Some Special Types of First-Order PDEs Solving Cauchy s problem for nonlinear PDEs. MA 201: Partial Differential Equations Lecture - 6

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1 MA 201: Partial Differential Equations Lecture - 6

2 Example Find a general solution of p 2 x +q 2 y = u. (1) 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. 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.

3 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 f p = dy = f q dx 2px = dy 2qy = du pf p +qf q = dp (f x +pf u ) = du 2(p 2 x +q 2 y) = dp p 2 +p = dq (f y +qf u ) dq q 2 +q From which it follows that = 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 log(p 2 x) = log(q 2 y)+loga q 2 dy +2qydq 2q 3 y +2q 2 y 2q 3 y = p 2 x = aq 2 y,where a is an arbitrary constant. (2)

4 Step 3: (Solving for p and q). Using (1) and (2), 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

5 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 which gives the general solution of equation (1).

6 Equations involving only p and q If the equation is of the form then Charpit s equations take the form f(p,q) = 0, (3) dx = dy du = = dp f p f q pf p +qf q 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 (3), we obtain a relation Thus, we have q = Q(a). u = ax +Q(a)y +b, (4) where b is a constant. Thus, (4) is a general solution of (3).

7 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 (4), we obtain a general solution as u = ax + y a +b = a 2 x +y au = b, where a and b are arbitrary constants.

8 Equations not involving the independent variables For equation of the type Charpit s equation becomes f(u,p,q) = 0, (5) 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, (6) where a is an arbitrary constant.solving (5) and (6) for p and q, we obtain q = Q(a,u) = p = aq(a,u).

9 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, (7) 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.

10 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,

11 Separable equations A first-order PDE is separable if it can be written in the form h(x,p) = g(y,q). (8) So that f(x,y,u,p,q) = h(x,p) g(y,q). For this type of equation, Charpit s equations become dx = dy du = = dp = dq. h p g q ph p qg q h x g y Consider two relations having only x and p dp h x = dx h p = dp dx + h x h p = 0. (9) Writing (9) in the form h p dp +h x dx = 0, we see that its solution is h(x,p) = a. Similarly, we get g(y,q) = a.determine p and q and solve equation du = pdx +qdy to determine an integral surface.

12 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 h(x,p) = p2 (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 = = g(y,q) (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 solution for the given PDE.

13 Clairaut s equation A first-order PDE is said to be in Clairaut form if it can be written as u = px +qy +g(p,q). (10) Charpit s equations take the form dx = dy = x +g p y +g q du px +qy +pg p +qg q = dp 0 = dq 0. Clearly p = a and q = b. Substituting the values of p and q in (10), we obtain the required general solution u = ax +by +g(a,b).

14 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, (11) which is of Clairaut s type. Putting p = a and q = b in (11), a general solution is given by u = ax +by + 1 a+b, where a and b are arbitrary constants.

15 Method of Characteristics Suppose, we need to find an integral surface for equation and equation f(x,y,u,u x,u y ) = 0. (12) g(x,y,u,p,q) = 0 (13) is compatible to the above equation. Then any point (x,y,u,p,q) on the surface g = 0 satisfies the PDE 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.(14) Thus, we have the following system of five ODEs x (t) = f p, y (t) = f q u (t) = pf p +qf q (15) p (t) = {f x +pf u } q (t) = {f y +qf u } These equations are known as the characteristic equations associated with PDE (12).

16 Let us consider the general Cauchy s problem for the first-order PDE F(x,y,u,u x,u y ) = 0 (16) subject to an appropriate initial condition given by an initial curve Γ C 1 : x(0) = x 0 (τ), y(0) = y 0 (τ), u(0) = u 0 (τ). (17) Note: We need initial conditions also for p and q in order to obtain a complete initial value problem for the system (16)-(17). We have p = u(x,y)/ x = p(x,y). For the transformation we have x = x(t), y = y(t), u = u(t), p = p(x(t),y(t)) = p(t), so that p(0) = u 0(τ) x 0 (τ) = p 0(τ) (say). Similarly, we use following notation q(0) = u 0(τ) y 0 (τ) = q 0(τ) (say).

17 Step-by-step method for solving Cauchy problem: Step 1: Find functions p 0 (τ) and q 0 (τ) (if possible) such that F(x 0 (τ),y 0 (τ),u 0 (τ),p 0 (τ),q 0 (τ)) = 0, u 0(τ) = p 0 (τ)x 0(τ)+q 0 (τ)y 0(τ) Here, we have used the fact that u(0) = u(x(0),y(0)) = u 0 (τ) Note that if p 0 (τ) and q 0 (τ) do not exist, then the Cauchy s problem for (16)-(17) has no solution. If there are several choices for (p 0 (τ),q 0 (τ)), then a solution for the Cauchy s problem (16)-(17) exists for each such choice. Step 2: Solve the characteristic system with the given initial conditions x(0) = x 0 (τ), y(0) = y 0 (τ), u(0) = u 0 (τ), p(0) = p 0 (τ), q(0) = q 0 (τ), where p 0 (τ) and q 0 (τ) are the functions given in Step 1.

18 Step 3: Step 2 will give solutions as x = x(t) = x(t,τ), y = y(t) = y(t,τ), u = u(t) = u(t,τ) (18) Here, check whether the transformation (x,y) (t,τ) is invertible or not. Consider the Jacobian J = F p (x 0 (τ),y 0 (τ),u 0 (τ),p 0 (τ),q 0 (τ))y 0(τ) F q (x 0 (τ),y 0 (τ),u 0 (τ),p 0 (τ),q 0 (τ))x 0 (τ) along given curve Γ. If J 0 (generalized transversality condition), the map (x,y) (t,τ) is invertible around Γ. Example Solve the PDE u x u y u = 0 subject to the condition u(x, x) = 1.

19 Solution. Here, we have f(x,y,u,p,q) = pq u. The characteristic system takes the form dx dt = f p = q(t), dp dt = [f x +p(t)f u ] = p(t), dy dt = f q = p(t), du dt = pf p +qf q = 2p(t)q(t), dq dt = [f y +q(t)f u ] = q(t). Note that dp dt = p(t) = p(t) = cet and dq dt = q(t) = q(t) = det, where c and d are arbitrary constants. From the given equation, we have u(t) = p(t)q(t) = cde 2t.

20 The equations for the characteristic curve are x(t) = de t +d 1, y(t) = ce t +c 1, u(t) = cde 2t, p(t) = ce t, q(t) = de t. Writing the initial condition in parametric form, we have x 0 (τ) = τ, y 0 (τ) = τ, u 0 (τ) = 1. Next, we must find p 0 (τ) and q 0 (τ) such that p 0 (τ)q 0 (τ) u 0 (τ) = 0 & 0 = u 0 (τ) = p 0(τ) q 0 (τ), Thus, we have two choices p 0 (τ) = 1 and q 0 (τ) = 1, or p 0 (τ) = 1 and q 0 (τ) = 1. For the choice p 0 (τ) = 1 = p(0) and q 0 (τ) = 1 = q(0), we obtain x = e t 1+τ, y = e t 1 τ, u = e 2t

21 Check generalized transversality condition J = q 0 (τ) p 0 (τ) 0. From the first two equations, we obtain e t = (x +y +2)/2. Thus, u = (x +y +2)2. 4 If we choose p 0 (τ) = 1 and q 0 (τ) = 1, the solution is given by u(x,y) = (x +y 2)2. 4