PDE Parabolic coefficient
조회 수: 2 (최근 30일)
이전 댓글 표시
Hi !
This is the first time I use matlab, so I have a bad level.
Parabolic solver has this form:
u1=parabolic(u0,tlist,b,p,e,t,c,a,f,d)
It is said on this page (<http://www-rohan.sdsu.edu/doc/matlab/toolbox/pde/assempde.html>) that "*c*" coefficient can be a 2by2 matrix.
Here is my little code :
[p,e,t]=initmesh('squareg');
[p,e,t]=refinemesh('squareg',p,e,t);
u0=zeros(size(p,2),1);
ix=find(sqrt(p(1,:).^2+p(2,:).^2)<0.4);
u0(ix)=ones(size(ix));
tlist=linspace(0,0.1,20);
c=[1 2; 3 4];
f=10;
u1=parabolic(u0,tlist,'squareb1',p,e,t,c,0,f,1);
pdeplot(p,e,t,'xydata',u1(:,end))
Impossible to run it. I get this error message :
??? Error using ==> times
Matrix dimensions must agree.
Error in ==> pdeasmc at 18
c3=((cc1.*g1x.*g2x+cc2.*g1y.*g2y).*ar);
Error in ==> assema at 112
ks1=pdeasmc(it1,it2,it3,np,ar,x,y,sd,u,ux,uy,time,g1x,g1y,g2x,g2y,g3x,g3y,c);
Error in ==> parabolic at 85
[K,M,F]=assema(p,t,c,a,f,time);
Error in ==> ex_parab at 9
u1=parabolic(u0,tlist,'squareb1',p,e,t,c,0,f,1);
Why is there a matrix dimension problem ? I work on D=2 space, so the gradient is just a column vector, with 2 lines, and 'c' is just a 2by2 matrix in my code.
Does some can help me to understand and fixe it ? Thanks in advance !!
댓글 수: 0
채택된 답변
Bill Greene
2012년 4월 18일
Hi,
PDE Toolbox accepts c-matrices in many different forms depending on whether the matrix is symmetric, unsymmetric or diagonal and also whether the coefficients vary as a function of x. This is discussed more on page 5-21 of the manual.
For your scalar case with a full, unsymmetric matrix, c would be: c=[1 3 2 4]'; Note, this is 4 rows with the coefficients entered by column.
Regards,
Bill
추가 답변 (0개)
참고 항목
카테고리
Help Center 및 File Exchange에서 Eigenvalue Problems에 대해 자세히 알아보기
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!