Filter löschen
Filter löschen

Where have I made mistakes in this code

1 Ansicht (letzte 30 Tage)
Lukgaf
Lukgaf am 22 Mär. 2018
Kommentiert: Walter Roberson am 22 Mär. 2018
%prob2.m
% u’’- 4u’ + 4u = exp(x)+ C;
% Subject to u(1) = 0; u(-1) = 0; C=(-4*exp)/(1+(exp)^2);
N = 40;
[D, x]=cheb(N); %%we call this function to get the Derivative matrix D
I = eye(N+1); % I is the identity matrix of size (N+1)
C=(-4*exp(x))/(1+(exp(x).^2)); %the value of the constant
A = D^2 - 4*D + 4*I; f = exp(x)+ C;
B = A(2:N,2:N); %to impose the boundary conditions
ff = f(2:N);
v = B\ff; %solve the BVP.
u = [0;v;0]; xx = -1:0.1:1;
u_exact = exp(xx)-(sinh(1)/sinh(2))*exp(2*xx)+ C/4;
plot(x,u); %plot the numerical solution
xlabel('x')
ylabel('u')
hold on
plot(xx,u_exact,'o'); % plot the exact solution
hold off
I have used the similar code to solve many questions, why is not working for this? Thanks. It shows this eror: Error
in prob2 (line 11)
v = B\ff; %solve the BVP.
  7 Kommentare
Lukgaf
Lukgaf am 22 Mär. 2018
>> prob2
Error using \
Matrix dimensions must agree.
Error in prob2 (line 11)
v = B\ff; %solve the BVP.
Jan
Jan am 22 Mär. 2018
Bearbeitet: Jan am 22 Mär. 2018
@Lukgaf: This is the line, which causes the error. But what is the message, which explains what the problem is?

Melden Sie sich an, um zu kommentieren.

Akzeptierte Antwort

Walter Roberson
Walter Roberson am 22 Mär. 2018
v = B\ff.'; %solve the BVP.
  2 Kommentare
Lukgaf
Lukgaf am 22 Mär. 2018
Thanks all. But why is the exact soluation has the similar problem.
Matrix dimensions must agree.
Error in prob2 (line 13)
u_exact = exp(xx)-(sinh(1)/sinh(2))* exp(2*xx)+ C/4;
Walter Roberson
Walter Roberson am 22 Mär. 2018
You would get that error if xx is a vector or array and C is a vector or array of different size than xx (though starting in R2016b some size combinations can be meaningfully computed as if you had used bsxfun)

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Linear Algebra finden Sie in Help Center und File Exchange

Community Treasure Hunt

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

Start Hunting!

Translated by