F (x). Thus (˜ct, ˜xt)t∈[0,∞)is feasible.10.2 The Shallow Lake ProblemShallow lakes are sen sitive to phosphorus inflow (loading) caused by fertilizeruse in the agricultural sector. For low amounts of phosphorous the ecologyof the shallow lake gives a positive feedback to an increase in phosphorousloading.1An increase in the phosphorous stock triggers the growth of algaeand can cause a loss of oxygen in the water. The water becomes turbid andthe ecosystem changes (to the distress of trouts, fishermen, and swimmers).Such a state is called eutrophic. A state of the lake with clear water isreferred to as oligotrophic. In small ponds the change from an oligotrophicstate to a euthrophic state can happen within a day, in lakes within weeks.The following analysis will explain the underlying dynamics.Definitions:• x(t): Stock of phosphorous (suspended in algae) in the lake• c(t): Loading=phosphorous inflow from watershed (agriculture)• α: (Self-) Purification rate of the lake• f(x): feedback, increases for small phosphorous stock → convex-concave.We assume limt→∞f′(x) = 0. An example that we will use for numericanalysis is: f (x) =x21+x2.Lake dynamics:˙xt= ct− αxt+ f(xt) ≡ X(ct, xt) (2)Thu s, a lake with a constant loading c is in equilibrium if˙xt= 0⇔ αxt− ct= f(xt) (3)⇔ ct= αxt− f(xt) ≡ h(x) (4)1Another way to think about it is that for low stocks and low flows of phosphoroussome mechanisms in the lake hold back part of the phosphorous from going into the waterand producing algae. This ‘holding back effect’ decreases as the stock increases (while thenatural purification rate increases as the stock of phosphorous increases).3Non-convex Control Problems ARE 263 - Fall 08Figure 1 depicts the two sides of equation (3) for different loadings c. Without0.51.01.52.0x0.20.40.60.81.0Αx-c , fHxLFigure 1 : Shallow lake model. Feedback f (x) curve and ‘purification net loading’ αx − ccurve for c = 0 (straight line), c = .05 (dashed line) and c = .1 (dotted line). Thepurification parameter determines the slope of the line, here it is α = .52. The feedbackfunction is f =x21+x2.any phosphorous inflow from the watershed we have a unique equilibrium ata zero phosphorous stock. With a small inflow (dashed line) we find threeequilibria and history determines which one the lake is in. Note that youcan read off t he phosphorous stock as the absolute of the intersection withthe y-axis. If we start with a zero phosphorous stock and slowly load thelake with a constant rate, the equilibrium stock moves up the f (x) curve aswe increase this constant rate until we reach x1. If we further increase theloading the equilibrium with the low phosphorous stock disappears and thelake will move all the way tox. This move in figure 1 corresponds to themove from a clear to a turbid lake or from an oligotrophic to a eutrophicstate. Note that not only a small difference in the loading can cause a hugedifference in the equilibrium phosphorous stock, but also a decrease of theloading does not immediately take us back to x1. If we start reducing theloading we slowly move down the f(x) curve until we reach x2. This pointcorresponds to a significantly lower loading than the one corresponding tox1, bu t still to a higher phosphorous stock. Only if we decrease the loadingfurther we ‘jump’ back to t he lower part of the f(x) curve and reach a newequilibrium at x. This effect is called hysteresis.4Non-convex Control Problems ARE 263 - Fall 08The scenario described above is the one that we would like to analyzein more detail and from an economic perspective. However, I would like topoint out that two other qualitatively different scenarios are possible if thepurification rate of the lake is