Sample 4 - Fisher's Equation PDE

3/26/2013

Fisher’s Equation: Analysis and numerical solutions Author:

Mandeep Singh

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Contents Introduction ............................................................................................................................................................................ 2

Stability Analysis ..................................................................................................................................................................... 2

Numerics ................................................................................................................................................................................. 3

Forward Euler Method (Results) ............................................................................................................................................. 5

Runge-Kutta Method, 4th Order (Results) ............................................................................................................................... 8

Trapezoidal Method (Results) ............................................................................................................................................... 11

Backwards Euler Method (Results) ....................................................................................................................................... 14

Forward Euler Method (MATLAB Code) ............................................................................................................................... 17

Runge-Kutta Method, 4th Order (MATLAB Code) .................................................................................................................. 18

Trapezoidal Method (MATLAB Code) ................................................................................................................................... 20

Backwards Euler Method (MATLAB Code) ............................................................................................................................ 23

References ............................................................................................................................................................................ 26

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Introduction The main equation: Fisher’s equation, also known as the Fisher–Kolmogorov equation and the Fisher–KPP

equation

𝑣𝜏 = 𝐷𝑣𝑦𝑦 + 𝑟𝑣 (1 −𝑣

𝑘)

𝑦 ∈ (−𝐿, 𝐿), 𝑡 > 0

𝑣𝑥(±𝐿, 𝑡) = 0

𝑣(𝑦, 0) = 𝑓(𝑦)

Fisher’s Equation governs the evolution of a population of density 𝑣, subject to diffusion and logistic growth

(given by the 𝑟𝑣 (1 −𝑣

𝑘) term. Discovered in 1937 by Ronald Fisher, this model was designed to answer

whether the gene has a typical speed of propagation. This model is a semilinear equation, where diffusion is

coupled with logistic growth through the term 𝑟𝑣 (1 −𝑣

𝑘). The parameter 𝑟 represents a biological potential

(net birth-death rate), 𝑘 is the carrying capacity of the habitat.

Non-dimensionalization:

𝑡 = 𝑟𝜏; 𝑥 = √𝑟

𝐷𝑦; 𝑢 =

𝑣

𝑘

𝑢𝑡 = 𝑢𝑥𝑥 + 𝑢(1 − 𝑢), 𝑡 > 0

𝑥 ∈ [−𝑀,𝑀], 𝑡 > 0

𝑢𝑥(±𝑀, 𝑡) = 0

𝑢(𝑥, 0) = 𝑔(𝑥)

Stability Analysis

We start with assuming a travelling wave solution for 𝑢(𝑥, 𝑡), i.e.:

𝑢(𝑥, 𝑡) = 𝑢(𝑥 − 𝑐𝑡); 𝑐 =speed of the wave

𝑢𝑡 = 𝑢𝑥𝑥 + 𝑢(1 − 𝑢) → (𝑢(𝑥 − 𝑐𝑡))𝑡= (𝑢(𝑥 − 𝑐𝑡))

𝑥𝑥+ 𝑢(𝑥 − 𝑐𝑡) − (𝑢(𝑥 − 𝑐𝑡))

2

Let 𝑥 − 𝑐𝑡 = 𝑧 → 𝑢(𝑧)𝑡 = 𝑢(𝑧)𝑥𝑥 + 𝑢(𝑧) − 𝑢2(𝑧)

→ −𝑐𝑢′(𝑧) = 𝑢′′(𝑧) + 𝑢(𝑧) − 𝑢2(𝑧)

→ 𝑢′′(𝑧) + 𝑐𝑢′(𝑧) + 𝑢(𝑧) − 𝑢2(𝑧) = 0

Let

𝑢′ = 𝑣 = 𝑓(𝑢, 𝑣)

𝑣′ = −𝑐𝑣 − 𝑢 + 𝑢2 = 𝑔(𝑢, 𝑣)

Critical points: (0,0) and (1,0)

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𝐽(𝑢, 𝑣) = (𝑓𝑢 𝑓𝑣𝑔𝑢 𝑔𝑣

) = (0 1

−1 + 2𝑢 −𝑐)

𝐽(0,0) = 𝐽1 = (0 1−1 −𝑐

) ; 𝐽(1,0) = 𝐽2 = (0 11 −𝑐

)

𝜆0,0 =−𝑐 ± √𝑐2 − 4

2

From above we get that the point (0,0) is a stable node for all values greater than or equal to 2 i.e. for 𝑐 ≥ 2

and for 𝑐 = 2, we have a degenerate node. For 𝑐2 < 4, the system is a stable spiral.

𝜆1,0 =−𝑐 ± √𝑐2 + 4

2

From above we get that the point(1,0) is a saddle point because both eigenvalues are real but one is positive

and one is negative.

Here is the phase portrait for Fisher’s equation for 𝑐 = 2:

Note that the two critical points are marked in red. It is easy to see both the separatrix and stable degenerate

node.

Numerics 𝑢𝑡 = 𝑢𝑥𝑥 + 𝑢(1 − 𝑢)

We start with computing the approximation for the spatial derivative for 𝑢(𝑥, 𝑡)

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𝑢𝑥𝑥(𝑥, 𝑡) ≈𝑢(𝑥 − Δ𝑥, 𝑡) − 2𝑢(𝑥, 𝑡) + 𝑢(𝑥 + Δ𝑥, 𝑡)

(Δ𝑥)2

The purpose of this approximation is to take our nonlinear PDE and approximate it as a nonlinear, time

dependent ODE.

∴, 𝑢𝑡 ≈1

(Δ𝑥)2[𝑢(𝑥 − Δ𝑥, 𝑡) − 2𝑢(𝑥, 𝑡) + 𝑢(𝑥 + Δ𝑥, 𝑡)] + 𝑢(𝑥, 𝑡)(1 − 𝑢(𝑥, 𝑡)) = 𝑓(𝑥, 𝑡) … 1

We can re-write the above equation using indices:

𝑢𝑛 ≈1

(Δ𝑥)2[𝑢(𝑥𝑖−1, 𝑡𝑛) − 2𝑢(𝑥𝑖, 𝑡𝑛) + 𝑢(𝑥𝑖+1, 𝑡𝑛)] + 𝑢(𝑥𝑖, 𝑡𝑛)(1 − 𝑢(𝑥𝑖, 𝑡𝑛)) = 𝑓(𝑥𝑖 , 𝑡𝑛)… 2

Note the subscript on 𝑢 changes from 𝑡 to 𝑛. As 𝑛 is our time index, the above equation is at the time step

𝑡 = 𝑛ℎ, where ℎ = Δ𝑡. In other words, for the given space point, 𝑥𝑖, we calculate the time evolution by

keeping the space index,𝑖, fixed and going to the next time index, 𝑛 + 1. We perform this calculation using

nested loops.

Also, in order to account for the no-flux boundary conditions 𝑢𝑥(±𝑀, 𝑡) = 0, the space points at 𝑖 = 1 and 𝑖 =

𝑁 need to be treated differently. These points are approximated by the following equations:

𝑢𝑛 ≈(−2𝑢(𝑥1, 𝑡𝑛) + 2𝑢(𝑥2, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥1, 𝑡𝑛)(1 − 𝑢(𝑥1, 𝑡𝑛))… 3

The above equation is at 𝑥 = −𝑀 (left boundary point), and as before, the equation represents the time

evolution of this boundary point at index 𝑖 = 1.

𝑢𝑛 ≈(−2𝑢(𝑥𝑁 , 𝑡𝑛) + 2𝑢(𝑥𝑁−1, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥𝑁 , 𝑡𝑛)(1 − 𝑢(𝑥𝑁 , 𝑡𝑛))… 4

i.e.:

𝑢𝑛 = 𝑓(𝑥𝑖, 𝑡𝑛) ≈

{

(−2𝑢(𝑥1, 𝑡𝑛) + 2𝑢(𝑥2, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥1, 𝑡𝑛)(1 − 𝑢(𝑥1, 𝑡𝑛)); 𝑖 = 1

1

(Δ𝑥)2[𝑢(𝑥𝑖−1, 𝑡𝑛) − 2𝑢(𝑥𝑖 , 𝑡𝑛) + 𝑢(𝑥𝑖+1, 𝑡𝑛)] + 𝑢(𝑥𝑖 , 𝑡𝑛)(1 − 𝑢(𝑥𝑖, 𝑡𝑛)); 𝑖 = 2,… ,𝑁 − 1

(−2𝑢(𝑥𝑁 , 𝑡𝑛) + 2𝑢(𝑥𝑁−1, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥𝑁 , 𝑡𝑛)(1 − 𝑢(𝑥𝑁 , 𝑡𝑛)); 𝑖 = 𝑁

The above equation is at 𝑥 = −𝑀 (left boundary point), and as before, the equation represents the time

evolution of this boundary point at index 𝑖 = 𝑛.

The solution technique employed in this assignment is as follows: we approximate the spatial derivative with a

G1 type grid, we assign the number of nodes to the said grid (N), and the solution to the time derivative is

obtained by simply “marching” technique i.e. we just approximate the integral for the time using different

methods (namely forward Euler, backwards Euler, RK4, and Trapezoidal). The following schematic shows a

pictorial description on this summary.

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Each column vector (red), is a time snapshot of the entire spatial domain, and each row vector (green) is the

evolution of the solution at that given point 𝑥𝑖 for the entire time from 0 to 𝑡𝑓.

In order to apply different numerical techniques to the Fisher’s equation, we take the spatial approximation

from equation 2 , i.e. 𝑓(𝑥𝑖, 𝑡𝑛), and apply whatever numerical method we desire.

Lastly, here are the two initial conditions given for the assignment:

𝑔1(𝑥) = {0.990

|𝑥| ≤ 15|𝑥| > 15

𝑔2(𝑥) = {1

4cos2

𝜋𝑥

100

|𝑥| ≤ 5|𝑥| > 5

Also, in order to apply the implicit methods, we will need to compute the Jacobian matrix for the given

system.

The Jacobian for Fisher’s Equation problem is given as:

𝐽(𝑓) =𝜕𝑓

𝜕�⃗⃗�=

{

−

2

(Δ𝑥)2+ 1 − 2𝑢𝑘; 𝑖𝑓 𝑘 = 𝑙

1

(Δ𝑥)2; 𝑖𝑓 |𝑘 − 𝑙| = 1

0; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

Forward Euler Method (Results)

The definition of the method is given as:

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑦𝑛, 𝑡𝑛)

∴,we solve: 𝑢(𝑥𝑖, 𝑡𝑛+1) = 𝑢(𝑥𝑖, 𝑡𝑛) + ℎ𝑓(𝑥𝑖, 𝑡𝑛)

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Results using IC = 𝑔1(𝑥):

2D plot

3D plot

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2D plot

3D plot

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Runge-Kutta Method, 4th Order (Results)

The definition of the method is given as:

𝑢𝑛+1 = 𝑢𝑛 +ℎ

6(𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4)

𝑘1 = 𝑓(𝑡𝑛, 𝑢𝑛)

𝑘2 = 𝑓 (𝑡𝑛 +ℎ

2, 𝑢𝑛 +

ℎ𝑘12)

𝑘3 = 𝑓 (𝑡𝑛 +ℎ

2, 𝑢𝑛 +

ℎ𝑘22)

𝑘4 = 𝑓(𝑡𝑛 + ℎ, 𝑢𝑛 + ℎ𝑘3)

As our discretized system is “time independent”, the 𝑘 values simply become:

𝑘1 = 𝑓(𝑢𝑛) = 𝑢𝑛

𝑘2 = 𝑓 (𝑢𝑛 +ℎ𝑘12) = 𝑢𝑛 +

ℎ𝑘12

𝑘3 = 𝑓 (𝑢𝑛 +ℎ𝑘22) = 𝑢𝑛 +

ℎ𝑘22

𝑘4 = 𝑓(𝑢𝑛 + ℎ𝑘3) = 𝑢𝑛 + ℎ𝑘3

And since:

𝑢𝑛 = 𝑓(𝑥𝑖, 𝑡𝑛) ≈

{

(−2𝑢(𝑥1, 𝑡𝑛) + 2𝑢(𝑥2, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥1, 𝑡𝑛)(1 − 𝑢(𝑥1, 𝑡𝑛)); 𝑖 = 1

1

(Δ𝑥)2[𝑢(𝑥𝑖−1, 𝑡𝑛) − 2𝑢(𝑥𝑖 , 𝑡𝑛) + 𝑢(𝑥𝑖+1, 𝑡𝑛)] + 𝑢(𝑥𝑖 , 𝑡𝑛)(1 − 𝑢(𝑥𝑖, 𝑡𝑛)); 𝑖 = 2,… ,𝑁 − 1

(−2𝑢(𝑥𝑁 , 𝑡𝑛) + 2𝑢(𝑥𝑁−1, 𝑡𝑛))

(Δ𝑥)2+ 𝑢(𝑥𝑁 , 𝑡𝑛)(1 − 𝑢(𝑥𝑁 , 𝑡𝑛)); 𝑖 = 𝑁

We are numerically solving the following system: 𝑢(𝑥𝑖, 𝑡𝑛+1 ) = 𝑢(𝑥𝑖, 𝑡𝑛) +ℎ

6(𝑘1 + 2𝑘2 + 2𝑘3 + 𝑘4)

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2D plot

3D plot

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2D plot

3D plot

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Trapezoidal Method (Results) The definition of the method is given as:

𝑦𝑛+1 = 𝑦𝑛 +1

2ℎ[𝑓(𝑦𝑛, 𝑡𝑛) + 𝑓(𝑦𝑛+1, 𝑡𝑛+1)]

However, since we have a nonlinear system, we need to solve for 𝑢𝑛+1 using Newton’s Method. For the n-

dimensional case, the Newton’s Method is given as:

�̃�𝑛+1 = 𝑢𝑛 − 𝐽−1(𝑓(𝑢𝑛))𝑓(𝑢𝑛)

The Jacobian is defined as:

𝐽(𝑓) =𝜕𝑓

𝜕�⃗⃗�=

{

−

2

(Δ𝑥)2+ 1 − 2𝑢𝑘; 𝑖𝑓 𝑘 = 𝑙

1

(Δ𝑥)2; 𝑖𝑓 |𝑘 − 𝑙| = 1

0; 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒

The computational challenge with such an approach is that we cannot choose ℎ and Δ𝑥 with the same degree

of freedom as we were allowed in the explicit methods. This is due to the fact that the Jacobian matrix must

be a square matrix in order for the inverse to exist. Therefore, the grid size for both 𝑡 and 𝑥 has to be the

same.

The following results were obtained, using Newton’s method with a tolerance of 10−4. The general shape of

the result matches that of the two explicit methods discussed above.

The approximation, �̃�𝑛+1, obtained from Newton’s Method is implemented in the following manner:

𝑢𝑛+1 = 𝑢𝑛 +1

2ℎ[𝑢𝑛 + �̃�𝑛+1]

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2D Plot

3D Plot

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2D Plot

3D Plot

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Backwards Euler Method (Results) The definition of the method is given as:

𝑦𝑛+1 = 𝑦𝑛 + ℎ𝑓(𝑦𝑛+1, 𝑡𝑛+1)

The same discussion of nonlinearity and accuracy from Trapezoidal method applies to Backwards Euler

method as well. The only different here is the manner of implementation of the approximation, �̃�𝑛+1. For

backwards Euler Method, the results are given as:

𝑢𝑛+1 = 𝑢𝑛 + ℎ�̃�𝑛+1


2D Plot

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3D Plot


2D Plot

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3D Plot

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Forward Euler Method (MATLAB Code) %Code for Forward Euler Method

clear

clc

%Initializing constants, parameters, and data structures

N = 200; %Number of space points

M = 50; %Space endpoint

t0 = 0; %Initial time

tf = 20; %Final time

h = (tf-t0)/(N-1); %Delta t

t = t0:h:tf; %Time vector

dx = (2*M)/(N-1); %Delta x

x = -M:dx:M; %Space vector

g1 = zeros(length(x),1); %IC 1 Vector


u = zeros(N,length(t)); %Solution Matrix

%Create IC g1

for i = 1:N

if abs(x(i))<= 15

g1(i,1) = 0.99;

end

end

%Create IC g2

for i = 1:N

if abs(x(i))<= 5

g2(i,1) = 0.25*(cos((pi*x(i))/10))^2;

end

end

%---------------Solving using Forward Euler and IC 1----------------------%

u(:,1) = g1(:,1);

%i = 1

for n = 1:length(t)-1

fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n));

u(1,n+1)=u(1,n)+h*(fn);

end

%i = 2,3,...,N-1


fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n));

u(2:N-1,n+1)=u(2:N-1,n)+h*(fn);

end

%i = N


fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n));

u(N,n+1)=u(N,n)+h*(fn);

end

%Plot solutions using IC g1

figure1 = figure;

plot(x,u(:,1),'black','linewidth',2)

hold on

%This part produces profile for each second starting at 1 to 20

for i = 2:length(t)

if rem(t(i),1) == 0

plot(x,u(:,i),'r')

end

end

legend('t = 0','t > 0','location','northeastoutside')

title('Solution using IC = g_1, plotted every 1 second','fontsize',16)

xlabel('x','fontsize',16)

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ylabel('u(x,t_{fixed})','fontsize',16)

grid on

axis tight

hold off

%------------Solving using Forward Euler and IC 2-------------------------%

u(:,1) = g2(:,1);

%i = 1


fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n));

u(1,n+1)=u(1,n)+h*(fn);

end

%i = 2,3,...,N-1


fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n));

u(2:N-1,n+1)=u(2:N-1,n)+h*(fn);

end

%i = N


fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n));

u(N,n+1)=u(N,n)+h*(fn);

end


figure2 = figure;


hold on


for i = 2:length(t)

if rem(t(i),1) == 0

plot(x,u(:,i),'r')

end

end





grid on

axis tight

hold off

Runge-Kutta Method, 4th Order (MATLAB Code) %Code for RK4 Method

clear

clc

%Initializing constants, parameters, and data structures

N = 200; %Number of space points

M = 50; %Space endpoint

t0 = 0; %Initial time

tf = 20; %Final time

h = (tf-t0)/(N-1); %Delta t

t = t0:h:tf; %Time vector

dx = (2*M)/(N-1); %Delta x

x = -M:dx:M; %Space vector



u = zeros(N,length(t)); %Solution Matrix

%Create IC g1

for i = 1:N

if abs(x(i))<= 15

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g1(i,1) = 0.99;

end

end

%Create IC g2

for i = 1:N

if abs(x(i))<= 5

g2(i,1) = 0.25*(cos((pi*x(i))/10))^2;

end

end

%----------------------Solving using RK4 and IC 2-------------------------%

u(:,1) = g1(:,1);

%i = 1


k1 = (1/dx^2)*(-2*u(1,n) + 2*u(2,n)) + u(1,n)*(1-u(1,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(1,n+1)=u(1,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end

%i = 2,3,...,N-1


for i = 2:N-1

k1 = (1/dx^2)*(u(i-1,n)-2*u(i,n)+u(i+1,n)) + u(i,n)*(1-u(i,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(i,n+1)=u(i,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end

end

%i = N


k1 = (1/dx^2)*(2*u(N-1,n)-2*u(N,n)) + u(N,n)*(1-u(N,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(N,n+1)=u(N,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end


figure1 = figure;


hold on


for i = 2:length(t)

if rem(t(i),1) == 0

plot(x,u(:,i),'r')

end

end





grid on

axis tight

hold off

k1 = 0; k2 = 0; k3 = 0; k4 = 0;

u = zeros(N,length(t));

%----------------------Solving using RK4 and IC 2-------------------------%

u(:,1) = g2(:,1);

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%i = 1


k1 = (1/dx^2)*(-2*u(1,n) + 2*u(2,n)) + u(1,n)*(1-u(1,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(1,n+1)=u(1,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end

%i = 2,3,...,N-1


for i = 2:N-1

k1 = (1/dx^2)*(u(i-1,n)-2*u(i,n)+u(i+1,n)) + u(i,n)*(1-u(i,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(i,n+1)=u(i,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end

end

%i = N


k1 = (1/dx^2)*(2*u(N-1,n)-2*u(N,n)) + u(N,n)*(1-u(N,n));

k2 = k1 + 0.5*h.*k1;

k3 = k2 + 0.5*h.*k2;

k4 = k3 + h.*k3;

u(N,n+1)=u(N,n) + h/6*(k1 + 2.*k2 + 2.*k3 + k4);

end


figure2 = figure;


hold on


for i = 2:length(t)

if rem(t(i),1) == 0

plot(x,u(:,i),'r')

end

end





grid on

axis tight

hold off

Trapezoidal Method (MATLAB Code) %Code for Trapezoidal Method clear clc

%Initializing constants, parameters, and data structures N = 200; %Number of space points M = 50; %Space endpoint t0 = 0; %Initial time tf = 20; %Final time tol = 10^-4; %Tolerance for Newton's Method h = (tf-t0)/(N-1); %Delta t t = t0:h:tf; %Time vector dx = (2*M)/(N-1); %Delta x x = -M:dx:M; %Space vector g1 = zeros(length(x),1); %IC 1 Vector g2 = zeros(length(x),1); %IC 2 Vector u = zeros(N,length(t)); %Solution Matrix J = zeros(N,length(t)); %Jacobian Matrix

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u1 = 0; u2 = 0;

%Create IC g1 for i = 1:N if abs(x(i))<= 15 g1(i,1) = 0.99; end end

%Create IC g2 for i = 1:N if abs(x(i))<= 5 g2(i,1) = 0.25*(cos((pi*x(i))/10))^2; end end

%-----------------Solving using Trapaziodal and IC 1----------------------% u(:,1) = g1(:,1);

%Deriving the Jacobian for n1 = 1:N for k1 = 1:length(t) if n1 == k1 J(n1,k1) = 1/dx^2 + 1 - 2*u(n1,k1); else if abs(k1 - n1) == 1 J(n1,k1) = 1/dx^2; else J(n1,k1) = 0; end end end end

Ji = inv(J);

%i = 1 for n = 1:length(t)-1 fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(2,n+1)*fn; u1 = u2; end u(1,n+1)=u(1,n)+(h/2)*(fn + u1); end

%i = 2,3,...,N-1 for n = 1:length(t)-1 fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(3:N,n+1).*fn; u1 = u2; end u(2:N-1,n+1)=u(2:N-1,n)+(h/2)*(fn + u1); end

%i = N for n = 1:length(t)-1 fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(N,n+1)*fn; u1 = u2; end u(N,n+1)=u(N,n)+(h/2)*(fn + u1);

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end

%Plot solutions using IC g1 figure1 = figure; plot(x,u(:,1),'black','linewidth',2) hold on %This part produces profile for each second starting at 1 to 20 for i = 2:length(t) if rem(i,10) == 0 plot(x,u(:,i),'r') end end legend('t = 0','t > 0','location','northeastoutside') title('Solution using IC = g_1, plotted every 1 second','fontsize',16) xlabel('x','fontsize',16) ylabel('u(x,t_{fixed})','fontsize',16) grid on axis tight hold off

%-----------------Solving using Trapezoidal and IC 2----------------------% u(:,1) = g2(:,1);


Ji = inv(J);

%i = 1 for n = 1:length(t)-1 fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(1,n+1)*fn; u1 = u2; end u(1,n+1)=u(1,n)+(h/2)*(fn + u1); end

%i = 2,3,...,N-1 for n = 1:length(t)-1 fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(2:N-1,n+1).*fn; u1 = u2; end u(2:N-1,n+1)=u(2:N-1,n)+(h/2)*(fn + u1); end

%i = N for n = 1:length(t)-1 fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n)); u1 = fn;

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while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(1,n+1)*fn; u1 = u2; end u(N,n+1)=u(N,n)+(h/2)*(fn + u1); end


Backwards Euler Method (MATLAB Code) %Code for Backwards Euler Method clear clc

%Initializing constants, parameters, and data structures N = 200; %Number of space points M = 50; %Space endpoint t0 = 0; %Initial time tf = 20; %Final time tol = 10^-4; %Tolerance for Newton's Method h = (tf-t0)/(N-1); %Delta t t = t0:h:tf; %Time vector dx = (2*M)/(N-1); %Delta x x = -M:dx:M; %Space vector g1 = zeros(length(x),1); %IC 1 Vector g2 = zeros(length(x),1); %IC 2 Vector u = zeros(N,length(t)); %Solution Matrix J = zeros(N,length(t)); %Jacobian Matrix u1 = 0; u2 = 0;

%Create IC g1 for i = 1:N if abs(x(i))<= 15 g1(i,1) = 0.99; end end

%Create IC g2 for i = 1:N if abs(x(i))<= 5 g2(i,1) = 0.25*(cos((pi*x(i))/10))^2; end end

%------------Solving using backward Euler and IC 1------------------------% u(:,1) = g1(:,1);

%Deriving the Jacobian for n1 = 1:N

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for k1 = 1:length(t) if n1 == k1 J(n1,k1) = 1/dx^2 + 1 - 2*u(n1,k1); else if abs(k1 - n1) == 1 J(n1,k1) = 1/dx^2; else J(n1,k1) = 0; end end end end

Ji = inv(J);

%i = 1 for n = 1:length(t)-1 fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(2,n+1)*fn; u1 = u2; end u(1,n+1)=u(1,n)+(h)*(u1); end

%i = 2,3,...,N-1 for n = 1:length(t)-1 fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(3:N,n+1).*fn; u1 = u2; end u(2:N-1,n+1)=u(2:N-1,n)+(h)*(u1); end

%i = N for n = 1:length(t)-1 fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(N,n+1)*fn; u1 = u2; end u(N,n+1)=u(N,n)+(h)*(u1); end


Page 25 of 26

%------------Solving using backward Euler and IC 2------------------------% u(:,1) = g2(:,1);


Ji = inv(J);

%i = 1 for n = 1:length(t)-1 fn = (1/dx^2)*(-2*u(1,n) + 2*u(2,n))+u(1,n)*(1-u(1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(1,n+1)*fn; u1 = u2; end u(1,n+1)=u(1,n)+(h)*(u1); end

%i = 2,3,...,N-1 for n = 1:length(t)-1 fn = (1/dx^2).*(u(1:N-2,n)-2.*u(2:N-1,n)+u(3:N,n))+u(2:N-1,n).*(1-u(2:N-1,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(2:N-1,n+1).*fn; u1 = u2; end u(2:N-1,n+1)=u(2:N-1,n)+(h)*(u1); end

%i = N for n = 1:length(t)-1 fn = (1/dx^2)*(2*u(N-1,n)-2*u(N,n))+u(N,n)*(1-u(N,n)); u1 = fn; while(abs(norm(u1)-norm(u2))) >= tol u2 = u1 - Ji(1,n+1)*fn; u1 = u2; end u(N,n+1)=u(N,n)+(h)*(u1); end

%Plot solutions using IC g2 figure1 = figure; plot(x,u(:,1),'black','linewidth',2) hold on %This part produces profile for each second starting at 1 to 20 for i = 2:length(t) if rem(i,10) == 0 plot(x,u(:,i),'r') end end legend('t = 0','t > 0','location','northeastoutside') title('Solution using IC = g_2, plotted every 1 second','fontsize',16) xlabel('x','fontsize',16) ylabel('u(x,t_{fixed})','fontsize',16)

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grid on axis tight hold off

References

Salsa, Sandro. Partial Differential Equations in Action: From Modelling to Theory. 1st ed. Milan: Springer, 2008.

Debnath, Lokenath. Nonlinear Partial Differential Equations for Scientists and Engineers. 3rd ed. New York,

2012.

Documents

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