MA347 Numerical Solution of PDE Problem: Scheme Comparison Report

This report addresses the numerical solution of a two-dimensional non-linear PDE problem, focusing on the application and comparison of implicit and explicit schemes. The solution begins with the formulation of the problem, including initial and boundary conditions, and introduces the Crank-Nicolson and Lax-Friedrichs methods for implicit and explicit schemes, respectively. The report details the computational domain, grid specifications, and the use of the Reynolds number to evaluate the schemes. Numerical results are presented and compared with exact solutions at different time steps, demonstrating the accuracy and stability of the schemes. The report also discusses the advantages and disadvantages of each scheme, highlighting their stability conditions, computational speed, and error characteristics. The conclusion emphasizes the implicit scheme's preference due to its efficiency and convenience in obtaining reliable solutions.

Numerical Solution PDE
Problem d
Consider a two dimensional non-linear equations
dk
dt +u dm
dm +v dk
dn - 1
ℜ ( d2 h
dm + d2 k
dn )=0
dh
dt +u dh
dm +v dk
dn - 1
ℜ ( d2 v
d m2 + d2 u
d n2 )=0
In reference to the original state of the derivatives;
k(m,n,0)=ѱ1(m,n);(m,n) ∈Ὡ
h(m,n,0)=ѱ2(m,n);(m,n)∈Ὡ
And the boundary conditions are;
k (m,n,t)=Ƹ(m,n,t);(m,n) ∈ dὩ , t>0 i.e. at time step t greater than 0
h(m,n,t)=Ƹ(m,n,t);(m,n) ∈dὩ ,t>0 i.e. at time step t greater than 0
Ὡ ={(m,n):a≤ m≤ b , c ≤ n ≤ d } and dὩ is boundary.
k (m,n) and h(m,n) being the components of velocity at the step time.
ѱ1,ѱ2 Ƹ1,Ƹ2 are the known factors to determine the velocity.
Re is the Reynolds number.

Problem e
Implicit Scheme
Kranck-Nicolson Scheme
Assume u=k,v=h
Let k(m,n,t)= 3
4 - 1
4 ¿ ¿
h(m,n,t)= 3
4 + 1
4 ¿ ¿
Consider the computational domain to be a square domain,i.e
Ὡ={m,n):0≤ m≤ 1 ,0≤ n ≤1 }
k(m,n,t) and h(m,n,t) are the conditions of the initial boundaries and are obtained from the solutions of the analysis.
Computation of numerical analysis is conducted on a uniform grid of mesh width of ∆ m=∆ n=0.05
The study case is in optimal agreement with the exact solutions when the Reynolds number is substituted with different values.
It is clear from the result to conclude that from the current scheme ,there is enough evidence to support that it is not conditionally
stable and accurate for any time step –size.
Realization

Re=500at , ∆ t=0.001, t=0.5
In comparison of numerical results V with the exact solution at t=0.5 and t=2 with ∆ t=0.001∧ℜ=10 we get values as per the given
below.
m, n) t=0.5 t=2
numerical exact numerical exact
(0.10, 0.10) 00.88475 00.88475 00.91284 00.91284
(0.50, 0.10) 00.91460 00.91460 00.93871 00.93873
(0.90, 0.10) 00.93938 00.94016 00.95670 00.95887
(0.30, 0.30) 00.88474 00.88475 00.91283 00.91284
(0.70, 0.30) 00.91445 00.91460 00.93812 00.93873
(0.10, 0.5o) 00.85372 00.85373 00.88279 00.88280
(0.50, 0.50) 00.88471 00.88475 00.91262 00.91284
(0.90, 0.50) 00.91293 00.91460 00.93346 00.93873
(0.30, 0.70) 00.85362 00.85373 00.88229 00.88280
(0.70, 0.70) 00.88437 00.88475 00.91101 00.91284
(0.10, 0.90) 00.82443 00.82519 00.84903 00.85183
(0.50, 0.90) 00.85227 00.85373 00.87736 00.88280
(0.90, 0.90) 00.88198 00.88475 00.90358 00.91284
Explicit Scheme
Lax –Friedrichs Scheme
Choose a function k.
Let k(m) =m2
The first order forward finite difference approximation is given as ;

km(3)
Use the step-size h=0.1
km(m0)≈ U ( m0+h ) −U (m0)
h
Substitute km(X0) ≈ ( m0+ h )2−m 02
h
Replacing X0 by 3 and h by 0.1
km(3)= ( 3+0.1 ) 2−32
0.1 =6.1
You can see that the approximation is 6.1-6 =0.1 i.e half the step-size.
Repeat the problem with h= 0.05
km(x)≈ ( 3+0.05 )2−32
0.05 =6.05
Error =6.05-6=0.05
It is indeed true that reducing the step-size to half results into halving the error.
Problem f
Comparison And Analysis of Implicit and Explicit Schemes
Explicit schemes are only stable for Courant Numbers <1, while implicit schemes are only stable for Courant Numbers >1

Implicit solutions are coded to run faster hence are more preferred, while explicit solutions are coded to run slow and as result not
often preferred.
Implicit formulation consumes less time and doesn’t need to be verified to stabilize any criteria for each time step, while explicit
formulation is time consuming as it requires more time to stabilize the criteria for each time step.
Explicit Schemes
Implement the use of central and forward difference.
Is an easy method i.e the estimation of derivative is already calculated.
The stability condition has to be satisfied i.e grid sizes of the scheme cannot be chosen freely.
Has got an order error ∆ t ∆ x2.
It is more accurate to get a solution as finer grids are used.
Implicit Schemes
Offer a free choice of grid sizes and there is no any stability conditions required.
Employ the use of central and backward difference.
Has got an order error ∆ t ∆ x2.
Even though both the explicit and implicit schemes have got the same order error, the implicit scheme involve less errors as
compared to the explicit schemes. Consequently ,it is better proposed for an implicit scheme than an explicit scheme as it is more
convenient and efficient in any problem formulation in getting a reliable solution and result.

REFERENCES
[1] L. Evans, Partial Differential Equations. American Mathematical Society (2006).