Chapter 2. First-Order Partial Differential Equations. Prof. D. C. Sanyal
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1 BY Prof. D. C. Sanyal Retired Professor Of Mathematics University Of Kalyani West Bengal, India dcs klyuniv@yahoo.com 1
2 Module-2: Quasi-Linear Equations of First Order 1. Introduction In this section, we propose to consider some characteristics and properties of quasilinear first-order partial differential equation of the form (5) of Module-1. As shown in Section 3 of Module-1 all these equations can be represented by Lagrange s equation (8) of Module-1, vi. P p + Qq = R, (p = x, q = y ) (1) where each of P,Q, R is a function of x, y, and they do not involve p or q. The equation (1) can be generalised to n independent variables x 1,x 2,,x n in the form P 1 p 1 + P 2 p P n p n = R, (2) in which P i = P i (x 1,x 2,,x n ), R = R(x 1,x 2,,x n ) and p i = x i, (i = 1,2,,n). 2. Lagrange s Auxiliary Equations Theorem 1: The quasi-linear first order partial differential equation P p + Qq = R has the general solution ϕ(u,v) = 0, where u(x,y,) = c 1 and v(x,y,) = c 2 are solutions of the equations P = Q = d R. (3) Proof: Since u(x,y,) = c 1 is a solution of equation (3), so the equations u x + u y + u d = 0 and P = Q = d R are compatible to each other and, therefore, we have P u x + Qu y + Ru = 0. Similarly, we get P v x + Qv y + Rv = 0. Solving these equations, we get P (u,v) (y,) = Q (u,v) (,x) = R (u,v) (x,y) (4) 2
3 Now, we have alrea seen in Section 3 of Module -1 that the relation ϕ(u,v) = 0 leads to the partial differential equation p (u,v) (y,) + q (u,v) (,x) = (u,v) (x, y) Substituting from (4) into this equation, we see that ϕ(u,v) = 0 is a solution of the equation P p + Qq = R. The equation (3) is known as Lagrange s auxiliary equation. (5) Geometrical interpretation of the equation P p + Qq = R Since the direction cosines of the normal to the surface = f (x,y) at a point are proportional to x, y, 1, i.e. to p,q, 1, so Lagrange s equation P p + Qq = R can be written as P p + Qq + R( 1) = 0. (6) Thus the normal at a point to a given surface is perpendicular to a straight line having direction cosines in the ratio P : Q : R. Also the equations /P = /Q = d/r represent a family of curves, the tangent at any point of which has direction cosines in the ratio P : Q : R. Again, the relation ϕ(u,v) = 0, where u(x,y,) = constant and v(x,y,) = constant are two particular integrals of the equations (3), represents a surface through this point. Now, a curve of the family through any point on the surfaces lies entirely on the surface. Hence, the normal to this surface at this point is at right angles to the tangent at that point to the curve. In other words, it is perpendicular to the straight line with direction cosines P : Q : R. Now, since the equations (3) and (6) are equivalent (as they define the same set of surfaces), so the relation ϕ(u,v) = 0 is an integral of the equation (6), provided that u(x,y,) = constant and v(x,y,) = constant are two independent solutions of (3) and the function ϕ is arbitrary. Example 1: Find the general integral of the quasi-linear equation px( 2y 2 ) = ( qy)( y 2 2x 3 ). Solution: The given equation can be written as x( 2y 2 )p + y( y 2 2x 3 )q = ( y 2 2x 3 ). 3
4 Lagrange s auxiliary equations are x( 2y 2 ) = y( y 2 2x 3 ) = d ( y 2 2x 3 ). (1) The last two equations of (1) give y constant. Thus u(x,y,) = y/ = c 1. Also, from (1) we get = d leading to the solution y = c 1, where c 1 is or x( 2y 2 ) = 2y d 2y 2 ( y 2 2x 3 ) ( y 2 2x 3 ) = d(y 2 ) ( y 2 2x 3 )(2y 2 ) x = d(y2 ) y 2 + 2x 3 = dw w + 2x 3, where w = y2 or 2x 3 + w xdw = 0, or xdw w 2x x 2 = 0 Integrating, we have x 2 w x = constant = c 2, i.e. v(x,y,) = x 2 + x y2 x 2 = c 2. Thus, the required general integral is ϕ( y,x2 + x y2 x 2 ) = 0. Theorem 2: If u i (x 1,x 2,,x n ) = c i, (i = 1,2,,n), be n independent solutions of the equations 1 = 2 = = n = d P 1 P 2 P n R (7) where c i are constant, P i = P i (x 1,x 2,,x n ), = (x 1,x 2,,x n ) and R = R(x 1,x 2,,x n ), then the relation Φ(u 1,u 2,,u n ) = 0, where Φ is arbitrary, is the general solution of the quasi-linear partial differential equation where p i = x i, (i = 1,2,,n). P 1 p 1 + P 2 p P n p n = R, (8) Proof: Since u i (x 1,x 2,,x n ) = c i, (i = 1,2,,n), are the solutions of the equation (7) so the n equations n u i x j + u i d = 0, (i = 1,2,,n), j j=1 must be compatible with equations (7) and we must have n j=1 u P i j + R u i x j = 0, (i = 1,2,,n), 4
5 Solving these equations for P i, we get where the Jacobian P i (u 1,u 2,,u n ) (x 1,x 2,,x i 1,,x i+1,,x n ) J = (u 1,u 2,,u n ) (x 1,x 2,,x n ) = = u 1 u 1 x 1 R (u 1,u 2,,u n ) (x 1,x 2,,x n ) x 2 u 2 x 1 u 2 x 2. u n x 1 u 1 x n u 2 x n. u n x 2 u n x n (9) Now, differentiation of the relation Φ(u 1,u 2,,u n ) = 0 partially with respect to x i leads to n j=1 ( Φ u j + u j u j x i ) = 0, (i = 1,2,,n). x i Eliminating the n quantities Φ u j, (j = 1,2,,n), from these equations, we get n j=1 (u 1,u 2,,u n ) (x 1,x 2,,x i 1,,x i+1,,x n ) p j + (u 1,u 2,,u n ) (x 1,x 2,,x n ) = 0. Substituting from (9) into these equations, if follows that P 1 p 1 + P 2 p P n p n = R. Hence, the function Φ(u 1,u 2,,u n ) = 0 satisfies the equation (8) whenever u 1,u 2,,u n are independent solutions of (7). Example 2: Solve the equation (u + y + ) u + (u + + x) u + (u + x + y) u x y = x + y +. Solution: Here the auxiliary equations are (u + y + ) = (u + + x) = d (u + x + y) d(x y) d(y ) d( x) d(x + y + + u) = = = x y y x 3(x + y + + u) Integration of first two equations gives x y y = constant, second and third equations lead to y u = constant, while the last two equations gives (x + y + + u) 3 1 ( u) = constant. Hence the required solution is ( ) x y Φ y, y u, (x + y + + u) 3 1 ( u) = 0. 5
6 3. Integral Surfaces To investigate the geometrical content of a first-order-partial differential equation, we consider the general quasi-linear equation P p + Qq = R, (1) where each of P, Q, R is a function of x, y, ; form = (x,y), p = x, q = y. We assume that the possible solution of (1) in the form = (x,y) or in implicit f (x,y,) (x,y) = 0 (2) represents a possible solution surface in (x, y, )-space. This is called an integral surface of the equation (1). A curve in (x,y,)-space, whose tangent at every point coincides with the characteristic direction field (P, Q, R) is called a characteristic curve. If the parametric equations of this characteristic curve are x = x(t), y = y(t), = (t), then the tangent vector to this curve is ( dt, dt, d dt ) which must be equal to (P,Q,R). Thus, the system of ordinary differential equations of the characteristic curve is given by dt = P (x,y,), dt = Q(x,y,), d = R(x,y,), (3) dt These are called the characteristic equations of the quasi-linear equation (1). Integral Surfaces Passing Through a Given Curve Suppose that u(x,y,) = c 1 and v(x,y,) = c 2 are the solutions of the Lagrange s auxiliary equations P = Q = d R (4) so that the solution of the corresponding quasi-linear equation (1) is ϕ(u,v) = 0 arising from a relation ϕ(c 1,c 2 ) = 0 between the constants c 1 and c 2. 6
7 To find the integral surface through the curve Γ having parametric equations x = x(t), y = y(t), = (t), t being a parameter, the particular solutions u(x,y,) = c 1 and v(x,y,) = c 2 must be such that u(x(t),y(t),(t)) = c 1, and v(x(t),y(t),(t)) = c 2 from which the parameter t can be eliminated to obtain a relation the form ϕ(c 1,c 2 ) = 0. The desired solution is then given as ϕ(u,v) = 0. Example 3: Find the equation of the integral surface of the differential equation 2y( 3)p + (2x )q = y(2x 3) which passes through the circle = 0, x 2 + y 2 = 2x. Solution: Here Lagrange s auxiliary equations are 2y( 3) = (2x ) = d y(2x 3) From the first and last equations, we get (1) 3 = 2d, i.e. (2x 3) 2( 3)d = 0 2x 3 whose solution is u(x,y,) = x 2 3x = c 1, where c 1 is constant. Also, from the equation (1), we derive + 2(y d) = 0 Integrating, v(x,y,) = x + y 2 2 = c 2, c 2 being constant. Hence the general solution of the given equation is ϕ(u,v) = 0, i.e. ϕ(x 2 3x 2 + 6,x + y 2 2) = 0. Now, the parametric equations of the given curve are x = 1 + cost, y = sint, = 0 and hence (1 + cost) 2 3(1 + cost) = c 1 and 1 + cost + sin 2 t = c 2 i.e. cos 2 t cost = 2 + c 1 and cost cos 2 t = c 2 so that 2 + c 1 + c 2 = 0 by elimination of t. Thus the required equation of the integral surface is 2 + x 2 3x x + y 2 2 = 0 i.e x 2 + y 2 2 2x + 4 = 0. 7
8 Surfaces Orthogonal to a Given System of Surfaces We now find a system of surfaces which cut orthogonally a one-parameter system of surfaces given by the equation F(x, y, ) = c, c being a parameter. (5) First, we note that the normal at any point (x,y,) to the surface (5) has direction ratios (P,Q,R) = ( F x, F y, F ). If the surface = f (x,y) cuts system of surfaces (5) orthogonally, then its normal at the point (x,y,) is in the direction ( x, y, 1) and is perpendicular to the direction (P,Q,R) of the normal to the surface (5) at the point. Thus, we have F x x + F y y + F ( 1) = 0, i.e. P p + Qq = R (6) Conversely, we note that the normal to any solution of (6) is perpendicular to that member of the system (5) which passes through the point and so any solution of (6) is orthogonal to every surface of the system (5) at that point. Thus, the equation (6) is the general partial differential equation which determines the surface orthogonal to the system (5), i.e. the surfaces orthogonal to the system (5) are generated by the integral curves of the equations F/ x = F/ y = d F/. (7) Example 4: Find the equation of the system of surfaces which cut orthogonally the cones of the system x 2 + y = cxy, c being a parameter. Solution: The given system of surfaces is F(x,y,) = x2 +y = c. The auxiliary equations are xy i.e. F/ x = F/ y = d F/ x x 2 y 2 2 = y x 2 + y 2 2 = d 2. 1 y y = 2 x 2 xy 2 y x + 1 x = 2 xy 2 d 2 xy It follows that leading to the solutions x + y + d = 0 and x y x 2 y 2 = d u(x,y,) = x 2 + y = c 1 and v(x,y,) = x2 y 2 2 = c 2, 8
9 respectively, where c 1 and c 2 are constants. Thus the general solution of the given ( equation is ϕ x 2 + y 2 + 2, x2 y 2 = 0 and the equation of the required system of surfaces is x 2 + y = f ) ( 2 x 2 y 2 2 ) Example 5: Find the surface which intersects the surfaces of the system (x + y) = c(3 + 1) orthogonally and passes through the circle x 2 + y 2 = 1, = 1. Solution: The given system of surfaces is F(x,y,) = (x+y) 3+1 auxiliary equations are = c, c being parameter. The i.e. F/ x = = F/ y = (3 + 1)d =. x + y d F/ 3+1 = 3+1 = d x+y (3+1) 2 It follows that = 0 and (x + y)d(x + y) ( )d = 0 whose solutions are u(x,y,) = x y = c 1 and v(x,y,) = (x y) (2 + 1) = c 2, where c 1 and c 2 are constants. Now the given circle has parametric equations x = cost, y = sint, = 1 so that cost sint = c 1 and (cost + sint) 2 6 = c 2, i.e. sin2t = 1 c1 2 as well as sin2t = 5 + c 2. Eliminating t between these two relations, we get c1 2 + c = 0 (x y) 2 + (x + y) (2 + 1) + 4 = 0, i.e. x 2 + y 2 = , which is the equation of the required surface. 9
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