Skiba without unstable equlibrium in a linear quadratic framework

Skiba without unstable equlibrium in a linear quaratic framework Richar F. Hartl Institute of Management, University of Vienna, Brünnerstraße 8, -0 Vienna, ustria Richar.Hartl@univie.ac.at Tel. +43--477-3809 Fax. +43--477-38094 Peter M. Kort Deartment of Econometrics an CentER, Tilburg University, P.O. Box 9053, 5000 LE Tilburg, The Netherlans Kort@kub.nl May 30, 00 bstract We consier a simle moel in a LQ framework where the revenue funvtion is concave (quaratic) but su ers a convex kink. It is shown that this yiels Skiba-tye behavior although no unstable equilibrium exists. JEL #:C6, C6 reliminary - o not quote

Introuction We consier a simle moel in a LQ framework where the revenue function is concave (quaratic) but su ers a convex kink. It is shown that this yiels Skiba-tye behavior although no unstable equilibrium exists. Moel formulation The moel we consier is the following: max u Z 0 e t [r(k) c(u)] t; () _k u ¹k; k(0) k 0 ; () where k enotes the caital stock an u is investment. Labor is assume to be roortional to caital stock so that it oes not nee to be exlicitly inclue. The revenue function is given by r(k) while the investment costs are c(u). The iscount rate is while ¹ enotes the ereciation rate. lthough Barucci i not imose this constraint, for economic reasons (e.g. rm or inustry seci city) we assume that investments are irreversible: u 0: (3) Following Barucci (998) we assume quaratic revenue an costs functions where - in aition - we assume that the revenue function has a convex kink: r(k) max a L k bk ;a H k bk g ª (4) c(u) cu + u : (5) We require all arameters a L, a H, b, c,, g, ¹, an to be ositive an a L <a H Revenue r(k) reaches its maximum for an there is a kink at oint k max a H b (6) g a H a L (7) We restrict ourselves to the scenario where revenue is increasing for all k<k max, imlying that the kink is before the maximum of the rst exression a L k bk in the revenue function: g < a L a H a L b g < a L (a H a L ) (8) b The relevant region of the roblem is 0 k k max since for larger k revenue ecreases in k, an it is never otimal for the rm to be there.

3 Mathematical nalysis First the necessary conitions for otimality are resente, after which the equilibria are etermine an their stability roerties are stuie. 3. Necessary conitions To etermine the necessary otimality conitions we rst write own the current value Hamiltonian: H a L k bk cu u + q[u ¹k] for k< (9) H a H k bk g cu u + q[u ¹k] for k> (0) Since H uu < 0, maximization of the Hamiltonian yiels a unique solution. It follows that u is continuous over time (see Feichtinger an Hartl 986, Corollary 6.). From the maximum rincile it is obtaine that @H@u 0 i.e. u q c : () If (3) is imose, then () hols only for u>0 i.e. for q>c.otherwisewehave u 0 if q c: () which has the usual economic interretation that it is only otimal to invest, if the shaow rice of caital excees the marginal cost of the rst unit. The ajoint equation is _q q @H@k ( + ¹)q a i +bk: (3) where i L for k< an i H for k> :From (), i.e., q u + c, an(3)we get: _u _q ( + ¹)u + ( + ¹)c a i + b k: (4) We will also use the transversality conition lim t! et q (t) 0 i.e. lim e t u (t) 0: (5) t! 3. Equilibria an their Stability We rst comute the isoclines. The _ k 0-isocline oes not een on whether k< or k> an is a straight line: u ¹k (6) an the _u 0-isocline oes een on k< or k> via the arameter a i.itisgiven by u ( + ¹) [a i ( + ¹)c bk]: (7) 3

Hence, the _u 0-isocline is linear an ecreasing in k with an uwar jum of size a H a L at k. (+¹) From (6) an (7) we erive that the steay states must satisfy: ¹k i a i ( + ¹) c ¹ ( + ¹)+b : (8) for i L; H. In orer to have multile steay states, we must assume because otherwise ¹ k L becomes negative. We must also verify that i.e., a L > ( + ¹) c (9) ¹k L < < ¹ k H ; (0) (a L ( + ¹) c)(a H a L ) ¹ ( + ¹)+b <g< (a H ( + ¹) c)(a H a L ) : () ¹ ( + ¹)+b This must be comare to (8), i.e. we nee that (a L(+¹)c)(a H a L ) < a L(a H a L ) which is ¹(+¹)+b b clearly the case. We conclue that uner the assumtion (8) an (9) we have multile steay states satisfying (0). The stability of the equilibria follows from the general observation that an equilibrium is a sale oint if an only if the erivative of the k _ 0-isocline is larger than the erivative of the _u 0-isocline there: 3 @ k _ 6 et 4 @k @ _u @k @ k _ @u @ _u @u 7 5 < 0 i u k _k0 @ _ k @k @ _ k @u @ _u @k > @ _u @u u k since @ k _ @ _u > 0 an @u @u + ¹>0: Since the k _ 0-isocline is increasing an the _u 0-isocline is ecreasing, both steay states are sale oints. Usually between two sale oints one woul exect an unstable steay state. Here, this is not the case an we will emonstrate that the kink in the revenue function at k lays the same role as an unstable steay state. 4 nalysis of the multile steay states scenario _u0 _k u ¹k; k(0) k 0 ; () _u _q ( + ¹)u + ( + ¹)c a i + b k: (3) 4

arently, in each of the two regions k< an k> the canonical system () an (4) is linear an can be solve analytically. Following Hartl an Kort (000), we obtain k k ¹ i + k 0 k ¹ i e t u ¹ k ¹ i + k 0 k ¹ i (¹ + ) e t (4) (5) where is the negative eigenvalue of the Jacobian of the ynamic system () an (4): q ( +¹) + 4b (6) Note that here we use the transversality conition (5) so that we can omit the term involving the ositive eigenvalue. From (4) an (5), we also obtain analytical exressions for the caniate olicy functions (sale oint aths), showing that the sale oint aths are ecreasing straight lines: u ¹ ¹ k i + k ¹ k i (¹ + ) (7) We comute the objective values when starting from k 0 < converging to equilibrium ¹k L an starting from k 0 > converging to equilibrium ¹ k H. Using the notation g L 0 an g H g, the instantaneous ro t rate is given by ¼(k; u) a i k bk g i cu u : (8) Now inserting the solutions (4) an (5) while using the notation we get i k 0 k ¹ i (9) ¹ + (30) ¼(k(t);u(t)) a ¹ki i + i e t b ki ¹ + i e t gi c ¹ k ¹ i + i e t ¹ k ¹ i + i e t k ¹ i (a i ¹c) k ¹ i b + ¹ g i i e t a i +b k ¹ i + c + ¹ k ¹ i i e t b + : Now, the objective functional value becomes (k 0 ) Z 0 e t ¼(k(t);u(t))t ¹ k i (a i ¹c) ¹ k i (b + ¹ ) g i a i +b ¹ k i + c + ¹ ¹ k i Reinserting i from (9) an collecting terms w.r.t. k 0,weget k (k 0 ) ¹ i (a i ¹c) k ¹ i (b + ¹ ) g i + a i +b k ¹ i + c + ¹ k ¹ i µ + b + ¹k i c a i +b ¹ k i + ¹ ¹ k i 5 i b + i ¹k i b + ¹k i k 0 b + k0 (3)

4. Value function an Hamiltonian We can also comute using the Hamiltonian, since H; see Feichtinger an Hartl (986,. 7). First, we have u qc, i.e., q u + c. Then, we can use (7) to obtain u ¹ k ¹ i + k k ¹ i (¹ + ) for i L; H. Now, for i L an the trajectories aroaching the smaller equilibrium k ¹ L we have: L (k) H a L k bk cu u +(u + c)(u ¹k) a L k bk + u u¹k c¹k a L k bk ¹ k + k k k ¹ L + k ¹ L c¹k a L c¹ ¹ k L k + ¹ b k + k ¹ L (3) while for i H an the trajectories aroaching the larger equilibrium k ¹ H we obtain in an analogous way: H (k) a H c¹ ¹ k H k + ¹ b k + k ¹ H g (33) In enix it is shown by teious comutations that formulas (3) for i L; H an (3) an (33) are inee equivalent. We note that these functions i from (3) an (33) nee not hol in the whole regions k< an k> as the DNS-oint usually will not coincie with. 5 DNS Threshol In orer to obtain the threshol between the regions of initial values where convergence to the two equilibria occurs, we must comute the oint, where the value functions for the two caniates intersect. For this, it turns out that it is much simler to use the exressions (3) an (33) rather than (3). When equating the two exressions L (k) a L c¹ ¹ k L k + ¹ b k + k ¹ L ; H (k) a H c¹ ¹ k H k + ¹ b k + k ¹ H g; we observe that the quaratic term in k isthesameanthatc¹k cancels. Thus, the intersection must satisfy: ¹kH ¹ k L (ah a L ) k ¹ k H ¹ k L g: Thus, we get k g ¹k H k ¹ L (a H a L ) ¹kH k ¹ (34) L We note that this is only an aroximation of the DNS oint, since the above formulas only hol on one sie of. On the other sie of ; the extension of this solution caniate yiels a higher value of the objective function since - because of the kink - revenue is, in fact, higher than assume in (3) an (33). Thus, we know that If k <,thenk DNS <k <, an (35a) if k >,thenk DNS >k > (35b) 6

Only if k haens to coincie with, thenk DNS k. This hairline can be characterize as follows: g ¹k H k ¹ L g (a H a L ) ¹kH k ¹ > L a H a L g ¹k H k ¹ L (ah a L ) > g (a H a L ) ¹kH k ¹ L ¹k H k ¹ L (ah a L ) < g ¹kH k ¹ L ¹kH + k ¹ L (ah a L ) < g so that (??) can also be written as If g < ¹kH + k L (ah a L ),thenk DNS <k < (36a) if g ¹kH + k L (ah a L ),thenk DNS k, an (36b) if g > ¹kH + k L (ah a L ),thenk DNS >k >. (36c) For reasons of comleteness, we rove that the enominator of (34) is ositive: r a H a L > ¹kH k ¹ L () a H a L > a H a L > (a H a L ) ¹ ( + ¹)+b () 0 @ 4 q ¹ ( + ¹)+b > ( +¹) + 4b ¹ ( + ¹)+b > µ ah ( + ¹) c ¹ ( + ¹)+b a L ( + ¹) c ¹ ( + ¹)+b q ( +¹) + 4b () + +4¹ +4¹ + 4b () 4 + + ¹ + ¹ + b () ( +¹) + 4b > : which is true. From (34) we can also conclue that can k can become negative, since k > 0 i g> ¹k H ¹ k L > 0: (37) Note that in (6) an ¹ k H an ¹ k L in (8) o not een on the arameter g which can be chosen ineenently. Thus, for any given set of the other arameters, values of g small enough to violate (37) lea to a situation that the smaller steay state, although strictly ositive, is no equilibrium since it is always otimal to converge to the larger steay state being the only long run otimal equilibrium here. It is also clear that a very large g will lea to the situation that k > ¹ k H, imlying that it is never otimal to converge to the larger steay state. In enix B we investigate, whether this will always haen for g close the uer boun (8). It seems that this nee not always be true. 7

6 Numerical Examles fter we have mae as many calculations as ossible analytically, we now ientify ifferent scenarios by choosing aroriate arameter values. Let us assume a L 5;a H 0;b0:; ;¹0:; 0:; an c : (38) From (8) we nee g< a L(a H a L ) 5. From the above analytical investigations we can b exect that for intermeiate values of g both steay states will be equilibria, for small values of g only the larger steay state will be a long run otimal equilibrium an for large values of g only the smaller steay state will be an equilibrium. Below we give arameter values for each of these situations to occur. 6. Two equilibria We choose g 0: This yiels the maximum relevant caital stock while the kink (7) at an the steay states (8) are k max a H b 50; g a H a L ; ¹k L a L ( + ¹) c ¹ ( + ¹)+b 4:688; ¹k H a H ( + ¹) c ¹ ( + ¹)+b 30:33: Finally, the oint of intersection (34) of the two value function exressions (3) an (33) is given by ¹k k H k ¹ L g ¹kH k ¹ L (ah a L ) 0:34: In Figure, we lot the revenue function for these arameters, where we see that the kink is inee at. The value functions can be obtaine using (3) or using (3) an (33). Both ways we get: al c¹ k L (k) ¹ L k + ¹ b k + k ¹ L 68:98 + :373k 0:53k (39) ah c¹ k H (k) ¹ H k + ¹ b k + k ¹ H g 45:699 + :407k 0:53k : (40) 8

revenue r(k) 00 50 0 0 0 κ 30 40 50 60 k -50-00 Figure : The revenue function r(k) maxf5k 0:k ; 0k 0:k 0g : These value function caniates are lotte in Figure, where we see that the inee intersect for k 0:34: We have also veri e (36a) since g 0< ¹kH + ¹ k L (ah a L ) :5: Next we raw the hase iagram. The k _ 0isocline is u k¹ 0:k while the _u 0isoclines are the arallel lines uj k< ( + ¹)c a L ( + ¹) uj k> ( + ¹)c a H ( + ¹) b ( + ¹) k 7:8 _3 0:_3k b ( + ¹) k 6: _6 0:_3k The caniates for the olicy functions (i.e. the sale oint aths) are the arallel lines u L ¹ k ¹ L + k k ¹ L (¹ + ) 5:863 0:53k (4) u H ¹ k ¹ H + k k ¹ H (¹ + ) 0:704 0:53k (4) while the unstable aths are the arallel lines u unstable L ¹ ¹ k L + k ¹ k L (¹ + ) 6:655 + 0:653k u unstable H ¹ ¹ k H + k ¹ k H (¹ + ) 3:735 + 0:653k 9

value function caniates 700 Π H (k) 600 500 400 Π L (k) 300 00 00 0 0 0 k* 30 40 50 k Figure : The two caniates for the value function. 0

u 0 8 x 0 6 u L 0 4 0 k L k LB k DNS k UB k H u H 0 κ 0 0 30 40 50 k Figure 3: Phase iagram with two equilibria. where + q ( +¹) + 4b is the ositive eigenvalue of the Jacobian ¹ b + ¹ 0:453 0: :0 0: 0:3 In Figure 3, we lot the hase iagram. t thejuminthe _u 0isocline occurs an thus the trajectories su er a kink there. The bol lines reresent the otimal olicy functions while the otte lines are caniates which are not otimal. The thin soli lines are the unstable branches an the ash-otte lines are the isoclines. From k 0:34; an (??) we can conclue that k DNS <k 0:34 an that it will be close to k. We can now comute this DNS-oint k DNS by rst obtaining the intersection of olicy function u H ¹ ¹ k H + k ¹ k H (¹ + ) with k as : u H ( ) ¹ ¹ k H + ¹ k H (¹ + ) 7:3353: Next, we can comute the hyerbola like trajectory belonging to the region k< an ening in the oint (k; u) ( ; u H ( )) (; 7:3353:). This trajectory is given by k k ¹ i + F e t + F e t u ¹ k ¹ i +(¹ + ) F e t +(¹ + ) F e t (43) (44)

with i L. If this trajectory ens in (k; u) ( ; u H ( )) (; 7:3353:) at time t 0 (we choose this time transformation for convenience), then we obtain the two arameters F an F from k ¹ L + F + F 4:688 + F + F ; u H ( ) ¹ k ¹ L +(¹ + ) F +(¹ + ) F :937 5 0:53 F +0:653 F 7:3353; which yiels F 0:4685 an F 6:8435. Now the hyerbola like trajectory in question is vertical when _ k F e t + F e t 0, which haens at time t 3:6353. tthattime, k LB ¹ k L + F e t + F e t 7:697 an u ¹ ¹ k L +(¹ + ) F e t +(¹ + ) F e t 3:539 3. This oint is the lower bounary of the overla region an is therefore a lower boun for k DNS. It is smaller than k 0:34 which is what we nee, since we must have k LB <k DNS <k <.Notethatk LB <k DNS follows from the fact, that the trajectory u L is farther away from the _ k 0isocline than this trajectory (which intersects _ k 0 here) an thus has a better value of the objective function; see Feichtinger an Hartl (986,. 8). Thus, we know that k LB 7:697 <k DNS <k 0:34: Forreasonofcomletenesswecanalsoeterminetheoverlaregionofthetwoolicy function caniates to the right of. Now we must rst etermine the intersection of olicy function u L ¹ ¹ k L + k ¹ k L (¹ + ) with k : u L ¹ ¹ k L + ¹ k L (¹ + ) :879: Next, we can comute the hyerbola like trajectory belonging to the region k > aneningintheoint(k; u) ( ; u L ( )) (; :879). This trajectory is again given by (43) an (44) with i H an must satisfy (k; u) (; :879) at time t 0. Now we obtain the two arameters F an F from ¹ k H + F + F an u L ( ) ¹ ¹ k H +(¹ + ) F +(¹ + ) F :879 an obtain F :469 an F 6:8434. Now the hyerbola like trajectory in question is vertical when _k F e t + F e t 0, which haens at time t :77. tthattime, k UB ¹ k H + F e t + F e t 4:59 an this is the uer bounary of the overla region. In Figure 3 we have also comute these two hyerbola-like trajectories which etermine the overla region aroun the kink. We can now try to comute the real value function H for k LB <k<,forwhichthe intersection with L gives k DNS. In a similar way to (3) we etermine the value function H (k) for k< but this time for the trajectory going to ¹ k H.Wehave H (k) H a L k bk cu u +(u + c)(u ¹k) a L k bk + u u¹k c¹k (45)

but now we must use the arametric reresentation for k an u from (43) an (44) with i L an F 0:4685, F 6:8435. Thus, this value function caniate is arametrically given by H (k) a Lk bk + u u¹k c¹k k k ¹ L + F e t + F e t u ¹ ¹ k L +(¹ + ) F e t +(¹ + ) F e t F 0:4685; F 6:8435: (46) On the one han we have from (3) that the value function w.r.t. the solution caniate going to k ¹ L satis es (k) al c¹ ¹ k L k + ¹ b k + k ¹ L : Equating these two value function caniate, we obtain ¹ k L k + ¹ k + k ¹ L u u¹k. fter inserting (43) an (44) with F 0:4685, F 6:8435, one can comute the arameter t 0:899 for the intersection an insert this in (43) to obtain k DNS 9:884: (47) In Figure 4, we lot all relevant value function caniates an illustrate that they intersectatthelevelk DNS from (47). Note that in the interval [k LB ; ] the value function caniate H reresenting solutions converging to k ¹ H is not given by (33) but by the (slightly) larger value (46). In Figure 4, this is reesentes by the bol otte curve in [k DNS ; ] an the thin otte curves in [k LB ;k DNS ]. We note that for g ¹kH + ¹ k L (ah a L ) :5 accoring to (36a) k k DNS :5, but we refrain from rawing the hase iagram an that value function since the result is obvious. 6. Only larger equilibrium We kee all arameters as in (38) but now choose a i erent g. From (39) an (40) evaluate at ¹ k L (an keeing g 0 for the moment) we have L ¹kL 403; H ¹kL 34:78 (48) an we see that at the smaller equilibrium ¹ k L the value function L is by 6:4 higher than the other caniate H. We also note from (33) that g enters H via the aitive term g. Thus, if we reuce g by more than 6:4 6:4 then H ¹kL will excee L ¹kL Let us e.g. choose g 00: 3

value function 700 Π H (k) 600 500 400 Π L (k) 300 00 00 k LB k DNS κ 0 0 0 30 40 50 k Figure 4: Value function an DNS oint. 4

u 0 8 x 0 6 u L 0 4 u H 0 0 k L k H κ 0 0 30 40 50 k Figure 5: Phase iagram where only ¹ k H is an equilibrium. While the maximum relevant caital stock k max 50an the two steay states ¹ k L 4:688 an ¹ k H 30:33 o not change (since the formulas o not een on g), the kink in revenue an thus the jum in the _u 0isocline occurs at g a H a L 0; We comare the value function caniates (3) an (33) at k ¹ k L 4:688 an inee obtain L (k) 403 < H (k) 44:8 which means that their oint of intersection (34) is k : 7 < ¹ k L 4:688: So even although the revenue of aroaching k ¹ H is unerestimate by H for all k<, this value still excees the value function caniate L at the smaller steay state ¹k L 4:688. Thus only k ¹ H is the equilibrium here an for all k the olicy function consists of the sale oint ath (4) associate with this equilibrium. When rawing this hase iagram in Figure 5, we note that all lines, namely the isoclines an the stable an unstable aths remain unchange comare to Figure 3. 5

6.3 Only smaller equilibrium We again kee all arameters as in (38) an evaluate (39) an (40) evaluate at ¹ k H (keeing g 0)wehave L ¹kH 473:0; H ¹kH 584:4 an we see that at the smaller equilibrium k ¹ H the value function L is by :3 lower than the other caniate H. Noting again that g enters H via the aitive term g, if we increase g by more than :3 :3 then ¹kH L will excee H ¹kH Let us e.g. choose g : gain maximum relevant caital stock k max 50an the two steay states k ¹ L 4:688 an k ¹ H 30:33 o not change while the kink in revenue an thus the jum in the _u 0isocline occurs at g a H a L 4:4 We comare the value function caniates (3) an (33) at k ¹ k H 30:33 an inee obtain L ¹kH 473:0 > H ¹kH 464:4 which means that their oint of intersection (34) is k 3:07 > ¹ k H 30:33 So even although the revenue of aroaching ¹ k L is unerestimate by L for all k>, this value still excees the value function caniate H at the larger steay state ¹ k H 30:33. Thus only ¹ k L is the equilibrium here an for all k the olicy function consists of the sale oint ath (4) associate with this equilibrium (as long as it is ositive; for k>33:873 it is u 0). We raw this hase iagram in Figure 6, an note again that all lines (isoclines, stable an unstable aths) remain unchange comare to Figure 3. 7 Conclusions an Extensions It shoul be note that the observations in this aer were obtaine in a LQ framework. The reason is that this enables us to comute everything (incluing solution trajectories, olicy functions an value function) analytically. However, the result that a kink in the revenue function can lea to DNS oints in its neighborhoo is clearly not restricte to this LQ framework. ny concave revenue function r (k) with a convex kink an any convex cost function c (u) can lea to this result More generally, this henomenon can occur ue to any jum in the costate isocline cause by a non-smoothness of a moel function. 6

u 0 8 x 0 6 u L 0 4 u H 0 0 k L k H κ 0 0 30 40 50 k Figure 6: Phase iagram where only ¹ k L is an equilibrium. References [] Barucci, E.: Otimal investments with increasing returns to scale, International Economic Review 39, 789-808 (998) [] Davison, R., Harris, R.: Non-convexities in continuous-time investment theory, Review of Economic Stuies 48, 35-53 (98) [3] Dechert, W.D.: Increasing returns to scale an the reverse exible accelerator, Economic Letters 3, 69-75 (983) [4] Dechert, W.D., Nishimura, K.: comlete characterization of otimal growth aths in an aggregate moel with a non-concave rouction function, Journal of Economic Theory 3, 33-354 (983) [5] Eisner, R., Strotz, R.H.: Determinants of Business Investments in Imacts of Monetary Policy. Englewoo Cli s, N.J.: Prentice Hall, 963 [6] Feichtinger, G., Hartl, R.F.: Otimale Kontrolle Ökonomischer Prozesse: nwenungen es Maximumrinzis in en Wirtschaftswissenschaften. Berlin: e Gruyter, 986 [7] Hartl, R.F., Kort, P.M.: Caital accumulation of a rm facing an emissions tax, Journal of Economics 63, -3 (996) 7

[8] Hartl, R.F., Kort, P.M.: Otimal investments with increasing returns to scale: a further analysis, in: E.J. Dockner et al. (es.) Otimization, Dynamics, an Economic nalysis: Essays in Honor of Gustav Feichtinger, 6-38 (000) [9] Jorgensen, S., Kort, P.M.: Otimal ynamic investment olicies uner concaveconvex ajustment costs, Journal of Economic Dynamics an Control 7, 53-80 (993) [0] Rothschil, M.: On the cost of ajustment, Quarterly Journal of Economics 85, 605-6 (97) 8 enix. Check whether the two formulas are comatible This material to be omitte later or to be shortene!! We can use the formulas (+¹) + 4b + (+¹) + 4b q ( +¹) + 4b ¹ + +¹ (+¹) + 4b in orer to simlify the exressions for (k 0 ). We o this searately by comaring the koecients of k0,ofk 0,antheconstantterm. 8.0. Quaratic term First, we simlify the coecient b+ +¹ an using the notation b+ b+ +¹ 4b+((+¹) +(+¹) ) 4 4b+ ((+¹) +((+¹) + 4b )) 4 4b+((+¹) + 4b ) 4 + (+¹) 8b+(+¹) 4 of k 0 in (3) by inserting an ( +¹) + 4b 4b+(+¹ ) + (+¹) 4b +(+¹) + (+¹) µ +¹ 4 + (+¹) 4 + (+¹) (+¹) (+¹) + 4b (49) 8

Thesameresultweshoulgetbysimlifyingthecoecientofk in (3): ¹ b b b µ ¹ µ 4 + (+¹) + 4b + 4 4 ¹ ( +(+¹) + 4¹ ) 4 + +4¹ + 4 4 which µ is same as above. Maybe the nicest formula for the coecient of k0 is: µ +¹ (+¹) + + 4b +¹ + (+¹) + 4b ( + ¹ ) 8.0. Linear term Next, we simlify the coecient of k 0 in (3) by inserting an +¹ ¹k i ca i+b k ¹ i + ¹ k ¹ i b+ b+ +¹ b+ (+¹ ) ¹ ki +¹ (c+¹ ¹ k i)a i +b ¹ k i + ¹ ki (+¹ )(c+¹ ¹ k i)a i +4b ¹ k i + : (+ ) b+ (+¹ ) ¹ki (+¹ )(c+¹ k ¹ i)+(a i 4b k ¹ i) (+ ) If we exan the numerator an collect terms with,, an resectively, we obtain that this numerator is: b + ( +¹) ki ¹ ( +¹) k ¹ i + b + ( +¹) ki ¹ ( +¹) c +¹ k ¹ i + ai b k ¹ i + ¹ k i ( +¹) k ¹ i + c +¹ k ¹ i + k ¹ i ¹ k i + ¹ k i ( +¹) c +a i + ¹ k i ( +¹) ¹ k i + c +¹ ¹ k i + ¹ k i k ¹ i + (a i ( +¹) c)+ ¹ k i + c c + (a i ( +¹) c) i Thus, the coecient of k 0 simli es to: c+ai c¹c c+ (a i (+¹)c) (+ ) + c + a i(+¹)c + c + a i (+¹)c + (+¹) + 4b ( )c+(a i ¹c) + c + a i(+¹)c ( +(+¹))c+a i + On he other han, we simlify the coecient of k in (3) by inserting an ¹k L a L(+¹)c : ¹(+¹)+b ) a L c¹ ¹ k L a ic¹ ( a i (+¹)c ¹(+¹)+b 9

Since we want te get an exression wher in the enominator we have ther term +, we multily by this exression: (+ )(a i c¹) ( )( ) a ¹(+¹)+b i(+¹)c (+ ) (+ )(a i c¹)(¹(+¹)+b) ( )( )(a i (+¹)c) (+ )(¹(+¹)+b) a i((+ )(¹(+¹)+b) ( )( ))c¹(+ )(¹(+¹)+b)+ ( )( )(+¹)c (+ )(¹(+¹)+b) In the numerator, the coecient of a i simli es: (¹ ( + ¹)+b) ( ) + (¹ ( + ¹)+b)+ ( ) (¹ ( + ¹)+b) ( +¹) + 4b + (¹ ( + ¹)+b)+ ( +¹) + 4b (¹ ( + ¹)+b) ( +¹) + 4b 4 (¹ ( + ¹)+b) The remainer in the numerator also simli es: c¹ + (¹ ( + ¹)+b)+ ( ) c¹ (¹ ( + ¹)+b)+ ( ) ( + ¹) c c¹ (¹ ( + ¹)+b)+ ( )( + ¹) c c (¹ ( + ¹)+b) ( +¹)+ Thus, the coecient of k in (3) nally becomes: 4(¹(+¹)+b)a i+c(¹(+¹)+b)((+¹)+ ) (+ )(¹(+¹)+b) a i + + c (¹+ + ) (+ ) a ic(+¹) + + c whichissameasabove. 8.0.3 Constant term + a i+b ¹ k i + c+ ¹ ¹ k i ( + ¹) c The most comlicate is the constant term. This constant term in (3) is ¹k i (a i ¹c) k ¹ i (b+¹ )g i ¹k i b+ ¹ k i while in in (3) it is ¹ k L Now let us simlify the former by collecting terms involving k ¹ i an k ¹ i. ¹k i (a i ¹c) ¹ k i (b+¹ )g i k ¹ (a i ¹c) i + k ¹ ca i i g i ¹ k i (b+¹ ) (b+ ¹) + a i+b ¹ k i + c+ ¹ ¹ k i g i + k ¹ i b+ ¹ k i (equals zero for i L) + k ¹ (ai ¹c) i + ca i + k ¹ i b+¹ + (b+ ¹) b+ The coecient of k ¹ i is ¹k i b+ ¹ k i 0

(a i ¹c) + ca i (a i¹c) (a i¹c) + (+¹ )ca i + The coecient of ¹ k i b+¹ b+¹ + b+¹ b+¹ b+¹ b+¹ b+¹ b+¹ b+¹ is + (b+ ¹) b+ b+ +¹ + ¹ + +¹ ca i + b+ + (b+¹(+¹ )) + ¹ +4 b+¹(+¹) + ¹ +4 b+¹(+¹) + ¹ +4 b+¹(+¹) + (a i¹c) +¹ c + (+¹)ca i + 4b+ (+¹ ) 4 4b+(+¹) (+¹) + 4 4b+(+¹) 4 4 + (+¹) 4 4b+(+¹) 4 + (+¹) ¹ +4 b+¹(+¹) + + (¹) +4 b+¹(+¹) + + (¹) +4 b+¹(+¹) + Thus, the constant term in (3) is: (+¹) +4b 4 (+¹) +4b 4 + (+¹) 4 4b+(+¹) 4 4b+(+¹) 4 9: 86 a i (+¹)c (ai ¹c) c + (+¹)ca i ¹(+¹)+b + + ai (+¹)c b+¹ + (¹) +4 b+¹(+¹) 4 ¹(+¹)+b + a i (+¹)c (ai ¹c) c + a i(+¹)c (+¹)ca i ¹(+¹)+b ¹(+¹)+b + + ai (+¹)c b+¹ + (¹) 4 ¹(+¹)+b + ai (+¹)c 4 b+¹(+¹) 4 ¹(+¹)+b + (+¹) +4b a i (+¹)c (ai ¹c) c + ai (+¹)c b+¹ + (¹) ¹(+¹)+b 4 ¹(+¹)+b (a i(+¹)c) ¹(+¹)+b + + ai (+¹)c b+¹(+¹) ¹(+¹)+b + + ai (+¹)c (+¹) +4b 4 ¹(+¹)+b a i (+¹)c (ai ¹c) c + ai (+¹)c b+¹ + (¹) ¹(+¹)+b 4 ¹(+¹)+b + ai (+¹)c (+¹) +4b 4 ¹(+¹)+b ai (+¹)c ¹(+¹)+b (ai ¹c) ¹(+¹)+b a i c + b+¹ + (¹) + (+¹)c 4 4 ai (+¹)c ¹(+¹)+b + b+¹ + (¹) + (+¹) +4b ¹(+¹)+b 4 4 µ ai (+¹)c 4¹(+¹)+4b(b+¹ )+(¹) + (+¹) +4b ¹(+¹)+b 8 4 ai (+¹)c 4¹+4¹¹+4bb¹ +¹ + (+¹) +4b ¹(+¹)+b 8 4 ai (+¹)c ¹+¹ +b+ + (+¹) +4b ¹(+¹)+b 8 4 (+¹) +4b 7: 04 9 0 (+¹) +4b

µ ai (+¹)c (¹+¹ + )+b + ¹(+¹)+b 8 4 We want to get: ¹ k L ai (+¹)c 4 ¹(+¹)+b So we must show that 4 (¹+¹ + )+b 8 + 4 (+¹) +4b (+¹) +4b (¹ +¹ + )+b + (+¹) +4b + (¹ +¹ + )+b (+¹) +4b + (¹ +¹ + )+b ( +¹) +4b + ¹ +¹ + +b ( +¹) 4b + ¹ ¹ b ( +¹) 4b + ( +¹) + 4b ¹ ¹ b ( +¹) + 4b ( +¹) 4b which is true, since both, l.h.s. an r.h.s. vanish. 9 enix B. We check whether g close the uer boun (8) leas to the situation that k > k ¹ H, imlying that it is never otimal to converge to the larger steay state an thus k ¹ L is the only equilibrium: for g a L(a H a L ) we have b a L (a H a L k ) ( k b ¹ H ¹ kl) (a H a L ) ( k ¹ H k ¹ > k ¹ L) H a H(+¹)c ¹(+¹)+b a L (a H a L ) ( k ¹ b H ¹ kl) (a H a L ) ( k ¹ H k ¹ > a H(+¹)c L) ¹(+¹)+b al (a H a L ) ¹k b H k ¹ L (¹ ( + ¹)+b) > (a H ( + ¹) c) (a H a L ) ¹kH k ¹ L a L (a H a L ) (¹ ( + ¹)+b) ¹k b H k ¹ L (¹ ( + ¹)+b) > (a H a L )(a H ( + ¹) c) ¹kH k ¹ L (ah ( + ¹) c) a L (a H a L ) (¹ ( + ¹)+b) (a b H a L )(a H ( + ¹) c) > ¹k H k ¹ L (¹ ( + ¹)+b) ¹kH k ¹ L (ah ( + ¹) c) (a H a L ) a Lb (¹ ( + ¹)+b) (a H ( + ¹) c) > ¹kH k ¹ ¹kH L + k ¹ L (¹ ( + ¹)+b) (ah ( + ¹) c) (a H a L ) a Lb ¹ ( + ¹)+( + ¹) c + a L a H a > H a L ah +a L (+¹)c (¹ ( + ¹)+b) (a ¹(+¹)+b ¹(+¹)+b H ( + ¹) c) (a H a L ) a Lb ¹ ( + ¹)+( + ¹) c + a L a H a > H a L ah +a L ( + ¹) c a ¹(+¹)+b H +( + ¹) c a Lb a ¹ ( + ¹)+( + ¹) c + a L a H > L a H ¹(+¹)+b a Lb ¹ ( + ¹)+( + ¹) c>(a H a L ) ¹(+¹)+b

(+¹)(a L ¹+bc) b(a H a L > ) ¹(+¹)+b (¹ ( + ¹)+b) (+¹)(a L¹+bc) b(a H a L ) (¹ ( + ¹)+b) (+¹)(a L¹+bc) b(a H a L ) (¹ ( + ¹)+b) (+¹)(a L¹+bc) b(a H a L ) >¹( + ¹)+b >¹( + ¹)+b + (+¹) + 4b µq > ( +¹) + 4b q (¹ ( + ¹)+b) (+¹)(a L¹+bc) b(a H a L + > ( +¹) + 4b ) Does not seem clear that this is always true, although for the arameters I trie it hols. So, maybe it coul be that even if the kink equals the maximum of the revenue function before the kink, the larger steay state is an equilibrium. For the arameters below, this is not the case. ¹ ¹ b 3