Two Dimensional Non-Linear Equations

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 ;