Second order ODE - BVP
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Hi,
I am trying to obtain the solution of the following second order ODE but I struggle.
and 
on the interval [0, 0.5]. It is my understanding that 1) transoformation to the system of first order ODE and a Matlab bvp4c solver should be used. I wrote therefore the code:function bvp5
xlow=0; xhigh=0.5;
solinit = bvpinit(linspace(xlow,xhigh,10),[0 1]);
sol = bvp4c(@bvp5ode,@bvp5bc,solinit);
xint = linspace(xlow,xhigh);
Sxint = deval(sol,xint);
plot(xint,Sxint(1,:))
% -----------------------------------------------
function dydx = bvp5ode(x,y)
dydx = [ y(2) 0.64*y(1)-(2/x)*y(2) ];
% -----------------------------------------------
function res = bvp5bc(ya,yb)
res = [ ya(1)-0.2 yb(2) ];
I obtain the folelowing error: Unable to solve the collocation equations -- a
singular Jacobian encountered.
1) How to form a guess ? I dont have any idea ...
2) What is the problem with my equation or my code?
Best wishes,
采纳的回答
Alan Stevens
2021-2-1
I used an alternative method that makes use of Matlab's fzero function. You can also see one way of avoiding the problem at x = 0:
xlow=0; xhigh=0.5;
xspan = [xlow xhigh];
dydx0 = 0;
y0 = 0; % Initial guess for y(x=0)
% Use fzero to get value of y0 that makes y(x=0.5) = 0.1;
y0 = fzero(@(y0) y0val(y0, xspan), y0);
disp(['y(x=0) = ' num2str(y0)])
% Now use ode45 to integrate from x=0 to x=0.5 with good starting value
% for y(x=0)
Y0 = [y0 0];
[x, Y] = ode45(@odefn, xspan, Y0);
plot(x,Y(:,1)),grid
xlabel('x'),ylabel('y')
function Z = y0val(y0, xspan)
Y0 = [y0, 0];
[~, Y] = ode45(@odefn, xspan, Y0);
yend = Y(end,1);
Z = yend - 0.1;
end
function dYdx = odefn(x, Y)
y = Y(1); dydx = Y(2);
d2ydx2 = 0.64*y;
if dydx>0
d2ydx2 = d2ydx2 - (2/x)*dydx;
end
dYdx = [dydx;
d2ydx2];
end
3 个评论
Alan Stevens
2021-2-2
编辑：Alan Stevens
2021-2-2
I think the problem with using bpv4c is that it expects values of y(x=0) and y(x=0.5), but you have y'(x=0), not y(x=0). I'm not very familiar with bvp4c (I rarely use it), so I can't really help you further with it.
It's true that the shooting method requires a reasonable initial guess. Presumably your actual problem is much more complicated if the method is no use to you.
(NB In your analytical solution you have k = 6.4 and D = 0.1. This doesn't result in k/D = 0.64 !!!)
sko
2021-2-2
更多回答（1 个）
Alan Stevens
2021-2-1
编辑：Alan Stevens
2021-2-1
1 个投票
You start at xlow = 0, but your function bvp5ode has an x in the denominator. bvp4c doesn't like it when this is zero! If you set xlow = 0.001, say, then it runs.
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