System of mixed PDE with 2 variables using FDM and RK4

조회 수: 5 (최근 30일)
University Glasgow
University Glasgow 2022년 8월 3일
답변: University Glasgow 2022년 8월 9일
I want solve the attached mixed pdes in 2 variables with FDM and R4K. See my trial below: I received and error that 'N' is uncrecognised.
% parameters.
k1 = 6*10^(-12);
eta1 = 0.0240;
apha3 = -0.001104;
gama1 = 0.1093;
d = 0.0002;
%thetab = 0.0001;
% activity parameter
xi = 0.1;
%%%% Simplify parameters
A = (apha3*k1)/gama1;
B = A/k1;
%%%% Discretize xspace
z_vec = linspace(0, d, N);
dz = z_vec(2) - z_vec(1);
%%%% Discretize t
dt = (dz^2)/(2);
%%%% Discretize t
dtrk4 = (dz^2)/(2);
trk4_vec = 0: dtrk4:10;
% Allocate memory for u
thetark4_mat = zeros(length(z_vec), length(trk4_vec));
vrk4_mat = zeros(length(z_vec), length(trk4_vec));
%%%% IC
thetark4_mat(1,:) = 0; % Left end of pipe
thetark4_mat(end,:) = d; % right end of the pipe
vrk4_mat(1,:) = 0; % Left end of pipe
vrk4_mat(end,:) = d; % right end of the pipe
for tdz = 1:length(trk4_vec)-1
k11 = Derivative(thetark4_mat(:, tdz));
k12 = Derivative(vrk4_mat(:, tdz));
k21 = Derivative(thetark4_mat(:, tdz) + k11*dtrk4/2);
k22 = Derivative(vrk4_mat(:, tdz) + k12*dtrk4/2);
k31 = Derivative(thetark4_mat(:, tdz) + k21*dtrk4/2);
k32 = Derivative(vrk4_mat(:, tdz) + k22*dtrk4/2);
k41 = Derivative(thetark4_mat(:, tdz) + k31*dtrk4);
k42 = Derivative(vrk4_mat(:, tdz) + k32*dtrk4);
phi = (1/6)*(k11+ 2*k21 + 2*k31 +k41);
phi1 = (1/6)*(k12+ 2*k22 + 2*k32 +k42);
thetark4_mat(:, tdz+1) = thetark4_mat(:, tdz) + phi*dtrk4;
vrk4_mat(:, tdz+1) = vrk4_mat(:, tdz) + phi*dtrk4;
end
%%%% Plot this
[tt, zz] = meshgrid(trk4_vec, z_vec);
mesh(zz, tt, thetark4_mat)
xlabel('z')
ylabel('time')
zlabel('theta')
figure()
mesh(zz, tt, vrk4_mat)
xlabel('z')
ylabel('time')
zlabel('v')
function dthetadt = Derivative(theta, v)
%global x_vec dx k
dthetadt = 0*theta;
dvdt = 0*theta;
z_vec = linspace(0,10,10);
% parameters.
k1 = 6*10^(-12);
eta1 = 0.0240;
apha3 = -0.001104;
gama1 = 0.1093;
d = 0.0002;
%thetab = 0.0001;
% activity parameter
xi = 0.1;
%%%% Simplify parameters
A = (apha3*k1)/gama1;
B = A/k1;
dz = z_vec(2) - z_vec(1);
for j = 2:length(z_vec)-1
dvdt(j) = 0;
dthetadt(j) = A/(apha3*dz^2)*(theta(j+1) -2*theta(j) + theta(j-1)) + ...
B/(2*dz)*(v(j+1) - v(j-1));
dvdt(j) = A/(2*dz^3)*(theta(j+2) +2*theta(j+1) -2*theta(j-1)- theta(j-2))+ ...
(eta1 - B)/(dz^2)*(v(j+1) -2*v(j) + v(j-1)) - z/(2*dz)*(theta(j+1) - theta(j-1));
end
end
  댓글 수: 20
Torsten
Torsten 2022년 8월 6일
theta and v both have only 5 elements.
But in these loops
for i=(N+4):(2*N)
rhsode(i,1)= C/(2*h^3)*(v(i+1) + 2*v(i) - 2*v(i-1))+ ...
+ G/(h^3)*(theta(i+2) -2*theta(i+1) + 2*theta(i-1) - theta(i-2))...
+xi/(2*h)*(theta(i+1) - theta(i-1)); %uses central difference for third derivative for theta
end
rhsode(2*N+1,1)=0;
for i = (5+N):(2*N + 1)
rhsode(i,1)= C/(2*h^3)*(v(i+1) + 2*v(i) - 2*v(i-1))+ ...
+ G/(h^3)*(-theta(i-2) +3*theta(i-1) -3*theta(i) + theta(i+1))...
+xi/(2*h)*(theta(i+1) - theta(i-1)); %involves special third derivative
end
almost every index of theta and v exceeds 5.
Use two arrays rhsode_theta and rhsode_v both arrays having 5 elements. Then indexing should be simple.
At the end, return
rhsode = [rhsode_theta;rhsode_v]

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답변 (2개)

University Glasgow
University Glasgow 2022년 8월 8일
% I still don't this error despite having 5 by 5 entries in both you theta and v:
Error in counts_4_IjuptilK_080822 (line 41)
[t,y] = ode15s(@lcode1,tspan,u0, options);
% parameters.
k1 = 6*10^(-12);
eta1 = 0.0240;
apha3 = -0.001104;
gama1 = 0.1093;
d = 0.0002;
N = 4;
% Boundary Conditions
Phi = 0;
% activity parameter
xi = 0.1;
h = d/N; %% step size
%%%% Simplify parameters
A = k1/gama1;
B = apha3/k1;
C = eta1 - B;
G = apha3*A;
Theta = 0.0001;
z=linspace(0,d,N+1);
theta0 = Theta*sin(pi*z/d)
v0 = zeros(1,N+1)
M1=eye(N+1,N+1);
M1(1,1)=0;
M1(N+1,N+1)=0;
M2=zeros(N+1,N+1);
M=[M1 M2;M2 M2];
u0 = [theta0'; v0'];
tspan = [0 10];
options = odeset('Mass',M,'RelTol',1e-4,'AbsTol',1e-6);
%[t,y] = ode15s(@lcode1,tspan,[u0;v0], options);
[t,y] = ode15s(@lcode1,tspan,u0, options);
%theta = y(:,1:5);
%v = y(:,1:5);
% Plot the solution.
function rhsode = lcode1(~,y, k1, eta1, apha3, d, Phi, xi, A, B, C, G)
theta = y(1:5);
v = y(6:10);
rhsode = [theta(1)-Phi
(A/(h^2))*(theta(3)+2*theta(2)-theta(1))+(B/(2*h))*(v(3)-v(1))
(A/(h^2))*(theta(4)+2*theta(3)-theta(2))+(B/(2*h))*(v(4)-v(2))
(A/(h^2))*(theta(5)+2*theta(4)-theta(3))+(B/(2*h))*(v(5)-v(3))
theta(5)-Phi
v(N+1)
(G/(h^3))*(-theta(5) + 3*theta(2)-3*theta(3)+theta(4))+(C/(2*h^2))*(v(3) +2*v(2) - v(1)) + (xi/(2*h))*(theta(3)-theta(1))
(G/(h^3))*(-theta(2) + 3*theta(3)-3*theta(4)+theta(5))+(C/(2*h^2))*(v(4) -2*v(3) - v(2)) + (xi/(2*h))*(theta(4)-theta(2))
(G/(h^3))*(-theta(3) + 3*theta(4)-3*theta(5)+theta(2))+(C/(2*h^2))*(v(5) -2*v(4) - v(3)) + (xi/(2*h))*(theta(5)-theta(3))
v(5)];
end
  댓글 수: 1
Torsten
Torsten 2022년 8월 8일
I don't understand why you deleted the call to "fsolve" in an earlier code.
Here is the error again I already commented on.
% parameters.
k1 = 6*10^(-12);
eta1 = 0.0240;
apha3 = -0.001104;
gama1 = 0.1093;
d = 0.0002;
N = 4;
% Boundary Conditions
Phi = 0;
% activity parameter
xi = 0.1;
h = d/N; %% step size
%%%% Simplify parameters
A = k1/gama1;
B = apha3/k1;
C = eta1 - B;
G = apha3*A;
Theta = 0.0001;
z=linspace(0,d,N+1);
theta0 = Theta*sin(pi*z/d)
v0 = zeros(1,N+1)
M1=eye(N+1,N+1);
M1(1,1)=0;
M1(N+1,N+1)=0;
M2=zeros(N+1,N+1);
M=[M1 M2;M2 M2];
u0 = [theta0'; v0'];
tspan = [0 10];
options = odeset('Mass',M,'RelTol',1e-4,'AbsTol',1e-6);
%[t,y] = ode15s(@lcode1,tspan,[u0;v0], options);
[t,y] = ode15s(@(t,y)lcode1(t,y,k1, eta1, apha3, d, Phi, h, xi, A, B, C, G),tspan,u0, options);
%theta = y(:,1:5);
%v = y(:,1:5);
% Plot the solution.
function rhsode = lcode1(~,y, k1, eta1, apha3, d, Phi, h, xi, A, B, C, G)
theta = y(1:5);
v = y(6:10);
rhsode = [theta(1)-Phi
(A/(h^2))*(theta(3)+2*theta(2)-theta(1))+(B/(2*h))*(v(3)-v(1))
(A/(h^2))*(theta(4)+2*theta(3)-theta(2))+(B/(2*h))*(v(4)-v(2))
(A/(h^2))*(theta(5)+2*theta(4)-theta(3))+(B/(2*h))*(v(5)-v(3))
theta(5)-Phi
v(1)
(G/(h^3))*(-theta(5) + 3*theta(2)-3*theta(3)+theta(4))+(C/(2*h^2))*(v(3) +2*v(2) - v(1)) + (xi/(2*h))*(theta(3)-theta(1))
(G/(h^3))*(-theta(2) + 3*theta(3)-3*theta(4)+theta(5))+(C/(2*h^2))*(v(4) -2*v(3) - v(2)) + (xi/(2*h))*(theta(4)-theta(2))
(G/(h^3))*(-theta(3) + 3*theta(4)-3*theta(5)+theta(2))+(C/(2*h^2))*(v(5) -2*v(4) - v(3)) + (xi/(2*h))*(theta(5)-theta(3))
v(5)];
end

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University Glasgow
University Glasgow 2022년 8월 9일
Hi, thank you for your help. The code is working well for 4 counts.
But when I tried to generalised, I received this error: Error using daeic12 This DAE appears to be of index greater than 1.
Below is my code:
clear
clc
close all
% parameters.
global k1 eta1 alpha3 gamma1 d N Phi xi h A B C G
k1 = 6*10^(-12);
eta1 = 0.0240;
xi = -0.1; % activity parameter
alpha3 = -0.001104;
gamma1 = 0.1093;
d = 0.0002;
N = 4;
% Boundary Conditions
Phi = 0;
%%%% Simplify parameters
A = k1/gamma1;
B = alpha3/k1;
C = eta1 - B;
G = alpha3*A;
% step size
h = d/N;
% range of z
z=linspace(0,d,N+1);
% initials
Theta = 0.0001;
theta0 = Theta*sin(pi*z/d);
v0 = zeros(1,N+1);
theta0_int=theta0(1:N);
v0_int=v0(1:N);
u0 = [theta0_int'; v0_int'];
% Matrix M
M1=eye(N,N);
M2=zeros(N,N);
M=[M1 M2;M2 M2];
% t span
tspan = [0 4];
% ode solver
options = odeset('Mass',M,'RelTol',1e-4,'AbsTol',1e-6);
[t,y] = ode15s(@lcode1,tspan,u0, options);
% Extract the solution for theta and v
%theta = [Phi*ones(length(t), 1) y(:,1:N-1) Phi*ones(length(t), 1)];
%v = [zeros(length(t), 1) y(:,N:2*(N-1)) zeros(length(t), 1)];
% Right hand side of the ODEs: F(U)
function rhsode = lcode1(t,y)
global Phi xi h A B C G N
% initialize theta and v
theta = y(1:N);
v = y(N+1:2*N);
% theta equation
% discretise equation:(A/(h^2))*(theta(i+1)+2*theta(i)-theta(i-1))+(B/(2*h))*(v(i+1)-v(i-1))
for i=2:N
theta(N+1) = 0;
v(N+1) = 0;
rhsode(i,1)=(A/(h^2))*(theta(i+1)+2*theta(i)-theta(i-1))+(B/(2*h))*(v(i+1)-v(i-1)); % internal nodes for 0<z<d
end
% v equation
% for i = N: corresponds to i=9-5 =4 =N
rhsode(N+2,1) = C/(h^2)*(v(N-1) + 2*v(N-2) - v(N-3))+ ...
+ G/(h^3)*(-theta(N-3) +3*theta(N-2) - 3*theta(N-1) + theta(N))...
+xi/(2*h)*(theta(N-1) - theta(N-3)); %involves special third derivative
% for i = 8, this corresponds i = 8-5 =3
% let j =N+4-(N+1) = 3
for j=3:N-1
theta(N+1) = 0;
v(N+1) = 0;
rhsode(j,1)= C/(h^2)*(v(j+1) + 2*v(j) - v(j-1))+ ...
+ G/(2*h^3)*(v(j+2) -2*theta(j+1) + 2*theta(j-1) - theta(j-2))...
+ xi/(2*h)*(theta(j+1) - theta(j-1)); %uses central difference for third derivative for theta
end
% for i = N: corresponds to i=9-5 =4 =N
theta(N+1) = 0;
v(N+1) = 0;
rhsode(2*N,1) = C/(h^2)*(v(N+1) - 2*v(N) + 2*v(N-1))+ ...
+ G/(h^3)*(-theta(N-2) +3*theta(N-1) -3*theta(N) + theta(N+1))...
+xi/(2*h)*(theta(N+1) - theta(N-1)); %involves special third derivative
end

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