NOTES ON BURGERS’S EQUATION - University Of Maryland

1 NOTES ON BURGERS S EQUATIONMARIA CAMERONC ontents1. Solution of the Burgers EQUATION with nonzero viscosity12. Shock speed33. Characteristics of the Burgers equation54. Weak solutions65. The Riemann Case 1:uL> Case 2:uL< uR96. Numerical methods for hyperbolic conservation Conservative methods for nonlinear Discrete Generalization of methods developed for the advection Godunov s Glimm s method17 References17 Burgers s EQUATION (1)ut+uux= uxxis a successful, though rather simplified, mathematical model of the motion of a viscouscompressible gas, where u= the speed of the gas, = the kinematic viscosity, x= the spatial coordinate, t= the of the Burgers EQUATION with nonzero viscosityLet us look for a solution of Eq.

2 (1) of the form of traveling wave [1], ,u(x,t) =w((x x0) st) w(y).12 MARIA CAMERONT henut= sw ,ux=w , anduxx=w . Plugging this into Eq. (1) we obtain sw +ww = w sw +(w22) = w sw+w22= w + also impose conditions at the :w( ) =uL,w( ) =uR, whereuL> uR, andw ( ) = 0. Then we have suL+u2L2=C= suR+ ,smust be (uL+uR)/2. Thus, the shock speed is the same as in the case of zeroviscosity. HenceC= uLuR/2. Then we continue. w =w22 uL+uR2w+uLuR2dy2 =dww2 (uL+uR)w+uLuRdy2 =dww2 (uL+uR)w+(uL+uR)24 (uL uR)24dy2 =dw(w uL+uR2)2 (uL uR)24 Integrating the both parts using dw(w a)2 b2=12blogw a bw a+bwe obtainy2 +C=1uL uRlogw uL+uR2 uL uR2w uL+uR2+uL uR2=1uL uRlogw uLw uR=1uL uRloguL ww the last equality we used the fact thatuL> w > uR. Hence,uL ww uR=ey(uL uR)2 +CuL w=weA uReA,whereA=y(uL uR)/(2 ) +Cw(eA+ 1) = (uL+uReA)w=uL+uReAeA+ 1=uR+uL uR22eA+ ON BURGERS S EQUATION3 Multiplying and dividing by exp( A/2) and using the identity2e A/2eA/2+e A/2= 1 eA/2 e A/2eA/2+e A/2= 1 tanhA2we getw(y) =uR+uL2 uL uR2tanh(y(uL uR)4 +C).

3 Hence,(2)u(x,t) =uR+uL2 uL uR2tanh(([x x0] st)(uL uR)4 ).The profilesw(y) for various values of are shown in Fig. 1. As 0,u(x,t) tends to astep function of the argumentx = = = profiles of the solution of the viscous Burgers EQUATION foruR= 0,uL= 1,x0= 0, and equal to , , and Note thatw(y)tends to a step function as speedIf the viscosity = 0, or neglected, Eq. (1) can be rewritten as(3)ut+ [12u2]x= 0Eq. (3) is easier to study theoretically and numerically than Eq. (1). From now on, unlessindicated otherwise, we will refer to Eq. (3) as the Burgers (3) has a solution in the form of the traveling wave [2](4)u(x,t) =V(x x0 st),whereV(y) is a step function:(5)V(y) ={uLy <0uRy >0},4 MARIA CAMERON whereuL> uR. This wave is called theshock the speed of propagation of theshock wave.

4 It can be obtained from the following reasoning. LetMbe some large the integral M Mu(x,t) the shock speed using the law of conservation of M Mu(x,t)dx= M M uuxdx= u22M M=u2L2 the other hand, M Mu(x,t)dx= (M+st)uL+ (M st) ,ddt M Mu(x,t)dx=s(uL uR).Hence, the speed of propagation of the wave is(6)s=(u2L2 u2R2)/(uL uR) =uL+ argument above is valid for a more general EQUATION of the form(7)ut+ [f(u)]x= ON BURGERS S EQUATION5 Such equations are calledhyperbolic conservation laws. The shock speed is given by(8)s=f(uL) f(uR)uL uR=jump inf(u)jump EQUATION is called theRankine-Hugoniot of the Burgers equationThe characteristics of Eq. (3) are given by(9)dxdt=u(x,t).Let us show thatuis constant along the characteristics. Let (x(t),t) be a (x(t),t) = u t+ u xdxdt=ut+uux= , the solution of Eq.

5 (9) is given by(10)x(t) =u(x(0),0)t+x(0) =u0(x0)t+x0,wherex0=x(0), u0(x) =u(x,0).Eq. (10) shows that the characteristics are straight lines, they may intersect, they do not necessarily cover the entire (x,t) is a new phenomenon in comparison with the linear first order equationsut+a(x,t)ux=0. For a linear first order EQUATION , there is a unique characteristic passing through everypoint of the (x,t) space. Thus, its characteristics never intersect and cover the entire , even for a smooth initial speed distributionu0(x) the solution of the Burgersequation may become discontinuous in a finite time. This happens whenu 0(x) is negativesomewhere. Then the characteristics intersect, , the wave breaks. Let us find the breaktime. Consider two characteristicsx(t) =u0(x1)t+x1andx(t) =u0(x2)t+x2.

6 Then weequatex(t) =u0(x1)t+x1=u0(x2)t+ the time at which they intersect ist= x2 x1u0(x2) u0(x1).Therefore, the break time isTb= minx1,x2 R( x2 x1u0(x2) u0(x1))= 1minx1,x2 R(u0(x2) u0(x1)x2 x1)= 1minx1,x2 R(1x2 x1 x2x1u 0(x)dx)= 1minx Ru 0(x).6 MARIA CAMERONAn example is shown in Figure 3. The initial datau0(x) = exp( 16x2) and the cor-recponding characteristics of the Burgers EQUATION are shown in Fig. 3 (a) and (b) respec-tively. The solution at timest= andt= obtained by the method of characteristicsis shown in Fig. 3 (c) and (e). The numerical solution computed by Godunov s method(see Section 6) is shown in Fig. 3 (d) and (f). The solution obtained by the method ofcharacteristis is triple-valied at some values ofxand non-physical in the sense that it isnot the vanishing viscosity solution (see Section 1).

7 In contrast, the solution computed byGodunov s method tends to the vanishing viscosity solution as we refine the solutionsWe have seen in the previous section that a solution to the Burgers EQUATION can becomediscontinuous even if the initial data are smooth. Then the discontinuity travels with acertain speed, the shock speeds, given by Eq. (6). In Section 2 we foundsusing theunderlying integral conservation law. However, at this point, we can only call the stepfunction given by Eqs (4), (5), and (6) a solution of the integral conservation law ratherthan the solution of Eq. (3). In order to validate discontinuous solutions for differentialequations, the concept of the weak solutions was introduced (see [3]). This extensionof the concept of the solution must satisfy the following requirements: a smooth function is a weak solution iff it is a regular solution, a discontinuous function can be a weak solution, only those discontinuous functions which satisfy the associated integral equationcan be weak (x,t) be an infinitely smooth function with a compact support, ,it is different from zero only within some compact subset of the space (x,t) =R [0,+ ).]

8 Letu(x,t) be a smooth solution of a hyperbolic conservation law given by Eq. (7). Then(11) 0 = 0 (ut+ [f(u)]x) dxdt=( udx) 0 0 tu+ xf(u) , here is the definition [3].Definition (x,t)is a weak solution of the conservation lawut+ [f(u)]x= 0if for anyinfinitely differentiable function (x,t)with a compact support(12) 0 tu+ xf(u)dxdt=( udx) 0= (x,0)u(x,0) a function (x,t) is called atest Riemann problemIn this section, we consider the following initial value problem for the Burgers EQUATION :u(x,0) ={uLx <0uRx >0}.This problem is called theRiemann problem. We will consider two ON BURGERS S EQUATION7(a)(b)(c)(d)(e)(f)Figure 3.(a) The initial datau0(x) = exp( 16x2). (b) The correspondingcharacteristics of the Burgers EQUATION .(c-d) The solution at timet= by the method of characteristics (c) and computed numerically(by Godunov s method) (d).

9 (e-f) The solution at timet= obtained bythe method of characteristics (e) and computed numerically (by Godunov smethod) (f).8 MARIA 1:uL> this case, the characteristics cover the entire (x,t) space but alsocross. Hence the construction of the solution using only the characteristics is us show that in this case there exists a unique weak solution given by(13)u(x,t) ={uLx < stuRx > st},wheres=uL+ characteristics for this solution are chown in Fig. 4(a).(b)Figure 4.(a) The characteristics for the shock wave. (b) Illustration forthe proof that the shock wave is the unique weak (x,t) be a test function. First suppose that supportUlies entirely in one ofthe sets{x < st}or{x > st}. Then sinceu(x,t) is constant in each of these sets, it satisfiesthe Burgers EQUATION on the support of.

10 Then using Eq. (11) we conclude that Eq. (12) suppose that the supportUof is divided by the linex=stinto two setsULandUR(Fig. 4(b)). Then we have 0 tu+ xf(u)dx= UL tu+ xf(u)dxdt+ UR tu+ xf(u)dxdtApplying the Green identity(14) D(Px Qt)dxdt= DPdt+QdxNOTES ON BURGERS S EQUATION9and noting thatuis constant withinULand withinUR, hence ( u)t= tuand ( f(u))x= xf(u), we continue= UL (u2L2dt uLdx)+ UR (u2R2dt uRdx)= x=st (u2L2s uL)dx x=st (u2R2s uR)dx 0 uLdx 0 uRdx= x=st(u2L u2R2s (uL uR)) dx (x,0)u(x,0)dxThe first integral in the last equality is zero for any test function iffs= (uL+uR) , the solution given by Eq. (13) is the unique weak solution for the Riemann problemin the caseuL> uR. The discussion in Section 1 indicates that the solution given by Eq.