two-point boundary value problem + plot

ateq alsaadi

2015 4 2

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Hello,

I am trying to solve this question: Solve the two-point boundary value problem utilizing LU factorization for tridiagonal matrices. Make two plots: one with h = 1/4 and the other with h = 1/16. On each plot, graph the numerical solution with circles connected by lines and the real solution. Include a legend on each graph.

−y'' = 25 sin(πx), 0 ≤ x ≤ 1, y(0) = 0, y(1) = 1

I am using this code and I let n=4. How can I fix my code to plot both n=4 and n=16 together.

 %-y''=25sin(pix), y_0=0, y_1=1
clc
r=inline('25.*sin(pi.*x)','x');
alpha=0; beta=1;
n=4;
h=1/n;
x=[alpha+h:h:(n-1)*h];
A=2*eye(n-1)-diag(ones(n-2,1),-1)-diag(ones(n-2,1),1);
A=(1/h^2)*A;
b=r(x);
b(n-1)=b(n-1)+beta/h^2;
b(1)=b(1)+alpha/h^2;
w=A\b'
soln=[alpha, w' beta]
xval=[alpha, x, beta]
plot(xval,soln,'o-')

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nhagen 2015-4-2

在 MATLAB Online 中打开

Hi, is that what you need ?

%-y''=25sin(pix), y_0=0, y_1=1
clc
r=inline('25.*sin(pi.*x)','x');
alpha=0; beta=1;
n=[4 16];
color='br'; %blue for n=4 and red for n=16
figure
hold on
for i=1:2
h=1/n(i);
x=[alpha+h:h:(n(i)-1)*h];
A=2*eye(n(i)-1)-diag(ones(n(i)-2,1),-1)-diag(ones(n(i)-2,1),1);
A=(1/h^2)*A;
b=r(x);
b(n(i)-1)=b(n(i)-1)+beta/h^2;
b(1)=b(1)+alpha/h^2;
w=A\b';
soln=[alpha, w' beta];
xval=[alpha, x, beta];
plot(xval,soln,'o-','Color',color(i))
legend_text{i}=['n= ',num2str(n(i))];
end
legend(legend_text);