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How do I solve a second order ordinary differential equation with boundary conditions?

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Dorothy  Carter
Dorothy Carter 2017 年 4 月 16 日
回答済み: Sachin Kumar 2017 年 4 月 19 日
The equation I am trying to solve is c"+2/a*c'=c/0.4(0.1+c); In my code c is 'x' and t is 'a'. The boundary conditions are dc(0)=0 and c(1)=0 % this function breaks down the second order function into two first order % eq function dxdt=F(t,x) dxdt=[x(2); -2./t.*x(1)+t./(0.04.*(0.01+t))]; end% This function applies the boundary conditions. function res=fb(xa,xb) % xa is the derivative boundary condition and xb the normal boundary % condition res=[xa(2); xb(1)-1]; end % This is the code to solve bme=bvpinit([0.1 1],[0.1 1]); sol=bvp4c(@F, @fb, bme); plot(sol.y(1,:), sol.x); xlabel('a'); ylabel('Ss/S'); title('Plot for S~Km');
Can anyone verify my initial set up is correct and if so, how to just have values from 0 to 1 for my x-axis?

回答 (1 件)

Sachin Kumar
Sachin Kumar 2017 年 4 月 19 日
Below link has some examples on using bvp4c, you can verify with these examples:

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