θ x = f ( x,t) could be written as

Transcription

1 9. Higher orer PDEs as systems of first-orer PDEs. Hyperbolic systems. For PDEs, as for ODEs, we may reuce the orer by efining new epenent variables. For example, in the case of the wave equation, (1) θ tt = c θ xx, the efinitions () u θ t an v θ x imply (3) u x = v t, while (10) itself may be written as u t = c v x. Thus the secon-orer equation (1) is equivalent to the first-orer system (4) u t c v x u x v t The motivation for this approach is our success with first-orer equations. We foun that (5) a( x,t) θ t + b( x,t) θ = f ( x,t) coul be written as (6) θ s = f where (7) s = a t + b is the irectional erivative along the curve (8) t s = a, x s = b. Can we o a similar trick for first-orer systems? Consier the general system with epenent variables, u(t,x) an v(t,x), in the inepenent variables x an t: (9) D 1 u + 1 v = f 1 D u + v = f 9-1

2 Here, (10) D 1 = A 1 ( x,t,u, v) t + B ( 1 x, t,u,v ) 1 = a 1 ( x,t,u,v) t + b ( 1 x,t, u,v ) (an similarly, with replacing 1) are the irectional erivatives that appear in the given system. We allow the coefficients to epen on u an v as well as x an t; such systems are calle quasilinear. The system (9) is the generalization of (5) to the case of epenent variables. The theory can be extene to n epenent variables in m inepenent variables (see Whitham chapter 5). However, success is rare when there are more than inepenent variables, an more than epenent variables complicates the notation; for these reasons we shall be content with (9). The primary ifference between (5) an (9) is that each equation in (9) contains irectional erivatives which generally point in ifferent irections. Thus it is generally impossible to integrate either of (9) in the same way as (5). But what about linear combinations of (9)? Multiplying (9a) by c 1 an (9b) by c, where c 1 an c are functions of (x,t,u,v) to be etermine, an aing the equations, we obtain (11) ( c 1 D 1 + c D )u + ( c c )v = c 1 f 1 + c f. We want the irectional erivative of u in (11) to be proportional to the irectional erivative of v in the same equation. That is, we want (1) c 1 D 1 + c D = α ( c c ) where α (another function of ( x,t, u,v) ) is the factor of proportionality. Rewriting (1) in the form (1a) ( L) t + ( L) an requiring the coefficients of / t an / to vanish, we obtain equations for c 1 an c : (13) A 1 α a 1 A α a B 1 α b 1 B α b c 1 c = 0 0. For nonzero (c 1,c ), the eterminant in (13) must vanish. This gives a quaratic equation for α. If the roots of this quaratic equation are real, an if the corresponing pairs (c 1,c ) are ifferent (so that the equations of form (11) are inepenent equations), then we have achieve our goal of writing (9) in the form of equations, each of which involves erivatives along a single family of lines in the xt-plane. In this case, the system is sai to 9-

3 be hyperbolic, an the families of lines are its characteristics. If we are very lucky, we might even succee in manipulating (11) into the form (14) R 1 s 1 = F 1, R s = F where / s 1 an / s are the irectional erivatives corresponing to the families of characteristics. Then, if it happens that F 1 = F, the variables R 1 ( t,x,u,v) an R ( t, x,u,v) are calle Riemann invariants. In exceptional cases, the families of characteristics may actually coincie. In working out particular problems, it is almost never worthwhile to use the general notation use above. It is almost always better to simply follow the foregoing strategy in each particular case. However, the general notation shows why the metho usually fails in the case of 3 or more inepenent variables. In the case of epenent variables in 3 inepenent variables, (13) becomes a set of 3 equations (corresponing to the 3 kins of first-erivative) in the unknowns (c 1,c ) an is thus generally overetermine. In the remainer of this section, we focus on examples. Example. In the case of system (4) the linear combination is so we want c 1 ( u x v t ) + c ( u t c v x ) ( c,c 1 ) = α ( c 1, c c ), that is, α 1 1 αc c 1 = 0 0. c Thus α c 1 α = ±1 / c. The choice α = +1/ c correspons to / s 1 = t c x an R 1 = u + cv. The choice α = 1/ c correspons to / s = t + c x an R = u cv. Example. For the heat equation, θ t κθ xx, we let u = θ an v =θ x to obtain the system u t κ v x v u x The secon equation fits the esire form, but no other linear combination oes, because there is no v t -term. Thus there is only one characteristic, an therefore the system is not hyperbolic (big surprise). Example. For Laplace s equation, θ xx +θ yy, the choices u = θ x an v =θ y yiel the system 9-3

4 u x + v y u y v x The general linear combination is Thus we want c 1 ( u x + v y ) + c ( u y v x ) = ( c 1 x + c y )u + ( c 1 y c x )v. ( c 1 x + c y ) = α( c 1 y c x ) 1 α c 1 α 1 = 0 0 α = ±i. c Thus there are no characteristics, an the system is not hyperbolic. Example. u t + u et v t + v + v u t u + x v t v v These equations are alreay in characteristic form; the irectional erivatives in each equation point in the same irection. Defining we have s = t + an r = t (*) u s v et s + v an u r + x v r v. The first of these implies u s s et v ( ) + v s et e t v s u et v ( ) since t / s = 1. Similarly, the secon of (*) implies r ( u + xv) v x r v r ( u + xv) since x / r = 1. Thus the Riemann invariants are u e t v an u + xv. The characteristics corresponing to / s are x t = const. Thus u e t v = F( x t) where F is an arbitrary function. Similarly, the characteristics corresponing to / r are 9-4

5 x + t = const. Thus u + xv = G( x + t) where G is another arbitrary function. The general solution can be written u = x F( x t) + et G( x + t), v = G( x + t) F( x t) x + e t x + e t In the remainer of this section we consier a far more interesting example, the oneimensional shallow-water equations: (15) u t + u u h = g h t + h u + u h Note that these equations are nonlinear. Before proceeing with the analysis of (15), we note that these equations may be viewe as a special case of the equations for a one-imensional polytrope, a homentropic gas with equation of state p = cρ γ, where c an γ are constants. The equations governing the gas are u t + uu x = p x / ρ an ρ t + uρ x + ρu x. Thus (15) correspon to ρ = h, c = g / an γ =. Using this analogy, many of the general results from gas ynamics (Whitham chapter 6) may be taken over to the shallowwater system. The general linear combination of (15) is (16) c 1 ( u t +u u x + g h x ) + c ( h t + hu x + u h x ) which is equivalent to (17) u + h s 1 s where (18) c s 1 1 t + c u 1 + c h an (19) s c 1 g + c t + c u. 9-5

6 To make the irectional erivatives proportional we set (0) = α s 1 s which implies (1) c 1 α c c 1 u + c h α gc 1 α c u Substituting (1a) into (1b) we obtain () c h α gc. If c 0 then α = h / g an we obtain α = ± h / g. Thus the system is hyperpolic if h>0. We choose c = g; then c 1 = ± gh ; an we have (3) ( ) g s = g s t + ( u ± gh) x an (4) = α = ± gh s 1 s s. Thus (17) becomes (5) ± gh u s + g h s ( s u ± gh) (6) t + ( u ± gh ) u ± gh ( ) where both signs are to be taken the same. The equations (6) fit the form of (14) with (7) R 1 = u + gh = s 1 t + ( u + gh) R = u gh = s 1 t + ( u gh ) an F 1 = F (where we have re-efine / s 1 an / s ). Suppose that u an h are given along some curve in the x-t plane, such as t=0. Then the solution elsewhere is obtaine by integrating along the curves efine by the irectional erivatives / s 1 an / s. Since these irectional erivatives epen on both u an h, each step in this integration requires information carrie by both sets of characteristics: 9-6

7 Thus the initial conitions in THIS interval etermine the solution below THESE characteristics. This situation iffers from the linear equation θ tt = c θ xx in that the characteristics have a slope that epens on u an h. Just as in the case of quasilinear firstorer equations, this may lea to multi-value solutions. For general hyperbolic systems the situation can become very complex. We consier the following initial-value problem for our shallow-water system: The flui is initially at rest (u=0) with uniform epth (h = h 0 ), an lies to the right of a wall at x=0. Beginning at t=0, the wall moves to the left along the trajectory x=x(t) as shown: What is the motion of the flui? From the analysis above we know that u + gh is constant along (+) characteristics with slope x / t = u + gh, while u gh is constant along (-) characteristics with slope x / t = u gh. If u < gh, then the (-) characteristics originating on the positive x-axis fill up the entire area of the x-t plane occupie by the flui. This implies that (8) u gh = gh 0 throughout the flui. On the (+) characteristics with slope (9) x t = u + gh we have (30) u + gh = const, 9-7

8 but the constant in (30) is ifferent along each particular characteristic. Nevertheless it follows from (30) an (8) that u an h are (ifferent) constants along each (+) characteristic. Hence by (9) each of the (+) characteristics is a straight line. For the (+) characteristics intersecting the x-axis, the slope is x / t = these characteristics the constant in (30) is gh 0. Thus we conclue that gh 0. On (31) u h = h 0 for all x > gh 0 t. To etermine the solution in the remaining part of the omain, we use the fact that u = X ( t) at the wall. By (8) an (9) the slope of the (+) characteristics may be written as (3) x t = 3 u + gh 0 Thus (33) x t = 3 X ( τ ) + gh 0 where τ is the time at which the characteristic intersects the wall. Since, as we alreay know, the right-han sie of (33) must be a constant, it follows from (33) that (34) x = X( τ) + 3 X ( τ ) + gh 0 ( t τ). Given any (x,t) in the region of the moving flui, we can etermine u(x,t) an h(x,t) as follows. First we solve (34) for τ, the time at which the (+) characteristic passing through (x,t) intersects the wall. Then (35) u( x,t) = X ( τ ) an by (8) (36) gh( x,t) = gh X ( τ ). Eqns (35) an (36) hol at all points along the characteristic. The solution will be singlevalue if (34) has only one solution for τ. 9-8

9 An interesting special case is (37) X = V (constant) corresponing to a wall moving steaily to the left at spee V>0. In this case the slope of the (+) characteristics intersecting the wall is (38) x t = 3 V + gh 0 On these characteristics the solution is (39) u = V, gh = gh 0 1 V Between this region an the region of quiescent flow lies a fan region, in which the (+) characteristics pass through the origin with every slope between (38) an gh 0, an the solution varies smoothly across the characteristics. As V gh 0, the fan extens right up to the wall, an the epth at the wall vanishes. When V > gh 0 the wall simply outruns the flui, an the problem reuces to the am break problem, in which the wall is mae to isappear at t=0. In the am-break problem, all the (+) characteristics to the left of the quiescent region on x > gh 0 t emanate from the origin x = t. Each (+) characteristic has the constant slope (9), an since it passes through the origin, we may write its equation as (40) x = ( u + gh)t. But then the solution in the entire fan region may be obtaine from (40) an (8). We fin (41) u = x 3 t gh 0 an gh = 3 gh x t 9-9

10 Once again, the rightwar ege of the fan region is x = h = h 0, matching the solution on x > gh 0 t. At this ege u=0 an gh 0 t. At the leftwar ege, x = gh 0 t, h=0, an u = gh 0. This is the spee at which the floo covers ry groun. The solution looks like this: Cases, like that just consiere, in which one of the Riemann invariants is uniform throughout the flow, are calle simple waves. In such cases one can usually use the uniformity of the Riemann invariant to eliminate one of the epenent variables a priori, an thereby reuce the entire problem to a first-orer equation. For example, in the am-break problem we may use (8) to eliminate u in (4) h t + ( hu) which then takes the form (43) h t + c( h) h with c( h) = ( 3 gh gh 0 ). It is in fact far simpler to solve the am break problem in the form (43) than by the previous metho; it follows easily from (43) that gh 0, x > gh 0 t (44) gh = 3 gh + 1 x 0 3 t, gh t < x < gh t

11 an u may then be obtaine from (8). Now, returning to the problem with the moving wall, we consier the case X = Vt in which the wall moves to the right at a uniform spee V. We shall fin that the fan appearing in the previous solution is replace by a shock. We begin by noting, from symmetry consierations, that, just as in the case of the leftwar moving wall, the solution must take the form u = u( x / t), h = h( x / t). The bounary conitions are u=0 an h = h 0 along the x-axis, an u=v along x = Vt, the trajectory of the wall. We let h = h w along the wall trajectory, where h w is a constant which must be etermine by our solution to the problem. Let ( x,t) be a point in the flui omain. If the characteristics through ( x,t) intersect the x-axis, then, as before, (45) u + gh + gh 0 u gh gh 0 imply that u an h = h 0 ; the flui is quiescent as in its initial state. On the other han, if the characteristics through x,t ( ) intersect the wall trajectory, then (46) u + gh = V + gh w u gh = V gh w imply that u = V an h = h w ; the flui moves at the same spee as the wall, but its uniform epth h w is as yet unetermine. These solutions obviously isagree, an the only conclusion can be that the quiescent solution hols in a wege-shape region near the x- axis, whereas the solution with u = V hols in a wege-shape region near the line x = Vt. These regions are separate by a shock along x = Ut, where U too must be etermine. To analyze the shock with complete physical correctness, we must a viscosity to the problem, an treat the shock as an internal bounary layer in which u an h vary rapily between the uniform values on either sie. However, as we realize from our stuy of 9-11

12 Burger s equation, we can moel the shock as a true iscontinuity if we are careful to apply jump conitions that express conservation laws that woul survive the inclusion of viscosity. (47) The shallow-water equations imply the conservation of mass, an momentum h t + ( hu) (48) ( hu) + 1 t gh + hu. Both of these fit the form (49) P t + Q consiere in Section 4. There we showe that (49) implies (50) U[ P] = [ Q] where U is the velocity of the shock, an [ ] enotes the jump across the shock. In the present case the jump conitions imply (51) U( h w h 0 ) = Vh w 0 an (5) U Vh w 0 ( ) = 1 gh w + h w V 1 gh 0 Eqns (51) an (5) etermine the values of h w an U, completing the solution. We fin that (53) U = gh w h 0 ( ) ( an V = U h h w 0) h 0 + h w (It is easier to regar U an V as functions of h w.) Reference. Whitham chapters 5 & 6 an pp h w. 9-1