One Explicit Scheme for the Linear Heat Conduction Equation and the Numerical Approximation

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One Explicit Scheme for the Linear Heat

Conduction Equation and the Numerical

Approximation

Hongxia Li

Zhejiang University of Finance and Economics , Hangzhou, Zhejiang, 310018, China Email: [email protected]

Abstract—In this paper, one explicit scheme for the linear he at conduction equation in one space dimension is conside re d. We will analysis the truncation error of the method and w ill use the super norm, Fourier series to analysis the stability of the scheme; at last the numerical experiments of initial va lue problem will be presented.

Index Terms—explicit scheme, heat conduction equation, nu merical experiments

I. INTRODUCTION

Many problems in science and engineering are modelled by special cases of the linear parabolic equation for the unknown u x t( , )

t xx

u =u (1) An initial condition will be needed; if this is given at t=0 it will take the form

0 ( ,0) ( )

u x =u x (2) where 0_{( )}

u x is a given function.

In this paper we consider one explicit scheme 1

1 1

2 2

n n n n n j j j j j

u u u u u

h τ

+

+ −

− − +

= (3) 1

1 1

( 2 )

n n n n n j j j j j

u + =u +γ u₊ − u +u₋ (4)

(where γ is the mesh ration, γ =τ_h2 , τ is the temporal spacing and h is the spatial spacing ) in[1], [2], [3] forthe linear heat conduction in one space dimension (1). In section 2 We will analysis the truncation error of the method. In section 3 we will use the super norm, Fourier series to analysis the stability of the scheme. In section 4 the numerical experiments of initial value problem will be presented.

II.THE TRUNCATION ERROR

The Truncation error R(x,t) of the numerical scheme (3) is the difference between the two sides of the equation, when the approximation n

j

u is replaced throughout by the exact solution u x t( , )j n

1

( , ) ( , ) ( , ) u x tj n u x tj n

R x t

τ

+ − =

1 1

2

( j , ) 2 ( , )n j n ( j , )n

u x t u x t u x t

h

+ − + −

− (5)

where

2

' ''

1 ( , ) ( , ) ( , )

( , ) ( , ) j n t j n ₂ tt j n

j n j n u x t u x t u x t

u x t u x t τ τ

τ τ

+ − ₌ + +

3

' '' 2

( ) ( , )

( , ) ( , ) ( )

2

j n

t j n tt j n

o u x t

u x t u x t o

τ τ _τ

τ

−

+ = + + (6)

1 1

2

( j , ) 2 ( , )n j n ( j , )n

u x t u x t u x t

h + − + − ₌ 2 3 ' '' ''' 2 ( , ) ( , ) ( , ) ( , ) 2 3!

j n x j n xx j n xxx j n

h h

u x t hu x t u x t u x t

h + + + + 4 (4) 5 2

( , ) ( ) 2 ( , )

4! x j n j n

h

u x t o h u x t

h + − 2 3 ' '' ''' 2 ( , ) ( , ) ( , ) ( , ) 2 3!

j n x j n xx j n xxx j n

h h

u x t hu x t u x t u x t

h − + − + 4 (4) 5 2 ( , ) ( ) 4! x j n

h

u x t o h

h − +

2

'' _{( , )} (4)_{( , )} _{( )}3 12

xx j n x j n

h

u x t u x t o h

= + + (7) substitute (6), (7) into (3), we get

' '' ''

( , ) ( , ) ( , ) ( , ) 2

t j n tt j n xx j n

R x t =u x t +τu x t −u x t

2

(4)_{( , )} _{( )}3 12 x j n

h

u x t o h

− + (8) since '_{( , )} '' _{( , ) 0}

t j n xx j n

u x t −u x t = , so 2

'' (4) 2 3

( , ) ( , ) ( , ) ( ) ( )

2 tt j n 12 x j n

h

R x t =τu x t − u x t +oτ +o h

₌_O₍_τ₊_h2₎₍₉₎

III. THE STABILITY OF THE SCHEME

We will use the super norm and Fourier series to analysis the stability of the scheme (4).

A. super norm

Definition(super norm): Assume u is a vector with N

components, then the super norm of u is defined by

u_∞= 1

max

_{≤ ≤} k

k N

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1

max

+ _≤ n n j j j e e 1

max

n+ _≤

max

n

j j

e e

1

n n

e + e

∞≤ ∞ (13)

we note that for stability of the scheme (4) we have required that 0≤ ≤γ 1/ 2. In this case we say that the scheme is conditionally stable(where the condition is

0≤ ≤γ 1/ 2).

B. Fourier series

We can show that a similar fourier mode = ikxj

n n j

u λ e (14) is an exact solution of the difference equation (4) (where

λ is the amplification factor). Suppose we substitute (14) into the difference equation (4):

1 1

1 − _{(1 2 )} +

+ ikxj₌ ikxj _{+ −} ikxj₊ ikxj

n n n n

e e e e

λ γλ γ λ γλ

we can then divide by n ikxj

e

λ to get (15)

(1 2 ) ( − )

= − + ikh+ ikh

e e

λ γ γ

1 2 (1 cos )

= − γ − kh 2 1 4 sin

2

= − γ kh (15) The difference scheme (4) is stable wwhenever

2

1 4 sin 1

2

− γ kh ≤ (16) is satisfied. Since (16) equivalent to

2

1 1 4 sin 1

2

− ≤ − γ kh≤ or

_{2 4 sin}2 2

≥ γ kh

which is true when 0≤ ≤γ 1/ 2. Thus in this case we also say that the scheme is conditionally stable(where the condition is 0≤ ≤γ 1/ 2).

VI. THE NUMERICAL EXPERIMENTS

Use the explicit scheme (4) to simulate the following examples.

Example: consider the initial and boundary problem

of (1)

(a) γ=2 / 5

(b) γ=3 / 5

Fig1: numerical solution of example at level=36 when k=0.5; (a) γ=2 / 5, (b)γ=3 / 5.

We can see from the Fig1 that the numerical solution is stable whenγ≤1/ 2and is unstable when γ>1/ 2. The explicit scheme (4) is conditional stable.

Case2: Takeγ =2 / 5 is less than 1/2 and γ=3 / 5 is

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(a) γ=2 / 5

(b) γ=3 / 5

Fig2: numerical solution of example at level=36 when k=1; (a) γ=2 / 5, (b)γ =3 / 5.

(a) γ=2 / 5

(b) γ=3 / 5

Fig3: numerical solution of example at level=36 when k=2; (a) γ=2 / 5, (b)γ=3 / 5.

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(b) γ=3 / 5

(a) γ=2 / 5

(b) γ=3 / 5

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(b) γ=3 / 5

(a) γ=2 / 5

(b) γ=3 / 5

(a) γ=2 / 5

(b) γ=3 / 5

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(b) γ=3 / 5

(a) γ=2 / 5

(b) γ=3 / 5

The graphs indicate When γ>1/ 2, the k is greater then the numerical solution is great unstable.

REFERENCES

[1] K.W.Morton; D.F.Mayers, ” Numerical solution of partial differential equations”. Posts Telecom Press (2006). [2] J.W.Thomas; “Numerical partial differential equations-

finite difference methods”. Spring-verlag New York Berlin Heidelberg (1995).

[3] Li likang, Yu chonghua, Zhu zhenghua “Numerical solution of differential equations (in chinese)”. Fudan university press (1999).

Hongxia Li Hebei Province, China. Birth