Hyperbolic Equation Solution by a Second Order Upwind Approximation by MATLAB

Let's assume that the temperature distribution inside a pipe is governed by the following one dimensional unsteady hyperbolic partial-differential-equation

T_t + u T_x = 0

Consider negligible heat diffusion and the initial velocity u is, 0.1 cm/s. The boundary conditions are described below:

T(x, 0) = 200 x;       0 ≤ x ≤ 0.5

T(x, 0) = 200 (1 – x);       0.5 ≤ x ≤ 1

Develop an algorithm in MATLAB to solve this problem using the second-order approximation in time and the second-order upwind approximation in space. Consider, Δx = 0.05 cm; Δt = 0.05. 

The following MATLAB program develops the second order upwind method to solve the given one-dimensional unsteady hyperbolic convection equation.

% Hyperbolic equation - Convection equation: Upwind Method

close all;

clc;

% Input Properties from Problem

L=1.0; % (dimension in cm)

Lmax=5.0;

alpha=0.01;

u=0.1; % (dimension in cm/s)

Tt=10; % Simulation Time Period

% Mesh/Grid size

dx=0.05;

dt=0.05;

c=u*dt/dx; % Depends on dt values

d=alpha*((dt/dx)^2.0);

% Matrix Parameters

imax=(Lmax/dx)+1; % Array dimension along the width

tmax=(Tt/dt); % Array dimension along the height

ndim=imax-2;

tdim=tmax-2;

% Total no of iterations

iteration=10000;

% Convergence Criterion

tolerance=0.000001;

% Array and Variables

% a = Matrix of Coefficients A(i,j)

% b = Right Side Vector, b(i)

% x = Solution Vector, x(i)

% Initial Condition

f(:,1)=0.0;

for i=1:imax-2

    x(i)=(i-1)*dx;

   

    if x(i)<=1.0

        f(i,1)=0.0;

    end

    if x(i)>1.0 && x(i)<=1.5

        f(i,1)=200*(x(i)-1.0);

    end

    if x(i)>1.5 && x(i)<=2.0

        f(i,1)=200*(L-x(i)+1.0);

    end

    if x(i)>2.0

        f(i,1)=0;

    end

end

fprintf('imax = %f\n',imax);

fprintf('c=%f\n',c);

% Upwind Method

for t=1:tmax

    time(t+1)=(t)*dt;

    for i=2:imax-2

        % Second Order Approximation

       f(i,t+1)=f(i,t)-c*(f(i,t)-f(i-1,t))-0.5*c*(1.0-c)*(f(i,t)-2.0*f(i-1,t)+f(i-2,t));

    end

end

hold on

plot(x,f(:,1))

plot(x,f(:,find(time==1)),'o-')

plot(x,f(:,find(time==5)),'o-')

plot(x,f(:,find(time==10)),'o-')

xlabel('x, cm')

ylabel('Temperature, C')

legend('t=0s','t=1s','t=5s','t=10s')

title ('c=0.1')

Program Output:

Second-order approximation in time and second-order upwind approximation in space:

Hyperbolic Equation Solution by a First Order Upwind Approximation by MATLAB

Let's assume that the temperature distribution inside a pipe is governed by the following one dimensional unsteady hyperbolic partial-differential-equation:

T_t + u T_x = 0

Consider negligible heat diffusion and the initial velocity u is, 0.1 cm/s. The boundary conditions are described below:

T(x, 0) = 200 x;       0 ≤ x ≤ 0.5

T(x, 0) = 200 (1 – x);       0.5 ≤ x ≤ 1

Develop an algorithm in MATLAB to solve this problem using the first-order approximation in time and the first-order upwind approximation in space. Consider, Δx = 0.05 cm; Δt = 0.05. 

The following MATLAB program develops the first-order upwind method to solve the given one-dimensional unsteady hyperbolic convection equation.

% Hyperbolic equation - Convection equation: Upwind Method

close all;

clc;

% Input Properties from Problem

L=1.0; % (dimension in cm)

Lmax=5.0;

alpha=0.01;

u=0.1; % (dimension in cm/s)

Tt=10; % Simulation Time Period

% Mesh/Grid size

dx=0.05;

dt=0.05;

c=u*dt/dx; % Depends on dt values

d=alpha*((dt/dx)^2.0);

% Matrix Parameters

imax=(Lmax/dx)+1; % Array dimension along the width

tmax=(Tt/dt); % Array dimension along the height

ndim=imax-2;

tdim=tmax-2;

% Total no of iterations

iteration=10000;

% Convergence Criterion

tolerance=0.000001;

% Array and Variables

% a = Matrix of Coefficients A(i,j)

% b = Right Side Vector, b(i)

% x = Solution Vector, x(i)

% Initial Condition

f(:,1)=0.0;

for i=1:imax-2

    x(i)=(i-1)*dx;

   

    if x(i)<=1.0

        f(i,1)=0.0;

    end

    if x(i)>1.0 && x(i)<=1.5

        f(i,1)=200*(x(i)-1.0);

    end

    if x(i)>1.5 && x(i)<=2.0

        f(i,1)=200*(L-x(i)+1.0);

    end

    if x(i)>2.0

        f(i,1)=0;

    end

end

fprintf('imax = %f\n',imax);

fprintf('c=%f\n',c);

% Upwind Method

for t=1:tmax

    time(t+1)=(t)*dt;

    for i=2:imax-2

        % First Order Approximation

        f(i,t+1)=f(i,t)-c*(f(i,t)-f(i-1,t));

    end

end

hold on

plot(x,f(:,1))

plot(x,f(:,find(time==1)),'o-')

plot(x,f(:,find(time==5)),'o-')

plot(x,f(:,find(time==10)),'o-')

xlabel('x, cm')

ylabel('Temperature, C')

legend('t=0s','t=1s','t=5s','t=10s')

title ('c=0.1')

Program Output:

Adams-Bashforth Method with RK2 as a Start-up Scheme in MATLAB

Let's consider the following differential equation:

y'' + 10 y' + 500y = 0

Where, the initial conditions are,

y(0) = -0.025, and y'(0) = -1

We will solve this differential equation using a multi-step method, Adams-Bashforth, where second order Runge-Kutta approach (RK2) is added as a start-up scheme in the algorithm. 

MATLAB Program:

The following MATLAB program implements the Adams-Bashforth method with initialization with RK2 method.

close all;

clc;

h = 0.02;     % Step size

Tmax = 0.5;    % Maximum time

N = Tmax / h;  % Maximum number of steps

alpha=0.5;

t = linspace(0,0.5,N+1);  % Time range

% Analytical solution of the differential equation

y_real = -(((9.*sqrt(19))/760).*exp(-5.*t).*sin(5.*sqrt(19).*t)) - ((1/40).*exp(-5.*t).*cos(5.*sqrt(19).*t));

plot(t,y_real);

hold on

%Numerical solution

f=@(t,y) [y(2); -500*y(1)-10*y(2)]; % Governing system of equations

% Initial Conditions

Y = [-0.025; -1];

% Initialization with second order Runge-Kutta method

k1 = h.*f(t(1),Y(:,1));

    k2 = h.*f(t(1)+alpha.*h, Y(:,1)+alpha.*k1);

    Y(:,2) = Y(:,1) + (1-1/2/alpha).*k1 + k2/2/alpha;

% Second Order Adams-Bashforth method steps

for i=2:N

    Y(:,i+1) = Y(:,i) + 3/2*h*f(t(i),Y(:,i)) - h/2*f(t(i-1),Y(:,i-1));

end

plot(t,Y(1,:),'o:')

legend('Exact Solution','Adams-Bashforth Solution','Location','NorthEast')

title('When h = 0.02')

xlabel('t')

ylabel('y')

Program Output:

The following plot shows the numerical and analytical solution of the differential equation with two different step sizes with respect to time.

Leapfrog Method with RK2 as a Start-up Scheme in MATLAB

Let's consider the following differential equation:

y'' + 10 y' + 500y = 0 Where, the initial conditions are,

y(0) = -0.025, and y'(0) = -1

We will solve this differential equation using a multi-step method, Leapfrog, where second order Runge-Kutta approach (RK2) is added as a start-up scheme in the algorithm. The following MATLAB program implements the Leapfrog method with initialization with RK2 method.

MATLAB Program:

close all;

clc;

h = 0.02;     % Step size

Tmax = 0.5;    % Maximum time

N = Tmax / h;  % Maximum number of steps

alpha=0.5;

t = linspace(0,0.5,N+1);  % Time range

% Analytical solution of the differential equation

y_real = -(((9.*sqrt(19))/760).*exp(-5.*t).*sin(5.*sqrt(19).*t)) - ((1/40).*exp(-5.*t).*cos(5.*sqrt(19).*t));

plot(t,y_real);

hold on

%Numerical solution

f=@(t,y) [y(2); -500*y(1)-10*y(2)]; % Governing system of equations

% Initial Conditions

Y = [-0.025; -1];

% Initialization with second order Runge-Kutta method

k1 = h.*f(t(1),Y(:,1));

    k2 = h.*f(t(1)+alpha.*h, Y(:,1)+alpha.*k1);

    Y(:,2) = Y(:,1) + (1-1/2/alpha).*k1 + k2/2/alpha;

% Leapfrog method steps

for i=2:N

    Y(:,i+1) = Y(:,i-1) + 2.*h.*f(t(i),Y(:,i));

end

plot(t,Y(1,:),'o:')

legend('Exact Solution','Leapfrog Solution','Location','NorthEast')

title('When h = 0.001')

xlabel('t')

ylabel('y')

Program Output:

The following plot shows the numerical and analytical solution of the differential equation with respect to time.

Implicit Euler Method by MATLAB to Solve an ODE

In this example, an implementation of the Implicit Euler approach by MATLAB program to solve an ordinary differential equation (ODE) is presented. Let's consider a differential equation, which is defined as,

dv/dt = p(t) v + q(t)

Where,

p(t) = 5(1+t)

and,

q(t) = (1+t)e-t

The initial value is, v(0) = 1; and the time period is 0 < t < 10.

Implicit Euler approach is unconditionally stable. The implementation of Implicit Euler scheme may be represented as,

v_n+1 = (v_n + hq_n+1) / (1 + hp_n+1)

The following MATLAB program implements this scheme:

MATLAB program:

close all;
clc;
y1(1) = 0; % Initial condition
h = 0.1; % Time step
t1(1) = 0;
i = 1;

% Implementation of the Implicit Euler approach
while (t1(i) <= 30)
    q=(1+t1(i))*exp(-t1(i));
    p= 5*(1+t1(i));
       
    y1(i+1) = (y1(i)+(h*q))/(1+h*p);
    t1(i+1) = t1(i) + h;
    i = i + 1;
end
plot(t1,y1,'-o')
hold on


Program Output:
The following plot shows the progression of the solution as the time increases.