bvp4c setup for second order ODE

2 次查看(过去 30 天)
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!

回答(1 个)

Torsten
Torsten 2018-6-18
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 个评论
Torsten
Torsten 2018-6-19
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
Utkarsh Tiwari 2020-10-22
Wow this was incredibly helpful as I faced a similar problem. Thank you!

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Ordinary Differential Equations 的更多信息

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by