1
? b2i ? ? b21 ? =? B = ? bi = ? .. .. .. .. b2n ? .. ? 2 CHAPTER 1. DIMENSIONAL ANALYSIS to deem dimensionless quantities as large or small. The last assumption of D.A. is that formula (1.1) acts as a dimensionless equation in the sense that (1.1) is invariant under arbitrary scaling of any fundamental dimension i.e. it is independent of the choice of a system of units.
Next, we see a particular example in order to clarify all these abstract ideas. For this purpose we may consider the well known law of Newton F = ma = m d2x dt2 , (1.4) where we are interested in calculating the velocity, v, of some body, so v is the quantity to calculate. If we rewrite the above equation (1.4) as Fm-1x-1t2 = 1, (1.5) we may choose as dimensional base the following one B = {x,m,t} that we rewrite as B = {L, M,T}, (1.6) where L stands for length, M for mass and T for time, where the elements belonging to the dimensional base are usually written in capital letters instead of small letters. Note that we have only one equation and four quantities, hence the multiplicity of the dimensional base is: (number of quantities)-(number of equations), in fact is the rank of the dimensional matrix (see below, the recipe, for clarifying this calculation). In the next chapter we will clarify this point. In this way we say that the dimensional equation of the force F is [F] = LMT-2, while the dimensional equation of the quantity v is [v] = LT-1. (1.7)
(1.8) Other bases can be obtained, but I am not interested in exploring this fact. We only stress that it is very important calculate the appropriate base for each problem as we will see in the following sections (see section of examples). Now we may check if our equation veri?es the principle of dimensional homogeneity, in such a way that [F] = [m][a], LMT-2 = M · LT-2, (1.9) since [a] = LT-2. i This example is trivial, but we are only trying to clarify these new concepts.
1.2 The Recipe.
The above concepts are the main assumptions of the Pi-Theorem. Pi-Theorem allows us to calculate the relation- ship between the quantity u and the set of quantities (Wi)n =1, the set of variables and constants, characteristic and universal. These assumptions bring us to the following conclusions:
1. Formula (1.1) can be expressed in term of dimensionless quantities. 2. The number of dimensionless quantities is (1.10) k + 1 = n + 1- rg(B) where rg(B) is the rank of the dimensional matrix. This matrix is formed by the column vectors bmi b1i b11 b1n ? ? ?
? ? ? ? . ? ? . b12 b22 . ··· ··· . bm1 bm2 ··· bmn ?
? . ? (1.11)
? x(i) = ? .. ? ?, ? a = ? .. ? ?, ? y = ? .. ? ?, p = uW11W22….Wnn, pi = W1 1iW2 2i….Wnni, u = W1 W2 ….Wn [Wi] = L11i…..Lm mi, 3 i 1.3. THE PI-THEOREM.
where bi is the dimension vector of Wi,i = 1,…,n. Precisely k of these dimensionless quantities depend on the measurable quantities (Wi)n =1. 3. Let ? ? ? x1i x2i . ? ? ? (1.12) xmi with i = 1,…,k, represent the k = n – rg(B) linearly independent solutions x of the system Bx = 0. Let ? ? ? a1 a2 . am ? ? ? (1.13) be the dimension vector of the quantity u, and let ? ? ? y1 y2 . ? ? ? (1.14) (1.15) i yn represent a solution of the system By = -a. Then formula (1.1) simpli?es to p = g(p1,….,pk),
where p,(pi)k=1 are dimensionless quantities given by y y y x x x (1.16) with i = 1,…k, and g is some unknown function of its arguments. In particular eq. (1.1) becomes -y1 -y2 -yn g(p1,….,pk). (1.17) 1.3 The Pi-Theorem.
Now we will give a brief sketch of the proof of this famous theorem. First of all a a [u] = L11…..Lm m, b b (1.18) with i = 1,…n. Next, we use assumption (vi) and consider the invariance of eq. (1.1) under arbitrary scaling of the fundamental dimensions by taking each fundamental dimensions in turn. We ?rst scale L1 by letting 1 L* = eeL1, e ? R, (1.19) in turn this induces the following scalings of the measurable quantities u* = eea1u, i W* = eeb1iWi, i = 1,…n. (1.20) Equations(1.20)de?neaone-parameter(e)Liegroupofscalingtransformationsofthe n+1quantities(u,W1,W2,…,Wn) with e = 0 corresponding to the identity transformation. This group is induced by the one-parameter group of scalings (1.19) of the fundamental dimension L1.
Xi-1 = WiW1 1i 11, i = 2,3,…n, (1.23) v = uW1 p = u?Wi i, xmj CHAPTER 1. DIMENSIONAL ANALYSIS 4
From the assumption (vi), eq. (1.1) holds iff 1 2 * u* = f (W*W*…..Wn), (1.21) i.e. eea1u = f eeb1iWi n i=1 , ?e ? R, (1.22) then two cases need to be distinguished: Case I. b11 = …. = b1n = 0 = a1. Here L1, is not a fundamental dimension of the problem, i.e. eq. (1.1) is dimensionless with respect to L1. and a1 = 0. Hence u = 0 is a trivial solution. b1i = 0 for some i = 1,…n. Without loss of generality we assume b11 = 0. We de?ne
-b /b Case II. b11 = …. = b1n = 0, Therefore it follows that ? new measurable quantities
and let We choose as the new unknown Xn = W1.
-a1/b11 , (1.24)
(1.25) i i i i i i the transformation given by eqs. (1.23)-(1.25) de?nes a one-to-one mapping of the quantities (Wi)n =1 -? (Xi)n =1 , and a one-to-one mapping of the quantities u,(Wi)n =1 -? v,(Xi)n =1 , in this way eq. (1.1) is equiv- alent to v = F (Xi)n =1 , (1.26) where F is a unknown function of (Xi)n =1 . Thus the group of transformations (1.20) becomes v* X* i = v, = Xi, i = 1,2,….n – 1, (1.27) (1.28) * Xn = e eb11 Xn, (1.29) i – i therefore v,(Xi)n =1 1 are invariants of (1.20). Moreover, the quantities v,(Xi)n =1 satisfy assumptions (iii), and eq. (1.26) satis?es assumption (vi). Therefore i – v = F (Xi)n =1 1 ,eeb11Xn , ?e ? R. (1.30) i – i i where F is independent of the quantity Xn. Moreover, the measurable quantities (Xi)n =1 1 are products of powers of (Wi)n =1 and v is a product of u and powers of (Wi)n =1 . Thus eq. (1.1) reduces to i – v = G (Xi)n =1 1 , (1.31) i – where v;(Xi)n =1 1 are dimensionless with respect to L1 and G is some function of (n – 1) arguments. If we repeat this tactic with the (m – 1) fundamental dimensions, we reduce eq. (1.1) to a dimensionless formula i (1.32) i p = g (pi)k=1 ,
where [p] = [pi] = 1, with i = 1,…,k, and g is some function of (pi)k=1 , being n
i=1 y pi = k ?Wm j=1 , (1.33) with m = 1,…,n.
? x = ? .. ? ?, uW11W22….Wnn = 1, 2 0 -1 -3 ? ? 2 0 -1 -3 ? x2 ? ? x3 ? = 0, -2 1 -2 0 ? x1 = -5 2×4 x2 = 5 6×4 x3 = -35×4 1.4. EXAMPLES. 5 Note that this theorem makes no assumptions about the continuity of the function f and hence of g with respect of any of their arguments. It is shown immediately that the number of measurable dimensionless quantities is k = n – rg(B). This follows since x iff ? ? ? x1 x2 . ? ? ? (1.35) xn satis?es Bx = 0. This equation has k = n – rg(B) linearly independent solutions x(i) given by (1.12). The real numbers (1.14) follow from setting y y y (1.36) leading to y satisfying By = -a.
1.4 Examples.
In this section we will try to explain how works DA through different examples and how we can avoid the cases which we obtain a result with an unknown function g. For this purpose we will follow two different tactics, the ?rst of them will consist in making a mathematical hypothesis on the behaviour of this unknown function and we will justify (as far as possible) such assumption. In the second of our tactics we will use the so called dimensional discrimination. Such tactic allows us to obtain a complete solution for our problems and it is not needed to make any previous assumption.
Ejemplo 1.4.1 The famous atomic explosion of 1945.
Solution. We are interested in knowing the radius R of the shock wave that is produced an atomic explosion. For this purpose we treat R as the unknown (u = R with the previous notation) and assume that R = f(W1,W2,W3,W4), (1.37) where W1 = E is the energy released by the explosion, W2 = ?0 is the ambient air density, W3 = P0 is the ambient air pressure, W4 = t is the elapsed time after the explosion takes place. For this problem we use B = {L, M,T} as dimensional base. therefore the corresponding dimensional matrix is given by L M T E t 2 0 1 0 -2 1 P0 -1 1 -2 ?0 -3 1 0 , / ? B = ? 1 0 -2 1 ? 1 1 ? -2 0 (1.38) since for example [E] = L2MT-2, etc… and as we can see the rg(B) = 3 therefore k = n – rg(B) = 4 – 3 = 1 is the number of p – monomia. Following the above explained procedure we ?nd that Bx = 0 i.e. x1
x4 ? ?
? 1 0 1 1 ?? ? =? ?
? , (1.39)
1 -1 1 ? -2 ? 5 ? 1 ? Et t E2?0 1/5 n 3n-1 ? = E 5 t 5 ? 5 P0 n, (1.46) 6 CHAPTER 1. DIMENSIONAL ANALYSIS where x4 is arbitrary. Setting x4 = 1 we obtain the measurable dimensionless quantity p1 = P0 t6 E2?30 1/5 . (1.40) Now the dimension vector of R is ? ? a = ? 0 ?, 0 [R] = L, (1.41) the general solution of By = -a is y = 0 ? ?
? ? + x, (1.42) where x is the general solution of Bx = 0. Setting x = 0, we obtain the dimensionless unknown p = R Et2 ?0 -1/5 , (1.43) therefore the solution that DA gives us is the following one R = Et2 ?0 1/5 g(p1), (1.44) where g is some function of p1. How to solve the big drawback of having an unknown function g? We may make the following assumption (called the power law hypothesis, from now on) p = (p1)n, n ? R, (1.45) i.e. we are supposing that the general solution follows a power law and we ?x the value of the numerical constant n through experiment. Therefore the solution that we suggest take the following form R = 2 ?0 1/5 ? ?P0 6 3 ? 1-2n 2+6n 0 basically we are assuming that R ˜ Ct 2+6n 5 near t = 0. (1.47) where C is a constant. Numerical values from the experiments (i.e. after many nuclear explosions) bring us to obtain n = 0, so the solution is R = E ?0 1 5 2 t 5 =? R = At2/5. (1.48) Observación 1.4.1 The short cut. The problem is easily reduced to: B = {L, M,T} (1.49) R = f (E,t,P0,?0), where the dimensional matrix is given by L M T R 1 0 0 E t 2 0 1 0 -2 1 P0 ?0 -1 -3 1 1 -2 0 . (1.50)
R = EaEtat?0 0P0 0, ?0 1.4. EXAMPLES. 7 Since the number of quantities is 5 and the multiplicity of the dimensional base is 3 therefore we will obtain 2 p – monomials, i.e. a solution like p = g(p1) where p = f(R;E,t,?0) and p1 = f1 (E,t,P0,?0). From the dimensional matrix (1.50) we see that = 0, = 0, = 0 (1.51) (1.52) (1.53) – 1+ 2aE – aP0 – 3a?0 aE + aP0 + a?0 -2aE + at – 2aP0 therefore the solution under the assumption p = (p1)n is a? aP i.e. R = E 1-2n 5 t 2+6n 5 3n-1 5 Pn 0 (1.54) as we can deduce in a trivial way from the above system of linear equations (1.51)-(1.52).
We need to avoid this kind of assumptions, although they work well, or at least to try to justify mathematically why we may expect this class of solutions i.e. power law solutions.
Ejemplo 1.4.2 The range of a projectile. A projectile of mass m weight W and with initial velocity v, is launched in horizontal direction from a height h. We are interested in knowing the range x of the projectile.
Solution. In this case we will explore two possibilities, in the ?rst of them the application of DA brings us to obtain an unsatisfactory solution since this depends on an unknown function. To solve this drawback we explore another possibility which consists in enlarge the dimensional base taking into account the spacial discrimination.
1. Brie?y the problem is reduced to: x = f (m,W,v,h), (1.55) that with respect to the dimensional base B = {L, M,T} it is obtained the following dimensional matrix L M T x 1 0 0 m 0 1 0 W 1 1 -2 v 1 0 -1 h 1 0 0 , =? p = x h , p1 = mv2 hW (1.56) therefore the solution will be p = ?(p2), i.e. x = h? mv2 hW . (1.57) As we have already commented this solution is very poor for a very simple problem, for this reason we will try another tactic. 2. As we can see there are two privileged spatial directions in this problem Lx, Ly . For this reason, under this new consideration, the problem is reduced to (as above): x = f (m,W,v,h), (1.58) but with respect to the dimensional base B = matrix x Lx, Ly, M,T
m W v it is obtained the following dimensional
h Lx Ly M T 1 0 0 0 0 0 1 0 0 1 1 -2 1 0 0 -1 0 1 , 0 0 (1.59)
8 CHAPTER 1. DIMENSIONAL ANALYSIS
which bring us to obtain only one p – monomia. Therefore the solution is in this case x = v hm W = v h g , (1.60) as it is expected.
With this example we have tried to show another tactic to solve physical problem and without the necessity of doing assumptions on the behaviour of the solution as in the previous example.
Ejemplo 1.4.3 The golfer and the moral. We are interested in knowing how long, t, must train a golfer of mass m and weight W, in order to hit a golf-ball to a distance l.
Solution. As in the above example we may see that the set of governing parameters is (1.61) P = P (t;m,W,l), that with respect to the dimensional base B = {L, M,T}, we obtain the following the dimensional matrix t m W l L M T 0 0 1 0 1 0 1 1 1 0 -2 0 , =? t = C l g , (1.62) from which we obtain a single monomial, as it is observed in a trivial way. C ? R.
Observación 1.4.2 Obviously this solution is a complete nonsense since, from our point of view, it is necessary that the problem must be considered within some physical theory (and this is not the case). DA is subordinate to one physical theory and not the other way round as we may deduce from this simple example although some times one may found some interesting relationship between the quantities i.e. with physical meaning. 1.5 Applications of dimensional analysis to ODEs and PDEs
In this section we will try to show how to apply D.A. to solve differential equations and partial differential equations. 1.5.1 Differential equations. The method of the Lie groups it has showed as a very useful tool in order to solve nonlinear equations (nl-ODE) as well as PDE. Nevertheless, when one is studying ode of ?rst order its application becomes very complicated (very tedious) if one has not a computer algebra package since if one decides to look for the possible symmetries of an ODE with pencil and paper this task may be turned very exhausting. However, if we know that an ODE admits a concrete symmetry, then it is a trivial issue to ?nd new variables which allows us to rewrite the ODE in quadratures or as we will see in this paper to obtain an ode with separating variables.
y' = Cx3y3 + Bxy2 – A , y' = a2Cx3y3 + aBxy2 – A , ln u + u + 1 + 1.5. APPLICATIONS OF DIMENSIONAL ANALYSIS TO ODES AND PDES 9 Our purpose in this work is to explain, through an example, how D.A. works in order to ?nd these changes of variables (c.v.) in a trivial way, i.e. without the knowledge of the symmetries of the ODE under study. The idea is as follows. When we are studying an ODE from the dimensional point of view, we must require that such ODE veri?es the principle of dimensional homogeneity (pdh) i.e. that each term within the equation have the same dimensions, for example, speed, or energy density. To clarify this concept we consider the following y x equation of the quantity ·. As we can see, this ODE does not verify the pdh, since the term x has dimensions of x, i.e. [x] = x. In order to do that this equation veri?es the pdh we need to introduce dimensional constants, in such a way that after rewrite the ODE with these constants the ODE now veri?es such principle i.e. in this example and as we can see easily if we consider only one constant a, such that [a] = yx-2, we make that the y x studying physical or engineering problems since, as it is supposed, such problems (equations) verify the pdh and we do not need to introduce new dimensional constants, that must have physical meaning, for example, the viscosity coef?cient, etc…. Precisely this dimensional constant suggests us the c.v. (x,y(x)) -? (t,u(t)) where t = x,u = yx-2 in such a way that rewriting the original ODE in these new variables yields a new ODE with separating variables: u + u't = 1. The reason is the following. We know from the Lie group theory that if it is known a symmetry of an ODE then this symmetry bring us through a c.v. to obtain a simpler ODE (a quadrature or a ODE with separating variables) and therefore the solution is found in a closed form. To ?nd this c.v. one uses the invariants that generate each symmetry, i.e. the ?rst principle is that it is useful to pass to new coordinates such that one of the coordinate functions is an invariant of the group. After such transformation it often (but not always) happens that the variables separate and the equation can be solved in closed form. We must stress that taking the new dependent variables to be an invariant of the group does not guarantee the separation of variables. The choice of the independent variable is also very important. One must note that D.A. gives us this invariant (or at least a particular solution). It is observed in our example that the solution y = x2, (suggested by D.A.) is a particular solution for this ODE, but if we study this ode form the Lie theory point of view it is obtained that such ODE has the scaling symmetry X = x?x +2y?y and therefore the solution y = x2, is furthermore an invariant solution (generated by X), for this reason such c.v. works well. As we will show in the following example, the c.v. that D.A. induces is the same as the one generated by the symmetries of the ODE. In order to make see this fact, we will solve the example by D.A. as well as by the Lie group technique, calculating the symmetries and their corresponding invariants and c.v.. Ejemplo 1.5.1 Solve the Abel ODE. y x (1.63) Solution. If we rewrite eq. (3.256) introducing the following dimensional constants, y x (1.64) where A,B,C ? R, and [a] = X-2Y-1. As we can see this ODE is scale invariant since we have needed to introduce only one constant. DA suggests us the following c.v. t = x, u(t) = ax2y =? x = t, y = u at2 , (1.65) in such a way that eq. (3.256) is written in the following form: tu' = u u2 + u + 1 , (1.66) and its solution is: lnt + 1 2 2 v 3 3 arctan 3 2 u + 1 3 v 3 – lnu + C1 = 0, (1.67)
ax y + ax y + 1 + y' = Cx3y3 + Bxy2 – A , y' = aCx3y3 + ßBxy2 – A , x CHAPTER 1. DIMENSIONAL ANALYSIS 10
therefore in the original variables it yields: lnx + 1 2 ln 2 2 2 v 3 3 arctan 3 2 ax2y + 1 3 v 3 – ln ax2y + C1 = 0. (1.68) In second place, we study eq. (3.256) y x (1.69) with respect to the dimensional base B = {T}. This ODE veri?es the principle of dimensional homogeneity with 1 H' rewriting the equation in a dimensionless way we ?nd that y ? x-2. This ODE has been obtained studying a cosmological model and therefore H is the usual Hubble parameter (see the next chapter to understand the meaning of this quantity) and therefore [H] = T-1 as it is usual and T stands for dimensions of time. But if we study this equation with respect to the dimensional base B = {X,Y}, we need to introduce new dimensional constants that make the equation verify the principle of dimensional homogeneity y x (1.70) where a1/2 = [ß] = X-2Y-1, hence X Y y 0 1 ß x -2 1 -1 0 =? y ? 1 ßx2 , (1.71) which is a particular solution of the original ode. Lie Method. In order to ?nd the symmetry generator, X = ?(x,y)?x + ?(x,y)?y, of a ODE y' = f (x,y), we need to solve the following PDE ?x + (?y – ?x)f – ?y f2 – ?(x,y)fx – ?(x,y)fy = 0, (1.72)
(1.73) where fx = df dx , fy = df dy . (1.74) see the appendix in order to clarify notation and concepts. In this case we have to solve ?x + (?y – ?x) Cx3y3 + Bxy2 – A y x – ?y Cx3y3 + Bxy2 – A y x 2 – – ?(x,y) 3Cx2y3 + By2 + A y 2 – ?(x,y) 3Cx3y2 + Bxy – A x = 0, (1.75) (1.76) This ODE admits the following symmetry X = x?x – 2y?y, which is a scaling symmetry and it induces the following change of variables, Xr = 0, Xs = 1, (1.77) r = x2y, s(r) = ln(x), =? x = es(r), y = r e2s(r) , (1.78)
v 2 2Cx y+B =? lnx = – lny =? 11 1.5. APPLICATIONS OF DIMENSIONAL ANALYSIS TO ODES AND PDES
which brings us to obtain the next ode in quadratures s' = 1 r(Cr2 + Br + 2- A) , (1.79) and whose solution is: s(r) = – lnr A – 2 + 1 ln Cr2 + Br + 2- A 2 A – 2 – 2 +4C(B Barctanh (A – 2) vB2 Cr+ A-2) B2 + 4C(A – 2) + C1, (1.80) and hence in the original variables (x,y): lnx = – ln x2y A – 2 + 1 ln Cx4y2 + Bx2y + 2- A 2 A – 2 – Barctanh (A – 2) 2 B +4C(A-2) B2 + 4C(A – 2) + C1, (1.81) which is the most general solution for this ODE. Now if we consider the invariants I that induce the symmetry X i.e. dx ? = dy ? , -? y' := dy dx = ? ? , (1.82) Hence, for example the symmetry X = x?x – 2y?y generates the following invariant: dx ? = dy ? =? dx x = – dy 2y 1 2 I = x2y -? y = ax-2, (1.83) this would be the solution that suggests us precisely the direct use ofthe Pitheorem (ifthe ODE is scale invariant, as in this case, the solution obtained applying by the Pi theorem coincides with the invariant solution that generates the scale symmetry, in this case X). We see that we only obtain a particular solution, but that this is invariant, in fact if we think about the ODE as a dynamical system we see that the ?xed point of such equation would be precisely the solution y = Kx-2.
Observación 1.5.1 Usually the most general solution of this kind of ODEs is unphysical i.e. it lacks of physical meaning. So, why the invariant solution has physical meaning and is stable from the dynamical point of view? 1.5.2 Partial differential equations. i i i i i l i Using the notation of the section 2, now we consider a boundary value problem for a partial differential equation (PDE) u,(Wi)n =1 where u is the unknown function (the dependent variable of the PDE) and (Wi)n =1 are the independent variables and constants. From the Pi-theorem it follows that the PDE may be re-expressed in dimensionless form p,(pi)k=1 , being p the dimensionless dependent variable and (pi)k=1 the dimensionless independent variables and dimensionless constants. Considering the set of quantities (Wi)n =1 we distinguish between (Wi)i=1 variables and (Wi)n =l+1 constants. Let B be the dimensional matrix such that B = (B1 | B2), (1.84) i l i of all (Wi)n =1 and B1 = M((Wi)i=1) the dimensional matrix of the quantities and B2 = M((Wi)n =l+1) the dimensional matrix of the constants.
12 CHAPTER 1. DIMENSIONAL ANALYSIS An important objective in applying D.A. to PDE is to reduce the number of independent variables. The rank of B2 i.e. r(B2) represents the reduction in the number of constants through DA. Consequently the reduction in the number of independent variables is ? = r(B) – r(B2). In particular the number of dimensionless measur- able quantities is k = n – r(B) = [(l – ?) + (n – l – r(B2))], where (l – ?) of the quantities are dimensionless independent variables and (n – l – r(B2)) are dimensionless constants. If ? = r(B) – r(B2) = 0, then DA reduces the given boundary value problem to a dimensionless boundary problem with (n – l – r(B2)) dimensionless constants. In this case the number of independent variables is not reduced. If l = 2, (l – ?) = 1, then the resulting solution of the boundary value problem is called self-similar.
Ejemplo 1.5.2 Stokes ?rst problem. Let us consider a motionless ?uid in the proximity of a plane wall. If the wall starts moving abruptly along its own plane (starting current) with a constant speed U0, the Navier-Stokes equation reduce to: ?u ?t = ? ?2u ?y2 . (1.85) Find the velocity distribution of the ?uid as function of y, the distance to the plane as well as function of, t :i.e. u = u(y,t) Solution. Let u be the ?uid velocity in the direction of U0 (x-direction), in this case the Navier-Stokes equations reduce to the viscous diffusion equation = ? , (1.86) ?t with the boundary conditions: u(y,0) = 0, ?y > 0, (1.87) u(8,t) = 0, and u(0,t) = U0, t > 0 (1.88) where ? is the “kinematic viscosity”. We are going to explore two dimensional tactics, the ?rst of them is the usual one while the second will consist in the spatial discrimination.
1. In this case the dimensional base isB = {L,T} and therefore we obtain the following dimensional matrix: L T u 1 -1 U0 1 -1 y 1 0 t 0 1 ? 2 -1 (1.89) y obtaining in this way 3 monomials and therefore we do not reduce the number of variables in the problem. We have insisted in outlining this tactic in order to show the useful of the spatial discrimination as follows. 2. In this case the dimensional base is B = Lx, Ly,T and therefore the new dimensional equations of each quantity are: [?] = L2T-1, [u] = [U0] = LxT-1, [y] = Ly, [t] = T (1.90) The most general solution is: f (y,t,u,U0,?) = 0, (1.91) we have the following dimensional matrix: Lx Ly T u 1 0 0 U0 1 0 0 y 0 1 0 t 0 0 1 ? 0 2 -1 , (1.92)
p1 = v ? = v d? z' = – sz 1.5. APPLICATIONS OF DIMENSIONAL ANALYSIS TO ODES AND PDES 13 obtaining two p – monomia : p = u U0 , y ? t , (1.93) then, new variables ? and U are de?ned as y ? t , U (?) = u U0 , i.e. u = U0 · U (?), (1.94) Now if we calculate ? u ? t and ?2u ? y2 as functions of U and ?, we obtain d2U 2 + ? dU 2 d? = 0 (1.95) with the boundary conditions: U(0) = 1, U(8) = 0, (1.96) where U = 0 for t = 0, y > 0 (as, ? ? 8) and U = 1 for t ? 8, y = 0 (as, ? = 0). Therefore the solution is: U = erf ? 2 . (1.97) For example, the change of variables U' = z, ? = s, (1.98) brings us to obtain the following ode 1 2 =? z = C1e-s 2/4 , (1.99) and making the inverse change of variables U' = C1e-? 2/4 , =? U = C1 e-? 2/4 dx = erf ? 2 , (1.100) it is obtained the solution (1.97).
In this way we can see how works the dimensional discrimination, showing us that is it a powerful tool to solve partial differential equations.
Chapter 2
Introduction to the symmetry methods.
What are the symmetries of an ODE?. Consider the simplest example y' = dy dx = 0, (2.1) ˆ ˆ whose solution is obviously the set of straight lines y = c, with c ? R. The ODE (2.1) is represented geometrically by the set of all solutions and so any symmetry of the ODE must necessarily map the solution set into it self. Formally, the condition that any symmetry maps the object into itself requires that the set of solutions curves in the (x,y)-plane must be indistinguishable from the image in the (x,y)-plane, so ˆ y' = 0 when y' = 0. (2.2) A smooth transformation of the plane is invertible if its Jacobian is nonzero so we impose the further condition ˆ ˆ ˆ ˆ (2.3) xxyy – xyyx = 0. A particular solution curve will be mapped to a (possibly different) solution curve, and so ˆ ˆ y(x,c) = c(c), ?c ? R. (2.4) ˆ ˆ ˆ ˆ Here x is regarded as a function of x and c that is obtained by inverting x = x(x,c). ODE (2.1) is very simple and so all of its symmetries can be found. Differentiating (2.4) with respect to x, we obtain yx(x,c) = 0, ?c ? R, therefore taking (2.3) into account, the symmetries of (2.1) are of the form ˆ ˆ (x,y) = (f(x,y), g(y)), fx = 0, gy = 0. (2.5) ˆ ˆ where f and g are assumed to be smooth functions of their arguments. The ODE has very large family of symmetries. We were able to use the know general solution of (2.1) to derive (2.2) which led to (2.5). However, we could also have obtained this result directly from (2.2) as follows. On the solution curves, y is a function of x, and hence x = (x,y) and y = (x,y) may be regarded as a function of x. Then by the chain rule (2.2) can be written as ˆ ˆ d y dx = ˆ ˆ Dxy Dxx = 0, when y' = 0, (2.6) ˆ ˆ where Dx denotes the total derivative with respect to x, i.e. Dx = ?x + y'?y + y''?y' + …, therefore (2.2) amounts to
xx + y'xy and therefore (2.5) holds.
15
16 CHAPTER 2. INTRODUCTION TO THE SYMMETRY METHODS. The advantage of using the symmetry condition in the form (2.2) is that one can obtain information about the symmetries without having to know the solution of the differential equation in advance. For an ODE y' = ?(x,y), (2.8) the symmetry condition reads ˆ ˆ ˆ y' = ?(x,y), when y' = ?(x,y). (2.9) As before, we regard y as function of x on the solution curves, therefore (2.9) yields ˆ ˆ Dxy Dxx = ˆ ˆ ˆ ˆ yx + y'yy xx + y'xy ˆ ˆ = ?(x,y), when y' = ?(x,y), (2.10) therefore the symmetry condition for (2.8) is equivalent to the constraint ˆ ˆ ˆ ˆ yx + ?(x,y)yy xx + ?(x,y)xy ˆ ˆ = ?(x,y), (2.11) together with the requirement that the mapping should be a diffeomorphism. Now, suppose that we are able to ?nd a nontrivial one-parameter Lie group (LG) of symmetries of the ODE (2.8) then the Lie group can be used to determine the general solution of the ODE. “Suppose” that the symmetry of (2.8) include the Lie group of transformations in the y-direction i.e. (0,1) ˆ ˆ (x,y) = (x,y + e),
then, the symmetry condition (2.11) reduces to
?(x,y) = ? (x,y + e), (2.12)
(2.13) for a very small values of e. Differentiating (2.13) with respect to e at e = 0, ?y(x,y) = 0, and therefore the most general ODE whose symmetries include the Lie group (2.12) is of the form y' = ?(x), =? y = ?(x)dx + c, (2.14) the articular solution corresponding to c = 0 is mapped by the translation to the solutions ˆ y = ?(x)dx + e = ˆ ˆ ?(x)dx + e, (2.15) ˆ ˆ which is the solution corresponding to c = e. So by using the one-parameter LG we obtain the general solution from one particular solution. Is the same true for ODEs with other one-parameter LG?. Yes, the same idea works for all one-parameter LG. In a suitable coordinate system, the symmetries parametrized by e suf?ciently close to zero are equivalent to translations. How can ?nd symmetries of an ODE?. One method is to use the symmetry condition (2.11). In general this is a complicated nonlinear PDE in (x,y). However, Lie symmetries can be derived from a much simpler condition on the tangent vector ?eld. By de?nition, the Lie symmetries of an ODE are of the form ˆ ˆ x = x + e? +O(e2),
if we substitute (2.16) into (2.11) we obtain
?(x,y) + e ?x + ?(x,y)?y +O(e2) 1+ e ?x + ?(x,y)?y +O(e2) y = y + e? +O(e2),
= ? x + e? +O(e2),y + e? +O(e2) , (2.16)
(2.17)
we now expand each side of (2.17) as a Taylor series about e = 0, assuming that each series converges
? + e ?x + ?(?y – ?x – ?y?2) +O(e2) = ? + e ??x + ??y +O(e2),
where ? is shorthand for ?(x,y). Equating the O(e) terms gives the linearized symmetry condition ?x + ?(?y – ?x) – ?y?2 = ??x + ??y. 17
(2.18)
(2.19) ˆ ˆ Canonical Coordinates. We know that every ODE whose symmetries include the translations (2.12) may be integrated directly. How can these coordinates be found?. All orbits of the symmetry (2.12) have the same tangent vector at every point (?,?) = (0,1). Given any one-parameter LG, we aim to introduce coordinates such that (r,s) = (r(x,y),s(x,y)), where (r,s) = (r,s + e). If this is possible then, in the new coordinates, the tangent vector at the point (r,s) is (0,1) i.e. ˆ d r de|e=0 = 0, ˆ d s de|e=0 = 1, (2.20) using the chain rule and (2.16) we get ?rx + ?ry = 0, ?sx + ?sy = 1, (2.21) i.e. Xr = 0, Xs = 1. Since the change of coordinates should be invertible in some neighborhood of (x,y) then we impose the condition (2.3). Suppose that we have been able to ?nd a nontrivial Lie symmetry (LS) of a given ODE (2.8) then this ODE can be reduce to a quadrature by rewriting it in terms of canonical coordinates as follows: ds dr = sx + ? (x,y)sy rx + ? (x,y)ry , (2.22) the right-hand side of (2.22) can be written as a function of r and s. For a general change of variables (x,y) ? (r,s) the transformed ODE would be of the form ds dr = ?(r,s), (2.23) ˆ ˆ for some function ?. However, (r,s) are canonical coordinates and so the ODE is invariant under the group the translations in the s-direction: (r,s) = (r,s + e). Therefore, as we already know, ODE (2.23) is of the from ds dr = ?(r), (2.24) the problem is now reduced to a quadrature. The general solution of (2.24) is s – ?(r)dr = c, c ? R, (2.25) and hence the general solution of (2.8) is s(x,y) – r(x,y) ?(r)dr = c. (2.26) In general, for an ODE y(n) = ?(x,y,y',…..,y(n-1)), y(k) = dky dxk , (2.27) ˆ ˆ where ? is a smooth function of its arguments. We begin imposing the symmetry condition. A symmetry of (2.27) is a diffeomorphism. that maps the set of solutions of the ODE to itself. Any diffeomorphism. G : (x,y) ? (x,y) maps smooth planar curves to smooth planar curves. This action of G on the plane induces an action on the
18 CHAPTER 2. INTRODUCTION TO THE SYMMETRY METHODS. ˆ ˆ ˆ ˆ ˆ derivatives y(k) which is the mapping G : ?(x,y,y',…..,y(n-1)) ? ?(x,y,y',…..,y(n-1)). This mapping is called the nth prolongation of G. The functions y(k) are calculated recursively as follows: ˆ y(k) = ˆ ˆ dy(k-1) dx = ˆ ˆ Dxy(k-1) Dxx , ˆ ˆ y(0) = y, (2.28) where Dx is the total derivative. The symmetry condition for the ODE (2.27) is ˆ ˆ ˆ ˆ ˆ y(n) = ?(x,y,y',…..,y(n-1)), when (2.27) holds. (2.29) The LS are obtained linearizing (2.29) about e = 0. The linearized symmetry condition is derived as above. For e suf?ciently close to zero, the prolonged LS are of the form: ˆ x = x + e? +O(e2), ˆ y = y + e? +O(e2), ˆ y(k) = y(k) + e?(k) +O(e2), (2.30) if we substitute (2.30) into the symmetry condition (2.29) the O(e) terms yield the linearized symmetry condition ?(k) = ??x + ??y + ?(1)?y' + …..+ ?(n-1)?y(n-1), when (2.27) holds. (2.31) The functions ?(k) are calculated recursively from (2.28) as follows. For k = 1, ˆ y(1) = ˆ ˆ Dxy Dxx = y' + eDx? +O(e2) 1+ eDx? +O(e2) = y' + e Dx? – y'Dx? +O(e2), (2.32) ?(1) = Dx? – y'Dx?. (2.33) therefore from (2.30)
Similarly ˆ y(k) = y(k) + eDx?(k-1) +O(e2) 1+ eDx? +O(e2) , (2.34) and hence ?(k) = Dx?(k-1) – y(k)Dx?. (2.35) To ?nd the Lie point symmetries of an ODE (2.27) we must ?rst calculate ?(k),k = 1,2,…,n. The functions ? and ? depend upon x and y only and therefore (2.33) and (2.35) give the following results: ?(1) = ?x + (?y – ?x)y' – ?yy'2, ?(2) = ?xx + (2?xy – ?xx)y' + (?yy – 2?xy)y'2 – ?yyy'3 + (?y – 2?x – 3?yy')y'',
?(3) = ?xxx + (3?xxy – ?xxx)y' + 3(?xyy – ?xxy)y'2 + ?yyy – ?yy y'3 – ?yyyy'4 + 3(?xy – ?xx+ ?yy – 3?xy y' – 2?yyy'2)y'' – 3?yy''2 + ?y – 3?x – 4?yy' y''', (2.36)
(2.37)
(2.38) etc…. For example, for an ODE y'' = ?(x,y,y') the linearized symmetry condition is obtained substituting (2.36) and (2.37) into (2.31) and then replacing y'' by ?(x,y,y'). This gives ?(2) = ??x + ??y + ?(1)?y', i.e. ?xx + (2?xy – ?xx)y' + (?yy – 2?xy)y'2 – ?yyy'3 + (?y – 2?x – 3?yy')? = ??x + ??y + ?x + (?y – ?x)y' – ?yy'2 ?y'. (2.39)
dx ˆ dy ˆ 19
For a ?rst-order ODEs, the right-hand side of the linearized symmetry condition (2.31) is X? where X is the in?nitesimal generator i.e. X = ??x + ??y. This in?nitesimal generator is associated with the tangent vector to the orbit passing through (x,y), namely (?,?) = , de de |e=0 . (2.40) To deal with the action of LS on derivatives of order n or smaller we introduce the prolonged in?nitesimal generator X(n) = ??x + ??y + ?(1)?y' + …..+ ?(n-1)?y(n-1), (2.41) the coef?cient of ?y(k) and so X(n) is associated with the tangent vector in the space variables (x,y,y',…..,y(n-1)). We can use the prolonged in?nitesimal generator to write the linearized symmetry condition (2.31) in a compact form: X(n)(y(n) – ?(x,y,y',…..,y(n-1))) = 0, when (2.27) holds. (2.42) Ejemplo 2.0.3 Consider a third-order ODE y''' = N(x,y,y',y''). (2.43) i A vector ?eld S = ?(x,y)? ? x + ?(x,y)? ? y, is called a symmetry of (2.43) if and only if S[3](y''' – N(x,y,y',y''))|(2.43) = 0, where S[3] is the third prolongation of S de?ned by: S[3] = S + ?3=1 ?(i)?y(i), where ?(1) = ?x + (?y – ?x)y' – ?yy'2, ?(2) = ?xx + (2?xy – ?xx)y' + (?yy – 2?xy)y'2 – ?yyy'3 + (?y – 2?x – 3?yy')y'', ?(3) = ?xxx + (3?xxy – ?xxx)y' + 3(?xyy – ?xxy)y'2 + ?yyy – ?yy y'3 – ?yyyy'4 + 3(?xy – ?xx+ ?yy – 3?xy y' – 2?yyy'2)y'' – 3?yy''2 + ?y – 3?x – 4?yy' y''',
and |(2.43) means evaluated on (2.43). Straightforward calculations indicate that (2.44) is equivalent to ?(3) = ?Nx + ?Ny + ?(1)Ny' + ?(2)Ny''. (2.44)
(2.45) Since ? and ? are independent of y', Equation (2.45) can be separated with respect to powers of y' to give an overdetermined system of linear partial differential equations for ? and ?. This system is easier to solve than the original equation (2.43): this is the main stength of the Lie method.
Reduction of order by using canonical coordinates. Suppose that X is an in?nitesimal generator of a one paramenter LG of symmetries of the ODE y(n) = ? x,y',y'',….,y(n-1) , n = 2. (2.46) (2.47) Let (r,s) be canonical coordinates for the group generated by X, so that X = ?s. If the ODE (2.46) is written in terms of canonical coordinates, it is of the form ? s(n) = ? r,s,s,…,s(n-1) , s(k) = dks drk , (2.48)
?x + 2 ?y, x2 – 1 2y x3 X5 = 2y2e-x y(x2 – 1)ex x3 x2 X3 = e-x ?y, x3 x2 e-x 2ye-x X6 = ?x – ?y, (x – 1)2 (x + 1)2 ex ?y, X8 = ?x – 20 CHAPTER 2. INTRODUCTION TO THE SYMMETRY METHODS. dr symmetry condition gives, ?s = 0. Therefore ? s(n) = ? r,s,…,s(n-1) . (2.49) ? By writing the ODE (2.46) in terms of the canonical coordinates, we have reduced it to an ODE of order n – 1, for v = s :
drk+1
Suppose that the reduced ODE has the general solution (2.51) v = f (r;c1,c2,….,cn-1), then the general solution of the original ODE is s(x,y) = r(x,y) f (r;c1,c2,….,cn-1)dr + cn. (2.52) ? ? ? ? ? More generally, if v is any function of s and r such that vs (r,s) = 0, the ODE (2.49) reduces to an ODE of the form . (2.53) drk Once the general solution of (2.53) has been found, the relationship s = s(r,v), gives the general solution of (2.46): s(x,y) = r(x,y) ? s(r;v(r;c1,c2,….,cn-1))dr + cn. (2.54) Ejemplo 2.0.4 Consider the linear ODE y'' = 3 x – 2x y' + 4y (2.55) where by using all the above theoretical results we get -2 3 x ?yy = 0, – 2x ?y + ?yy – 2?xy = 0, -12y?y – ?xx + 3 x2 + 2x ? – 3 x – 2x ?x + 2?xy = 0, 3 x – 2x ?x + ?xx = 0, -8y?x – 4? + 4y?y –
and therefore we obtaing the following symmetries: 2 y 2y2 2 2 x3 x2 X1 = y?y, X2 = x2 – 1 ?y,
x 2 2 X7 = ?x – X4 = ?x –
2 x3 2 ?y,
2y(x – 1)(x + 1)ex x2 ?y.
2r C1e-r – 1 C1e-r – 1+ r2 2x C1e-x – 1 C1e-x – 1+ x2 = ', v' = – – r + 21 We consider the the ?rst of them, namely X1 = y?y, where the simplest canonical coordinates are r = x, s = lny, which prolong to v = ds dr = y' y , d2s dr2 = y'' y – y' y 2 , hence our ODE reduces to the Riccati equation v' = 3 r – 2r v + 4- v2, where its solution is v = – 2 2 , all that remains it to carry out the c.v. y' y = – 2 2 , so the solution for our original ODE yields 2 Ejemplo 2.0.5 For the ODE y'' = y'2 y + y – 1 y y' we ?nd the following symmetry X = ?x which leads us to get the following c.v. r = y, s(r) = x, thus (this is a central step in all the work) v = ds dr 1 y v' = – y'' (y')3 , and therefore we get: v r 1 r v2.
Bibliography
[1] G. W. Bluman and J. D. Cole. “Similarity Methods for Differentials Equations”. (1974) Appl. Math. Sci. N13 Springer-Verlang New York. G. W. Blumann and S.W. Kumei “Symmetries and Differential Equations”. Springer Verlang (1989). G. W. Blumann and S.C. Anco. “Symmetries and Integration Methods for Differen- tial Equations". Springer Verlang (2002). [2] L. V. Ovsiannikov, Group Analysis of Differential Equations (Academic Press 1982). [3] N. H. Ibragimov. “Transformation Groups Applied to Mathematical Physics”. D. Reidel (1985). N. H. Ibragimov. “Elementary Lie Group Analysis and Ordinary Differential Equations” (John Wiley & Sons, 1999). [4] R. Seshadri and T. Y. Na. “Group Invariance in Engineering Boundary” Value Problems. Springer-Verlang. NY, 1985. [5] H. Stephani, “Differential Equations: Their Solutions Using Symmetries” (Cambridge University Press 1989). [6] P. T. Olver, “Applications of Lie Groups to Differential Equations” (Springer-Verlang, 1993). [7] L. Dresner, “Applications of Lie´s Theory of Ordinry and Partial Differential Equations”. IOP 1999 [8] G. Bauman, “Symmetry Analysis of Differential Equations”. Springer Verlang (2000). [9] P. E. Hydon, “Symmetry Methods for Differential Equations” (Cambridge University Press, 2000). [10] B. J. Cantwell, “Introduction to Symmetry Analysis” (Cambridge University Press, Cambridge, 2002). 23
+ x, where each term must have dimensions of y' i.e. [y'] = yx-1, where [·] stands for dimensional y' = as we can see easily if we consider only one constant a, such that [a] = yx-2, we make that the ode y' = + ax, Chapter 3 Ecuaciones de Primer Orden 3.1 Introducción D.A. has usually been employed in different areas (?elds) such as engineering problems, ?uid mechanics etc… and these problems are always described by partial differential equations (pde) (see [1]-[10]). This method (“tac- tic") helps us to reduce the number of quantities that appear into an equation and to obtain ordinary differential equations (ode). We would like to point out (emphasize) that this tool is more effective if one practices the spatial discrimination, such tactic allows us to obtain better results than with the standard application of D.A. (see [11]). Knowing that D.A. works well in pde we would like to extend this method to the study of ode (the ?rst order in this case) in a systematic way. There are in the literature previous work in this direction for ode of ?rst order (see [12]-[13]. The method of the Lie groups it has showed as a very useful tool in order to solve nonlinear equations (nl-ode) as well as pde (see [14]-[27] ). Nevertheless, when one is studying ode of ?rst order its application results very complicated (very tedious) if one has not a computer algebra package since if one decides to look for the possible symmetries of an ode with pencil and paper this task may be turned very exhausting. However, if we know that an ode admits a concrete symmetry, then it is a trivial issue to ?nd new variables which allows us to rewrite the ode in quadratures or as we will see in this paper to obtain an ode with separating variables. Our purpose in this work is to explain, through examples, how D.A. works in order to ?nd these changes of variables (c.v.) in a trivial way, i.e. without the knowledge of the symmetries of the ode under study. The idea is as follows. When we are studying an ode form the dimensional point of view, we must require that such ode veri?es the principle of dimensional homogeneity (pdh) i.e. that each term within the equation have the same dimensions, for example, speed, or energy density. To clarify this concept we consider the following ode, y x equation of the quantity ·. As we can see, this ode does not verify the pdh, since the term x has dimensions of x, i.e. [x] = x. In order to do that this equation veri?es the pdh we need to introduce dimensional constants, in such a way that after rewrite the ode with these constants the ode now veri?es such principle i.e. in this example and y x veri?es the pdh. We would like to emphasize that this situation does not appear (arise) when one is studying physical or engineering problems since (as it is supposed) that such problems (equations) verify the pdh and we do not need to introduce new dimensional constants, that must have physical meaning, for example, the viscosity coef?cient, etc…. Precisely this dimensional constant suggests us the c.v. (x,y(x)) -? (t,u(t)) where t = x,u = yx-2 in such a way that rewriting the original ode in these new variables it is obtained a new ode with separating variables: u + u't = 1. The reason is the following. We know from the Lie group theory that if it is known a symmetry of an ode then this symmetry bring us through a c.v. to obtain a simpler ode (a quadrature or a ode with separating 25
26 CHAPTER 3. ECUACIONES DE PRIMER ORDEN variables) and therefore the solution is found in a closed form. To ?nd this c.v. it is used the invariants that generate each symmetry, i.e. the ?rst principle is that it is useful to pass to new coordinates such that one of the coordinate functions is an invariant of the group. After such transformation it often (but not always) happens that the variables separate and the equation can be solved in closed form. We must stress that taking the new dependent variables to be an invariant of the group does not guarantee the separation of variables. The choice of the independent variable is also very important. One must note that D.A. gives us this invariant (or at least a particular solution). It is observed in our example that the solution y = x2, (suggested by D.A.) is a particular solution for this ode, but if we study this ode form the Lie theory point of view it is obtained that such ode has the scaling symmetry X = x?x + 2y?y and therefore the solution y = x2, is furthermore an invariant solution (generated by X), for this reason such c.v. works well. As we will show in the paper, the c.v. that D.A. induces is the same than the generated one by the symmetries of the ode. In order to make see this fact, we will solve each example by D.A. as well as by the Lie group technique (abusing of the trivial calculations), calculating the symmetries and their corresponding invariants and c.v.. However D.A. has a little (or rather big) drawback, it only gives us relationship of type power i.e. y = xn, with n ? R, and we cannot obtain relationship as y = (x + 1)n. Nevertheless, and as we will see in the examples, although D.A. does not give us an invariant solution it will be suf?cient that it provides us a particular solution in order to obtain a c.v. which brings us to a simpler ode than the original one. This task will be very useful in the case of studying Riccati and Abel odes. Furthermore, this “tactic" is valid even in the case in which the ode has no symmetries but unfortunately one always founds examples in which any tactic does not work. The paper is organized as follows: In section 2 we describe (with all the tedious and super?uous calculations), through two examples, how the proposed method works by comparing it with the Lie method in order to show which are the symmetries and their corresponding invariants. In section 3 we shall show several kind of ODEs paying special emphasis in for example Bernoulli, Riccati and Abel ODEs (see [28]-[32]). We think that these kind of ode are the most dif?cult to solve and therefore any simpli?cation is welcome. We show two Abel odes that do not admit symmetries (see [33]-[37]). In the ?rst of them we use the D.A. to obtain a particular solution that allows us to obtain a Bernoulli ode but in the second one we are not able to ?nd any particular solution and therefore a c.v. which brings us to obtain a simpler ode. We end with some conclusion as well as pointing out some of the limitations of the proposed tactic. We would like to emphasize the pedagogical character of this chapter (at least we try) for this reason we have abused of super?uous calculations and we have omitted some technical details since the supposed audience are mainly engineers and/or physicists but not for mathematicians.
3.2 The method
In this section we will explain how the D.A. works in order to solve odes through two simple examples. The main idea is to introduce dimensional constants that make the ode under study verify the principle of dimen- sional homogeneity. These dimensional constants help us to ?nd c.v. which bring us to obtain ode with sepa- rating variables. These c.v. obtained through D.A. correspond to invariant solution (or at least to a particular solution) and therefore they are induced by one symmetry. We compare our tactic with the Lie one.
Ejemplo 3.2.1 Solve the homogeneous equation x2 + y2 dx = 2xydy. (3.1) Solution. In order to solve it we will use three different ways, the traditional, the dimensional and the Lie method.
1- u x + C1x, 27 3.2. THE METHOD
Traditional method. Making the c.v. u = y/x we have: u' = 1- u2 2ux =? 2udu 2 = dx x =? ln(1- u2) = lnx =? y2 = x2 + x, (3.2) Dimensional Analysis. We go next to consider eq. (3.1), written as follows y' = a2x2 + y2 2xy , (3.3) where the dimensional constant a, makes the ode verify the dimensional principle of homogeneity (d.p.h.) if [a] = y x = X-1Y, (3.4) where [·] stands for the dimensional equation of the quantity ·. Applying the Pi theorem we obtain the dimensionless variables that help us to simplify the original ode. There- fore taking into account the following dimensional matrix we take x y y 0 1 x 1 0 a -1 =? p1 = 1 ax y , =? y = ax, (3.5) as we will see later this solution is a particular solution of this ode. We would like to emphasize that as we have only needed one constant then the equation is scale-invariant in such a way that the generator of this group is X = x?x + y?y as it is observed from eq. (3.5). This fact will be probed studying this equation under the Lie group tactic, see below. In this way the new variables are (t,u(t)) : t = x, u(t) = a x y , =? x = t, y = a t u(t) , (3.6) this change of variables brings us to rewrite eq. (3.3) as follows: u' u(1- u 2) = 1 2t =? v u 1- u2 = 1 2 lnt, (3.7) and hence y2 = a2x2 + C1x. (3.8) As we can see in this trivial example, the D.A. induces a c.v. which helps us to obtain an ode simpler than the original one. We can also think in the following way ay' = b x 2y + a y 2x , (3.9) where [a] = x, [b] = y2x-1, (3.10) and hence t = x a , u(t) = y2 bx =? x = at, y = abtu(t) , (3.11) u' = 1 =? u = t + C1, (3.12) therefore eq. (3.9) yields
in this way we obtain the solution y2 = b 2 a (3.13)
u(t) = v 1 (?y – ?x)(x + y ) 1 ?y(x + y ) – ?(x,y)( – ) – ?(x,y)( – 2x?y t = v ,u(t) = x 28 CHAPTER 3. ECUACIONES DE PRIMER ORDEN once we have obtained the solution, the constants a,b are makig equal to 1, i.e. a = b = 1. In the same way we may consider the following change of variables t = x a , y bx =? x = at, v y = u(t) abt , (3.14) which brings us to the following ode 2u'u = 1 =? u2 – t + C1 = 0, (3.15) and therefore we obtain again the solution (3.13). As we will see below all these c.v. are generated by their corresponding (respective) symmetry. As we can see, this last change of variables is better than the ?rst and the tactic is the same: to introduce dimen- sional constants that make the equation dimensional homogeneous. Lie Method. In order to ?nd the symmetry generator of a ode y' = f (x,y),
we need to solve the following pde ?x + (?y – ?x)f – ?y f2 – ?(x,y)fx – ?(x,y)fy = 0, (3.16)
(3.17) where fx = df dx , fy = df dy . (3.18) In this case we have to solve ?x + 2 2 2 xy – 2 2 2 4 x2 y2 1 y x2 + y2 2x2y 1 x x2 + y2 2x y2 ) = 0, (3.19) we ?nd that X = ??x + ??y : X1 = x 2y ?y, X2 = – x2 2y + y 2 ?y, X3 = x?x + y?y, X4 = ?x + y 2x ?y, X5 = x2 + y2 ?x + 2xy?y, (3.20) observing that X5 is a trivial symmetry. Each symmetry induces a change of variable (canonical variables) which are obtained through the following formula Xt = 0, Xu = 1. (3.21) In this way the change of variables that induces the ?eld X4 = ?x + y is: y x =? x = u(t),y = t u(t) , (3.22) in such a way that eq. (3.75) yields: u' = 2t =? u(t) = t2 + C1, (3.23) and therefore x = y2 x + C1, (3.24) which is the same solution. The c.v. induced by this symmetry is similar to the obtained one in (3.14).
2y?y X4 -? I4 = v . 3.2. THE METHOD 29 The c.v. that induces the ?eld X1 = x is the following one t = x, u(t) = y2 x =? x = t, y = u(t)t , (3.25) therefore eq.(3.1) yields u' = b2, (3.26) this c.v. is the same than the obtained one in (3.11). For example the c.v. that induces the ?eld X3 = x?x + y?y is t = y x ,u(t) = lnx =? x = eu(t),y = teu(t) , (3.27) which brings us to the following ode : u' = – t2 2t – 1 =? u(t) = -ln(-1+ t) – ln(1+ t) + C1, (3.28) now writing the solution in the original variables we take lnx = C1 + ln x (y – x)(y + x) . (3.29) Now if we consider the invariants that induce each symmetry dx ? = dy ? , -? y' := dy dx = ? ? , (3.30) we take that: X1 -? I1 = x, X2 -? I2 = x, X3 -? I3 = y x , y x (3.31) For example the symmetry X3 = x?x + y?y generates the following invariant: dx ? = dy ? =? dx x = dy y =? lnx = lny =? I3 = y x -? y = ax, (3.32) this would be the solution that suggests us precisely the direct use of the Pi theorem (if the ode is scale invariant, as in this case, the solution obtained applying by the Pi theorem coincides with the invariant solution that generates the scale symmetry, in this case X3). We see that we only obtain a particular solution, but that this is invariant, in fact if we think about the ode as a dynamical system we see that the ?xed point of such equation would be precisely the solution y = ±x.
Ejemplo 3.2.2 Solve the linear ode 1- x2 y' + xy = 1. (3.33) Solution. Solution through D.A. Our ?rst step will be to introduce dimensional constants in such a way that eq. (3.33) veri?es the principle of dimensional homogeneity. In this way we rewrite the equation as follows A – x2 y' + xy = B, (3.34) where [A] = X2, [B] = XY, (3.35)
30 CHAPTER 3. ECUACIONES DE PRIMER ORDEN i.e. p1 = x2 A p2 = xy B , (3.36) where y = 1 Bx ? x2 A , (3.37) being ? a unknown function. It is possible to make the following assumption y = 1 Bx x2 A n , n ? R. (3.38) Since y = x-1 is not a particular solution of (3.33) we need to combine the p – monomias in order to ?nd a particular solution, the simplest way is as follows: t = x2 A , u(t) = A y B x =? x = v At, y = B A v u(t) At (3.39) observing that 1 u(t) = p2 · p-1, (3.40) 1 where y = cx, derived from p2 · p-1 is a particular solution of (3.33).
Observación 3.2.1 (Recipe) When we have two p -monomia, we have to check if they induce any particular (or invari- ant) solution. If they are not particular solutions then we must combine them in order to obtain such solution, this solution usually is obtained by combining them in a very simple way.
In this way our ode is written now as follows: 2u' u – 1 = 1 t(t – 1) =? u = 1+ C1 v t + 1 t , (3.41) and in the original variables the solution of eq. (3.33) yields: y = A B x ± C1 A – x2 , (3.42) where C1 is an integration constant. Now making A = B = 1, we have the ordinary solution to eq. (3.33). Lie method. Following the standard procedure we have to solve the pde: ?x + (?y – ?x)(1- x y) 1- x2 – ?y (1- x y)2 (1- x2)2 – ?(x, y)(- y 1- x2 + 2(1- x y) x (1- x2)2 )+ ?(x, y) x 1- x2 = 0, (3.43) which solutions are: X1 = -1+ x2?y, X2 = (-y + x)?y, X3 = (2xy – y2 – 1)(-y + x) (x – 1)(x + 1) ?y, (3.44) and their corresponding invariants are: X1 -? I1 = x, X2 -? I2 = x, X3 -? I3 = x. (3.45) For example the symmetry X1 generates the following c.v.: t = x, u(t) = v y -1+ x2 =? x = t, y = u(t) -1+ t2 , (3.46)
u = – -1+ t2 -1+ t2 C1(x2 – 1) + x x2 – 1 u(t) = – ln(t – 1) – – ln(-y + x) = – ln(x – 1) – 3.3. CUADRATURAS. 31 therefore eq. (3.33) is written as: ' v -1+ t2 1- 2t2 + t4 , (3.47) which solution is: u(t) = – 3/2 4(t + 1)2 + 3/2 4(t – 1)2 + C1, (3.48) and hence v y x2 – 1 = v x2 – 1 v v =? y = x ± c x – 1 x + 1. (3.49) Now, if for example we consider the symmetry X2 then we have: (t = x, u(t) = -ln(-y + x)) =? x = t, y = teu(t) – 1 eu(t) , (3.50) therefore: u' = – t -1+ t2 , (3.51) ?nding in this way that its solution is: 1 2 1 2 ln(t + 1) + C1, (3.52) hence in the original variables the solution yields: 1 2 1 2 ln(x + 1) + C1, (3.53) which after a simple simpli?cation it yields: v v y = x ± c x – 1 x + 1, (3.54) as we already know.
With these two simple examples we have tried to show how the D.A. works in order to introduce c.v. which brings us to obtain simpler odes than the original one. As we have emphasized in the recipe, the trick is to look for a particular solution. Sometimes this particular solution will be furthermore invariant solution (induced by a concrete symmetry). If this is the case, then our ode will be reduced to an ode with separating variables but if this solution is only a particular then, as we will see below, we have not any guarantee of reducing our ode to an ode with separate variables, nevertheless we will obtain a simpler ode than the original one. In the next section we will study some examples.
3.3 Cuadraturas.
Decimos que una ecuación diferencial está en forma de cuadraturas cuando sucede alguna de las siguientes situaciones:
1. la ecuación es de primer orden y el lado derecho de la ecuación depende sólo de x y' = F(x) (3.55) la solución es por lo tanto de la forma y(x) = F(x)dx + C1. (3.56)
-x2 + x + arcsin(2x – 1) + C1, x 32 CHAPTER 3. ECUACIONES DE PRIMER ORDEN
2. la ecuación es de primer orden y el lado derecho de la ecuación depende sólo de y(x): y' = F(y(x)) (3.57) la solución es por lo tanto de la forma y(x) 1 F(a) da + C1 = x (3.58) 3.3.1 Simetrías. Desde el punto de vista de las simetrías vemos que la ecuación (3.55) sólo puede tener las siguientes simetrías ?x + (?y – ?x) F(x) – ?y F(x)2 – ?(x, y) F' = 0 encontrando que: [? = 0, ? = 1] mientras que la ecuación (3.57) sólo puede tener las simetrías. ?x + (?y – ?x) F(y) – ?y F(y)2 – ?(x, y) Fy = 0 encontrando que: [? = 1, ? = 0] 3.3.2 Ejemplos Veremos unos sencillos e inmediatos ejemplo de cada tipo de ecuació descrito en la anterior subsección.
Ejemplo 3.3.1 Resolver la ecuación en cuadraturas x + x y'2 = 1
Solución. Vemos que y = –
y = -x2 + x – 1 2 1 2 arcsin(2x – 1) + C1 ya que ?x + (?y – ?x) -x(x – 1) x + ?y (x – 1) x – ?(x, y)(- -x(x – 1) 2 + 2x -2x + 1 -x(x – 1) )= 0 y por lo tanto [? = 0, ? = 1] como ya sabíamos. Intentaremos obtener alguna solución particular a la ecuación y' = a – bx x
3.4. ECUACIONES EN VARIABLES SEPARABLES Y REDUCIBLES A ELLAS. 33 donde las constantes dimensionales a y b tienen las siguientes dimensiones: [a] = y2x-1,[b] = y2x-2. En este caso la solución que sugiere el AD es de la forma y2 = ax · ? bx2 y2 , observando que semejante sugerencia está muy lejos de ser una solución a la ODE propuesta. Intentaremos otra táctica dimensional, forzando a que la función desconocida ? sea de la forma pnb, i.e. y2 = ax · bx2 y2 n =? y2n+2 = abnx2n+1 de esta forma y = abnx2n+1 1 2n+2 que tampoco funciona.
Ejemplo 3.3.2 Resolver la ecuación en cuadraturas y' = y2 Solución. Vemos que y = – 1 x – C1 ya que
y por lo tanto ?x + (?y – ?x)y2 – ?y y4 – 2?(x, y)y = 0
[? = 1, ? = 0] como ya sabíamos. En este ejemplo la solución que nos sugiere el AD es la siguiente. Introducimos la constante a para hacer que la EDO sea d.h. i.e. y' = ay2, con [a] = y-1x-1, de esta forma llegamos a y x y 1 0 a x -1 0 =? y = -1 1 1 ax -1 observando que sólo después de haber sustituido esta solución en la ODE original nos daríamos cuenta del signo i.e. y = ax , ya que el AD “no entiende” ni de signos ni de constantes numéricas, sólo de órdenes de magnitud.
Claim 3.3.1 Resaltamos que a pesar de no tener simetrías de escala hemos podido obtener una solución por AD. 3.4 Ecuaciones en variables separables y reducibles a ellas. Son de la forma y' = f(x)g(y) (3.59)
y g(y) = y. el AD) sii f(x) = 34 CHAPTER 3. ECUACIONES DE PRIMER ORDEN = f(x) (3.60) 3.4.1 Método tradicional.
Tenemos que separar, viendo de forma inmediata que: y' g(y) por lo tanto dy g(y) = f(x)dx =? dy g(y) = f(x)dx + C1 (3.61) 3.4.2 Método dimensional. En primer lugar deberemos reescribir la ecuación introduciendo cuantas constantes dimensionales sena nece- sariaas para hacer que la ecuación veri?que el principio de homogeneidad de tal forma que dy g(y) = [A1f(x)dx] (3.62) donde la constante A1 hace que la anterior ecuación sea dimensionalmente homogénea. 3.4.3 Método de Lie. En esta ocasión el método de Lie nos lleva a obtener las siguientes simetrías para la ecuación (3.59) al resolver la siguiente ecuación: ?x + (?y – ?x) f(x)g(y) – ?y f(x)2g(y)2 – ?(x, y) f 'g(y) – ?(x, y) f(x)g' = 0 (3.63) cuyas soluciones son: [? = 0,? = g(y)], [? = 1 f(x) ,? = 0],[? = 1 f(x) ,? = g(y)] que conducen a la siguiente solución: dx ?(x,y) = dy ?(x,y) =? dy g(y) = f(x)dx + C1 1 f(x) ,? = g(y), podremos tener simetrías de escala (las que genera Resaltamos que de las simetrías obtenidas ? = 1 x
3.4.4 Ejemplo. Ejemplo 3.4.1 Vamos a estudiar la siguente ode: y' = -sin(x)y
Solución. Resolveremos la ecuación planteada por los tres métodos, intentando mostar las ventajas y desventa- jas de cada método.
35 3.4. ECUACIONES EN VARIABLES SEPARABLES Y REDUCIBLES A ELLAS.
1. Método tradicional. Vemos de forma inmediata que la solución es: dy y = -sinxdx =? y = C1ecos(x) 2. Método dimensional. Tenemos que introdir constantes para hacer que la ecuación sea dimensionalmente homogénea, encon- trando así que: y' = -A1sin(x)y donde [A1] = x-1 y donde estamos considerendo que [sin(A1x)] = x0y0, i.e. es adimensional. Por lo tanto la soluión que sugiere el AD es: y = y0?(sin(A1x)) donde ? hace referencia a una unción desconocida de sin(A1x). Con este ejemplo vemos que en este caso el AD no nos acerceca ni de lejos a una solución a una ecuación tan sencilla como la planteada. 3. Método de Lie. Veamos ahora las simetrías de la ecuación al resolver la ecuación: ?x – (?y – ?x)ysin(x) – ?yy2 sin(x)2 + ?(x, y)y cos(x) + ?(x,y)sin(x) = 0 todas las posibles simetrías son: [? = 0,? = y],[? = 0,? = ecos(x)],[? = 1 sin(x) ,? = 0],[? = 1 sin(x) ,? = -y]. por lo tanto: -sinxdx = dy y llegando así a la solución: y = C1ecos(x) Las variables canónicas generadas por [? = 0,? = y] nos llevan al siguiente resultado:
(t = x,u(t) = lny) -? x = t,y = eu(t)
por lo tanto la ecuación diferencial original queda reducida a: u' = -sint =? u = cost + C1 por lo tanto la solución en las variables originales es: lny = cosx + C1 =? y = C1ecos(x). Podemos observar las ventajas y desventajas que presentan cada método. Ejemplo 3.4.2 Vamos a estudiar la siguente ode: y' = xy Solución. Resolveremos la ecuación planteada por los tres métodos, intentando mostar las ventajas y desventa- jas de cada método.
xdx =? y(x) = C1e 2x [? = 0,? = y],[? = 0,? = e- 2x ],[? = y(x) = C1e 2x 1 2 2 1 2 2 y(x) = C1e 2x . 36 CHAPTER 3. ECUACIONES DE PRIMER ORDEN 1. Método tradicional. Vemos de forma inmediata que la solución es: dy y = 1 2 2. Método dimensional. Como en l ejemplo anterior reescribimos la ecuación de tal forma que y' = A1xy con [A1] = x-2 y = y0? A1x2 bastante pobre. 3. Método de Lie. Veamos ahora las simetrías de la ecuación al resolver la ecuación: ?x – (?y – ?x)yx – ?yy2x2 + ?(x, y)y + ?(x,y)x = 0 una de las posibles simetrías es: 1 2 1 x ,? = 0],[? = 1 x ,? = y]. por lo tanto: xdx = dy y llegando así a la solución: 1 2 El empleo de las variables canónicas (v.c) que generan la simetría [? = 0,? = y] nos lleva a los siguientes resultados: (t = x,u(t) = lny) -? x = t,y = eu(t) por lo tanto la ecuación diferencial original queda reducida a: t + C1 u' = t =? u =
por lo tanto la solución en las variables originales es: lny = x + C1 =? 1 2 3.5 Ecuación homogénea
En primer lugar deberemos distinguir los siguientes casos:
1. Homogénea clase A dy dx = f y x (3.64)
37 3.5. ECUACIÓN HOMOGÉNEA
2. Homogénea clase B F y', y x (3.65) 3. Homogénea clase C y' = F ax + by + c rx + sy + t (3.66) 4. Homogénea clase D y' = y x + g(x)f y x (3.67) 5. Homogénea clase G y' = yf y xa x (3.68) 3.5.1 Método tradicional. Describos a continuación los métodos de resolución de cada tipo de ecuación.
1. Clase A. La forma de resolver la ecuación (3.64) es la de hacer el típico c.v. u = y x (3.69) de tal forma que la ecuación a resolver es: xu' = f(u) – u =? du f(u) – u = dx x (3.70) que es de variables separables y por lo tnato facilmente integrable.
2. Clase B. La forma de resolver la ecuación (3.65) es la de hacer el típico c.v. ?(t) = y x ,y' = ?(t) (3.71) de esta forma derivando y = x?(t) respecto de x aparece una ecuación en variables separadas.
3. Clase C. La forma de resolver la ecuación (3.66) deberemos tener en cuenta las siguientes posibilidades:
(a) Que las dos rectas ax + by + c y rx + sy + t se corten en un punto (x0,y0). En este caso se efectua el c.v. X = x – x0, Y = y- y0. De esta forma la ecuación resultante se reduce a una homogénea de clase A. (b) Que las rectas ax + by + c y rx + sy + t sean paralelas. En este caso se efectua el c.v. z = ax – by. De esta forma la ecuación resultante es de variables separadas.
4. Clase D. La forma de resolver la ecuación (3.67) es la de hacer el típico c.v.
5. Clase G. La forma de resolver la ecuación (3.68) es la de hacer el típico c.v.
?x + (?y – ?x)f( ) – ?y f( )2 + ?(x,y)D(f)( x)y x ?(x,y)D(f)( x) [? = 0,? = -y + x f( )],[? = x, ? = y] F( x,a) + 38 CHAPTER 3. ECUACIONES DE PRIMER ORDEN 3.5.2 Método dimensional. La ecuación (3.64) será homogénea dimensionalmente si no se discrimina entre x e y, en cuyo caso p = y A1x = ? (3.72) es adimensional, [A1] = yx-1. Expresándola en función de ? y de x, tendremos y = A1x?, y' = ? + A1x d? dx , (3.73) eliminando y', y en (3.64) obtenemos ? + A1x d? dx = f (?) =? dx x = d? f (?) – 1 , (3.74) consiguiéndose así la separación de variables que permite su integración. Como se observa es el mísmo método que el clásico. 3.5.3 Método de Lie. Como en el caso del método tradicional distinguiremos cada tipo:
1. Clase A. Tenemos que resolver la ecuación y y x x y 2 – y x = 0 que nos conduce a: y x y por lo tanto la osolución general ela ecuación es: y/x 1 a – f(a) da + ln(x) + C1 = 0 Resaltamos que esta tipo de ecuaciones admite simetrías de escala [? = x, ? = y] (AD) y que por lo tanto la menos seremos capaces de encontrar soluciones particulares (o la general comleta en el mejor de los casos) por medio de AD. 2. Clase B. En este caso las simetrías que genera la ecuación son:
[? = x,? = y]
y la solución general es del tipo: y/x 1 y a da + ln(x) + C1 = 0 3. Clase C. La ecuación a resolver es: ?x + (?y – ?x)F ax + by + c rx + sy + t – ?yF ax + by + c rx + sy + t 2 +
– 2 + g'(x)f [? = 0,? = x f( )], y2Df xa a f xa y yDf xa 3.5. ECUACIÓN HOMOGÉNEA 39 +?(x,y)D(f) ax + by + c rx + sy + t a rx + sy + t – (ax + by + c)r (rx + sy + t)2 – -?(x,y)D(f) ax + by + c rx + sy + t b rx + sy + t – (ax + by + c)s (rx + sy + t)2 = 0 encontrando que las simetrías en este caso son: [? = x(br – sa) + bo – sc br – sa ,? = rc – ao + (rb – as)y br – sa ] que conducen a la siguiente solución (invariantes): dx ?(x,y) = dy ?(x,y) =? dx x(br – sa) + bo – sc = dy rc – ao + (rb – as)y , 4. Clase D. El sistema a resolver es: ?x + (?y – ?x) y x + g(x)f y x – ?y y x + g(x)f y x 2 – -?(x,y) y x y x – y x y gD(f) x2 – ?(x,y) 1 x + y x gD(f) x = 0 encontrando que las simetrías en este caso son: y x [? = x g(x) ,? = y g(x) ] que conducen a la siguiente solución (invariantes): dx ?(x,y) = dy ?(x,y) =? g(x)dx x = g(x)dy y , sin embargo, si efectuamos el siguiente c.v. cv := {x = t, y = u(t)t} encontramos que ut t + u(t) = u(t) + g(t) f(u(t)) cuya solución es g(t) t dt – u(t) 1 f(a) da + C1 = 0 y por lo tanto – y/x 1 f(a) da + g(x) x dx + C1 = 0 5. Clase G. El sistema a resolver es: ?x + (?y – ?x) yf y xa x – ?y yf y xa x 2 – -?(x,y) y xax2 – y x2 – ?(x,y) Df y xa x – y xax2 = 0. Las simetrías que encontramos están generadas por: [? = 0, ? = -y(-f( y xn ) + n)], [? = x, ? = yn],
? = b , y = b , = -b 2 40 CHAPTER 3. ECUACIONES DE PRIMER ORDEN
si efectuamos el siguiente c.v. cv := {x = t, y = u(t)t}, encontramos una nueva ecuación diferencial ut t + u(t) = u(t) f( u(t)t tn ), cuya solución es: ln(t) – C1 + u(t)/n-1 1 a(n – f(a)) da, volviendo a las variables originale sencontramos que: ln(x) + C1 + y/xn 1 a(f(a) – n) da = 0. 3.5.4 Ejemplos. (3.75) Ejemplo 3.5.1 Queremos integrar la ecuación homogénea
x2 + y2 dx = 2xydy
Solución. Utilizaremos tres tácticas.
1. Método tradicional. Haciendo el c.v. u = y/x entonces u' = 1- u2 2ux =? 2udu 1- u2 = dx x =? ln(1- u2) = lnx =? y2 = x2 + x (3.76) 2. Solución mediante Análisis dimensional. Consideramos, en vez de (3.75), la ecuación
b2x2 + y2 dx = 2xydy
que será discriminadamente homogénea (d.h.) si [b] = y x (3.77) Obtendremos como variable que reemplaza a y la adimensional ?, obtenida obviamente de la siguiente matriz (ec. 3.78 ) b x y Lx Ly -1 1 0 1 0 1 (3.78) De manera que podemos escribir x y x ? dy dx x d? ? dx + b ? = b ? 1- x d? ? dx . (3.79)
? ln 1- ?2 =? A x = c1/2 = 3.5. ECUACIÓN HOMOGÉNEA 41 Sustituyendo y, dy dx en (3.76) b2 x2 + x2 2 = 2bx2 b ? ? 1- x d? ? dx (3.80) de donde resulta – ?2 + 1 = 2x d? ? dx . (3.81) Separando variables dx 2x = d? ? 1- ?2 = d? ? + ?d? 1- ?2 =? (3.82) 1 2 lnx + ln A = ln? – 1 2 v ? 1- ?2 , (3.83) x y sustituyendo ? = b y resulta tras sencillas operaciones y2 = b2 x2 + x A2 (3.84) que es la solución de (3.76). La solución de (3.75) corresponde a b = 1 y será y2 = x2 + cx (3.85) siendo c una constante arbitraria.
También podemos seguir este otro método dimensional alternativo al anterior. Reescribimos la ecuación (3.75) como y' = c1x2 + y2 2xy , (3.86) donde c1 esciertaconstantecondimensiones[c1] = y2x-2.Deestaformaintentamosencontarunasolución aplicando directamente el teorema Pi a las magnitudes (y,x,c1), obteniendo: x y y 0 1 x 1 0 c1 -2 =? y = c1/2x, 2 (3.87) viendo de esta forma que hemos obtenido una solución particular a nuestra ecuación diferencial por un método pedestre, comprobamos que se veri?ca . 1 c1x2 + c1x2 2xc1/2x = 2c1x2 2c1/2×2 . (3.88) Por lo tanto podemos tomar las sigientes variables t = x, u(t) = y x , (3.89) obteniendo así la siguiente ODE u't = 1 2 1 u + u – u, =? u2 – 1 = C1 t , (3.90) que en las variables originales y x 2 – 1 = C1 x , (3.91) es la solución anterior.
1 (?y – ?x)(x + y ) 1 ?y (x + y ) – ?(x, y)( – ) – ?(x, y)( – 2y?y t = v ,u(t) = x 2y?y, x . 42 CHAPTER 3. ECUACIONES DE PRIMER ORDEN 3. Solución mediante grupos de Lie. La solución: y = ± x2 + x_C1, Cálculo de las simetrías. Tenemos que resolver la siguiente ecuación en derivadas parciales: ?x + 2 2 2 xy – 2 2 2 4 x2 y2 1 y x2 + y2 2×2 y 1 x x2 + y2 2x y2 )= 0 las posibles soluciones a esta ecuación son: ? = 0,? = x 2y , ? = 0,? = – x2 2y + y 2 ,[? = x,? = y], ? = 1,? = y 2x , ? = x2 + y2,?(x, y) = 2xy Por ejemplo si seguimos la táctica de los invariantes encontramos que el campo X = x?x + y?y genera la siguiente solución:: dx ? = dy ? =? dx x = dy y =? lnx = lny =? y = ax esta sería la solución que nos sugiere precisamente el uso directo del teorema Pi. Vemos que sólo obten- emos una solución particular, pero que ésta es invariante, de hecho si pensamos en la ode como un sistema dinámico vemos que el punto ?jo de dicho sistema sería precisamente la solución y = ±x. De esta forma vemos el alcance de cada método. Veamos ahora la solución que sugiere las variables canónicas (v.c.). Las variables canónicas que obtenemos con el campo X = ?x + x son: y x =? x = u(t),y = t u(t) por lo tanto la ode original queda expresada en estas nuevas variables como: u' = 2t =? u(t) = t2 deshacemos el c.v. obteniendo x = y2 x + C1 que es precisamente la solución que hemos ido obteniendo hasta ahora. Veamos por último las variables canónicas que induce la simetría X = x?x + y?y, t = y x ,u(t) = lnx =? x = eu(t),y = teu(t) por lo tanto la ode original queda expresada en estas nuevas variables como: 2t t2 – 1 =? u(t) = -ln(-1+ t) – ln(1+ t) + C1 u' = –
deshacemos el c.v. obteniendo lnx = -ln y x – 1 – ln y x + 1 + C1 lnx = C1 + ln x (y – x)(y + x) . El cv que induce la simetría X = x es parecido al obtenido mediante el MD i.e. t = x,u(t) = y2
u = v v = – = ln 2 d(1- ?) 1- ? (1- ?) ± = – dx =? v = ln 2 = ln 2 x x 3.5. ECUACIÓN HOMOGÉNEA 43 Ejemplo 3.5.2 Resolver la ecuación: y + xy' 2 = 4x2y' (3.92) Solución. Atacaremos el problema por tres métodos:
1. Método tradicional. Haciendo el c.v. u = y/x entonces ' v 1- u) 2(1- u ± x =? du (1- u ± 1- u) = 2dx x efectuando el c.v.1- u = z, entonces: 2dx x z – 1 z + 1 c x dz z ± z deshaciendo todos los c.v. obtenemos: v =? ln|z – 1| ± ln v
C2 y = ±2C – x x 2. Solución mediante AD. es una ecuación no lineal que no es homogénea dimensionalmente si se discrimina. Sí lo será si introduci- mos una constante [a] = L-1Ly 2
Con lo que, desde el A.D., podemos plantear un cambio de variable a través del monomio, eviden- y resultando 2
Resolviendo la ecuación de segundo grado en ?' que hemos obtenido ?' = 2 x (1- ?) ± 1- ? (3.95) y una vez más se obtiene la separación de variables, que permite la integración: v 2 x dz c z ± z x (3.96) v ln|z – 1| ± ln v z – 1 z + 1 c x (3.97) En de?nitiva se obtienen 2 soluciones con sentido físico, para los dos signos, c 2 = v z – 1 2 -? z = 1+ v c x 2 ,? = 1- 1+ C x 2 =? y = -2aC – aC2 x (3.98) c 2 = v z + 1 2 -? z = v c x – 1 2 ,? = 1- C x – 1 2 =? y = 2aC – aC2 x (3.99) Las soluciones puramente matemáticas son y = ±2C – C2 x . (3.100)
x2 ?x + – ? ? ?(x,y)(-1+ v 2- v 2 ) ? x -y x 2x – y – 2 x2 – yx? x2-yx -?(x,y)? – ? – y = – 1 44 CHAPTER 3. ECUACIONES DE PRIMER ORDEN Vemos la alternativa pedestre: y + xy' 2 = 4ax2y' esta ecuación es d.h. sii [a] = x-1y, por lo tanto la solución que sugiere el AD (método pedestre(m.p.)) es: x y y 0 1 a x -1 1 =? y = ax, 1 0 comprobamos que se trata de una solución, como y' = 2ax entonces: (ax + xa)2 = 4ax2(a) por lo tanto es solución. Este monomio induce el siguiente cv. t = x, u(t) = y x , entonces 2u + tu' 2 = 4 u + tu' , = 0 i.e. lo mismo pero más sencillo y claro. 3. Método de Lie. La ecuación es homogenea ,D’alembert .Tenemos que resolver la siguiente ecuación (?y – ?x)(2x – y – 2 x2 – y x) ?y (2x – y – 2 x2 – y x)2 x 2x-y x
x x2 x
cuyas soluciones son: [? = 0, ? = x(x – y) x ], [? = 0, _? = x – y + x(x – y)], [? = 0, ? = -2×2 + 2y x – 2 x(x – y) x + y x(x – y)], y x 2 2 y la soluciones son: y = x, y = C1(-C1 + 2x) x , C (-C1 + 2x) x Vemos que una de las posibles simetrías de la ecuación son las de escala generadas por [? = x, ? = y], esta simetría genera la solución que hemos obtenido mediante AD (m.p.). Vemos las diferencias entre ambos v x(x-y) x ?y induce el siguiente métodos. Las variables canónicas nos llevan a los siguientes resultados: Elcampo X = cambio de variables; t = x,u(t) = -2 x(x – y) =? x = t,y = 1 4t2 – u(t)2 4 t u' = ±2 =? u(t) = 2t + C1 la ode queda:
si deshacemos el c.v. obtenemos -2 x(x – y) = 2x + C1
u = x – 1 + 1y(x). 1 A1/2 ' 2 ?3/2 1- ? =? v = ln 2 + ln = – 2 2 x2c A 3.5. ECUACIÓN HOMOGÉNEA 45 Ejemplo 3.5.3 Estudiar la ecuación y' = y x + y2 x2 – 1 (3.101) Solución.
1. Método tradicional. Haciendo el c.v. u = y/x entonces ' v u2 – 1 x =? v du u2 – 1 = dx x cuya solución es: lnx – ln(u + u2 – 1) + C1 = 0 si expresamos esta solución en función de las variables originales, entonces obtenemos: x = C1 y2 (x) 2 C x x y 2. Solución mediante AD. Esta ode no será d.h. si entendemos que el número 1 dentro de la ecuación es precisamente un número puro. Conlaexperienciaanteriorpodemosdarlelasdimensionesdeunaconstante[A] = L-2L2 , obtenerel x2A y2 y = x A ? 1/2 , y' = A ? 1/2 – ? (3.102) sustituimos y simpli?camos A ? 1/2 – x A1/2 ?' 2 ?1/2 ? = A ? 1/2 + A ? – A =? (3.103) – x ?' 2 ? = d? ? 1- ? = -2 dx x (3.104) que integrando v ln v 1- ? – 1 1- ? + 1 1 x -1 c2 (3.105) es decir v v 1- ? – 1 1- ? + 1 1 c x (3.106) y obtenemos la solución de (3.101): ? = 1- 1- x2c2 1+ x2c2 , (3.107) y por lo tanto: y = ± v A 2c + v 2 i.e. y = ± 1 2c + x2c 2 (3.108)
-?(x, y)?- 2 – ? – ?(x, y)?1 + 46 CHAPTER 3. ECUACIONES DE PRIMER ORDEN
El m.p. en este caso nos conduce a la siguiente solución particular: x y y 0 1 a x -2 1 =? y2 = ax2 =? y = a'x, 2 se comprueba con facilicidad que esta solución particular veri?ca la ecuación diferencial original. El c.v. que induce este monomio es el siguiente: t = x, u(t) = y x , y por lo tanto obtenemos la siguiente ODE u't = u2 – 1, =? lnx – ln y x + y2 x2 – 1 + c1 = 0. 3. Solución mediante GL. La ecuacion que debemos resolver es la siguiente: ?x + (?y – ?x)( y x + y2 x2 – 1) – ?y y x + y2 x2 – 1 2 – ? y x y2 x2 y2 – 1x 3 ? ? x y2 x2 y – 1×2 ? ? = 0 donde las posibles solucione son: [? = 0,? = y2 x2 – 1x], [? = 0, ? = -x2 + y2 – y2 x2 – 1yx], [? = x, ? = y], simetrías de escala [? = -x y, ? =
| Página siguiente |