Compatible Systems and Charpit s Method Charpit s Method Some Special Types of First-Order PDEs. MA 201: Partial Differential Equations Lecture - 5

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1 Compatible Systems and MA 201: Partial Differential Equations Lecture - 5

2 Compatible Systems and Definition (Compatible systems of first-order PDEs) A system of two first-order PDEs and f(x,y,u,p,q) 0 (1) g(x,y,u,p,q) 0 (2) are said to be compatible if they have a common solution. Theorem Equations (1) and (2) are compatible on a domain D if (i) J (f,g) (p,q) f p f q g p g q 0 on D. (ii) p and q can be explicitly solved from (1) and (2) as p φ(x,y,u) and q ψ(x,y,u). Further, the equation is integrable. du φ(x,y,u)dx +ψ(x,y,u)dy

3 Compatible Systems and Theorem A necessary and sufficient condition for the integrability of the equation du φ(x,y,u)dx +ψ(x,y,u)dy is [f,g] (f,g) (x,p) + (f,g) (y,q) +p (f,g) (u,p) +q (f,g) 0. (3) (u,q) In other words, equations (1) and (2) are compatible iff (3) holds. Example Show that the equations xp yq 0, xup +yuq 2xy are compatible and solve them. Solution. Take f xp yq 0, g u(xp +yq) 2xy 0. Then f x p, f y q, f u 0, f p x, f q y, g x up 2y, g y uq 2x, g u xp +yq, g p ux, g q uy.

4 Compatible Systems and Compute J (f,g) (p, q) f p g p f q g q x y ux uy uxy +uxy 2uxy 0 for x 0, y 0, u 0. Further, (f,g) (x, p) f x f p g x g p p x up 2y ux uxp x(up 2y) 2xy (f,g) (u, p) f u f p g u g p 0 x xp +yq ux 0 x(xp +yq) x2 p xyq (f,g) (y,q) f y f q g y g q q y uq 2x uy quy +y(uq 2x) 2xy (f,g) (u, q) f u f q g u g q 0 y xp +yq zy y(xp +yq) y2 q +xyp.

5 Compatible Systems and It is an easy exercise to verify that [f,g] (f,g) (x,p) + (f,g) (y,q) +p (f,g) (u,p) +q (f,g) (u, q) 2xy x 2 p 2 xypq 2xy +y 2 q 2 +xypq y 2 q 2 x 2 p 2 0. So the equations are compatible. Next step is to determine p and q from the two equations xp yq 0, u(xp+yq) 2xy. Using these two equations, we have uxp +uyq 2xy 0 xp +yq 2xy u 2xp 2xy u p y u φ(x,y,u). and xp yq 0 q xp y xy yu q x u ψ(x,y,u).

6 Compatible Systems and Substituting p and q in du pdx +qdy, we get and hence integrating, we obtain udu ydx +xdy d(xy), u 2 2xy +k, where k is a constant. NOTE: For the compatibility of f(x,y,u,p,q) 0 and g(x,y,u,p,q) 0, it is not necessary that every solution of f(x,y,u,p,q) 0 be a solution of g(x,y,u,p,q) 0 or vice-versa. For instance, the equations f xp yq x 0 (4) g x 2 p +q xu 0 (5) are compatible. They have common solutions u x +c(1+xy), where c is an arbitrary constant. Note that u x(y +1) is a solution of (4) but not of (5).

7 Compatible Systems and Charpit s method It is a general method for finding the general solution of a nonlinear PDE of first-order of the form f(x,y,u,p,q) 0. (6) Basic Idea: To introduce another partial differential equation of the first order g(x,y,u,p,q,a) 0 (7) which contains an arbitrary constant a and is such that (i) equations (6) and (7) can be solved for p and q to obtain p p(x,y,u,a), q q(x,y,u,a). (ii) the equation du p(x,y,u,a)dx +q(x,y,u,a)dy (8) is integrable.

8 Compatible Systems and When such a function g is found, the solution F(x,y,u,a,b) 0 of (8) containing two arbitrary constants a and b will be the solution of (6). The compatibility of equations (6) and (7) yields [f,g] (f,g) (x,p) + (f,g) (y,q) +p (f,g) (u,p) +q (f,g) (u,q) 0. Expanding it, we are led to the following linear PDE in g(x,y,u,p,q): g f p x +f g q y +(pf p +qf q ) g u (f x +pf u ) g p (f y +qf u ) g q 0. (9)

9 Compatible Systems and Now solve (9) to determine g by finding the integrals of the following auxiliary equations: dx dy f p f q du pf p +qf q dp (f x +pf u ) dq (f y +qf u ) (10) These equations are known as Charpit s equations. Once an integral g(x,y,u,p,q,a) of this kind has been found, the problem reduces to solving for p and q, and then integrating equation (8). Remarks. For finding integrals, all of Charpit s equations (10) need not be used. p or q must occur in the solution obtained from (10).

10 Compatible Systems and Example Find a general solution of p 2 x +q 2 y u. (11) Solution. To find a general solution, we proceed as follows: Step 1: (Computing f x, f y, f u, f p, f q ). Set f p 2 x +q 2 y u 0. Then and hence, f x p 2, f y q 2, f u 1, f p 2px, f q 2qy, pf p +qf q 2p 2 x +2q 2 y, (f x +pf u ) p 2 +p, (f y +qf u ) q 2 +q.

11 Compatible Systems and Step 2: (Writing Charpit s equations and finding a solution g(x,y,u,p,q,a)). The Charpit s equations (or auxiliary) equations are: dx dy f p f q dx 2px dy 2qy From which it follows that du pf p +qf q dp (f x +pf u ) du 2(p 2 x +q 2 y) dp p 2 +p p 2 dx +2pxdp 2p 3 x +2p 2 x 2p 3 x p 2 dx +2pxdp p 2 x On integrating, we obtain q2 dy +2qydq q 2 y dq (f y +qf u ) dq q 2 +q q 2 dy +2qydq 2q 3 y +2q 2 y 2q 3 y log(p 2 x) log(q 2 y)+loga p 2 x aq 2 y, (12) where a is an arbitrary constant.

12 Compatible Systems and Step 3: (Solving for p and q). Using (11) and (12), we find that p 2 x +q 2 y u, p 2 x aq 2 y (aq 2 y)+q 2 y u q 2 y(1+a) u [ ] 1/2 q 2 u (1+a)y q u. (1+a)y and p 2 aq 2y x a u (1+a)y [ ] 1/2 au p. (1+a)x y x au (1+a)x

13 Compatible Systems and Step 4: (Writing du p(x,y,u,a)dx +q(x,y,u,a)dy and finding its solution). Writing Integrate to have [ ] 1/2 ] 1/2 au u du dx +[ dy (1+a)x (1+a)y ( ) 1/2 1+a ( a ) ( ) 1/2dx 1/2 1 du + dy. u x y [(1+a)u] 1/2 (ax) 1/2 +(y) 1/2 +b the general solution of equation (11).

14 Compatible Systems and Equations involving only p and q If the equation is of the form then Charpit s equations take the form dx f p dy f q f(p,q) 0, (13) du pf p +qf q dp 0 dq 0 the last two are actually equivalent to dp dt 0, dq 0 and hence dt an immediate solution is given by p a, where a is an arbitrary constant. Substituting p a in (13), we obtain a relation Then, integrating the expression we obtain q Q(a). du adx +Q(a)dy u ax +Q(a)y +b, (14) where b is a constant. Thus, (14) is a general solution of (13).

15 Compatible Systems and Note: Instead of taking dp dq dt 0, we can take dt 0 q a. In some problems, taking dq 0 the amount of computation involved may be reduced considerably. Example Find a general solution of the equation pq 1. Solution. If p a then pq 1 q 1 a. In this case, Q(a) 1/a. From (14), we obtain a general solution as u ax + y a +b a 2 x +y au b, where a and b are arbitrary constants.

16 Compatible Systems and Equations not involving the independent variables For equation of the type Charpit s equation becomes f(u,p,q) 0, (15) dx dy du dp dq. f p f q pf p +qf q pf u qf u From the last two relations, we have dp dq dp pf u qf u p dq q p aq, (16) where a is an arbitrary constant. Solving (15) and (16) for p and q, we obtain q Q(a,u) p aq(a,u).

17 Compatible Systems and Now It gives general solution as where b is an arbitrary constant. du pdx +qdy du aq(a,u)dx +Q(a,u)dy du Q(a,u)[adx +dy]. du ax +y +b, (17) Q(a,u) Example Find a general solution of the PDE p 2 u 2 +q 2 1. Solution. Putting p aq in the given PDE, we obtain a 2 q 2 u 2 +q 2 1 q 2 (1+a 2 u 2 ) 1 q (1+a 2 u 2 ) 1/2.

18 Compatible Systems and Now, p 2 (1 q 2 )/u 2 p 2 a 2 1+a 2 u 2 p a(1+a 2 u 2 ) 1/2. ( )( ) (1+a 2 u 2 ) u 2 Substituting p and q in du pdx +qdy, we obtain du a(1+a 2 u 2 ) 1/2 dx +(1+a 2 u 2 ) 1/2 dy (1+a 2 u 2 ) 1/2 du adx +dy 1 { } au(1+a 2 u 2 ) 1/2 log[au +(1+a 2 u 2 ) 1/2 ] 2a which is the general solution of the given PDE. ax +y +b,

19 Compatible Systems and Separable equations A first-order PDE is separable if it can be written in the form f(x,p) g(y,q). (18) For this type of equation, Charpit s equations become dx f p dy g q du dp dq. pf p qg q f x g y From the last two relations, we obtain an ODE dp dx dp f x f p dx + f x 0 (19) f p which may be solved to yield p as a function of x and an arbitrary constant a. Writing (19) in the form f p dp +f x dx 0, we see that its solution is f(x,p) a. Similarly, we get g(y,q) a. Determine p and q from the equation f(x,p) a, g(y,q) a and then use the relation du pdx +qdy to determine a complete integral.

20 Compatible Systems and Example Find a general solution of p 2 y(1+x 2 ) qx 2. Solution. First we write the given PDE in the form It follows that p 2 (1+x 2 ) x 2 q y p 2 (1+x 2 ) x 2 a 2 p where a is an arbitrary constant. Similarly, q y a2 q a 2 y. Now, the relation du pdx +qdy yields du (separable equation) ax 1+x 2, ax dx 1+x 2 +a2 ydy u a 1+x 2 + a2 y 2 +b, 2 where a and b are arbitrary constants, a general solutionfor the given PDE.

21 Compatible Systems and Clairaut s equation A first-order PDE is said to be in Clairaut form if it can be written as Charpit s equations take the form u px +qy +f(p,q). (20) dx dy x +f p y +f q du px +qy +pf p +qf q dp 0 dq 0. Now, equivalently considering dp dt 0 p a, where a is an arbitrary constant. dq dt 0 q b, where b is an arbitrary constant. Substituting the values of p and q in (20), we obtain the required general solution u ax +by +f(a,b).

22 Compatible Systems and Example Find a general solution of (p +q)(u xp yq) 1. Solution. The given PDE can be put in the form u xp +yq + 1 p +q, (21) which is of Clairaut s type. Putting p a and q b in (21), a general solution is given by u ax +by + 1 a+b, where a and b are arbitrary constants.