Nodal basis function 1D

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mohamad hoseini
mohamad hoseini am 25 Okt. 2018
Kommentiert: KSSV am 25 Okt. 2018
Hello all, I coded a nodal basis function for 1D element from [-1,1]. the code is below:
close all; clc; clearvars;
n=10;
x = linspace(-1,1,n);
for i=1:n
a= x(i);
for j=1:n
b(j)=a.^(j-1);
end
v(i,:)=b';
end
vinv=inv(v);
for i=1:n
k=zeros(1,n);
k(i)=1;
f=vinv*k'
p(:,i)=f;
end
for i=1:n
g=@(x) p(1,i)+p(2,i).*x+p(3,i).*x.^2+p(4,i).*x.^3+p(5,i).*x.^4+p(6,i).*x.^5+p(7,i).*x.^6+p(8,i).*x.^7+p(9,i).*x.^8+ ....
p(10,i).*x.^9;
legendInfo{i} = ['Phi ' num2str(i)];
fplot(g, [-1 1])
legend(legendInfo)
hold on;
end
The code works already but my problem is in last "for loop" where I calculated "g" as a function handle. I want to instead of adding the terms from 1 to 10, use an automated calculation. Now, if I want to change number of nodes (n) from 10 to 20 I have to add 10 additional terms by hand. Moreover, Does somebody knows a better way to calculate nodal basis function for 1D element? Great thanks,

Akzeptierte Antwort

KSSV
KSSV am 25 Okt. 2018
Read about polyval
n=10;
x = linspace(-1,1,n);
for i=1:n
a= x(i);
for j=1:n
b(j)=a.^(j-1);
end
v(i,:)=b';
end
vinv=inv(v);
for i=1:n
k=zeros(1,n);
k(i)=1;
f=vinv*k'
p(:,i)=f;
end
x = linspace(-1,1,1000) ;
for i=1:n
r = flipud(p(:,i)) ;
y = polyval(r,x) ;
legendInfo{i} = ['Phi ' num2str(i)];
plot(x,y)
legend(legendInfo)
hold on;
end
  4 Kommentare
mohamad hoseini
mohamad hoseini am 25 Okt. 2018
Thanks
KSSV
KSSV am 25 Okt. 2018
Thanks is accepting and voting the answer..:)

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