MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 16 Lecture 3 Quasilinear First-Order PDEs A first order quasilinear PDE is of the form a(x, y, z) + b(x, y, z) x y = c(x, y, z). (1) Such equations occur in a variety of nonlinear wave propagation problems. Let us assume that an integral surface z = z(x, y) of (1) can be found. Writing this integral surface in implicit form as F (x, y, z) = z(x, y) z = 0. Note that the gradient vector F = (z x, z y, 1) is normal to the integral surface F (x, y, z) = 0. The equation (1) may be written as az x + bz y c = (a, b, c) (z x, z y, 1) = 0. (2) This shows that the vector (a, b, c) and the gradient vector F are orthogonal. In other words, the vector (a, b, c) lies in the tangent plane of the integral surface z = z(x, y) at each point in the (x, y, z)-space where F 0. At each point (x, y, z), the vector (a, b, c) determines a direction in (x, y, z)-space is called the characteristic direction. We can construct a family of curves that have the characteristic direction at each point. If the parametric form of these curves is then we must have x = x(t), y = y(t), and z = z(t), (3) = a(x(t), y(t), z(t)), = b(x(t), y(t), z(t)), = c(x(t), y(t), z(t)), (4) because (/, /, /) is the tangent vector along the curves. The solutions of (4) are called the characteristic curves of the quasilinear equation (1). We assume that a(x, y, z), b(x, y, z), and c(x, y, z) are sufficiently smooth and do not all vanish at the same point. Then, the theory of ordinary differential equations ensures that a unique characteristic curve passes through each point (x 0, y 0, z 0 ). The IVP for (1) requires that z(x, y) be specified on a given curve in (x, y)-space which determines a curve C in (x, y, z)-space referred to as the initial curve. To solve this IVP, we pass a characteristic curve through each point of the initial curve C. If these curves generate a surface known as integral surface. This integral surface is the solution of the IVP.
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 17 REMARK 1. (i) The characteristic equations (4) for x and y are not, in general, uncoupled from the equation for z and hence differ from those in the linear case (cf. Eq. (3) of Lecture 2). (ii) The characteristics equations (4) can be expressed in the nonparametric form as a = b = c. (5) Below, we shall describe a method for finding the general solution of (1). This method is due to Lagrange hence it is usually referred to as the method of characteristics or the method of Lagrange. 1 The method of characteristics It is a method of solution of quasi-linear PDE which is stated in the following result. THEOREM 2. The general solution of the quasi-linear PDE (1) is F (u, v) = 0, (6) where F is an arbitrary function and u(x, y, z) = c 1 and v(x, y, z) = c 2 form a solution of the equations a = b = c. (7) Proof. If u(x, y, z) = c 1 and v(x, y, z) = c 2 satisfy the equations (1) then the equations are compatible with (7). Thus, we must have u x + u y + u z = 0, v x + v y + v z = 0 au x + bu y + cu z = 0, av x + bv y + cv z = 0. Solving these equations for a, b and c, we obtain a (u,v) (y,z) = b (u,v) (z,x) = c (u,v) (x,y) Differentiate F (u, v) = 0 with respect to x and y, respectively, to have { F u u x + u } + F { v x v x + v } = 0 x { F u u y + u } + F { v y v y + v } = 0. y. (8)
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 18 Eliminating F u and F v from these equations, we obtain (u, v) x (y, z) + (u, v) (u, v) = y (z, x) (x, y) (9) In view of (8), the equation (9) yields a x + b y = c. Thus, we find that F (u, v) = 0 is a solution of the equation (1). proof.. REMARK 3. This completes the All integral surfaces of the equation (1) are generated by the integral curves of the equations (4). All surfaces generated by integral curves of the equations (4) are integral surfaces of the equation (1). EXAMPLE 4. Find the general integral of xz x + yz y = z. Solution. The associated system of equations are From the first two relation we have Similarly, x = y = z. x = y = ln x = ln y + ln c 1 = x y = c 1. z = y = z y = c 2. Take u 1 = x y and u 2 = z y. The general integral is given by F ( x y, z y ) = 0. EXAMPLE 5. Find the general integral of the equation z(x + y)z x + z(x y)z y = x 2 + y 2. Solution. The characteristic equations are z(x + y) = Each of these ratio is equivalent to y + x z 0 z(x y) = = x 2 + y 2. x y z. 0
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 19 Consequently, we have Integrating we obtain two integrals d{xy z2 2 } = 0 and d{1 2 (x2 y 2 z 2 )} = 0. 2xy z 2 = c 1 and x 2 y 2 z 2 = c 2, where c 1 and c 2 are arbitrary constants. Thus, the general solution is where F is an arbitrary function. F (2xy z 2, x 2 y 2 z 2 ) = 0, Next, we shall discuss a method for solving a Cauchy problem for the first-order quasilinear PDE (1). The following theorem gives conditions under which a unique solution of the initial value problem for (1) can be obtained. THEOREM 6. Let a(x, y, z), b(x, y, z) and c(x, y, z) in (1) have continuous partial derivatives with respect to x, y and z variables. Let the initial curve C be described parametrically as x = x(s), y = y(s), and z = z(x(s), y(s)). The initial curve C has a continuous tangent vector and J(s) = a[x(s), y(s), z(s)] b(x(s), y(s), z(s)] 0 (10) ds ds on C. Then, there exists a unique solution z = z(x, y), defined in some neighborhood of the initial curve C, satisfies (1) and the initial condition z(x(s), y(s)) = z(s). Proof. The characteristic system (4) with initial conditions at t = 0 given as x = x(s), y = y(s), and z = z(s) has a unique solution of the form with continuous derivatives in s and t, and x = x(s, t), y = y(s, t), z = z(s, t), x(s, 0) = x(s), y(s, 0) = y(s), z(s, 0) = z(s). This follows from the existence and uniqueness theory for ODEs. The Jacobian of the transformation x = x(s, t), y = y(s, t) at t = 0 is x x s t J(s) = J(s, t) t=0 = = t=0 y s y t [ y t a x ] t b 0. (11) t=0
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 20 in view of (10). By the continuity assumption, the Jacobian J 0 in a neighborhood of the initial curve. Thus, by the implicit function theorem, we can solve for s and t as functions of x and y near the initial curve. Then z(s, t) = z(s(x, y), t(x, y)) = Z(x, y). a solution of (1), which can be easily seen as c = = x + y = a x + b y, where we have used (4). The uniqueness of the solution follows from the fact that any two integral surfaces that contain the same initial curve must coincide along all the characteristic curves passing through the initial curve. This is a consequence of the uniqueness theorem for the IVP for (4). This completes our proof.. EXAMPLE 7. Consider the IVP: where f(x) is a given smooth function. y + z x = 0 z(x, 0) = f(x), Solution. We solve this problem using the following steps. Step 1. (Finding characteristic curves) To solve the IVP, we parameterize the initial curve as The characteristic equations are x = s, y = 0, z = f(s). = z, = 1, = 0. Let the solutions be denoted as x(s, t), t(s, t), and z(s, t). We immediately find that x(s, t) = zt + c 1 (s), y(s, t) = t + c 2 (s), z(s, t) = c 3 (s), where c i, i = 1, 2, 3 are constants to be determined using IC. Step 2. (Applying IC) The initial conditions at s = 0 are given by x(s, 0) = s, y(s, 0) = 0, z(s, 0) = f(s).
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 21 Using these condition, we obtain x(s, t) = zt + s, y(s, t) = t, z(s, t) = f(s). Step 3. (Writing the parametric form of the solution) The solutions are thus given by x(s, t) = zt + s = f(s)t + s, y(s, t) = t, z(s, t) = f(s). Step 4. (Expressing z(s, t) in terms of z(x, y)) Applying the condition (10), we find that J(s) = 1 0, along the entire initial curve. We can immediately solve for s(x, y) and t(x, y) to obtain s(x, y) = x tf(s), t(x, y) = y. Since t = y and s = x tf(s) = x yz, the solution can also be given in implicit form as z = f(x yz). EXAMPLE 8. Solve the following quasi-linear PDE: zz x + yz y = x, (x, y) R 2 subject to the initial condition z(x, 1) = 2x, x R. Solution. Here a(x, y, z) = z, b(x, y, z) = y, c(x, y, z) = x. The characteristics equations are On solving the above ODEs, we obtain = z, x(s, 0) = s, = y, y(s, 0) = 1, = x, z(s, 0) = 2s. x(s, t) = s 2 (3et e t ), y(s, t) = e t, z(s, t) = s 2 (3et + e t ). Solving for (s, t) in terms of (x, y), we obtain s(x, y) = 2xy 3y 2 1, t(x, y) = ln(y), z(x, y) = z(s(x, y), t(x, y)) = (3y2 + 1)x (3y 2 1).
MODULE 2: FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 22 Note that the characteristics variables imply that y must be positive (y = e t ). In fact, the solution z is valid only for 3y 2 1 > 0, i.e., for y > 1 3 > 0. Observe that the change of variables is valid only where x s (s, t) x t (s, t) y s (s, t) y t (s, t) 0. It is easy to verify that this condition leads to y 1/ 3. Practice Problems 1. Find a solution of the PDE z x + zz y = 6x satisfying the condition z(0, y) = 3y. 2. Find the general integral of the PDE (2xy 1)z x + (z 2x 2 )z y = 2(x yz) and also the particular integral which passes through the line x = 1, y = 0. 3. Solve z x + zz y = 2x, z(0, y) = f(y). 4. Find the solution of the equation z x + zz y = 1 with the data x(s, 0) = 2s, y(s, 0) = s 2, z(0, s 2 ) = s. 5. Find the characteristics of the equation z x z y = z, and determine the integral surface which passes through the parabola x = 0, y 2 = z.