Maths Solutions: Differential Equations and Numerical Methods

This assignment provides detailed solutions to two mathematical problems. The first problem focuses on the Lorenz system, a system of three first-order differential equations. It explores the use of Maclaurin series to develop an equivalent non-linear system of equations and utilizes initial conditions and parameters to derive a system of difference equations for the coefficients. The second problem addresses the Dirichlet problem for Poisson's equation, involving the discretization of the equation and the application of finite difference approximations to solve for a scalar potential. This includes defining mesh points, approximating derivatives, and formulating a matrix equation to solve the linear system using Gaussian elimination. The solutions are presented step-by-step, including the centered difference approximation and the final matrix equation, along with a justification using Gaussian elimination.

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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

Running head: MATHS SOLUTIONS
-1 1 0 0 0 ….. 0 u1 h2f1/2
+hu᾿a
1 -2 1 0 0…… 0 u2
h2f2
0 1 -2 1 0……… 0 x u3
h2f3
0 0 1 -2 1……… 0 u4 =
h2f4
0 0 0 1 -2……….
0 0 0 0……. 1 uN-1
h2fN-1
0 0 0 0 0 1 -2 uN
h2fN -ub
The centered difference approximation leads to an 3x 3 matrix A=(aij) that is diagonally
dominant and tridiagonal
To solve the linear system with justification using gaussian Elimination
Let us consider the general mxn linear system in the variables x1,x2,……….,xn which is of the form
A11x1 +a12x2 +…+a1nxn =b1
A21x1 +a22x2 +….+a2nxn=b2
Am1x1 +am2x2+…..+amnxn=bn
A solution of this linear system is an n- tuple .We first express the linear system in the form
augmented matrix,we then use elementary row operatioins to reduce the to an echelon form
We then solve the linear system of the echelon form using back substitution

Running head: MATHS SOLUTIONS
Reference
Hirsch, M. W., Smale, S., & Devaney, R. L. (2012). Differential equations, dynamical systems,
and an introduction to chaos. Academic press.
Heydari, M. H., Hooshmandasl, M. R., Ghaini, F. M., & Fereidouni, F. (2013). Two-dimensional
Legendre wavelets for solving fractional Poisson equation with Dirichlet boundary conditions.
Engineering Analysis with Boundary Elements, 37(11), 1331-1338.

Running head: MATHS SOLUTIONS

Running head: MATHS SOLUTIONS