Implicit and Explicit scheme

Raju SharmaAssistant ProfessorDepartment of Civil EngineeringChandigarh university

Implicit and Explicit scheme Parabolic EquationsThe one dimensional heat conduction equation u/ t = c2 (2u/ x2 ) is a well known example of parabolic partial differential equation

In general, the study of pressure waves in a fluid, propagation of heat and unsteady state problem lead to parabolic type of equations.Solution of One Dimensional Heat Equationu/ t = c2 (2u/ x2 )Where c2 =k/sp is the diffusivity of the substance (cm2/sec.)

Schmidit MethodCrank-Nicolson MethodIterative Methods of solutionDu Fort and Frankal Method Finite Difference Approximation to Partial DerivativesXYX=hx-h, yi-1, jx, y+ki, j+1x, yi, jx+h, yi+1, jx, y-ki, j-1y=koR=RegionMesh Pointu/ x = [(u(x+h,y) u (x,y))/h]+o(h) (f) = [u ((x,y)-u (x-h,y))/h] + o (h) = [(u(x+h,y) -2u (x,y) +u ((x+h,y)]/o(h2)Writing u(x, y) = u(ih, jk) as simply ui,j

Schmidt Methodui,j+1=ui-1,j +(1-2 ) ui,j + ui+1,j= kc2/h2 is the mesh ratio parameterThis formula enables us to determine the value of u at the (i, j+1)th mesh point in terms of the known function values at the points xi-1, xi and xi+1 at the instant tjtkhjth level(j+1)th leveli, j

i, j+1

i+1, j

i-1, j

xIt is a relation between the function values at the two time levels j+1 and j and is therefore, called a 2 level formula.ui,j+1=ui-1,j +(1-2 ) ui,j + ui+1,jThis equation is called the schmidit explicit formula which is valid only for 0< Put = in above equation ui,j+1= (ui-1,j + ui+1,j )Which shows that value of u at xi at time tj+1 is the mean of the u-values at xi-1 and xi+1 at time tj. This relation, known as Bendre-Schmidt recurrence relation, gives the values of u at the internal mesh points with the help of boundary conditions.Crank-Nicolson MethodSchmidt scheme is computationally simple and for convergent results i.e. k h2/2c2To obtain more accurate results, h should be small i.e. k is necessarily very small. This makes the computations exceptionally lengthy as more timelevels would be required to cover the region. A method that does not restrict and also reduces the volume of calculations was proposed by crank and nicolson in 1947.According to this - ui-1,j+1 +(2+2 ) ui, j+1 - ui+1,j+1 = ui-1, j +(2-2 ) ui, j + ui+1, j ..(1)Where = kc2/h2Left side of above equation contains three unknown values of u at the (j+1)th level while all the three values on the right are known values at the jth level.Thus above equation is a 2-level implicit relation and is known as crank-nicolson formula. It is convergent for all finite values of If there are n internal mesh Points on each row, then the equation gives n simultaneous equations for the n unknown values in terms of the known boundary Values khjth level(j+1)th leveli, j

i, j+1

i+1, j

i-1, j

xti-1, j +1

i+1, j +1

Iterative Method of solution for an implicit schemeui (n+1) = /2(1+ ) (ui-1 (n) +u i+1 (n) ) +bi/(1+ )

bi = ui , j + /2 (ui-1 , j 2 ui , j + u i+1 , j)

As the latest value of ui-1 i.e. ui-1 (n+1) is already available, the convergence of the iteration formula can be improved by replacing ui-1 (n) by ui-1 (n+1) .The above equation becomes

ui (n+1) = /2(1+ ) (ui-1 (n+1) +u i+1 (n+1) ) +bi/(1+ )

Which is known as gauss-seidal iteration formula

Du Fort and Frankal MethodThis method is based on central difference method

u/ t = ui , j+1 ui, j-1/2k2u/ x2 = ui-1 , j 2 u i, j-1+ u i +1, j /h2Put above values in equation given below u/ t = c2 (2u/ x2 )

We obtain ui , j+1 ui, j-1 =2k c2 /h2 [ui-1 , j 2 u i, j-1+ u i +1, j ]

Solve the equation u/t = 2u/x2 subject to the conditions u(x,0)=sinx, 0x1; u(0, t)= u(1, t)=0, using Schmidt method. Carryout computation for two levels, h=1/3, k = 1/36Here C2 =1, h = 1/3, k = 1/36.So that = kc2/h2 = Also u 1,0 = sin /3=3/2u 2,0= sin 2 /3= 3/2 and all boundary values are zeroSchmidt formulaui,j+1=ui-1,j +(1-2 ) ui,j + ui+1,jui,j+1= [ui-1,j +2 ui,j + ui+1,j] K=1/3h=1/3jth level(j+1)th level3/2

1, 2

3/20,0

t02, 2

x 001, 12, 1 0 0For i =1,2; j=0:u1,1 = [u 0,0 +2u1,0+u2,0]= (0+2 x 3/2+ 3/2)= 0.65u2,1 = [u 1,0 +2u2,0+u3,0]= (3/2 +2 x 3/2+ 0)= 0.65For i =1,2; j=1u1,2 = [u 0,1 +2u1,1+u2,1]= (0+2 0.65+0.65)=0.49u2,2 = [u 1,1 +2u2,1+u3,1]= (0.65+2 0.65+0)=0.49