VIEWS: 87 PAGES: 31 CATEGORY: Financial Models POSTED ON: 1/13/2010 Public Domain
A General Solution to the Bensoussan-Crouhy-Galai Equation appearing in Finance Solomon M. Antoniou SKEMSYS Scientific Knowledge Engineering and Management Systems 37 oliatsou Street, Corinthos 20100, Greece solomon_antoniou@yahoo.com Abstract We find a general solution to the non-linear PDE introduced by Bensoussan, Crouhy and Galai. This equation models the stochastic equity volatility as a function of the stock price and time. Our solution relies on the exploitation of the underlying Lie symmetries of the given equation. We also determine the classical Lie symmetries of a similar in nature non-linear PDE which models the volatility of the stock price as a function of the total asset value and time. Keywords: Stochastic equity volatility, Lie Symmetries, Nonlinear PDEs, Exact Solutions. 1 1. Introduction The Black-Scholes option pricing model (ref. [1]) was derived on the assumption of constant volatility. However this assumption does not appear to be true in everyday market transactions (volatility smile, see for example ref. [2]). The volatility dynamics of the stochastic volatility models (or the stochastic volatility models combined with jump diffusion processes, [5]-[9]) appears to be exogenous and cannot be justified on economic grounds. It is the Bensoussan-Crouhy-Galai (BCG) approach introduced some time ago (refs [3]-[4]) that considers the stochastic nature of volatility to be endogenous and comes from the impact of a change in the value of the firm’s assets on the financial leverage. The core of the BCG model relies on a PDE (see equation (11) of [3]) that models the stochastic equity volatility as a function of the stock price and the time. This equation is a nonlinear PDE, where the unknown function is the stochastic equity volatility function given by ∂ ∂ 1 2 2 ∂2 + + (r + 2 ) S =0 S 2 ∂t 2 ∂S ∂S There is a second equation (see equation (12) of [3]) which is also a nonlinear PDE, where the unknown function is the function ~ (V, t ) , V being the total asset (1.1) value and t the time variable. It is given by ~ ∂ ~ 1 2 2 ∂ 2~ ~) V ∂ = 0 (1.2) + + (r + V ∂t 2 ∂V ∂ V2 We have found a general solution to the Bensoussan-Crouhy-Galai equation (1.1), long believed to be as one not having an analytical solution. (S, t ) , S being the stock price and t the time variable. It is 2 In solving this equation, we have found the Lie point symmetries of the equation. We then determined two invariants, which allowed us to convert the equation into an ordinary, nonlinear, second order DE. Under two successive transformations, we transformed this last DE into an Emden-Fowler equation, which admits an analytical solution given by the error function erfi( x ) of imaginary argument. We have also determined the Lie point symmetries of the second equation (1.2) for further investigation. Despite the fact that the BCG approach is – to the best of our knowledge – one of the most serious alternative candidates to the Black-Scholes approach, it has not received proper attention. That might come from the non-linear nature of the equations involved and the lack of analytical solutions. A further reformulation of the BCG approach, using the closed-form solution of the above-mentioned equations is under way. The paper is organized as follows: In Section 2 we determine the Lie symmetries of differential equation (1.1). In Section 3 we find a general solution of the BCG equation, taking into account the generators of the Lie symmetry. In Appendix A we prove an auxiliary Theorem that helps to convert an ordinary DE into an Emden-Fowler equation. In Appendix B we solve the Emden-Fowler equation. In Appendix C we give a brief description on how to find the inverse of the error function. In Appendix D we consider another general solution of the equation, which is based on a generator not considered in deriving the general solution in Section 3, using a paraphrasis of the so-called Tanh-Method. In Appendix E we derive the Lie symmetries of the second equation, which can be used for the determination of a general, closedform solution. 3 2. Lie Symmetries of the BCG Equation (1.1) Lie’s method of symmetry is a well-established method initiated by S. Lie (see references [10]-[18]). In principle, this method can be used both for ODEs and PDEs (or systems of these equations), linear or nonlinear, in many areas of science and engineering (and even in finance). Theorem 1. Equation (1.1) can be transformed to the equation (2.1) where x = ln S . Proof. We consider the BCG equation (1.1). We make the substitution (2.2) Since S 2 ∂ 2 2 ∂ u 1 2 ∂ 2u 1 2 ∂ u + u + r + u =0 ∂ t 2 ∂ x2 2 ∂ x x = ln S ∂S = ∂ 2u ∂x 2 − ∂ ∂u ∂u and S , = ∂S ∂ x ∂x the BCG equation (1.1) takes the form (2.1), where (S, t ) = u ( x , t ) with x = ln S . Theorem 2. The generators of the Lie Symmetries of the equation (2.1) are given by (2.3) X1 = ∂ ∂t ∂ ∂ 1 ∂ +t − u ∂x ∂t 2 ∂u ∂ ∂ + e −(x − r t) u ∂x ∂u (2.4) X2 = r t (2.5) X3 = − e −(x − r t) X4 = ∂ ∂x (2.6) 4 Proof. Let (2.7) ( x , t , u ( 2) ) be defined by 1 1 ( x , t , u ( 2) ) = u t + u 2 u xx + r + u 2 u x 2 2 ∂ ∂ ∂ + (x, t, u ) + (x, t, u ) ∂x ∂t ∂u Introduce the vector field X (the generator of the symmetries) by (2.8) X = (x, t, u ) The second prolongation of the vector field is defined in our case by the equation ∂ ∂ ∂ ∂ ∂ + t + xx + xt + tt pr ( 2) X = X + x ∂ux ∂ut ∂ u xx ∂ u xt ∂ u tt Invariance of the equation under the symmetry of the vector field V is implemented by the equation pr ( 2) X [ ( x , t , u ( 2) )] = 0 , as long as ( x , t , u ( 2) ) = 0 . We have pr ( 2) X [ ( x , t , u ( 2) )] = 0 1 1 ⇔ pr ( 2) X u t + u 2 u xx + r + u 2 u x = 0 2 2 or (2.9) [ u u xx + u u x ] + 1 1 + x r + u 2 + t + xx u 2 = 0 2 2 The coefficients x t x , t and xx are given by = x + ( u − x )u x − u u 2 − x u t − u u x u t x = t − t u x + ( u − t )u t − u u x u t − u u 2 t = xx + (2 xu − xx )u x − xx u t + ( uu − 2 xu ) u 2 − x − 2 xu u x u t − uu u 3 − uu u 2 u t + ( u − 2 x ) u xx − x x xx 5 − 2 x u xt − 3 u u x u xx − u u t u xx − 2 u u x u xt For the functions , and we have = (x, t, u ) x = (t) , = (x, t ) Thus, using the previous expressions for (2.10) , t and xx , equation (2.9) becomes 1 [ u u xx + u u x ] + r + u 2 [ x + ( u − x )u x ] + 2 + t − t u x + ( u − t )u t + 1 + u 2 { xx + (2 xu − xx )u x + 2 + uu u 2 + ( u − 2 x )u xx } = 0 x 1 1 We substitute u t by − u 2 u xx − r + u 2 u x into (2.10): 2 2 [ u u xx + u u x ] + 1 + r + u 2 [ x + ( u − x )u x ] + t − t u x + 2 1 1 + ( u − t ) − u 2 u xx − r + u 2 u x + 2 2 1 + u 2 { xx + (2 xu − xx )u x + 2 + uu u 2 + ( u − 2 x ) u xx } = 0 x which can be written in the equivalent form (2.11) 1 1 2 r + u x + t + u 2 xx + 2 2 1 + u + r + u2 ( u − x )− t − 2 6 1 1 − ( u − t ) r + u 2 + u 2 (2 xu − xx ) u x + 2 2 1 1 + [ u − ( u − t ) u 2 + ( u − 2 x )u 2 ] u xx + 2 2 1 + u 2 uu u 2 = 0 x 2 All the coefficients of the partial derivatives of the function u are set equal to zero. We then have a system of partial differential equations: • Coefficient of the zero-th order derivative of u: (2.12) 1 2 1 2 r + u x + t + u xx = 0 2 2 • Coefficient of u x (2.13) 1 u + r + u2 ( u − x )− t − 2 1 1 − ( u − t ) r + u 2 + u 2 (2 xu − xx ) = 0 2 2 • Coefficient of u xx : (2.14) 1 1 u − ( u − t )u 2 + ( u − 2 x )u 2 = 0 2 2 • Coefficient of u 2 : x (2.15) 1 2 u uu = 0 2 uu = 0 , which means that We now have to solve the system of equations (2.12)-(2.15). From (2.15) we get to u: (2.16) ( x, t, u ) = (x, t ) ⋅ u + ( x, t ) is a linear function with respect From (2.14), after canceling opposite terms and dividing by u, we get the equation 7 + We substitute 1 tu − xu = 0 2 in the previous equation by the expression (2.16), to obtain ⋅u + + 1 tu − xu = 0 2 It follows that (2.17) and (2.18) + 1 t − x =0 2 by (2.16), we get =0 From equation (2.12), and substituting 1 2 r + u ( x ⋅ u + x ) + ( t ⋅ u + t ) + 2 1 + u 2 ( xx ⋅ u + xx ) = 0 2 which can be written in equivalent form 1 r ( x + t ) + (r x + t ) u + ( x + xx ) u 2 + 2 1 + ( x + xx ) u 3 = 0 2 From the previous equation we get (since (2.19) (2.20) r x + t =0 x + xx = 0 = 0) From (2.13), canceling opposite terms and substituting 1 u ( ⋅ u + ) + r + u2 ( t − x )− t + 2 1 + u 2 (2 x − xx ) = 0 2 The previous equation can be written in equivalent form by (2.16), we get 8 {r ( t − x ) − t } + u + 1 1 + { + ( t − x ) + (2 x − xx )} u 2 = 0 2 2 from which we get (2.21) and (2.22) 1 1 + ( t − x ) + (2 x − xx ) = 0 2 2 r( t − x )− t = 0 Since, (as it follows from (2.18)), x = + 1 t and 2 xx = x equation (2.22) can be written as 1 1 1 + t − − t + (2 x − x ) = 0 2 2 2 which is equivalent to (2.23) x+ =− 1 t 2 The previous equation has the general solution (2.24) (x, t ) = − 1 ′( t ) + g ( t ) e − x 2 We can prove that the solution in (2.24) is compatible with (2.20). After substitution of (2.24), equation (2.19) takes the form − r g( t ) e − x − which is equivalent to − 1 ′′( t ) + { g ′( t ) − r g( t )}e − x = 0 2 1 ′′( t ) + g ′( t ) e − x = 0 2 From this equation we get 9 (2.25) and (2.26) − 1 ′′( t ) = 0 2 g ′( t ) − r g ( t ) = 0 ( t ) = a1 + a 2 t Equation (2.25) conveys that (2.27) and (2.26) that (2.28) g( t ) = a 3 e r t We therefore get from (2.24) that (2.29) 1 ( x, t ) = − a 2 + a 3 e − ( x − r t ) 2 From (2.16), (2.17) and (2.29), we determine ( x , t , u ) : (2.30) 1 (x, t, u ) = − a 2 u + a 3 e − ( x − r t ) u 2 We can now find x from (2.18) : x = a3 e −(x − r t) Integrating with respect to x we get (2.31) = −a 3 e − ( x − r t ) + f ( t ) Using (2.21), (2.27) and (2.31) we can determine the unknown function f ( t ) . The differential equation we get for the function f ( t ) is f ′( t ) = r a 2 Integrating with respect to t, we obtain (2.32) f (t) = a 2 r t + a 4 The function ( x , t ) can be derived from (2.31): (2.33) (x, t ) = a 2 r t − a 3 e − ( x − r t ) + a 4 The vector X, the generator of the symmetries, can be expressed as 10 X = (a 2 r t − a 3 e − ( x − r t ) + a 4 ) ∂ ∂ + (a 1 + a 2 t ) + ∂x ∂t 1 ∂ + − a 2 + a 3 e −(x − r t) u 2 ∂u The previous expression can be written as X = a1 ∂ ∂ ∂ 1 ∂ + + a2 r t +t − u ∂x ∂t ∂t 2 ∂u ∂ ∂ ∂ + a4 + a 3 − e −(x − r t) + e −(x − r t) u ∂x ∂u ∂x From the previous expression we get the generators of the symmetries given by (2.3)-(2.6). 3. A General Solution of the BCG Equation (1.1) After determining the symmetries of the BCG equation, we are now in a position to find its general solution. 3.1 Some preliminary Theorems We state and prove the following Theorem 3. Equation (2.1) takes the form yy + y =1 under the substitutions u (x, t ) = y = x−rt Proof. We consider a linear combination of the generators X1 , X 2 and X 4 : (3.1) where and X= 1+ 2+ 4 r ( y) 1+ r t are constants to be determined. We therefore consider the generator 11 (3.2) X = ( + t) ∂ ∂ 1 ∂ + ( r t + 1) − u ∂t ∂x 2 ∂u Another linear combination of the generators might be possible. Any other choice depends on the boundary or/and initial conditions. We can now determine the invariants corresponding to X given by (3.2). For this purpose, we are going to solve the system (3.3) dt dx = = + t rt + 1 du 1 − u 2 The equation dt dx = can be written as + t rt + 1 1− r ⋅ dx = r + t+ dt and by integration, x = r t + k ⋅ ln( t + ) + C1 where k = (3.4) 1− r . Therefore, one invariant of the equation is y = x − r t − k ⋅ ln( t + ) dt = + t du we get 1 − u 2 Integrating the equation 1 ln u = − ln( + t ) + C 2 or ln u = ln[( + t ) −1 / 2 A] , or 2 u = ( + t ) −1 / 2 A where A is a constant. Therefore, another invariant is given by (3.5) where = ( y) . = + tu 12 Using (3.4) and (3.5), we find ut = − 1 k ( t + ) −3 / 2 − r + 2 t+ ( t + ) −1 / 2 y u x = ( t + ) −1 / 2 y u xx = ( t + ) −1 / 2 yy Thus, under the transformations (3.4) and (3.5), the non-linear partial differential equation ∂ u 1 2 ∂ 2u 1 2 ∂ u + u + r + u =0 ∂ t 2 ∂ x2 2 ∂ x becomes the ordinary second order (nonlinear) differential equation − 1 2 −k y+ 1 2 1 2 yy + y =0 2 2 Under the choice k = 0 , i.e. (3.6) yy + y =1 1 = , the previous equation takes the form r where we have taken = 1 (any other choice might be possible). y = ln , Theorem 4. Under the transformation (3.7) equation (3.6) can be transformed to the equation (3.8) 2 +2 = 1 Proof. Introducing the substitution (3.7), since y = and yy = 2 + equation (3.6) takes the form (3.8). Theorem 5. Equation (3.8) admits a similarity transformation (3.9) = , = −1 w , w = w ( ) 13 and takes the final form (3.10) w = 1 w which is an Emden-Fowler equation. Proof. The proof is in Appendix A. Theorem 6. (I) The general solution of equation (3.10) is given by (3.11) = ±∫ dw ln w + 2A 2 +B where A and B are arbitrary constants. (II) The integral appearing in (3.11), can be expressed in terms of the erfi( x ) , (3.12) = ± e−A erfi 2 2 A + ln w 2 +B 2 the error function of complex argument (Appendix C). Proof. (I) Appendix B. (II) Under the substitution 2 ln w 2 + 2 A = 2 z , i.e. w = e − A e z , the integral in (3.11) takes the form 2 = ± 2 e − A ∫ e z dz + B from which, taking into account (C.3) and readjustment of the constants A and B, we arrive at (3.12). Note. The “Integrator” by Wolfram Research (integrals.wolfram.com) gives the following result for the integral in (3.11): e−a log( x 2 ) x 2 erfi a + 2 2 x ∫ 1 2a + log( x ) 2 dx = 14 which, for x > 0 , is essentially the result we have found in (3.12). The following Theorem provides the general solution of the equation (3.10). Theorem 7. The function w ≡ w ( ) , which is a solution to (3.10), is given by (3.13) where (3.14) =± 2 e− ( − ) w ( ) = exp{[(erfi) −1 ( )]2 − } The function (erf i) −1 ( x ) is the inverse error function of complex argument. Proof. From (3.12) we get (3.15) where erfi ( A + ln w ) = is given by (3.14). Inverting (3.15) we find A + ln w = (erfi) −1 ( ) Solving the previous equation with respect to w, we find w ( ) given by (3.13). 3.2 A General Solution Let us now put everything together.We see that the final equation depends on the variable variables. We have that = (Theorem 5), = e y (Theorem 4) and y = x − r t (Theorem 3) . Therefore we have to find the relation between and the original where x = ln S . Therefore = e y = e x − r t = e ln S − r t or (3.16) = S ⋅ e−r t (S, t ) = u ( x , t ) (Theorem 1), Concerning the unknown function, we see that where u ( x , t ) = 1 r ( y) (Theorem 3) and ( y) = w ( ) (Theorem 4). 1+ r t 15 Function w ( ) is given by w ( ) = exp{[(erfi) −1 ( )]2 − } (Theorem 7). Therefore ( y) = 1 S⋅e −r t exp{[(erfi) −1 ( )]2 − } , or (3.17) er t ( y) = exp {[(erfi) −1 ( )]2 − } S We thus arrive at the following Theorem 8. The general solution of equation (1.1) is given by (3.18) (S, t ) = = S ⋅ e − rt and r er t exp{[(erfi) −1 ( )]2 − } 1+ r t S =± e− ( − ) . where 2 This is the main result of our paper. 3.3 Inverting the Error Function We now have to find an analytical expression for the inverse of the error function of imaginary argument. In Appendix C we prove the relation (C.10): (3.19) (erfi) −1 ( ) = − i (erf ) −1 ( i ) where (erf ) −1 ( x ) is the inverse error function given by the series expansion (C.4) and implemented in many Symbolic Languages, like “Mathematica”. Using (C.4) and (C.10), we can prove that (erfi) −1 ( ) has the series expansion (C.11). Appendix A. Proof of Theorem 5. We try a transformation of the form (A.1) = , = w 16 under which equation (3.8) is invariant. Under this transformation we find that the partial derivatives are transformed to 2 and 2 = 2 −1 1 w+ 2 1 w ( − 1) 1 = + − 1 −1 1 −2 1 w+ 2 1 11 2 + + − 1 2 w + 1 2 w We therefore find that under the transformation (A.1) the differential equation (3.8) takes the form ( − 1) 1 + + 1 1 11 2 + + + 1 2 −2 1 w+ 1 −1 w + + 1 2 2 w = − w −1 The coefficients of w and w of the previous equation vanish simultaneously under the choice (A.2) = 1, = −1 = −1 w Equation (3.8) therefore under the transformation = , takes the form (A.3) w = w −1 17 This is an Emden-Fowler (type of) equation. Appendix B In this appendix we shall solve the Emden-Fowler equation (3.10). Consider the equation d2y d x2 = 1 y dy we get the equation dx Multiplying the previous equation by 2 2 dy d 2 y dy 1 ⋅ =2 ⋅ dx d x 2 dx y d dy dy 1 =2 ⋅ dx dx dx y 2 2 which can be written as dy dy Integrating with respect to x we find = 2 ∫ from which we get y dx dy 2 = ln y + 2 A , where A is an arbitrary constant. dx Therefore we have dy = ± ln y 2 + 2 A dx and by a further integration with respect to x, ±∫ dy ln y + 2A 2 2 +B=x where B is another arbitrary constant. Note 1. Another method of finding the general solution is by the substitution dy d2y dp = p and therefore = p⋅ . 2 dx dy dx 18 The original equation can then be converted into p dp = integrated, etc. dy which can be y Note 2. The general form of the Emden-Fowler equation d2y dx 2 = a x m yn can be solved for a variety of choices for m and n using Lie Symmetry Analysis. Appendix C The error function is denoted by erf ( x ) and it is defined by the integral (C.1) erf (z) = 2 z 0 ∫e −x2 dx The error function of imaginary argument is denoted by erfi( x ) and it is defined by (C.2) erfi( x ) = −i erf (i x ) We can prove, using (C.2) and (C.1), that (C.3) erfi(z) = 2 z 0 ∫ e x dx 2 which means that erfi( x ) is a real-valued function. The inverse of the error function is a real-valued function given by (C.4) (C.5) (C.6) erf −1 ck (x) = ∑ x k = 0 2k + 1 2 ∞ 2 k +1 c0 = 1 ck = c m ⋅ c k −1− m m = 0 ( m + 1)(2m + 1) k −1 ∑ 19 and is being implemented in many of the well-known Symbolic Languages like “Mathematica”. The first few terms of the previous expansion are given by erf −1 2 3 4 x + x 3 + 7 x 5 + 127 x 7 + 4369 (x) = x9 + 2 12 480 40320 5806080 In order to determine the function w ( ) we have to invert the erfi( x ) function. Let us explain briefly how to invert the erfi( x ) function. Suppose we want to invert the function (C.7) erfi( x ) = z Since, according to (C.2), erfi( x ) = −i erf (i x ) , equation (C.7) can be written as − i erf (i x ) = z and therefore erf (i x ) = i z , from which we get i x = erf −1 (i z) or (C.8) x = − i erf −1 (i z) Inverting directly (C.7), we have (C.9) x = (erfi) −1 ( z) (erfi) −1 (z) = − i erf −1 (i z) Comparing (C.8) and (C.9), we get the formula (C.10) Using the series expansion (C.4), we get from (C.8) ck x = −i ∑ iz k = 0 2k + 1 2 ∞ 2 k +1 Since − i ⋅ i 2k +1 = i 2k + 4 = i 2k = (−1) k , the previous relation can be written as (C.11) (erfi) (z) = ∑ z k = 0 2k + 1 2 −1 ∞ ( −1) k c k 2 k +1 Details about the error function and its inverse can be found for example in References [19]-[21]. 20 Appendix D In trying to find a general solution of the BCG equation (2.2), we have used a linear combination of the generators X1 , X 2 and X 4 . It is obvious that the generator X 3 corresponds to the so-called Tanh-method of traveling-wave solutions (see for example [22] and references therein). In this case we introduce a new variable by = x − r t . Using the fact that u t = − r U ′( ) , u x = U ′( ) and u xx = U ′′( ) equation (2.2) transforms into 1 1 − r U ′( ) + U 2 U ′′ + r + U 2 U ′ = 0 2 2 which is equivalent to U ′′ + U ′ = 0 and has a general solution U = C1 + C 2 e − This solution is not of any practical use (see also [3]). Appendix E. Lie Symmetries of the BCG Equation (1.2) We consider the BCG equation (1.2) given by ∂ ~ 1 2 2 ∂ 2~ ∂~ + + (r + ~) V =0 V ∂t 2 ∂V ∂ V2 where ~ ≡ ~(V, t ) . We make the substitution x = ln V . Since V2 ∂ 2~ ∂ V2 = ∂ 2u ∂ x2 − ∂u ∂x and V ∂~ ∂u = ∂V ∂x the BCG equation takes the form ∂ u 1 2 ∂ 2u 1 2 ∂u + + r − + u =0 ∂t 2 ∂x ∂ x2 2 where ~(V, t ) = u ( x, t ) with x = ln V . The Lie Symmetries of the previous equation are determined next. 21 Let ( x , t , u ( 2) ) be given by ( x , t , u ( 2) ) = u t + 1 2 1 2 + uux u xx + r − 2 2 Introduce the vector field X (the generator of the symmetries) by (E.1) X = (x, t, u ) ∂ ∂ ∂ + (x, t, u ) + (x, t, u ) ∂x ∂t ∂u The second prolongation is defined in our case by ∂ ∂ ∂ ∂ ∂ + t + xx + xt + tt pr ( 2) X = X + x ∂ux ∂ut ∂ u xx ∂ u xt ∂ u tt We have now to implement the equation pr ( 2) X [ ( x , t , u ( 2) )] = 0 . We have pr ( 2) X [ ( x , t , u ( 2) )] = 0 1 2 1 2 + uux = 0 pr ( 2) X u t + u xx + r − 2 2 1 2 ⇔ [ ux ]+ xr − + u + 2 1 2 + t + xx =0 2 For the functions , and we have = (x, t, u ) x = (t) , = (x, t ) , Thus, using the same expressions for get that 1 2 + u + [ ux ]+ xr − 2 t and xx as in the equation (1.1), we 1 2 + t + xx =0 2 or 22 (E.2) 1 2 + u [ x + ( u − x )u x ] + [ ux ]+ r − 2 + t − t u x + ( u − t )u t + + 1 2 { xx + (2 xu − xx )u x + 2 + uu u 2 + ( u − 2 x )u xx } = 0 x We substitute u t by − 1 2 1 2 + u u x into (E.2): u xx − r − 2 2 1 2 + u [ x + ( u − x )u x ] + [ ux ]+ r − 2 1 2 1 2 + t − t u x + ( u − t ) − + uux + u xx − r − 2 2 + 1 2 { xx + ( 2 xu − xx )u x + uu u 2 + ( u − 2 x )u xx } = 0 x 2 which can be written in the equivalent form (E.3) 1 2 1 2 + u x + t + r − xx + 2 2 + 1 2 + r − + u( u − x )− t − 2 1 2 1 2 − ( u − t ) r − + u + (2 xu − xx ) u x + 2 2 1 1 + [ − ( u − t ) 2 + ( u − 2 x ) 2 ] u xx + 2 2 + 1 2 2 uu u x = 0 2 Equate all the coefficients of the partial derivatives of the function u to zero. We then have a system of partial differential equations: 23 • Coefficient of the zero-th order derivative of u: (E.4) 1 2 1 2 + u x + t + r − xx = 0 2 2 • Coefficient of u x (E.5) 1 2 + r − + u( u − x )− t − 2 1 2 1 2 − ( u − t ) r − + u + (2 xu − xx ) = 0 2 2 • Coefficient of u xx : (E.6) 1 1 − ( u − t) 2 + ( u −2 x) 2 = 0 2 2 • Coefficient of u 2 : x (E.7) 1 2 uu = 0 2 uu = 0 , which means that We are to solve the system of equations (E.4)-(E.7) From (E.7) we get to u: (E.8) ( x, t, u ) = (x, t ) ⋅ u + ( x, t ) is a linear function with respect From (E.6), after canceling opposite terms and simplifying, we get the equation (E.9) x = 1 ′( t ) 2 which has the general solution (E.10) (x, t ) = 1 ′( t ) ⋅ x + ( t ) 2 From the previous equation we get (E.11) (E.12) t = 1 ′′( t ) ⋅ x + ′( t ) 2 xx = 0 24 From equation (E.5), after canceling opposite terms and making the substitutions (E.8)-(E.12), we get the equation (E.13) ( ⋅u + ) + − 1 1 2 ′( t ) r − + u − 2 2 1 ′′( t ) ⋅ x − ′( t ) + 2 x = 0 2 The previous equation can be written as 1 ′ ( t ) u + + 2 + + 1 1 2 1 2 ′( t ) r − − ′′( t ) ⋅ x − ′( t ) + x=0 2 2 2 From the previous equation we derive the two equations (E.14) (E.15) + 1 ′( t ) = 0 2 + 1 1 2 1 2 ′( t ) r − − ′′( t ) ⋅ x − ′( t ) + x =0 2 2 2 by (E.8), we get From equation (E.4), and substituting 1 2 + u( x ⋅ u + x ) + ( t ⋅ u + t ) + r − 2 + 1 2 ( xx ⋅ u + xx ) = 0 2 which can be written in equivalent form 1 2 1 2 r − x+ t+ xx + 2 2 1 2 + r − x+ 2 x+ t+ 1 2 xx u + 2 xu 2 =0 From the previous equation we get 25 (E.16) 1 2 1 2 r − x+ t+ xx = 0 2 2 1 2 r − x+ 2 x =0 x+ t+ (E.17) (E.18) 1 2 xx = 0 2 From (E.14) we get (E.19) (x, t ) = − 1 ′( t ) 2 which is compatible with (E.18). Since xx = 0 , we get from (E.17) that (E.20) x + t =0 and therefore, using (E.19) and the above equation, we find (E.21) x = 1 2 ′′( t ) and so integrating the previous equation we obtain (E.22) (x, t ) = 1 x ⋅ ′′( t ) + ( t ) 2 From (E.16), using (E.21) and the fact (as it follows from (E.22)) t = 1 x ⋅ ′′′( t ) + ′( t ) 2 we derive the equation (E.23) 1 2 1 r − 2 2 ′′( t ) + 1 x ⋅ ′′′( t ) + ′( t ) = 0 2 The previous equation holds for any x if (E.24) and (E.25) ′′′( t ) = 0 1 2 1 ′′ ( t ) + ′( t ) = 0 r − 2 2 Finally, equation (E.15) can be written, using the previous expressions, 26 1 1 1 2 1 ′′ x ⋅ ′′( t ) + ( t ) + ′( t ) r − ( t ) ⋅ x − ′( t ) = 0 − 2 2 2 2 which is equivalent to the equation (E.26) (t) + 1 1 2 ′( t ) r − − ′( t ) = 0 2 2 Equation (E.25) gives (E.27) ( t ) = a1 + a 2 t + a 3 t 2 We can now determine ( t ) from equation (E.24). In fact, using (E.27) we find ′( t ) = − and by integration (E.28) 1 1 2 (t ) = − r − a 3t + a 4 2 1 1 2 r − 2a 3 2 2 Finally we can determine ( t ) from equation (E.26). We find, using (E.27) and (E.28) 1 2 1 1 1 2 ′( t ) = − r − a 3t + a 4 + r − ( a 2 + 2a 3 t ) 2 2 2 and by integration, (E.29) 1 1 2 (t) = a 4 t + r − a 2t + a5 2 2 From (E.19) we can determine ( x , t ) : (E.30) 1 ( x, t ) = − (a 2 + 2a 3 t ) 2 and from (E.22) the function ( x , t ) (E.31) ( x, t ) = 1 x ⋅ a3 − 1 1 2 r − a 3t + a 4 2 Using (E.30) and (E.31) we can determine the function ( x, t ) from (E.8): 27 (E.32) 1 1 1 1 2 ( x , t ) = − (a 2 + 2a 3 t ) u + x ⋅ a 3 − r − a 3t + a 4 2 2 The function ( x , t ) can be derived from (E.10), using (E.27) and (E.29): (E.33) 1 1 1 2 ( x , t ) = − (a 2 + 2a 3 t ) x + a 4 t + r − a 2t + a5 2 2 2 1 1 2 1 ∂ + X = − (a 2 + 2a 3 t ) x + a 4 t + r − a 2 t + a5 2 2 2 ∂x + (a 1 + a 2 t + a 3 t 2 ) ∂ + ∂t The vector X, the generator of the symmetries, can be expressed as 1 1 1 2 1 ∂ + − (a 2 + 2a 3 t ) u + x ⋅ a 3 − r − a3 t + a 4 2 2 ∂u The previous expression can be written as X = a1 ∂ + ∂t 1 ∂ 1 ∂ 1 1 2 ∂ + a 2 − x + r − + t − u + t ∂t 2 ∂u 2 2 ∂x 2 ∂ ∂ 1 1 1 2 ∂ + a 3 − t x + t2 + − tu + x − r − t + ∂x ∂t 2 ∂u + a4 ∂ ∂ + a5 ∂u ∂x Therefore, the generators of the symmetries are: X1 = ∂ ∂ ∂ ∂ , X 2 = (x − t) − 2t + u ∂x ∂t ∂u ∂t ∂ ∂ ∂ + t2 + (− t u + x − t ) ∂x ∂t ∂u X3 = − t x 28 X4 = where ∂ ∂ , X5 = ∂u ∂x 1 2 2 =r− The above generators can be used in finding a solution of the BCG equation (1.2) along the lines of Section 3, once the initial conditions are given. 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Bensoussan for kindly reading a preliminary draft of this report. 31