bvp4c setup for second order ODE
Mostrar comentarios más antiguos

Hello, I'm trying to solve for this second order ODE in steady state using bvp4c with the boundary conditions where at x=0, C_L=1 and x=100, C_L=0.
function bvp4c_test
%Diffusion coefficient
D_ij= 1*10^-6;
%Rate constants
k_1 = 0.00193;
k_2 = 0.00255;
k_3 = 4.09;
d_1 = 0.007;
d_2 = 3.95*10^-5;
d_3 = 2.26;
%Equilibrium constants
K_1 = 3.63;
K_2 = 0.0155;
K_3 = 0.553;
K_4 = 9.01;
K_i = 0.139;
%Total Receptors
N_T =1.7;
solinit = bvpinit(linspace(0,10,11),[1 0]);
sol = bvp4c(@odefcn,@twobc,solinit);
plot(sol.x,sol.y(1,:))
function dC_Ldx = odefcn(C_L,D_ij,k_1,k_2,k_3,d_1,d_2,d_3,K_1,K_2,K_3,K_4,K_i,N_T)
dC_Ldx = zeros(2,1);
dC_Ldx(1) = C_L(2);
N_r = (-(1+(C_L./K_2))+...
sqrt(((1+(C_L./K_2)).^2)-4.*((2./K_4).*(1+(C_L./K_2).*(1+(1./K_i).*(1+(C_L./K_3))))).*(-N_T)))./...
(2.*(2./K_4).*(1+(C_L./K_2).*(1+(1./K_i).*(1+(C_L./K_3)))));
N_R = (C_L./K_1).*N_r;
N_r2 = (1./K_4).*(N_r).^2;
N_rR = (C_L./(2.*K_2.*K_4)).*(N_r).^2;
N_pR = (C_L./(2.*K_2.*K_4.*K_i)).*(N_r).^2;
N_R2 = ((C_L.^2)./(K_2.*K_3.*K_4.*K_i)).*(N_r).^2;
R_L = ((2.*d_1.*(N_T-N_r-N_r2-N_rR-N_pR-N_R2))-(2.*k_1.*C_L.*N_r))+...
((2.*d_2.*(N_T-N_r-N_R-N_r2-N_pR-N_R2)))-((k_2.*C_L.*(N_T-N_r-N_R-N_rR-N_pR-N_R2)))+...
(d_3.*(N_T-N_r-N_R-N_r2-N_rR-N_pR))-(2.*k_3.*C_L.*(N_T-N_r-N_R-N_r2-N_rR-N_R2));
dC_Ldx(2) = -R_L./D_ij;
end
function res = twobc(ya,yb)
res = [ ya(1)-1; yb(1) ];
end
end
When I run it, I encounter the error stating Attempted to access C_L(2); index out of bounds because numel(C_L)=1.
| | _Error in bvp4c_test/odefcn dC_Ldx(1) = C_L(2);
Error in bvparguments testODE = ode(x1,y1,odeExtras{:});
Error in bvp4c [n,npar,nregions,atol,rtol,Nmax,xyVectorized,printstats] = ...
Error in bvp4c_test sol = bvp4c(@odefcn,@twobc,solinit);_||
Where am I going wrong? I am still very new to MATLAB and appreciate any help that I can get. Thank you!
Respuestas (1)
Torsten
el 18 de Jun. de 2018
sol = bvp4c(@(x)odefcn(x,D_ij,k_1,k_2,k_3,d_1,d_2,d_3,K_1,K_2,K_3,K_4,K_i,N_T),@twobc,solinit);
9 comentarios
Jennifer Yang
el 18 de Jun. de 2018
Torsten
el 18 de Jun. de 2018
Sorry, should be
sol = bvp4c(@(x,y)odefcn(x,y,D_ij,k_1,k_2,k_3,d_1,d_2,d_3,K_1,K_2,K_3,K_4,K_i,N_T),@twobc,solinit);
and
function dC_Ldx = odefcn(x,C_L,D_ij,k_1,k_2,k_3,d_1,d_2,d_3,K_1,K_2,K_3,K_4,K_i,N_T)
Jennifer Yang
el 18 de Jun. de 2018
Torsten
el 19 de Jun. de 2018
R_L is a 2-element vector. Thus the assignment
dC_Ldx(2) = -R_L./D_ij;
does not make sense.
Jennifer Yang
el 19 de Jun. de 2018
Torsten
el 19 de Jun. de 2018
Following your formula for N_r and R_L, you will have to replace C_L in these expressions by C_L(1).
Jennifer Yang
el 19 de Jun. de 2018
Torsten
el 19 de Jun. de 2018
function dy = odefcn(x,y,D_ij,k_1,k_2,k_3,d_1,d_2,d_3,K_1,K_2,K_3,K_4,K_i,N_T)
C_L = y(1);
dC_Ldx = y(2);
N_r = (-(1+(C_L./K_2))+...
sqrt(((1+(C_L./K_2)).^2)-4.*((2./K_4).*(1+(C_L./K_2).*(1+(1./K_i).*(1+(C_L./K_3))))).*(-N_T)))./...
(2.*(2./K_4).*(1+(C_L./K_2).*(1+(1./K_i).*(1+(C_L./K_3)))));
N_R = (C_L./K_1).*N_r;
N_r2 = (1./K_4).*(N_r).^2;
N_rR = (C_L./(2.*K_2.*K_4)).*(N_r).^2;
N_pR = (C_L./(2.*K_2.*K_4.*K_i)).*(N_r).^2;
N_R2 = ((C_L.^2)./(K_2.*K_3.*K_4.*K_i)).*(N_r).^2;
R_L = ((2.*d_1.*(N_T-N_r-N_r2-N_rR-N_pR-N_R2))-(2.*k_1.*C_L.*N_r))+...
((2.*d_2.*(N_T-N_r-N_R-N_r2-N_pR-N_R2)))-((k_2.*C_L.*(N_T-N_r-N_R-N_rR-N_pR-N_R2)))+...
(d_3.*(N_T-N_r-N_R-N_r2-N_rR-N_pR))-(2.*k_3.*C_L.*(N_T-N_r-N_R-N_r2-N_rR-N_R2));
dy = zeros(2,1);
dy(1) = dC_Ldx;
dy(2) = -R_L/D_ij;
end
Utkarsh Tiwari
el 22 de Oct. de 2020
Wow this was incredibly helpful as I faced a similar problem. Thank you!
Categorías
Más información sobre Boundary Value Problems en Centro de ayuda y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!