Why do I receive error using BVP5C "The boundary condition function BCFUN should return a column vector of length 1"?

My code is as following, aiming to solve a first order differential equation.
xmesh = linspace(0,8.6,10);
solinit = bvpinit(xmesh,@guess);
sol = bvp5c(@bvpfcn, @bcfcn, solinit); % Recived the error warning here!!!
plot(sol.x, sol.y, '-o')
function dydx = bvpfcn(x,y) % equation to solve
np0 = 1.3861*10^8; a = 10.2041; ni = 5*10^6; Mx = 10; ne = 5388520.4418; n = np0+ne/a;
Y0 = 2.19537; Ym = -0.87074;
A=2*ni/n; B=a*Mx^2; C=np0/n; D=a*ne/n; E=2*np0/n; F=exp(Ym-Y0); G=exp(Ym/a); % parameters
dydx = -(A*B*((1-2*y/B)^0.5-(1-2*Ym/B)^0.5)+ ...
C*(exp(y-Y0)*(1-erf((y-Ym)^0.5))+F*(2/pi^0.5*(y-Ym)^0.5-1))+ ...
D*(exp(y/a)*(1-erf(((y-Ym)/a)^0.5)+G*(2/pi^0.5*((y-Ym)/a)^0.5-1)))+ ...
E*(exp(y-Y0)*erf((y-Ym)^0.5)-2/pi^0.5*F*(y-Ym)^0.5))^0.5;
end
%--------------------------------
function res = bcfcn(ya,yb) % boundary conditions Y(0)=2 Y(zm)=-0.8
res = [ya(1)-2
yb(1)+0.8];
end
%--------------------------------
function g = guess(x) % initial guess for y
g = -sin(x);
end

 採用された回答

Torsten
Torsten 2024 年 11 月 20 日
移動済み: Torsten 2024 年 11 月 20 日
You have one first-order ODE, but two boundary conditions. That's mathematically impossible - you can impose only one condition. After having done this, switch to "ode45" as solver instead of "bvp5c".

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