Analytic solutions for the Burgers equation with source terms
AANALYTIC SOLUTIONS FOR THE BURGERS EQUATIONWITH SOURCE TERMS
G. I. MONTECINOS
Abstract.
Analytic solutions for Burgers equations with source terms, possi-bly stiff, represent an important element to assess numerical schemes. Here wepresent a procedure, based on the characteristic technique to obtain analyticsolutions for these equations with smooth initial conditions. Introduction
Exact solutions of hyperbolic balance laws are very useful to assess the perfor-mance of numerical schemes. Non-linearity of partial differential equation, as wellas, stiffness of source terms are desirable features to be recovered by numericalschemes. The analytic solutions presented here contain these elements.Exact solutions for Burgers equations are generally obtained by separation ofvariables [1], regularization techniques [3], expansion methods [5], to mention but afew. Here, the strategy to solve these equations is based on the characteristic curvemethod, see [4, 2]. The solution is obtained in two steps. First, the solution ofan Ordinary Differential Equation (ODE) called here, equivalent ODE , is obtained.This ODE is constructed by following the conventional characteristic curve methodand contains the influence of the source term. Second, the solution of an ODE,called here characteristic ODE is obtained. This equation is defined in the x − t plane and the solution of the equivalent ODE is included. In this way the influenceof the source term is present in the definition of the characteristic curve.To ensure the solvableness of the equivalent ODE , the existence of a primitivefunction for the reciprocal of the source term is required. Subsequently, if thesolution of the equivalent ODE has a primitive function then the characteristic ODE is solvable. If these requirements are satisfied, analytic solutions can be obtained.However, a non-linear algebraic equation has to be solved in the general case, whichrequire the initial condition to be a continuous function.This work is organized as follows. In section 2 the procedure is presented. Insection 3, the procedure is applied to the homogeneous Burgers equation. In section4 the solution for a linear source term is obtained. In section 5 the analytic solutionfor a quadratic source term is obtained. In section 6, the methodology is appliedto solve Burgers’s equation with a more general non-linear source term. Finally, insection 7 the main results of this work are summarized.2. Analytic solutions for Burgers equations with a special class ofsource terms
To start, let us consider a balance law in the form ∂ t q ( x, t ) + ∂ x (cid:18) q ( x,t ) (cid:19) = s ( q ( x, t )) ,q ( x,
0) = h ( x ) , (1)where h ( x ) is a continuous initial condition and s ( q ) is the source term which hasto satisfy the properties of lemma 2.2 shown below. a r X i v : . [ m a t h . A P ] M a r G. I. MONTECINOS
By following the characteristic curve method we define a curve x in the x − t plane, which satisfies ddt x ( t ) = q ( x ( t ) , t ) ,x (0) = y , (cid:27) (2)with y a constant value. This ODE will receive the name of characteristic ODE .On the other hand, we defineˆ q ( t ) = q ( x ( t ) , t ) , h (0) := q ( x (0) ,
0) = h ( y ) , (3)where y is that given in (2). With these definitions, (1) becomes an ODE given by ddt ˆ q ( t ) = s (ˆ q ( t )) , ˆ q (0) = h (0) . (cid:27) (4)This ODE is called here, equivalent ODE .Now, let us see the following definition in order to call for ensuring the existenceof solutions. Definition 2.1.
The primitive of a function f ( x ) is a differentiable function F ( x ) such that ddx F ( x ) = f ( x ) . (5)Next lemma sets the type of source terms which are considered in this work. Lemma 2.2. If s (ˆ q ) − contains a primitive with respect to ˆ q , then the equivalentODE is solvable. Additionally, there exists a function E ( t, h (0)) such that ∂∂t E ( t, h (0)) = s (ˆ q ( t )) , E (0 , h (0)) = h (0) (cid:27) (6) and ˆ q ( t ) = E ( t, h (0)) . Proof.
We integrate (2) as follows (cid:90) ˆ q ( t ) h (0) s ( q ) − dq = (cid:90) t dt , (7)as s ( q ) − has a primitive function, there exists G ( q ) such that ddq G ( q ) = s ( q ) − . (8)Therefore, E ( t, h (0)) is the solution to G ( E ) − G ( h (0)) − t = 0 . So, if there existsthe inverse function of G ( q ), which is denoted here by G − ( q ), the function E isexplicitly given by E ( t, h (0)) = G − ( t + G ( h (0))) . (9)In any case the exact solution is obtained asˆ q ( t ) = E ( t, h (0)) . (10) (cid:3) Once the solution to the equivalent ODE is available and observing that h (0) := h ( y ), the characteristic ODE takes the form ddt x ( t ) = E ( t, h ( y )) ,x (0) = y . (cid:27) (11)The following lemma deals with the existence of solution for this ODE. NALYTIC SOLUTIONS FOR THE BURGERS EQUATION WITH SOURCE TERMS 3
Lemma 2.3. If E ( t, h ( y ) has a primitive function F ( t, h ( y )) , such that ddt F ( t, h ( y )) = E ( t, h ( y )) and F (0 , h ( y )) = 0 . Then, the characteristic ODEhas the exact solution x = y + F ( t, h ( y )) . (12) Proof.
Integrating the ODE (11), we obtain (cid:82) xy dx = (cid:82) t E ( t, h ( y )) dt . (13)So, by using the properties of F ( t, h ( y )), the result holds. (cid:3) Remark 1.
The value y , is a constant for the characteristic ODE (11). However,if values x and t are set in (12), there exist a constant y satisfying (12). Thereforewe can identify a such constant by y = y ( x, t ) . Proposition 2.4. If s ( q ) − and E ( t, h ( y )) have their respective primitive func-tions. The problem (1) has the exact solution q ( x, t ) = E ( t, h ( y )) , (14) where y satisfies x = y + F ( t, h ( y )) , (15) with F ( t, h ( y )) the primitive of E ( t, h ( y )) with respect to t . Proof.
The construction of this function is given by lemmas (2.2) and (2.3). Now,we are going to probe that q ( x, t ) solves (1).By the chain rule ∂ t q = ∂∂y E ∂y∂t + ∂∂t E ,∂ x q = ∂∂y E ∂y∂x . (cid:41) (16)Then ∂ t q + q∂ x q = ∂∂y E ( ∂y∂t + q ∂y∂x ) + ∂∂t E . (17)On the other hand, we have0 = ∂y∂t + ∂∂t F + ∂∂ ( h ) F h ( y ) (cid:48) ∂y∂t , ∂y∂x + ∂∂ ( h ) F h ( y ) (cid:48) ∂y∂x . (18)Therefore, as F is the primitive of E and from (14), we have ∂y∂t + q ∂y∂x = (cid:18) ∂∂h F h (cid:48) (cid:19) − ( − ∂∂t F + q ) = 0 . (19)Finally, we note that ∂∂t E ( t, h ( y )) = s ( q ( x, t ))(20)and so, the result holds. (cid:3) Remark 2.
Note that in the general case, (15) is an algebraic equation and thebisection method is used to solve them. The regularity requirements of the bisectionmethod is that, functions have to be at least continuous ones, which is ensured bytaking h ( x ) a continuous function. In the following sections we will present some applications of the strategy describeabove.
G. I. MONTECINOS Homogeneous Burgers equations
In this section we apply the methodology seen in section 2, to solve the problem ∂ t q ( x, t ) + ∂ x (cid:18) q ( x,t ) (cid:19) = 0 ,q ( x,
0) = h ( x ) . (21)Then, the equivalent ODE has the form ddt ˆ q ( t ) = 0 , ˆ q (0) = h (0) . (cid:27) (22)This ODE is solvable and the exact solution has the form ˆ q ( t ) = E ( t, h (0)) = h (0) . Therefore, the characteristic ODE takes the form ddt x ( t ) = h ( y ) ,x (0) = y , (cid:27) (23)as E ( t, h (0)) has the primitive function F ( t, h (0)) = th (0) , we obtain x = y + th ( y ) . (24)However, h ( y ) = h ( x (0)) = q ( x (0) ,
0) = q ( x, t ). Therefore, (24) becomes x = y + tq ( x, t ) . (25)Hence the exact solution to (21), is computed as q ( x, t ) = h ( x − tq ( x, t )) , (26)which is the conventional solution of the homogeneous Burgers equation for generalsmooth initial conditions. Example 1.
Let us consider the problem (21) with h ( x ) = x . A simple inspectionshows that, u ( x, t ) = xt +1 , is the exact solution. Following the present methodologywe are going to obtain this exact solution.To start, let us note that equivalent ODE has the exact solution ˆ q ( t ) = h (0) . (27) Therefore, the exact solution to (21) is given by q ( x, t ) = h ( y ) , (28) with y satisfying the relationship obtained from the characteristic ODE, which byusing the form of h ( y ) has the form x = y + th ( y ) = y (1 + t ) . (29) So, by solving for y the sought solution is obtained q ( x, t ) = h ( y ) = y = xt + 1 . (30) 4. Burgers’s equation with a linear-source term
Now, let us consider the PDE given by ∂ t q ( x, t ) + ∂ x (cid:18) q ( x,t ) (cid:19) = βq ( x, t ) ,q ( x,
0) = h ( x ) . (31)Then, the equivalent ODE has the form ddt ˆ q ( t ) = β ˆ q ( t ) , ˆ q (0) = h (0) . (cid:27) (32) NALYTIC SOLUTIONS FOR THE BURGERS EQUATION WITH SOURCE TERMS 5
This ODE is solvable and the exact solution has the form ˆ q ( t ) = E ( t, h (0)) = h (0) exp ( βt ) . Therefore, the characteristic ODE takes the form ddt x ( t ) = h ( y ) e βt ,x (0) = y , (cid:27) (33)as E ( t, h (0)) has the primitive function F ( t, h (0)) = ( e βt − β ) h (0) , we obtain x = y + e βt − β h ( y ) . (34)Therefore, the exact solution to (31) is given by q ( x, t ) = h ( y ) e βt , (35)with y satisfying (34). Example 2.
The function u ( x, t ) = βxe βt β − e βt is the exact solution of (31) with h ( x ) = βxβ − , with β (cid:54) = 1 .To find this solution by using the present methodology we note that the equivalentODE has the exact solution ˆ q ( t ) = h (0) e βt . (36) On the other hand, the characteristic ODE due to the form of h ( y ) , has the exactsolution x = y + h ( y ) e βt − β = y (1 + e βt − β − . (37) Therefore, the exact solution to (31) is given by q ( x, t ) = h ( y ) e βt , (38) with y satisfying (37), which finally provides q ( x, t ) = h ( y ) e βt = βxe βt β − e βt . (39) Example 3.
Let us consider the problem (31) in the interval [0 , , with the initialcondition h ( x ) , given by h ( x ) = , x ≤ . , ( x − . . , . ≤ x ≤ . , , . ≤ x ≤ . , − ( x − . . , . ≤ x ≤ . , , x ≥ . . (40) In that case, the equation (34) is not explicitly solvable for y . In this work we usethe bisection method to obtain y .Figure 1, shows initial condition (40) and the solution of (31) at time t out = 0 . for β = − . Burgers’s equation with a quadratic source term
Let us consider the partial differential equation ∂ t q ( x, t ) + ∂ x (cid:18) q ( x,t ) (cid:19) = βq ( x, t ) ,q ( x, t ) = h ( x ) . (41) G. I. MONTECINOS out )q(x,0)
Figure 1.
Burgers’s equation with linear source term. Initial con-dition (dash line) and the solution at time t out = 0 . β = − equivalent ODE has the form d ˆ q ( t ) dt = β ˆ q ( t ) , ˆ q (0) = h (0) , (cid:27) (42)which is solvable and the exact solution isˆ q ( t ) = E ( t, h (0)) = h (0)1 − βth (0) . (43)On the other hand, we note that the characteristic ODE ddt x ( t ) = q ( x, t ) ,x (0) = y , (cid:27) (44)for q ( x, t ) satisfying (43) has the solution x = y − ln (1 − βth ( y )) β . (45)Therefore, the solution of (41) is given by q ( x, t ) = h ( y )1 − βth ( y ) , (46)with y satisfying (45). Example 4.
Let us consider (41) in the interval [0 , , with initial condition h ( x ) = sin(2 πx ) . (47) With this choice, y is not explicitly obtained from (45). Therefore we use thebisection method. Figure 2, shows the initial condition (dash line ) and the solutionat t out = 0 . and for β = − . NALYTIC SOLUTIONS FOR THE BURGERS EQUATION WITH SOURCE TERMS 7 q ( x , t ou t ) q(x,t out )q(x,0) Figure 2.
Burgers’s equation with quadratic source term. Initialcondition (dash line) and the exact solution at time t out = 0 . β = − . Burgers’s equation with a non-linear source term
Let us consider the partial differential equation ∂ t q ( x, t ) + ∂ x (cid:18) q ( x,t ) (cid:19) = e βq ( x,t ) ,q ( x, t ) = h ( x ) . (48)So, the equivalent ODE d ˆ q ( t ) dt = e β ˆ q ( t ) , ˆ q (0) = h (0) , (cid:27) (49)has the exact solutionˆ q ( t ) = E ( t, h (0)) = − ln ( e − βh (0) − βt ) β . (50)On the other hand, the characteristic ODE ddt x ( t ) = − ln ( e − βh (0) − βt ) β ,x (0) = y . (cid:41) (51)with ˆ q ( t ) = q ( x ( t ) , t ) satisfying (51) has the exact solution x = y + ( e − βh ( y ) − βt ) ln ( e − βh ( y ) − βt ) + βt + βh ( y ) e − βh ( y ) β . (52)Therefore, the solution to (48) is given by q ( x, t ) = − ln ( e − βh ( y ) − βt ) β , (53)with y a solution of (52). G. I. MONTECINOS q(x,t out )q(x,0)
Figure 3.
Bueger’s equation with exponential source term. Ini-tial condition (dash line ) and solution at t out = 0 .
26 (full line).Parameters β = − ε = 10 − . Example 5.
Let us consider (48) in the interval [0 , , with the initial condition h ( x ) = 2 (cid:18) − w ( x )2 (cid:19) + (cid:18) w ( x )2 (cid:19) . (54) with w ( x ) = 0 . − x (cid:112) (0 . − x ) + ε . Figure 3 shows the initial condition for ε = 10 − and the respective solution at time t out = 0 . for β = − . Conclusions
In this work we have presented a procedure to obtain analytic solutions to theBurgers equation with a family of source terms aimed to provide a set of testswhich is suitable for the assessment of numerical schemes. The family of sourceterms is that formed by functions containing primitive functions with respect to itsarguments. By following the characteristic method, we have defined two ODE’s,which were called equivalent ODE and characteristic ODE . This family of sourceterms allows us the solvableness of these ODE’s and as a consequence the soughtsolution of the original PDE is found. In the solution of the characteristic ODE an algebraic equation, which depends on the initial condition, is obtained. Therequirement of continuity for the initial condition allows us to find the root of thisalgebraic equation through the bisection method.
References [1] P. G. Est´evez, C. Qu, and S. Zhang. Separation of variables of a generalized porous mediumequation with nonlinear source.
Journal of Mathematical Analysis and Applications , 275(1):44– 59, 2002.[2] R.J. LeVeque.
Numerical Methods for Conservation Laws . Lectures in Mathematics ETHZ¨urich, Department of Mathematics Research Institute of Mathematics. Springer, 1992.[3] G. Norgard and K. Mohseni. A regularization of the Burgers equation using a filtered convectivevelocity.
Journal of Physics A: Mathematical and Theoretical , 41(344016):21 pp, 2008.[4] J.W. Thomas.
Numerical Partial Differential Equations: Finite Difference Methods . Numberv. 1 in Graduate Texts in Mathematics. Springer, 1995.
NALYTIC SOLUTIONS FOR THE BURGERS EQUATION WITH SOURCE TERMS 9 [5] M. Wang, X. Li, and J. Zhang. The (G’/G)-expansion method and travelling wave solutionsof nonlinear evolution equations in mathematical physics.
Physics Letters A , 372(4):417 – 423,2008.
Center for Mathematical Modeling (CMM), Universidad de Chile, Beauchef 851,Edificio Norte, Piso 7, Santiago - Chile
E-mail address ::