Unable to convert expression containing symbolic variables into double array.
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I am following this algorithm (pic) and in Step 4, in the line "wij(i,2)=(1-lambda^2)*f(i*h)+(lambda^2/2)*(f((i+1)*h)+f((i-1)*h))+k*g(i*h);" I am getting the error "Unable to convert expression containing symbolic variables into double array. Apply 'subs' function first to substitute values for variables." I tried applying subs to f(i*h) and g(i*h), so it was wij(i,2)=(1-lambda^2)*(subs(f(x),x,(i*h)))+(lambda^2/2)*(subs(f(x),x,((i+1)*h)))+(subs(f(x),x,((i-1)*h)))+k*(subs(g(x),x,(i*h))); but I still get the same error. Please help
This is my whole program:
%Problem:
l=1; %endpoint
T=1; %maximum time
alpha = 2; %constant
m=10;
N=20;
syms f(x) g(x,t)
f(x)=sin(pi*x);
g(x)=(sin(pi*x))*(cos(2*pi*t));
%STEP1
h=l/m;
k=T/N;
lambda=k*alpha/h;
%STEP2
for j=2:N
wij(1,j)=0; %wij(0,j)
wij(m,j)=0;
end
%STEP3
wij(1,1)=f(0); %wij(0,0)
wij(m,1)=f(l);
%STEP4
for i=2:m-1
wij(i,1)=subs(f(x),x,(i*h)); %wij(i,0)
wij(i,2)=(1-lambda^2)*(subs(f(x),x,(i*h)))+(lambda^2/2)*(subs(f(x),x,((i+1)*h)))+(subs(f(x),x,((i-1)*h)))+k*(subs(g(x),x,(i*h)));
end
%STEP5
for j=2:N-1
for i=2:m-1
wij(i,j+1)=2*(1-lambda^2)*w(i,j)+lambda^2*(wij(i+1,j)+wij(i-1,j))-wij(i,j-1);
end
end
%STEP6
for j=1:N
t=j*k;
for i=1:m
x=i*h;
fprintf( '%.1f %.1f %.10f',x,t,wij(i,j))
end
end
Akzeptierte Antwort
%Problem:
l=1; %endpoint
T=1; %maximum time
alpha = 2; %constant
m=10;
N=20;
syms f(x) g(x,t)
f(x)=sin(pi*x);
g(x)=(sin(pi*x))*(cos(2*pi*t));
%STEP1
h=l/m;
k=T/N;
lambda=k*alpha/h;
%STEP2
wij = zeros(m,N,'sym');
for j=2:N
wij(1,j)=0; %wij(0,j)
wij(m,j)=0;
end
%STEP3
wij(1,1)=f(0); %wij(0,0)
wij(m,1)=f(l);
%STEP4
for i=2:m-1
wij(i,1)=subs(f(x),x,(i*h)); %wij(i,0)
wij(i,2)=(1-lambda^2)*(subs(f(x),x,(i*h)))+(lambda^2/2)*(subs(f(x),x,((i+1)*h)))+(subs(f(x),x,((i-1)*h)))+k*(subs(g(x),x,(i*h)));
end
%STEP5
for j=2:N-1
for i=2:m-1
wij(i,j+1)=2*(1-lambda^2)*wij(i,j)+lambda^2*(wij(i+1,j)+wij(i-1,j))-wij(i,j-1);
end
end
%STEP6
for j=1:N
t=j*k;
for i=1:m
x=i*h;
fprintf( '%.1f %.1f %s\n',x, t, char(vpa(wij(i,j),10)));
end
end
2 Kommentare
LightFury Yeji
am 7 Feb. 2021
Walter Roberson
am 7 Feb. 2021
%STEP6
stopping_loop = false;
for j=1:N
t=j*k;
for i=1:m
x=i*h;
if ~isempty(symvar(wij(i,j))
stopping_loop = true;
break;
end
fprintf( '%.1f %.1f %.10f\n',x, t, double(wij(i,j)));
end
if stopping_loop; break; end
end
Weitere Antworten (1)
toka hakim
am 21 Mai 2021
syms x;
f=cos(x)-x;
x1=0.5;
x2=pi/4;
for i=3:6
x(i)=x(i-1)-(f(x(i-1))*(x(i-1)-x(i-2)))/f(x(i-1)-f(x(i-2)));
vpa(x(i),10);
end
1 Kommentar
Walter Roberson
am 21 Mai 2021
It is not clear how this is an Answer to the question that was asked?
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am 7 Feb. 2021
am 21 Mai 2021
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