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Introduction to Partial Differential Equations 1 Preliminaries A partial differential equation is an equation involving a function u of several vari- ables, say, u = u(x, y, z, t) and the partial derivatives of u. For example, in the study of thermal effects in a solid body the temperature u may vary from point to point in the solid as well as from time to time, and, the derivatives ∂u ∂u ∂u ∂u , , , ∂x ∂y ∂z ∂t Prepared by Swaroop Nandan Bora will, in general, be nonzero. Further in some other practical problems like wave prop- agation, potential ﬂow, steady-state heat conduction, higher derivatives of the type ∂2u ∂ 2u ∂ 2u ∂3u , , , etc. ∂x2 ∂x∂y ∂y 2 ∂t∂x2 may occur. When the laws of physics are applied to this kind of problems, we obtain some relations between the derivatives of the kind ∂u ∂ 2u ∂ 2u F x, t, u, ,··· , 2,··· , ,··· =0 (1) ∂x ∂x ∂x∂t Such an equation relating partial derivatives is called a partial differential equation (pde). As in the case of ODEs, we deﬁne the order of a pde to be the order of the derivative of the highest order occurring in the equations. Examples: ∂u ∂ 2u • = k 2 , k ∈ R+ : 2nd order equation in two variables x and t. ∂t ∂x 3 ∂u ∂u • + = 0: 1st order third degree equation in two variables x and t. ∂x ∂t ∂u ∂u ∂u • x +y + = 0: 1st order equation in three variables x, y and t. ∂x ∂y ∂t ∂3u ∂ 3u ∂u • 3 + 2 − = f (x, y, t): 3rd order equation in three variables x, y and ∂x ∂x ∂y ∂t t. ∂2u ∂ 2u ∂ 2u • + + = 0: 2nd order equation in three variables x, y and z. ∂x2 ∂y 2 ∂z 2 1 2 First order PDE in two independent variables A ﬁrst order partial differential equation in two independent variables in its most gen- eral form is given by F (x, y, u, ux , uy ) = 0 (2) where F is a known function of its arguments. When the function F is not a linear expression in ux and uy , equation (2) is said to be a nonlinear equation which will look like: a(x, y, u)(ux )2 + b(x, y, u)(uy )2 = c(x, y, u) (3) When F is a linear expression in ux and uy , but not necessarily in u, equation (2) is Prepared by Swaroop Nandan Bora a(x, y, u)ux + b(x, y, u)uy = c(x, y, u) (4) where a and b depend on u also. This equation is called a quasilinear equation. A ﬁrst order semilinear equation is an equation of the form a(x, y)ux + b(x, y)uy = c(x, y, u) (5) where the coefﬁcients of ux and uy do not depend on u and the nonlinearity in the equation is present only in the inhomogenous term on the right hand side of (5). A ﬁrst order linear equation is of the form a(x, y)ux + b(x, y)uy = c1 (x, y)u + c2 (x, y) (6) where the dependent variable u and its partial derivatives ux , uy all appear linearly with a, b, c1 and c2 as functions of x and y only. The solution u = u(x, y) represents a surface in (x, y, u) space. This surface is called an integral surface of the partial differential equation. Examples: 1. Nonlinear equation 3 2 2 ∂u ∂u ∂u ∂u + = 0, + =1 ∂x ∂t ∂x ∂y 2. Quasilinear equation ∂u ∂u ∂u ∂u u +k = u2 , (x2 + u2 ) − xy = u3 x + y 2 ∂x ∂y ∂x ∂y 2 3. Semilinear equation ∂u ∂u ∂u ∂u x +y = u2 , x −y = xu2 ∂x ∂y ∂x ∂y 4. Linear equation ∂u ∂u ∂u ∂u x +y = xu + y, y −x = xyu + x ∂x ∂y ∂x ∂y 2.1 Origin of ﬁrst order PDEs Consider the equation Prepared by Swaroop Nandan Bora x2 + y 2 + (z − c)2 = a2 (7) where a and c are arbitrary constants. Equation (7) represents the set of all spheres whose centres lie along z-axis. Differentiating (7) with respect to x and y separately, then we get x + p(z − c) = 0, y + q(z − c) = 0. ∂z ∂z where we use the notations p = and q = . Eliminating the arbitrary constant c ∂x ∂y from these two equations, we obtain the partial differential equation yp − xq = 0 (8) which is of ﬁrst order. In some sense, the set of all spheres with centres on the z-axis is characterized by the PDE (8). Now consider the equation x2 + y 2 = (z − c)2 tan2 α, (9) where c and α are arbitrary constants, which represents the set of all right circular cones whose axes coincide with the line Oz. Differentiating (9) with respect to x and y separately, we obtain p(z − c) tan2 α = x, q(z − c) tan2 α = y. (10) After eliminating the constants c and α, we get yp − xq = 0, (11) which is same as equation (8). It is evident what the spheres and the cones have in common is that they are surfaces of revolution which have the line Oz as the axis 3 of symmetry. All surfaces of revolution with this property are characterized by an equation of the form z = f (x2 + y 2 ), (12) where f is an arbitrary function. Let u = x2 + y 2 and differentiate (12) w.r.t. x and y respectively to get p = zx = 2xf (u), q = zy = 2yf (u), where f (u) = df /dx from which we obtain (11) after eliminating the arbitrary func- tion f (u). Thus, we see that the function deﬁned by each of (7), (9) and (12) is, in some sense, a solution of the pde (11). Prepared by Swaroop Nandan Bora Let us generalize this argument as: The relations (7) and (9) are both of the type F (x, y, z, a, b) = 0, (13) where a and b are arbitrary constants. Differentiating (13) w.r.t. x and y respectively, we get ∂F ∂F ∂F ∂F +p = 0, +q = 0. (14) ∂x ∂z ∂y ∂z (13) and (14) together constitute three equations involving two arbitrary constants a and b, and in the general case, it will be possible to eliminate a and b from these equations to obtain a relation of the kind f (x, y, z, p, q) = 0, (15) showing that the system of surfaces (7) or (9) or (12) gives rise to a pde (15) of ﬁrst- order. The obvious generalization of (12) is a relation between x, y and z of the type F (u, v) = 0, (16) where u and v are known functions of x, y and z, and F is an arbitrary function of u and v. Differentiating (16) w.r.t. x and y respectively, we get ∂F ∂u ∂u ∂F ∂v ∂v +p + +p = 0, ∂u ∂x ∂z ∂v ∂x ∂z ∂F ∂u ∂u ∂F ∂v ∂v +q + +q = 0. ∂u ∂y ∂z ∂v ∂y ∂z If we, now, eliminate ∂F/∂u and ∂F/∂v from these equations, we obtain the equation ∂(u, v) ∂(u, v) ∂(u, v) p +q = , (17) ∂(y, z) ∂(z, x) ∂(x, y) which is a pde of the type (15). Equation (17) is a linear ﬁrst-order pde. 4 2.2 Cauchy’s problem for ﬁrst-order partial differential equations The existence and uniqueness of solutions to ﬁrst-order pdes are discussed below in the form of Cauchy’s Problem. Cauchy’s Problem. If (a) x0 (µ), y0 (µ) and z0 (µ) are functions which, together with their ﬁrst derivatives, are continuous in the interval M deﬁned by µ1 < µ < µ2 ; and (b) if F (x, y, z, p, q) is a continuous function of x, y, z, p and q in a certain region U of the xyzpq space, then it is required to establish the existence of a function φ(x, y) with the following properties: Prepared by Swaroop Nandan Bora (1) φ(x, y) and its partial derivatives with respect to x and y are continuous functions of x and y in a region R of the xy space. (2) For all values of x and y lying in R, the point {x, y, φ(x, y), φx (x, y), φy (x, y)} lies in U and F [x, y, φ(x, y), φx (x, y), φy (x, y)] = 0. (3) For all µ belonging to the interval M , the point {x0 (µ), y0 (µ)} belongs to the region R, and φ{x0 (µ), y0 (µ)} = z0 . From the point of view of geometry, what we want to establish is that there exists a surface z = φ(x, y) which passes through the curve Γ whose parametric equations are x = x0 (µ), y = y0 (µ), z = z0 (µ) (18) and at every point of which the direction (p, q, −1) of the normal is such that F (x, y, z, p, q) = 0. (19) This is one of the various forms of Cauchy’s problem. To prove the existence of a solution of (19) passing through a curve with equations (18) it is necessary to make some further assumptions about the form of F and the nature of the curve Γ. Theorem 2.1 If g(y) and all its derivatives are continuous for |y − y0 | < δ, if x0 is a given number, and z0 = g(y0 ), z0 = g (y0 ), and if f (x, y, z, q) and all its partial derivatives are continuous in a region S deﬁned by |x − x0 | < δ, |y − y0 | < δ, |q − q0 | < δ, 5 then there exists a unique function φ(x, y) such that (a) φ(x, y) and all its partial derivatives are continuous in a region R deﬁned by |x − x0 | < δ1 , |y − y0 | < δ1 ; (b) For all (x, y) ∈ R, z = φ(x, y) is a solution of the equation ∂z ∂z =f x, y, z, ; ∂x ∂y (c) For all values of y in the interval |y − y0 | < δ1 , φ(x0 , y) = g(y). The above theorem is due to Sonia Kowalewski and is also known as Cauchy-Kowalewski theorem. Prepared by Swaroop Nandan Bora We have seen earlier that relations of the type F (x, y, z, a, b) = 0 (20) led to pdes of ﬁrst-order. Any such relations which contains two arbitrary constants a and b is a solution of a pde of ﬁrst-order, and is said to be a complete solution or a complete integral of that equation. On the other hand, any relation of the type F (u, v) = 0 (21) involving an arbitrary function F connecting a solution of a ﬁrst-order pde is called a general solution or a general integral of that equation. It is obvious that in some sense a general integral provides a much broader set of solutions of the pde than does a complete integral. 2.3 Linear equations of the ﬁrst order Let us consider the linear pde P p + Qq = R, (22) where P, Q and R are given functions of x, y and z ( which do not involve any deriva- tives), p and q respectively denote ∂z/∂x and ∂z/∂y and we are interested in ﬁnding a relation between x, y and z involving an arbitrary function. The ﬁrst systematic theory of this type was given by Lagrange and for this reason (22) is frequently referred to as Lagrange’s equation. Theorem 2.2 The general solution of the linear pde P p + Qq = R (23) 6 is F (u, v) = 0, (24) where F is an arbitrary function and u(x, y, z) = c1 and v(x, y, z) = c2 form a solution of the equation dx dy dz = = . (25) P Q R Proof: If equations u(x, y, z) = c1 , v(x, y, z) = c2 (26) Prepared by Swaroop Nandan Bora satisfy equations (25), then the equations ux dx + uy dy + uz dz = 0 and dx dy dz = = P Q R must be compatible, i.e., we must have P ux + Quy + Ruz = 0. (27) Similarly we have P vx + Qvy + Rvz = 0. (28) Solving equations (27) and (28) for P, Q and R, we have P Q R = = . (29) ∂(u, v)/∂(y, z) ∂(u, v)/∂(z, x) ∂(u, v)/∂(x, y) Previously we saw that the relation F (u, v) = 0 leads to the pde ∂(u, v) ∂(u, v) ∂(u, v) p +q = . (30) ∂(y, z) ∂(z, x) ∂(x, y) Substituting (29) into (30), we see that (24) is a solution of (23) if u and v are given by (26). Example: Find the general solution of ∂z ∂z x2 + y2 = (x + y)z. ∂x ∂y 7 Solution: The integral surfaces of this equation are generated by the integral curves of the equations dx dy dz 2 = 2 = . (31) x y (x + y)z Taking the ﬁrst two equations dx dy 2 = 2, x y we get 1/x − 1/y = c1 , (32) and it follows from the equations that Prepared by Swaroop Nandan Bora dx − dy dz 2 − y2 = x (x + y)z giving us x−y = c2 . (33) z Combining (32) and (33), we see that the integrals curves of equation (31) are given by (33) and the equation xy = c2 (34) z and that the curves given by these equations generate the surface xy x − y F , = 0, (35) z z where F is an arbitrary function. It should be observed that this surface can be ex- pressed by equations such as z = xy f (x − y/z), or z = xy g((x − y)/xy). We can generalize theorem 2.2 as follows: Theorem 2.3 If ui (x1 , x2 , x3 , . . . , xn , z) = ci , i = 1, 2, 3, . . . , n are independent solu- tions of the equations dx1 dx2 dxn dz = = ··· = = , P1 P2 Pn R then the relation φ(u1 , u2 , . . . , un ) = 0, in which the function φ is arbitrary, is a general solution of the linear pde ∂z ∂z ∂z P1 + P2 + · · · + Pn = R. ∂x1 ∂x2 ∂xn 8 2.4 Integral surfaces passing through a given curve Here we will discuss how a general solution may be used to determine the integral surface which passes through a given curve. Suppose we have found two solutions u(x, y, z) = c1 , v(x, y, z) = c2 (36) of the auxiliary equations dx dy dz = = . P Q R Then any solution of the corresponding linear equation is of the form Prepared by Swaroop Nandan Bora F (u, v) = 0 (37) arising from a relation F (c1 , c2 ) = 0 (38) between the constants c1 and c2 . The problem we have to consider is that of deter- mining the function F in special circumstances. If we want to ﬁnd the integral surface which passes through the curve C whose parametric equations are x = x(t), y = y(t), z = z(t), where t is a parameter, then the particular solution (36) must be such that u{x(t), y(t), z(t)} = c1 , v{x(t), y(t), z(t)} = c2 . Therefore, we have two equations from which we may eliminate the single variable t to obtain a relation of the type (38). The solution we seek is then given by (37). Example: Find the integral surface of the linear pde x(y 2 + z)p − y(x2 + z)q = (x2 − y 2 )z which contains the straight line x + y = 0, z = 1. Solution: The auxiliary equations are dx dy dz 2 + z) = 2 + z) = 2 . x(y −y(x (x − y 2 )z Taking y dx + x dy dz = 2 xy 3 + xyz − yx 3 − yxz (x − y 2 )z which will ultimately give us d(xy) dz =− , xy z 9 whose solution is xyz = c1 . (39) Similarly taking x dx + y dy dz = 2 , x2 y 2 +x 2 z − y 2 x2 − y 2 z (x − y 2 )z gives us x2 + y 2 − 2z = c2 . (40) For the curve x + y = 0, z = 1, we have the equations x = t, y = −t, z = 1. Substituting these values in (39) and (40), we have Prepared by Swaroop Nandan Bora −t2 = c1 , 2t2 − 2 = c2 , and eliminating t from them, we obtain the relation 2c1 + c2 + 2 = 0, showing that the desired integral surface is x2 + y 2 + 2xyz − 2z + 2 = 0. 2.5 Surfaces orthogonal to a given system of surfaces An interesting application of the theory of linear pdes of ﬁrst-order is to determine the systems of surfaces. Suppose we are given a one-parameter family of surfaces characterized by the equation f (x, y, z) = c1 (41) and we want to ﬁnd a system of surfaces which cut each of these given surfaces at right angles. The normal at the point (x, y, z) to the surface of the system (41) which passes through that point is the direction given by the direction ratios ∂f ∂f ∂f (P, Q, R) = , , . (42) ∂x ∂y ∂z If the surface with equation z = φ(x, y) (43) cuts each surface of the given system orthogonally, then its normal at the point (x, y, z) which is in the direction ∂z ∂z , , −1 ∂x ∂y 10 is perpendicular to the direction (P, Q, R) of the normal to the surface of the set (41) at that point. Therefore, we have the linear pde ∂z ∂z P +Q =R (44) ∂x ∂y for the determination of the surfaces (43). Substituting from (42), we see that this equation is equivalent to ∂f ∂z ∂f ∂z ∂f + = . ∂x ∂x ∂y ∂y ∂z Conversely, any solution of the linear pde (44) is orthogonal to every surface of the system characterized by (41), for (44) simply states that the normal to any solution of Prepared by Swaroop Nandan Bora (44) is perpendicular to the normal to that member of the system (41) which passes through the same point. The linear pde (44) is therefore the general pde determining the surfaces orthogonal to members of the system (41); i.e., the surfaces orthogonal to the system (41) are the surfaces generated by the integral curves of the equations dx dy dz = = . (45) ∂f /∂x ∂f /∂y ∂f /∂z Example: Find the surface which intersects the surfaces of the system z(x + y) = c(3z + 1) orthogonally and which passes through the circle x2 + y 2 = 1, z = 1. z(x + y) Solution: Here f = and hence we have 3z + 1 ∂f z ∂f z ∂f x+y = , = , = . ∂x 3z + 1 ∂y 3z + 1 ∂z (3z + 1)2 The integral curves are given by dx dy dz = = z/(3z + 1) z/(3z + 1) (x + y)/(3z + 1)2 giving us dx dy dz = = . z(3z + 1) z(3z + 1) x+y Taking the ﬁrst two, we get x − y = c1 and taking (x + y)dx + (x + y)dy = z(3z + 1)dz gives us x2 + y 2 − 2z 3 − z 2 = c2 . 11 Thus any surface which is orthogonal to the given surfaces has equations of the form x2 + y 2 − 2z 3 − z 2 = f (x − y). For the particular surface passing through the circle x2 + y 2 = 1, z = 1, we have f = −2. Thus the required surface is x2 + y 2 = 2z 3 + z 2 − 2. 2.6 Compatible system of ﬁrst-order equations Prepared by Swaroop Nandan Bora n Deﬁnition 2.1 An equation of the form Fi dxi = 0 is called a Pfafﬁan equation. i=1 Deﬁnition 2.2 A necessary condition that Pfafﬁan differential equation X · dr = 0 should be integrable is that X · curl X = 0 If every solution of the ﬁrst-order pde f (x, y, z, p, q) = 0 (46) is also a solution of the pde g(x, y, z, p, q) = 0 (47) then the equations are said to be compatible. We need to ﬁnd the conditions in order that (46) and (47) are compatible. If ∂(f, g) J≡ = 0, (48) ∂(p, q) we can solve equations (46) and (47) to obtain the explicit expressions p = φ(x, y, z), q = ψ(x, y, z) (49) for p and q. The condition that the pair of equations (46) and (47) should be compatible reduces then to the conditions that the system of equations (49) should be completely integrable, i.e., the equation φ dx + ψ dy − dz = 0 (50) should be integrable. We already know that the condition that (50) is integrable is φ(−ψz ) + ψ(φz ) − (ψx − φy ) = 0 12 which is equivalent to ψx + φψz = φy + ψφz . (51) Substituting from equations (49) into (46) and differentiating w.r.t. x and z respec- tively, we get fx + fp φx + fq ψx = 0, fz + fp φz + fq ψz = 0, from which it is deduced that fx + φfz + fp (φx + φφz ) + fq (ψx + φψz ) = 0. Prepared by Swaroop Nandan Bora Similarly we can deduce from (47) that gx + φgz + gp (φx + φφz ) + gq (ψx + φψz ) = 0. Solving these two equations, we ﬁnd that 1 ∂(f, g) ∂(f, g) ψx + φψz = +φ (52) J ∂(x, p) ∂(z, p) where J is as deﬁned as in (48). If we differentiate the given pair of equations w.r.t. y and z respectively we will obtain 1 ∂(f, g) ∂(f, g) φy + ψφz = − +ψ (53) J ∂(y, q) ∂(z, q) so that, substituting from (52) and (53) and replacing φ, ψ by p, q respectively, we see that the condition that the two equations should be compatible is [f, g] = 0, (54) where [f, g] is ∂(f, g) ∂(f, g) ∂(f, g) ∂(f, g) [f, g] = +p + +q . (55) ∂(x, p) ∂(z, p) ∂(y, q) ∂(z, q) Example: Show that the equations xp = yq, z(xp + yq) = 2xy are compatible and solve them. Solution: Here f = xp − yq, g = z(xp + yq) − 2xy 13 so that ∂(f, g) ∂(f, g) ∂(f, g) ∂(f, g) = 2xy, = −x2 p − xyq, = −2xy, = xyp + y 2 q ∂(x, p) ∂(z, p) ∂(y, q) ∂(z, q) which imply that [f, g] = y 2 q 2 − p2 x2 = 0. Hence the given equations are compatible. From the ﬁrst equations, we see that p = (y/x)q, and hence we can take p = y/z, q = x/z. Then z dz = y dx + x dy Prepared by Swaroop Nandan Bora giving us z 2 = 2xy + c1 . 2.7 Charpit’s Method Let us consider solving the pde f (x, y, z, p, q) = 0 (56) In Charpit’s method the fundamental idea is to introduce a second ﬁrst-order partial differential equation g(x, y, z, p, q, a) = 0 (57) containing an arbitrary constant a and which is such that: • Equations (56) and (57) can be solved to give p = p(x, y, z, a), q = q(x, y, z, a) • The equation dz = p(x, y, z, a)dx + q(x, y, z, a)dy (58) is integrable. When such a function g can be found, the solution of equation (58) F (x, y, z, a, b) = 0 (59) containing two arbitrary constants a and b, will be a solution of (56). 14 The main problem is then the determination of the second equation (57) which is not much difﬁcult as we need an equation g = 0 compatible with the given equation f = 0. That is, determining g from the following equation ∂g ∂g ∂g ∂g ∂g fp + fq + (pfp + qfq ) − (fx + pfz ) − (fy + qfz ) = 0. (60) ∂x ∂y ∂z ∂p ∂q Now the problem is to ﬁnd a solution of this equation, and it is equivalent to ﬁnding an integral of the subsidiary equations dx dy dz dp dq = = = = . (61) fp fq pfp + qfq −(fx + pfz ) −(fy + qfz ) Prepared by Swaroop Nandan Bora These equations are known as Charpit’s equations. Once an integral g(x, y, z, p, q, a) of this kind has been found, the problem reduces to solving for p, q and then integrating the equation (58). Example: Find a complete integral of the equation p2 x + q 2 y = z. (62) Solution: The auxiliary equations are: dx dy dz dp dq = = = = 2px 2qy 2(p2 x + q 2 y p − p2 q − q2 from which it follows that p2 dx + 2pxdp q 2 dy + 2qydq = p2 x q2y ⇒ p2 x = aq 2 y, a is a constant. (63) Solving (62) and (63) for p and q, we get 1/2 1/2 az z p= , q= (1 + a)x) (1 + a)y) So, equation (58) in this case becomes 1/2 1/2 1+a a 1/2 1 dz = dx + dy z x y ⇒ {(1 + a)z}1/2 = (ax)1/2 + y 1/2 + b which is the required complete integral of (62). 15 2.8 Special types of ﬁrst-order PDEs 2.8.1 Equations involving only p and q: For equations of the type f (p, q) = 0 (64) Charpit’s equations reduce to dx dy dz dp dq = = = = (65) fp fq pfp + qfq 0 0 An obvious solution of this equation is Prepared by Swaroop Nandan Bora p=a (66) the corresponding value of q being obtained from (64) as f (a, q) = 0 (67) so that q = Q(a), a a constant. The solution of the equation is z = ax + Q(a)y + b (68) where b is a constant. It is to be noted that we took dp = 0 here to provide our second equation. In some problems, the amount of work involved is considerably reduced if we take dq = 0, which leads to q = a. Example: Find the complete integral of pq = 1. Solution: We know that dp = 0 ⇒ p = a, and then aq = 1 ⇒ q = 1/a. The complete integral is y +b z = ax + a ⇒ a2 x + y − az = c, where a, c are arbitrary constants. 16 2.8.2 Equations not involving the independent variable: Let the pde be of the form f (z, p, q) = 0 (69) Then the Charpit’s equations take the form dx dy dz dp dq = = = = . (70) fp fq pfp + qfq −pfz −qfz which leads to the relation p = aq (71) Prepared by Swaroop Nandan Bora Solving (69) and (71), we obtain expressions for p, q from which a complete integral follows immediately. Example: Find a complete integral of p2 z 2 + q 2 = 1. Solution: Putting p = aq, we have q 2 (1 + a2 z 2 ) = 1 ⇒ q = (1 + a2 z 2 )−1/2 , p = a(1 + a2 z 2 )−1/2 Hence dz = pdx + qdy ⇒ (1 + a2 z 2 )1/2 dz = adx + dy. which leads to the complete integral az(1 + a2 z)1/2 − log[az + (1 + a2 z 2 )1/2 ] = 2a(ax + y + b). 2.8.3 Separable equations: We say that a ﬁrst-order pde is separable, if it can be written in the form f (x, p) = g(y, q) (72) For such equations, Charpit’s equations become dx dy dz dp dq = = = = . (73) fp −gq pfp − qgq −fx −gy so that we have an ODE dp fx + = 0. dx fp in x and p which may be solved to give p as a function of x, and an arbitrary constant. 17 Writing this equation in the form fp dp + fx dx = 0, we see that its solution is f (x, p) = a. Hence we determine p and q from the relations f (x, p) = a; g(y, q) = a (74) and then proceed as in the general theory. Example: Determine the complete integral of p2 y(1 + x2 ) = qx2 . Prepared by Swaroop Nandan Bora Solution: Given that p2 y(1 + x2 ) = qx2 . p2 (1 + x2 ) q ⇒ 2 = , x y ax so that p = √ , q = a2 y. 1+x 2 Hence a complete integral is √ 1 z = a 1 + x2 + a2 y 2 + b, 2 where a and b are constants. 2.8.4 Clairaut’s Equation A ﬁrst-order PDE is said to be of Clairaut type if it can be written in the form z = px + qy + g(p, q) (75) The corresponding Charpit’s equations are dx dy dz dp dq = = = = x + gp y + gq px + qy + pgp + qgq 0 0 ⇒ p = a, q = b. Using these values in (75), we get the complete integral z = ax + by + g(a, b) (76) as is readily veriﬁed by direct differentiation. Example: Find a complete integral of (p + q)(z − xp − yq) = 1. 18 Solution: The given equation can be written in the form 1 z = xp + yq + , p+q we see that the complete integral is 1 z = ax + by + . a+b Prepared by Swaroop Nandan Bora 19