An example for a closed-form solution for a system of linear partial di erential equations
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1 An example for a closed-form solution for a system of linear partial di erential equations Tobias Nagel (a) and Klaus Wälde (a,b), (a) University of Mainz (a,b) CESifo, Université Catholique de Louvain, University of Bristol June We analyse a system consisting of two linear partial di erential equations. We present a closed-form solution which has a gamma-structure, i.e. it is of form xe x. Keywords: linear partial di erential equation system, analytical solution Introduction Consider a function p(a z t) : R fw bg R +! R: We require that this function satis es the following system of partial di erential equations (PDEs), PDE w p (a w t) t m PDE b p (a b t) t m w b p (a w t) + sp (a w t) a p (a b t) = (a) p (a b t) + p (a b t) a sp (a w t) = (b) where s m w and m b are known real constants. Further we assume that m w 6= m b. Those equations are motivated in economic papers by Bayer and Wälde ( a,b,c). The authors analyse the distribution of wealth a in a world of uncertain labour income, i.e. labour income z that moves stochastically between a low state b and a high state w. This stochastic movement is modeled using two mutually independent Poisson processes. Arrival rates are s for the transition from state w to b and for the opposite direction. The state w can be called employment and state b unemployment. Individuals can save in this setup implying a distribution of wealth that is a function of labour market history. The above linear system results from Fokker-Planck equations describing the joint density of a and z under the assumption of a utility function characterised by so-called constant absolute risk aversion. As far as we can tell, general closed form solutions for a linear PDE system () do not seem to exist. A posting by Serre () supports this opinion by stating that a general closed-form solution for a linear PDE system does not exist. Special solutions, i.e. solutions for special initial functions, are well-known. An example are exponential-type solutions of the form e x e t (Israel, ). After having consulted, inter alia, Polyanin (, ), Polyanin et al. (), Evans (), Farlow (993) and Zachmanoglou and Thoe (986), we concluded that solutions of the gamma-type suggested here are not as widely known as one would expect. 3 Both authors are at the Mainz School of Mangement and Economics, University of Mainz, Jakob-Welder- Weg 4, 553 Mainz, Germany. Contact: nageltuni-mainz.de, klaus.waeldeuni-mainz.de, We would like to thank Christian Bayer (Department of Mathematics at the University of Vienna) for comments and discussions. In parallel work, ten Thije Boonkkamp and ourselves solve this system using the method of characteristics as presented e.g. in Mattheji, Rienstra and ten Thije Boonkkamp (5). We obtain a system of integral equations or ordinary di erential equations on characteristic lines that do not suggest an obvious closed-form solution. 3 We would call a solution widely known if it appears in textbooks. We would be very happy to receive hints to the journal literature if closed-form solutions are available there.
2 We suggest and prove that there are closed-form solutions of the type p(a w t) = (c w + a + t) e c t c a (a) p(a b t) = (c b + a + t) e c t c a : (b) where c w c b, c and c are parameters which are to be determined. These solutions could be called gamma-type solutions, given that their structure reminds of the gammadistribution. The remaining parts of this paper is structured as follows. Chapter gives a theorem showing that for certain parameter restrictions, our guess in () is indeed a solution to (). We also give a short sketch of the proof - which can be found in full detail in the appendix. Chapter 3 presents a numerical illustration of our solution. A gamma-type solution We try to keep this chapter as short as possible and at the same time include the key steps of the proof of our main theorem. Interested readers can nd the complete proof in the appendix.. Main result Proposition Consider the function p(a z t) = (c z + c az a + c tz t) e c t c a z fw bg (3) where c w c b c c Rnfg are constants and de ne c + c m b + (4a) c + c m w + s: (4b) Then, p(a w t) p(a b t) are solutions to the system of PDEs given in () if and only if all of the following ve conditions hold s = (5a) = (5b) = (5c) C w sm w + Cwm b caw = Cw + s (5d) m w = c w + c b (5e) where s m b and m w are given parameters as described after (). Proof. (sketch - see appendix for complete version) The proof is completely constructive. We start by substituting our (3) and its derivatives into the PDE system (). This yields = m w + c w c b + ( ) t + ( ) a = m b + c b c w s + ( s) t + ( s) a: Taking into consideration, that a R and t R + these equations can be satis ed only if the terms in front of a and t are zero. Imposing this on parameters, we also need the remaining
3 terms to equal zero. Imposing this as well, we end up with a system of six equations to determine the eight parameters c w c b c c : Analysing those equations, we nd that one of these six equations is redundant and we end up with the ve conditions in (5). We then choose c c b as free parameters. Once we have chosen c we can determine c by solving the quadratic equation (5a). As we can choose c, we always nd at least one real solution for c. Subsequently, xing and the previous results, we use (5b) to compute and (5d) for : The next step is given by (5c) an the computation of : Finally, given all previous results and xing c b we can use (5e) to determine the missing c w :. A rst example For a rst application, we set m w = s = and m b = = i.e. we want to solve One solution is t p (a w t) + p (a w t) p (a w t) p (a b t) = (6a) a p (a b t) p (a b t) + p (a b t) + p (a w t) = : t a (6b) p(a w t) = te a p(a b t) = ( + t) e a (7) i.e. with the notation of theorem : c = c w = = = c b = = = and c = : 4 To verify that, we compute partial derivatives, t p (a w t) = ea t p (a b t) = ea a p (a w t) = tea a p (a b t) = ( + t)ea substitute them plus our guess (7) into the initial system (6) and see that (6) holds. 3 Numerical illustration We now provide numerical examples for solutions. This is of importance for applications as it illustrates which values the parameters can take. 5 The rst subsection provides two equations that determine m w and m b : This is an outcome of the economic system in the background. Someone interested in the numerical illustration per se can skip this section and take values for m w and m b as given. Parameters for our numerical illustrations are given by r = :98 s = :4 = : w = b = = :79 = :: (8) While the values can be motivated from the economic model as well, they are relatively arbitrary at this point. 4 To get this solution, we chose c = and = c b = : The value of c together with the values of m w m b and s imply that we only have one solution for c i.e. c = : 5 All codes are available at right next to this paper. 3
4 3. Values of m w and m b The optimal consumption path under a utility function with constant absolute risk aversion is given by c(a z) = ra + z + m z z fb wg provided that r [ + m w ] s + se [w b+mw m b] = r [ + m b ] + e [w b+mw m b] = hold. Here, r is the constant interest rate, is the CARA-parameter, is the discount factor and s and are the arrival rates of the Poisson processes. These two equations result from the Bellman equations of the maximization problems (one for each state). Using our parameter values from (8), we get m w = :76 < m b = :747 > : 3. Determination of p(a z t) Having three parameters we can choose to our liking, we split up this section into three parts. In each part, we x two out of the three free parameters and the third parameter is allowed to vary within an interval. Using those values we compute the parameters xed by our conditions (5) and analyse their behaviour. As the proof of theorem has shown, we can have up to two real values for c given a xed value of c : We denote these two solutions by c + and c here. 3.. Fix c and For our st illustration we x c = and = : We let the remaining free parameter c b vary between and as plotted on the horizontal axis of the next gure. fixed c = and =, using c + fixed c = and =, using c 3 c + c c w.5.5 c b c b c w 3 3 c b c b Figure Parameter values for c = = and and c in the right panel. c b, using c + in the left panel The parameter values for c + are illustrated in the left panels, those for c in the right panels. In the case of c +, we see that all parameters are independent of c b apart from c w. Looking at the conditions in (5) immediately shows that this obviously needs to hold true. As the condition do not distinguish between di erent values of c we observe the same behaviour for c but with obviously di erent absolute values. 4
5 3.. Fix and c b Let = and c b = : fixed = and c b =, using c + fixed = and c b =, using c 5 5 c + c 5 c 5 c 3 3 c w c w c c Figure Parameter values for = c b = and di erent values of c, using c + in the left panel and c in the right panel. Here, as c is not xed, we have no constant parameters. Instead there exists a pole for c b and as shown in the second row in the gure. The value of that pole depends on : 6 c + is a at, opened up and positive valued parabola, whereas c is also at but open down with both positive and negative values. The left and right panel seems to be connected by a symmetry issue, which is not further analysed here Fix c and c b fixed c = and c b =, using c + fixed c = and c b =, using c 4 c + c w.5 c c w Figure 3 Parameter values for c = c b = and panel and c in the right panel. :5 :5, using c + in the left 6 Looking e.g. at in (5d) the denominator equals if = p (sm w ) =m b : 5
6 p(a,w,) p(a,b,) p(a,w,t) p(a,b,t) For both, c + in the left panel and c in the right panel, parameters c w and depend on : The remaining three parameters are constant. Obviously the absolute values in both cases are di erent Looking at the shape of p(a z t) This section looks at one example to analyse the structure of the p(a z t): Let We then plot [c c c b c w ] = [:83 :3 3:9 :68 :5 ] : p(a w t) = (:3 + :68a + t) e :83t a p(a b t) = ( 3:9a :5t) e :83t a : 4 a 4 5 t 6 8 a 4 5 t Figure 4 p(a w t) and p(a b t) for :5 a 5 and t : To get some impression, we solve the system for :5 a 5 and t : We observe that the peak moves to the left as time goes by. Taking a bigger interval for a, this movement would of course become even more visible a a Figure 5 Initial conditions p(a w ) and p(a b ) for :5 a 5: As any solution of a PDE is a function of the intitial condition, nding a closed-form solution implies in our case an explicit functional form for the initial condition (i.e. for t = ). These initial conditions are shown in g. 5. 6
7 4 Conclusion We showed that for a special two-dimensional linear system of PDEs, a solution of the form p (a w t) t m p (a b t) t m w b p (a w t) + sp (a w t) a p (a b t) = p (a b t) + p (a b t) a sp (a w t) = p(a z t) = (c z + c az a + c tz t) e c t c a z fw bg exist, if the parameters ful l certain conditions. In total we get ve conditions leaving us a high degree of freedom by choosing three parameters. References Bayer, C., and K. Wälde (a): Matching and Saving in Continuous Time : Theory, CESifo Working Paper 36. (b): Matching and Saving in Continuous Time: Proofs, CESifo Working Paper 36-A. (c): Matching and Saving in Continuous Time: Stability, mimeo - available at Evans, L. (): Partial Di erential Equations. American Mathematical Society, nd edition. Farlow, S. J. (993): Partial Di erential Equations for Scientists and Engineers. Dover Publications. Israel, R. (): Analytical solution to a Linear advection-reaction PDE: Answered 4.3., ow.net/questions/5844/analytical-solution-to-a-linearadvection-reaction-pde (5.6.). Mattheij, R., S. W. Rienstra, and J. H. M. ten Thije Boonkkamp (5): Partial di erential equations. Modeling, Analysis, Computation. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA. Polyanin, A. (): Handbook of Linear Partial Di erential Equations for Engineers and Scientists. Chapman and Hall. Polyanin, A., V. Zaitsev, and A. Moussiaux (): Handbook of First Order Partial Di erential Equations. Taylor and Francis. Polyanin, A. D. (): EqWorld - The World of Mathematical Equations, Serre, D. (): Analytical solution to a Linear advection-reaction PDE: Answered 5.3., ow.net/questions/5844/analytical-solution-to-a-linearadvection-reaction-pde (5.6.). Zachmanoglou, E., and W. Dale (986): Introduction to Partial Di erential Equations with Applications. Dover Pubn Inc, nd edition. 7
8 A Appendix - Proof of the main result This section proves proposition. Computing the partials derivatives of p(a z t) = (c z + c az a + c tz t) e c t c a z fw bg and substituting those derivatives back into () yields = m w + c w ( c + c m w + s) c b + ( ( c + c m w + s) )t + a( ( c + c m w + s) ) (9) = m b + c b ( c + c m b + ) c w s + ( ( c + c m b + ) s)t + a( ( c + c m b + ) s) () where we already used the fact that e c t c a >. If we can ensure that those two equations hold, p(a z t) are solutions. Due to the fact that (9) and () have to hold for all t [ ) and a R, we need that m w + c w = c b (a) = = (b) (c) m b + c b = c w s (d) = s = s where we simpli ed the expressions using = c + c m w + s and = c + c m b + : Substitution of the second and third equation into the last two equations yields m w + c w = c b = = m b + c b = c w s s = s = : and hence we can eliminate one equation. The remaining system is (e) (f) m w + c w = c b = = m b + c b = c w s s = : () Therefore we have ve equations for eight unknowns, yielding an underdetermined system. 8
9 Assume c is free, i.e. let c Rnfg: Now c can be chosen such that the last equation, i.e. c + c m b + = s c + c m w + s () c c ( + s + c (m b + m w )) + c (c m b m w + m w + sm b ) = holds. Hence c can be determined by solving a quadratic equation and we could end up with up to two solutions for c :Nevertheless we assume, that we can nd at least one value for c (maybe we have to change the chosen value of c ) and therefore we have xed c, c and of course. The remaining four equations (3) m w + c w = c b (4a) = (4b) = (4c) m b + c b = c w s (4d) can be used to determine four out of the remaining six parameters, c w c b and. To start with, we look at the rst and last equation. This system can be written in matrix notation as ctw m w Cw cb cb = M : m b s c w c w Looking at det(m) we see that det(m) = s or in nite solutions. Looking back at we can write those two equations as = by (). Hence, we have either none m w = c w + c b (5) m b = c w s c b (6) c w = m w + c b (7) c w = m b + s s c b: (8) Finding a solution of a system of linear equations is equivalent with nding an intersection of those two lines. By comparing those two lines with each other, we see, that they have to be parallel. as = s according to (). We therefore either have no solution or an in nite amount of solutions for (5) and (6). The second case - the one we need for our proof to work - arises for m w = m b : (9) s We therefore need to add this condition to our list of the remaining four equations in (4). Doing so, we can remove one of the two redundant equations (given that we impose them to be identical in (9)) and obtain = = c w = m b + s s c b m w = m b : s 9
10 We start by solving the rst two equations for and respectively and insert those expression into the third equation. This yields = () = () s + sm w = C w C wm b where the last equation is equivalent to sm w + m b C = c w tw + s () = C w + s smw + C wm b C () = c w + s tw sm w + Cwm b : (3) Let Rnfg be xed but arbitrary, then (3) determines : Also and can now be computed using () and (). At this stage we are left with only one equation, c w = m w + c b (4) to determine c w and c b : Without loss of generality, we assume c b to be the third and last free parameter and with (4) we have the condition on the missing parameter c w :Therefore the ve conditions of theorem are (), (), (), (3) and (4). q.e.d.
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