Solving Nonlinear System of Equations and Poisson's Equation with Dirichlet Boundary Conditions

[Type here]
MATHS SOLUTIONS
QUESTION 1
Lets consider the Lorenz system, which is a system of three first-order differential equations of
the form (Hirsch, Smale & Devaney, 2012).
dx/dt =yz-βx
dy/dt = σ(z-y)
dz/dt =y(-ρ-x)-βz
Parameters σ=10, ρ =28, β =8/3
Initial conditions x(0) =27, y(0) = -8, c = 8
So to develop an equivalent non linear system of equations of the Lorenz system of the form
F(x) = 0
If we assume that the solution of the system are expanded as Maclaurin series such that we have
X(t) = a0 +a1t +a2t2/2! +.....+antn/n! +...

Running head: MATHS SOLUTIONS
Y(t) = b0 +b1t + b2t2/2! +....+bntn/n! +...
Z(t) = c0 + c1t + c2t2/2! +....+cntn/n! +...
By consecutive differentiation, of the coefficients an, bn, cn we obtain the following system of
difference equation
an =σ(bn-1-an-1)
bn = ρan-1 -bn-1-
⅀i=0n-1 n-1 ai cn-i-1
i
cn = -bcn-1 + ⅀i=0n-1 n-1 aibn-i-1
i )
so using the initial values x(0), y(0), z(0) and parameters σ=10, ρ = 28, β =8/3 it is obtained the
system of difference equations for an(≈),bn(≈),cn(≈)
an(≈)=σ(bn-1(≈)-an-1(≈)) +Aσ(bn-2(≈)-an-2(≈))-Bσ(bn-3(≈)-an-3(≈))+cσ(bn-4(≈)-an-
4(≈))-Dσ(bn-5(≈)-an-5(≈))-Aan-1(≈) +Ban-2(≈), n>7
bn(≈)= (ρan-1(≈) -bn-1(≈)) +A(ρan-2(≈)) -bn-2(≈)) -B(ρan-3(≈)-bn-3(≈)),
n>6
cn(≈) = -Acn-1 (≈) +Bcn-2 (≈) -Ccn-3 (≈) + Dcn-4(≈), n>5
where A = 1+σ +b, B= σρ-a02, C=σa0b0, D= -σ2b02
QUESTION 2
Consider the Dirichlet problem for Poisson’s equation (Heydari et al. 2013)
-∆u(x)=f(x) in Ω
u=0 on ∂Ω

Running head: MATHS SOLUTIONS
where
Ω={(x1,x2) ∈ (0,1)x(0,1)}
f(x) =Afexp(- (x1-c1)/2s12 –(x2-c2)/2s22)1/𝛼(x)
𝛼(x)=1+Aexp(- (x1-c1)/2s12 –(x2-c2)/2s22)
We consider a scalar potential u(x) which satisfies the Poisson equation -∆u(x) =f(x)
Where f(x) is a specification u(x) fulfil the Neumann- Dirichlet boundary conditions
u ̓(a)=u ̓a and u(b)=ub.Then an appropriate discretization is chosen. The mesh is composed of
four discrete points belonging to the interval Ω
∆x=b-a/3 =h
The mesh points (xi) are defined by the following relation xi=a +(i-1)h, i=0,1,.......4
We denote by ui the approximate value of the desired potential at point xi:ui≈u(xi)
For each point xi in the interval Ω, the value of the function fi =f(xi)
ui ̓ =u ̓(xi) and u”i =u”(xi) are the first and second derivatives of the potential function u
At point xi with the centered difference approximation
(0(h2)), the first derivative
ui ̓ =ui+1-ui-1/2h +0(h2)
and the second derivative
u”i=ui-1-2ui +ui+1 /h2+0(h2) i=2,3
-u1 +u2 =h2f1/2 + hua ̓
We can introduce the vector F where elements Fi are defined by
F1 =h2f1/2 +hua ̓, FN= h2fN -ub and Fi=h2fi, i=2
We then obtain the following matrix equation