A second-order convergent and linearized difference scheme for the initial-boundary value problem of the Korteweg-de Vries equation

Wang Xuping Sun Zhizhong

(School of Mathematics, Southeast University, Nanjing 210096, China)

Abstract：To numerically solve the initial-boundary value problem of the Korteweg-de Vries equation, an equivalent coupled system of nonlinear equations is obtained by the method of reduction of order.Then, a difference scheme is constructed for the system.The new variable introduced can be separated from the difference scheme to obtain another difference scheme containing only the original variable.The energy method is applied to the theoretical analysis of the difference scheme.Results show that the difference scheme is uniquely solvable and satisfies the energy conservation law corresponding to the original problem.Moreover, the difference scheme converges when the step ratio satisfies a constraint condition, and the temporal and spatial convergence orders are both two.Numerical examples verify the convergence order and the invariant of the difference scheme.Furthermore, the step ratio constraint is unnecessary for the convergence of the difference scheme.Compared with a known two-level nonlinear difference scheme, the proposed difference scheme has more advantages in numerical calculation.

Key words：Korteweg-de Vries(KdV)equation; linearized difference scheme; conservation; convergence

The Korteweg-de Vries(KdV)equation was derived by Korteweg and de Vries[1]in 1895 and has a history of more than 130 years.The KdV equation is one of the classical mathematical physics equations.The KdV equation plays an important role in nonlinear dispersive waves and has a wide range of applications.Many scholars have investigated its solutions from the point of view of analysis and numerical value.Moreover, the general solutions of the KdV equation are difficult to obtain.Therefore, many numerical methods, such as the finite difference method[2-6], finite element method[7-10], spectral method[11-12], and meshless method[13-16], have been applied to solve the KdV equation.Among them, the finite difference method is simple and easy to implement on computers.The finite difference method is an important method for solving nonlinear evolution equations.Meanwhile, the theoretical analysis of the difference scheme is relatively difficult, particularly for the initial-boundary value problem.Consequently, we will use the finite difference method to solve the initial-boundary value problem of the KdV equation in this study.

When solving nonlinear evolution equations, we need to consider the corresponding initial and boundary value conditions.The three main types of problems are the initial, periodic boundary, and initial-boundary value problems.Currently, many studies on solving nonlinear evolution equations using the finite difference method have been conducted.Some of them analyzed the initial value problems.For this class of problems, difference schemes were conveniently established by adding homogeneous boundary conditions to the boundary in the practical computation, which is a Dirichlet boundary value problem.If the highest order of the spatial derivative of the equation for the space variablexis two, then the addition of homogeneous boundary conditions will not affect the construction and analysis of difference schemes, such as Burgers’ equation:

ut+uux=νuxx

whereνis a positive constant.The difference scheme has the same form at all inner grid points.If the highest order of the spatial derivatives is over two, then the difference will be significant because of the presence of derivative boundary conditions, such as the KdV equation:

ut+γuux+uxxx=0

whereγis a constant whose boundary conditions[17]are

u(0,t)=u(L,t)=ux(L,t)=0t≥0

The boundary condition is asymmetric, which means that the difference scheme will also be asymmetric.Therefore, for the initial-boundary value problem of the KdV equation, the construction and analysis of difference schemes need to be more detailed.Studies of the periodic boundary problems of the KdV equation using the finite difference method have also been conducted.However, because of the existence of derivative boundary conditions, similarly generalizing the method for the periodic boundary value problems to the initial-boundary value problems of the KdV equation is difficult.

Recently, we established two finite difference schemes for the KdV equation with the initial-boundary value conditions in Ref.[5].One was a nonlinear difference scheme and the other was a linearized difference scheme.The nonlinear difference scheme was proven to be unconditionally convergent, whereas the linearized difference scheme was conditionally convergent.The accuracy of these two finite difference schemes was the first order in space.Subsequently, in Ref.[6], we established a nonlinear difference scheme and proved that its spatial convergence order was two.In the numerical examples, we solved the nonlinear difference scheme using the Newton iterative method, which increased the computational cost at each time level.To improve computational efficiency and keep the convergence order unchanged, we consider establishing a linearized difference scheme for solving the initial-boundary problem of the KdV equation.

In this study, we construct a three-level linearized difference scheme for the following problem:

ut+γuux+uxxx=0 0

(1a)

u(x,0)=φ(x) 0

(1b)

u(0,t)=u(L,t)=ux(L,t)=0 0≤t≤T

(1c)

whereφ(0)=φ(L)=φ′(L)=0,γis a constant.We also establish the difference scheme and illustrate the truncation errors in detail.Then, we will present its conservation law, prove its unique solvability and conditional convergence, and provide some numerical simulations to verify our theoretical results and compare them with those of the nonlinear difference scheme in Ref.[6].

1 Difference Scheme

In this section, we will use the method of reduction of order to establish the difference scheme for Problem(1)and illustrate the truncation errors in detail.

1.1 Notation

Before presenting the difference scheme, we introduce the notations used.

We take two positive integersmandn.Then, we leth=L/m,xj=jh, 0≤j≤m;τ=T/n,tk=kτ, 0≤k≤n;Ωh={xj|0≤j≤m}.Ωτ={tk|0≤k≤n}.Moreover, we let

Vh={v|v∈Uhandv0=vm=0}

For anyu,v∈Uh, we introduce the following notations:

Then, we let

For anyw∈Sτ, we introduce the following notations:

It is easy to know that

1.2 Derivation of the difference scheme

We construct the difference scheme using the method of reduction of order.

Let

v=ux0≤x≤L, 0≤t≤T

Then Problem(1)is equal to the following problem of coupled equations:

ut+γuux+vxx=0 0

(2a)

v=ux0

(2b)

u(x,0)=φ(x) 0≤x≤L

(2c)

u(0,t)=u(L,t)=0 0

(2d)

v(L,t)=0 0≤t≤T

(2e)

Then, we denote

0≤j≤m, 0≤k≤n

Considering Eq.(2a)at points(xj,t1/2)and(xj,tk)and using the Taylor expansion, we obtain

1≤j≤m-1

(3)

and

1≤j≤m-1, 1≤k≤n-1

(4)

The constantc1>0 exists, such that

1≤j≤m-1, 1≤k≤n-1

(5)

We consider Eq.(2b)at points(xj+1/2,tk)and use the Taylor expansion to obtain

Lemma1[6]We denote

0≤j≤m-2, 0≤k≤n

The constantc2>0 exists, such that

Considering the initial and boundary value conditions expressed in Eqs.(2c)to(2e), we obtain

(7)

(8)

(9)

By omitting the small terms in Eqs.(3),(4), and(6)and combining them with Eqs.(7)to(9), we construct a three-level linearized difference scheme for Problem(2), as follows:

(10a)

1≤j≤m-1, 1≤k≤n-1

(10b)

(10c)

(10d)

(10e)

(10f)

1.3 Calculation of the difference scheme

For ease of calculation, we separate the variables for Eq.(10).

Theorem1The difference scheme expressed in Eq.(10)is equivalent to the following system of equations:

1≤j≤m-2

(11a)

(11b)

1≤j≤m-2, 1≤k≤n-1

(11c)

1≤k≤n-1

(11d)

(11e)

(11f)

and

(12a)

j=m-1,m-2,…,0; 0≤k≤n

(12b)

ProofⅠ)Based on Eqs.(10)to(12), first, we determine that Eqs.(10d)and(10e)are equivalent to Eqs.(11e)and(11f), respectively, and Eq.(10f)is equivalent to Eq.(12a).Moreover, Eq.(10c)is equivalent to Eq.(12b), and Eq.(10a)is equivalent to

(13a)

(13b)

Based on Eq.(10c), we derive

(14)

By substituting Eq.(14)into Eq.(13a), we derive Eq.(11a).

Similarly,based on Eq.(10f), we derive

Then, we obtain

By substituting the obtained equality into Eq.(13b)and using Eq.(14), we derive Eq.(11b).

Similarly, we obtain Eqs.(11c)and(11d)from Eqs.(10b),(10c), and(10f).

Ⅱ)Based on Eqs.(10)to(12), we determine that Eqs.(11e)and(11f)are equivalent to Eqs.(10d)and(10e), respectively, and Eq.(12a)is equivalent to Eq.(10f).Based on Eq.(12), we derive Eq.(10c)and

(15a)

(15b)

Based on Eq.(15), we obtain

By substituting the obtained equation into Eq.(11b), we derive

(16)

which is Eq.(10a)withj=m-1.By substituting Eq.(15b)into Eq.(11a), we obtain

that is

By combining the obtained equation with Eq.(16), we derive

which is Eq.(10a)with 1≤j≤m-2.

Similarly, we obtain Eq.(10b)from Eqs.(11c),(11d), and(12).

The proof is completed.

We denote

Based on Eq.(11e), we obtainu0.Then, we computeu1using Eqs.(11a),(11b), and(11f).We letw=u1/2.If we determinew, then we can obtainu1using the following expression:

u1=2w-u0

Based on Eqs.(11a),(11b), and(11f), we can obtain the system of equations forw, as follows:

The value ofwcan be obtained by solving the aforementioned system of quatic diagonal linear equations using the double-sweep method.

uk+1=2w-uk-1

Based on Eqs.(11c),(11d), and(11f), we can obtain the system of equations forw, as follows:

The value ofwcan be obtained by solving the aforementioned system of quadratic diagonal linear equations using the sweep method.

Furthermore,Theorem 1 illustrates that analyzing Eq.(10)is equivalent to analyzing Eq.(11).

2 Theoretical Analysis

In this section, we will analyze the conservation, unique solvability, and convergence of the difference scheme expressed in Eq.(10).

2.1 Conservation and Unique solvability

First, we provide several lemmas, which will be subsequently used.

Lemma2[18]For ?u∈Uhandv∈Vh, we have

(ψ(u,v),v)=0

Lemma3[6]We letv∈Uhandu∈Vhsatisfy

vm=0,vj+1/2=δxuj+1/20≤j≤m-1

Then, we derive

For the continuous problem expressed in Eq.(1), conservation exists.

Theorem2[5]Supposing thatu(x,t)is the solution to Problem(1), we denote

Then, we derive

Similarly,the difference scheme expressed in Eq.(10)has an invariant.

1≤k≤n

Then, we derive

Ek=‖u0‖21≤k≤n

ProofⅠ)Taking the inner product of Eq.(10a)withu1/2, we obtain

Together with Lemmas 2 and 3, we derive

That is,

(17)

1≤k≤n-1

Together with Lemmas 2 and 3, we derive

1≤k≤n-1

That is,

1≤k≤n-1

By adding the equality expressed in Eq.(17)to the obtained equality, we derive

0≤k≤n-1

This completes the proof.

Then, we prove the unique solvability.

Theorem4The difference scheme expressed in Eq.(10)has a unique solution.

ProofWe obtainu0using Eq.(10d)andv0using Eqs.(10c)and(10f).

Based on Eqs.(10a),(10c),(10e), and(10f), we deriveu1andv1.Considering the system of homogeneous equations, we obtain

(18a)

(18b)

(18c)

(18d)

Taking the inner product of Eq.(18a)withu1, we derive

Together with Lemmas 2 and 3, we obtain

Then, we have

‖u1‖=0

which follows

From Eq.(18b)and Eq.(18d), we can get

That is, Eqs.(10a),(10c),(10e)and(10f)have the unique solutionsu1andv1.

We suppose thatuk,uk-1andvk,vk-1are known.Based on Eqs.(10b),(10c),(10e)and(10f), we determineuk+1andvk+1.Considering the system of homogeneous equations, we obtain

1≤j≤m-1

(19a)

(19b)

(19c)

(19d)

Taking an inner product of Eq.(19a)with 2uk+1, we get

Using Lemmas 2 and 3, we obtain

Then, we have

‖uk+1‖=0

which follows

Based on Eqs.(19b)and(19d), we obtain

That is, Eqs.(10b), Eq.(10c), Eq.(10e), and Eq.(10f)have the unique solutionsuk+1andvk+1.

This completes the proof.

2.2 Convergence

0≤j≤m, 0≤k≤n

By subtracting Scheme(10)from Eqs.(3),(4), and(6)to(9), we derive the following system of error equations:

(20a)

1≤j≤m-1

(20b)

(20c)

(20d)

(20e)

(20f)

Before obtaining the convergence result, we present the following two lemmas.

Lemma4From the proof of Theorem 4.3 in Ref.[5], it follows that the following equality holds:

Lemma5From the proof of Theorem 4.1 in Ref.[6], it follows that the following equality holds:

We denote

Theorem5We let

Ifλ<1, then we have

‖ek‖≤c4(τ2+h2) 0≤k≤n

ProofⅠ)It follows from Eq.(20d)that

‖e0‖=0

(21)

Taking the inner product of Eq.(20a)with 2e1/2, we derive

That is,

(22)

It follows from Lemma 5 that

By substituting the previously derived equality into Eq.(22)and combining it with Lemma 1, Lemma 5, and the truncation error expressed in(5), we obtain

That is,

(23)

That is,

1≤k≤n-1

(24)

It follows from Lemma 5 that

By substituting the previously derived equality into Eq.(24)and combining it with Lemmas 4 and 1, we obtain

Then, we derive

Using the truncation error expressed in Eq.(5), we obtain

(25)

We let

It follows that

Ifλ<1, then we derive

It follows from inequality(25)that

That is,

(1-c3τ)Ek+1≤(1+c3τ)Ek+

1≤k≤n-1

Using the Gronwall inequality, we obtain

By substituting Eqs.(21)and(23)into the previously derived inequality, we obtain

It is easy to know that

Consequently,

‖ek‖≤c4(τ2+h2) 1≤k≤n

This completes the proof.

3 Numerical Examples

In this section, we present two numerical examples.The numerical results illustrate the efficiency of the difference scheme expressed in Eq.(11).

In Ref.[6], we presented the following two-level nonlinear difference scheme:

(26a)

(26b)

(26c)

(26d)

and solved it using the Newton iterative method.

We denote the error as follows:

Whenτis sufficiently small, the spatial convergence order is defined as follows:

Whenhis sufficiently small, the temporal convergence order is defined as follows:

Example1In Problem(1), we takeT=1,L=1,γ=-6,φ(x)=x(x-1)2(x3-2x2+2).The exact solution is unknown.

The difference scheme expressed in Eq.(11)will be employed to numerically solve this problem.The numerical accuracy of the difference scheme in space and time will be verified.

By varying step sizehwith the sufficiently small step sizeτ=1/12 800 and varying step sizeτwith the sufficiently small step sizeh=1/12 800, the numerical errors and convergence orders for Scheme(11)are listed in Tabs.1 and 2.From these tables, we determine that the numerical convergence orders of Scheme(11)can achieveO(τ2+h2), which is consistent with Theorem 5.

Tab.1 Errors and space convergence orders of Example 1(τ=1/12 800)

Tab.2 Errors and time convergence orders of Example 1(h=1/12 800)

Furthermore, we observe that the difference scheme expressed in Eq.(11)is more computationally efficient than the nonlinear difference scheme expressed in Eq.(26).

Fig.1 indicates that the energy of Scheme(11)is conserved for Example 1.

Fig.1 Energy conservation law of Example 1

We denote

and

Example2We conduct a numerical experiment with the exact solution of a solitary wave[19], as follows:

u(x,t)=4sech2(x-4t-4)

0≤x≤20, 0≤t≤1

The corresponding parameter isγ=3.

Similarly, we compute the example using the difference scheme expressed in Eq.(11)and observe the convergence.First, we fix the time step size as 1/3 200 and calculate the errors and convergence orders in the space directions, as shown in Tab.3.Then, we fix the space step size as 1/3 200 and calculate the errors and convergence orders in the time directions, as shown in Tab.4.

Tab.3 Errors and space convergence orders of Example 2

Tab.4 Errors and time convergence orders of Example 2

Notably, the numerical results are consistent with the theoretical analysis.In Fig.2, the numerical and exact solution curves are plotted fort=0.25,0.5,0.75,1.0.Fig.2 shows that the numerical solutions are consistent with the exact solutions.

(a)

4 Conclusions

In this study, we consider the numerical solution to the initial-boundary value problem of the KdV equation using the finite difference method.With the use of the method of reduction of order, we establish a three-level linearized difference scheme and show that it is more computationally efficient than the two-level nonlinear difference scheme through numerical simulations while retaining the same convergence orders.Using the energy analysis method, we prove the conservation, unique solvability, and conditional convergence of the difference scheme.

In Theorem 5, the convergence of the difference scheme was proven under the constraint of the step size ratio.This difference scheme may be unconditionally convergent.Indeed, we found from Example 1 that the restriction of step size ratio is unnecessary for ensuring that the convergence result holds.However, we have not yet discovered a better method to prove the unconditional convergence and will continue our research in future work.