Numerical solution of conservation laws applied to the Shallow Water Wave Equations

Transcription

1 Numerical solution of conservation laws applie to the Shallow Water Wave Equations Stephen G Roberts Mathematical Sciences Institute, Australian National University Upate January 17, 2013 (base on notes by SGR an Chris Zoppou, an report by Angus Griffith) AMSI Summer School,

2 Contents 1 First orer PDE s in 2 variables Definition Very Simple Example Constant coefficient Example Variable coefficient Example More complicate Example An an even more complicate Example General Semi-linear Case Alternative Formulation Problems

3 1 First orer PDE s in 2 variables 1.1 Definition In this section we show how to solve the most general semi-linear 1st orer PDE in two variables, i.e. a PDE of the form a(x, y)u x + b(x, y)u y + c(x, y, u) = 0. (1) [If a = a(x, y, u), an b = b(x, y, u) then the equation is calle quasi-linear.] More precisely, we will show how we can obtain the solution to (1) provie we can solve an associate 1st orer orinary ifferential equation (ODE). Let us first iscuss some examples to illustrate the metho. 1.2 Very Simple Example Consier the PDE u x = 0. Clearly the most general solution is given by u(x, y) = f(y), where f is an arbitrary function. Thus solutions to this simple PDE are constant along lines parallel to the x-axis, i.e. lines of the form y = C, for C R. The lines are calle the characteristic lines of the PDE u x = 0. To fin the value of u(x, y) on a characteristic line y = C we only have to know the value of u(x, y) in one particular point on the line. Thus, a solution to the PDE u x = 0 is completely etermine by giving for example the values of u on the y-axis, i.e. u(0, y). Alternatively, we might give as initial conitions the value of u on any line y = ax + b for a 0, or even on monotonically increasing, or ecreasing, curves y = g(x). However, for generic curves the problem only amits a solution if at all intersection points of the curve with all lines y = C the prescribe value of u(x, y) is the same. The treatment of the generic 1st orer PDE in (1) procees along the same lines, in fact can be reuce to the problem above. Before we get to that let us iscuss a few, more complicate, examples. 1.3 Constant coefficient Example Consier au x + bu y = 0, with a, b R, an a 2 + b 2 0. We will iscuss two ways of solving this PDE. Coorinate Metho: Upon efining new coorinates ξ = ax + by, η = bx ay, we have u x = u ξ ξ x + u η η x = au ξ + bu η, u y = u ξ ξ y + u η η y = bu ξ au η. 3

4 Thus au x + bu y = (a 2 + b 2 )u ξ, hence the PDE in new coorinates reas u ξ = 0. The general solution, as in the example before, is given by u(x, y) = f(η(x, y)) = f(bx ay). Geometric Metho: We can think ( ) of the equation as the statement that the erivative of u in a the irection of the vector vanishes. Recall that the irectional erivative D n u of a b multivariable function u(x) in the irection of a vector n, is given by u(x + tn) u(x) D n u(x) lim = t 0 t i n i i u(x) = n u, where the graient u is the vector with components (u x, u y,...). Thus we conclue that solutions ( to ) the PDE are precisely those functions which are constant in the irection of the a vector, i.e. functions which are constant on the lines bx ay = C, parametrize by C. b The value of u on these so-calle characteristic lines is thus etermine by the value of C. As before we can then introuce a characteristic variable η(x, y) = C = bx ay. Summarizing, as before, we fin the most general solution to be of the form u(x, y) = f(η(x, y)) = f(bx ay). As a concrete example, suppose we want to solve the PDE an initial conition 4u x 3u y = 0, u(0, y) = y 3. As we have seen the most general solution of the PDE is given by u(x, y) = f( 3x 4y). Setting x = 0 gives the conition f( 4y) = y 3. Hence, by setting w = 4y we fin f(w) = w3 64, an thus the solution is given by u(x, y) = 1 64 (3x + 4y) Variable coefficient Example Consier the transport equation u x + yu y = 0. (2) Again, geometrically this equation tells us that the function u(x, y) is constant along curves (x, y(x)) which at every point has a tangent vector in the irection (1, y), i.e. has slope y/1 = y. Such curves are etermine as solutions of the ODE y(x) = y(x), (3) x 4

5 which, by separation of variables, can easily be seen to have as a general solution y(x) = Ce x, (4) where C is an arbitrary constant. I.e. ye x = η(x, y). efines the characteristic variable η. As η varies, these curves sweep out the entire two-imensional (x, y)-plane. Thus, we conclue that the most general solution to the PDE (2) is given by u(x, y) = f(η(x, y)) = f(ye x ). (5) A particular solution, e.g. for initial conition u(0, y) = y 3 is easily foun by putting x = 0 in (5), i.e. u(0, y) = f(y) = y 3, hence u(x, y) = (ye x ) 3 = y 3 e 3x. 1.5 More complicate Example Consier the PDE u x + 2xy 2 u y = 0. Supposing that the characteristic curves can be parameterise as (x, y(x)), then y(x) satisfies the ODE Separating the variables leas to the solution y x (x) = 2xy(x)2. y y 2 = 2xx, 1 y = x2 C, where C is an arbitrary constant. Hence along a characteristic curve, x y = η, (C = η is the characterisitic variable). Hence the general solution of the PDE is given by u(x, y) = f(η(x, y)) = f(x y ). It can be easily verifie that this is inee a solution (check!). 5

6 1.6 An an even more complicate Example Now consier the non-homogeneous PDE u x + 2xy 2 u y = 2xy. Now u will NOT be constant along a characteristic, but will satisfy an ODE. have The characteristic curves can be parameterise as (x, y(x)), to obtain a characteristic variable x y = η. Now, from the equation for the characteristic curve, an the fact that u satisfies the PDE, we x u(x, y(x)) = u x(x, y(x)) + x y(x)u y(x, y(x)) = u x (x, y(x)) + 2xy(x) 2 u y (x, y(x)) = 2xy(x). So on a characteristic curve (η constant), u(x, y(x)) satisfies u(x, y(x)) = 2xy(x). x We have an expression for y(x) (for a given value of η) So u(x, y(x) satisfies (for η constant) y(x) = 1 η x 2. u(x, y(x)) = 2x x η x 2. Integrating this ODE we obtain (the constant of integration can be written as a function of η) u(x, y(x)) = ln η x 2 + f(η) = ln x2 + 1 y(x) x2 + f(x2 + 1 y(x) ) = ln y(x) + f(x y(x) ). Hence the general solution of the PDE is given by It can be verifie that this is a solution (check!). u(x, y) = ln y + f(x y ). Note that it is a particular solution of the non-homogeneous PDE, plus the general solution of the homogeneous PDE. 6

7 1.7 General Semi-linear Case Now let us return to the problem of solving the most general semi-linear first orer PDE a(x, y)u x + b(x, y)u y + c(x, y, u) = 0. (6) Motivate by the examples, let us efine a characteristic curve as a curve along which the solution of the PDE is given by an ODE (in fact, the ODE in the most of the examples was just a zeroth orer ODE stating that u was constant along those curves). First let us parameterise the characteristic curve as a function of x, i.e. (x, y(x)). In this case it is useful to consier the (essentially) equivalent PDE u x + The equation for the characteristic curves are then given by b(x, y) a(x, y) u c(x, y, u) y + = 0. (7) a(x, y) y b(x, y) (x) = x a(x, y). which shoul provie a solution for y(x) which epens on a constant. This is often written as which efines a characteristic variable η. η(x, y(x)) = C, The function u(x, y(x)) (which is just a function of x) satisfies the ODE y(x), u(x, y(x))) u(x, y(x)) = c(x,. x a(x, y(x)) The solution of this ODE will introuce a constant of integration which will be constant for a specific value of η. So the constant of integration can be ifferent for each value of η. So it is a function of η. Example Let us review this for the equation The equation is in form (6). Put in form (7). xu x + 2x 2 y 2 u y = 2x 2 y u. u x + 2xy 2 u y = 2xy u. This is similar to equation we have alreay investigate. We look for the characteristic curve of the form (x, y(x)). The equation for the characteristic curve is y x (x) = 2x y(x)2. 7

8 The characteristic curves can be parameterise as (x, y(x)), to obtain a characteristic variable have x y = η. Now, from the equation for the characteristic curve, an the fact that u satisfies the PDE, we x u(x, y(x)) = u x(x, y(x) + x y(x)u y(x, y(x)) = u x (x, y(x) + 2xy(x) 2 u y (x, y(x)) = 2xy(x) u(x, y(x)). So on a characteristic curve (η constant), u(x, y(x)) satisfies u(x, y(x)) = 2xy(x) u(x, y(x)). x We have an expression for y(x) (for a given value of η) So u(x, y(x) satisfies (for η constant) u(x, y(x)) = 2x x y(x) = 1 η x 2. u(x, y(x)). η x2 Integrating this ODE we obtain (the constant of integration can be written as a function of η) ln u(x, y(x)) = ln η x 2 + f(η) = ln x ( y x2 + f x ) y(x) ( = ln y(x) + f x ). y(x) Hence the general solution of the PDE is given by ln u(x, y) = ln y + f ( x ). y Leas to the solution ( u(x, y) = yg x ), y where the minus signs from the absolute values have been absorbe into the function g. As usual we shoul check that this function oes inee satisfy the PDE. ( u x + 2xy 2 u y = yg x ) ( ( 2x + 2xy 2 g x ) ( + yg x ) ) 1 y y y y ( 2 = 2xy 2 g x ) y = 2xyu. 8

9 1.8 Alternative Formulation Often it is useful to parametrize a characteristic curve in the two-imensional plane by the more general form (x(s), y(s)), an let u(s) = u(x(s), y(s)) be the solution of (6) along the curve. Since, u s = u x x s + u y y s, (8) it is clear that the characteristic curves, an the solution u(s) along the curve, are etermine by the following set of characteristic equations x = a(x, y), s y = b(x, y), s u = c(x, y, u). (9) s This set of ODE s has a unique solution, at least for s in a neighborhoo of s = 0, for given initial conitions (x(0), y(0), u(0) = u(x(0), y(0))). To make contact with the previous examples, note that if we eliminate the parameter s, an escribe the characteristic curve by an relation y(x) we have I.e. y s = y x x s. (10) y b(x, y) = x a(x, y). (11) This reuces the problem of solving the PDE (6) to a system of ODE s. Sometimes it is useful to aopt a ifferent, but equivalent, point of view. Namely, in case we are not only given a PDE (6), but also a specific initial conition on a curve γ, parametrize by γ(r) = (x(r), y(r)), i.e. u(x(r), y(r)) = u(r). In that case it is useful to consier replacing the coorinates (x, y) by coorinates (s, r), where s is a coorinate along the characteristic curves, an r a coorinate along the initial curve, normalize such that s = 0 on the curve γ(r). I.e. we get the following system of equations x(s, r) = a(x(s, r), y(s, r)), s y(s, r) = b(x(s, r), y(s, r)), s u(s, r) = c(x(s, r), y(s, r), u(s, r)), s x(0, r) = x(r), y(0, r) = y(r), u(0, r) = u(r). (12) 9

10 We fin solutions x(s, r), y(s, r) an u(s, r), an provie the transformation x = x(s, r), y = y(s, r) is invertible, i.e. we can solve s = s(x, y), r = r(x, y) we fin the solution to our original problem as u(x, y) = u(s(x, y), r(x, y)). 1.9 Problems Problem 1.1 (Systems of ODEs.). Solve the following systems of ODEs Hint: Try using ifferent methos, e.g, for (a) an (b) one might try turning the systems into secon orer equations an solve via the characteristic equation, or by the metho of matrix exponentiation. In (c) one might try an integrating factor or perhaps the metho of variation of parameters. (a) (b) (c) where: u t x s = y y s = x x s = y y s = x. + Au(t) = f(t) (i) u an f are real value functions an A a constant. (ii) u an f are vector value functions (in R n ) an A is a constant n n matrix. Problem 1.2 (Constant coefficient PDE). Solve the first-orer equation 3u x + 4u y = 0 with the auxiliary conition u = sin y when x = 0. Problem 1.3 (First orer linear PDE with constant coefficients.). Consier the equation au x + bu y + cu = 0, where a, b, c R. (a) First, use the metho characteristics to fin the general solution of this equation in terms of some given function. 10

11 (b) Secon, use the coorinate transformation ξ(x, y) = ax + by, η(x, y) = bx ay, to fin the general solution of this equation. (c) Show that both methos provie the same solution (though on first glance they look quite ifferent). Problem 1.4 (Homogeneous variable coefficient PDE). Solve the equation yu x + xu y = 0. Problem 1.5 (Inhomogeneous variable coefficient PDE). Obtain the general solution of the first orer PDE yu x + xu y yu = xe x. (13) If we prescribe u(x, y) = φ(x, y) on the upper portion of the hyperbola y 2 x 2 = 1, y 1, show that no solution exists unless φ(x, y) is of a special form. Fin this form an show that in such a case there are infinitely many solutions. Problem 1.6 (Inhomogeneous variable coefficient PDE). Consier the First Orer PDE yu x + xu y yu = y. (a) What are the characteristic equations for this equation? (b) Solve for the characteristic curves (for x an y). Either provie a parameterisation of the characteristic curves, or as expressions constant on the curve. (c) Fin the general solution of the PDE. () Fin the solution when the conition u(x, 1) = e x is specifie. (e) For what region of the x-y plane oes the solution in part () cover. Problem 1.7 (First orer inhomogeneous linear PDE). Solve u x +u y +u = e x+2y, with u(x, 0) = 0. Problem 1.8 (First orer linear PDE.). u(0, y) = e y2 (a) Solve the equation yu x +xu y = 0 with the conition (b) In which region of the xy-plane is the solution uniquely etermine? Problem 1.9 (First orer inhomogeneous linear PDE.). Solve u x +u y +u = e x+2y, with u(x, 0) = 0. Problem 1.10 (Heuristic metho for first orer PDEs). (a) Consier the most general 1st orer PDE in two variables F (x, y, u, u x, u y ) = 0 (14) with initial conition u(x, 0). Assuming we can solve for u y, give a heuristic erivation of the solution to (14) using Taylor series. 11

12 (b) Apply your formula to fin the solution to the equation au x + bu y = 0 with initial conition u(x, 0) = x 3. (c) Consier the semi-linear 1st orer PDE a(x, y)u x + b(x, y)u y + c(x, y, u) = 0. Show that uner a coorinate transformation ξ = ξ(x, y), η = η(x, y), the equation is transforme to an equation of a similar form where, in particular, the transforme b is given by β(x, y) = a(x, y)η x + b(x, y)η y, an erive the equations for a characteristic curve in analogy with the iscussion in the lectures on 2n orer PDEs. Show that this notion is equivalent to the notion of a characteristic curve introuce in the lectures on first orer PDEs. Problem 1.11 (A Counterexample.). Are all solutions of the PDE x u x = y u y (15) functions of xy? 12

13 References 13

14 Inex characteristic curve, 7 characteristic equations, 9 characteristic lines, 3 irectional erivative, 4 linear quasi-linear, 3 semi-linear, 3 14