Z Z Z The method of characteristics applied to quasi-linear PDEs 18.303 Linear Partial Differential Equations Matthew J. Hancock Fall 2006 1 Motivation [Oct 26, 2005] Most of the methods discussed in this course: separation of variables, Fourier Series, Green’s functions (later) can only be applied to linear PDEs. However, the method of characteristics can be applied to a form of nonlinear PDE. 1.1 Traffic flow Ref: Myint-U & Debnath §12.6 Consider the idealized flow of traffic along a one-lane highway. Let ρ (x, t) be the traffic density at (x, t). The total number of cars in x1 ≤ x ≤ x2 at time t is x2 N (t) = ρ (x, t) dx (1) x1 Assume the number of cars is conserved, i.e. no exits. Then the rate of change of the number of cars in x1 ≤ x ≤ x2 is given by dN = rate in at x1 − rate out at x2dt = ρ (x1, t) V (x1, t) − ρ (x2, t) V (x2, t) x2 ∂ = − (ρV ) dx (2) ∂x x1 where V (x, t) is the velocity of the cars at (x, t). Combining (1) and (2) gives x2 ∂ρ ∂ + (ρV ) dx = 0 ∂t ∂x x1 1 and since x1, x2 are arbitrary, the integrand must be zero at all x, ∂ρ ∂ + (ρV ) = 0 (3) ∂t ∂x We assume, for simplicity, that velocity V depends on den sity ρ, via ρ V (ρ) = c 1 − ρmax where c = max velocity, ρ = ρmax indicates a traffic jam (V = 0 since everyone is stopped), ρ = 0 indicates open r oad and cars travel at c, the speed limit (yeah right). The PDE (3) becomes ∂ρ ∂t + c 1 − 2ρ ρmax ∂ρ ∂x = 0 (4) We introduce the following normalized variables ρ u = , t˜= ct ρmax into the PDE (4) to obtain (dropping tildes), ut + (1 − 2u) ux = 0 (5) The PDE (5) is called quasi-linear because it is linear in the derivatives of u. It is NOT linear in u (x, t), though, and this will lead to interesting outcomes. 2 General first-order quasi-linear PDEs Ref: Guenther & Lee §2.1, Myint-U & Debnath §12.1, 12.2 The general form of quasi-linear PDEs is ∂u ∂u A + B = C (6) ∂x ∂t where A, B, C are functions of u, x, t. The initial condition u (x, 0) is sp ecified at t = 0, u (x, 0) = f (x) (7) We will convert the PDE to a sequence of ODEs, drastically simplifying its solu-tion. This general technique is known as the method of characteristics and is useful for finding analytic and numerical solutions. To solve the PDE (6), we note that (A, B, C) · (ux, ut, −1) = 0. (8) 2 Recall from vector calculus that the normal to the surface f (x, y, z) = 0 is ∇f. To make the analogy here, t replaces y, f (x, t, z) = u (x, t) −z and ∇f = (ut, ux, −1). Thus, a plot of z = u (x, t) gives the surface f (x, t, z) = 0. The vector (ux, ut, −1) is the normal to the solution surface z = u (x, t). From (8), the vector (A, B, C) is the tangent to this solution surface. The IC u (x, 0) = f (x) is a curve in the u − x plane. For any point on the initial curve, we follow the vector (A, B, C) to generate a curve on the solution surface, called a characteristic curve of the PDE. Once we find all the characteristic curves, we have a complete description of the solution u (x, t). 2.1 Method of characteristics We represent the characteristic curves parametrically, x = x (r; s) , t = t (r; s) , u = u (r ; s) , where s labels where we start on the initial curve (i.e. the initial value of x at t = 0). The parameter r tells us how far along the characteristic curve. Thus (x, t, u) are now thought of as trajectories parametrized by r and s. The semi-colon indicates that s is a parameter to label different characteristic curves, while r governs the evolution of the solution along a particular characteristic. From th e PDE (8), at each point (x, t), a particular tangent vector to the solution surface z = u (x, t) is (A (x, t, u) , B (x, t, u) , C (x, t, u)) . Given any curve (x (r; s) , t (r; s) , u (r; s)) parametrized by r (s acts as a label only), the tangent vector is ∂x ∂t ∂u , , . ∂r ∂r ∂r For a general curve on the surface z = u (x, t), the tangent vector (A, B, C) will be different than the tangent vecto (xr, tr, ur). However, we choose our curves (x (r; s) , t (r; s) , u (r; s)) so that they h ave tangents equal to (A, B, C), ∂x ∂t ∂u = A, = B, = C (9) ∂r ∂r ∂r where (A, B, C) depend on (x, t, u), in general. We have written partial derivatives to denote differentiation with respect to r, since x, t, u are functions of both r and s. However, since only derivatives in r are present in (9), these equations are ODEs! This has greatly simplified our solution method: we have reduced the solution of a PDE to solving a sequence of ODEs. 3( f(x) 2 1.5 1 0.5 0 −3 −2 −1 0 1 2 3 x Figure 1: Plot of f(x). The ODEs (9) in conjunction with some initial conditions specified at r = 0. We are free to choose the value of r at t = 0; for simplicity we take r = 0 at t = 0. Thus t (0; s) = 0. Since x changes with r, we choose s to denote the initial value of x (r; s) along the x-axis (when t = 0) in the space-time domain. Thus the initial values (at r = 0) are x (0; s) = s, t (0; s) = 0, u (0; s) = f (s) . (10) 3 Example problem [Oct 28, 2005] Consider the following quasi-linear PDE, ∂u ∂u + (1 + cu) = 0, u (x, 0) = f (x)∂t ∂x where c = ±1 and the initial condition f (x) is 1, x < −1 1, |x| > 1 2 + x, −1 ≤ x ≤ 0 f (x) = = 2 − |x| , |x| ≤ 1 2 − x, 0 < x ≤ 1 1, x > 1 The function f (x) is sketched in Figure 1. To find the parametric solution, we can write the PDE as ∂u ∂u (1, 1 + cu, 0) · , , −1 = 0 ∂t ∂xThus the parametric solution …
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