how can i run this code need help?

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Akhtar Jan 2022-3-9

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https://ww2.mathworks.cn/matlabcentral/answers/1667274-how-can-i-run-this-code-need-help

评论： Jan 2022-3-10

%its about Solving PDEs by Asymmetric MQ Collocation  the above code is already given in the book which %i have 
%pasted below i havn't much experience in MATALAB thats why i shared it here...
%Neumann Boundary Conditions  
%The function f is set to f(x, y) = -2(2y3 - 3y2 + 1) + 6(1 - x2)(2y - 1) and Dirichlet boundary conditions
%of u(0, y) = 2y2 - 3y2 + 1 and u(1, y) = 0 are applied as well as Neumann boundary conditions of
%∂u/∂y = 0, along y = 0 and y = 1.
%A shape parameter of ε = 2.5 results in an evaluation matrix to be inverted that has a very high condition %number 
%κ(H) ≈ 4.6e19.   
clc
Nb = 22; % number of boundary point s
Np = 60; % t o t a l number of c ent e r , i n t e r i o r and boundary
centers = dlmread(centersUnitCircle60.txt ,'_');
x = centers(:,1);y = centers(:,2);
u = 65./(65+(x-0.2).^2+(y+0.1).^2 );% e x ac t s o l u t i o n
f = 130./(65+(x-0.2).^2+(y+0.1).^2 ).^3.*(2.*x-0.4).^2 - ...
    260./(65+(x-0.2).^2+(y+0.1).^2).^2+...
    130./(65+(x-0.2).^2+(y+0.1).^2).^3.*(2.*y+0.2).^2;
H = zeros (Np,Np ); rx = zeros (Np,Np ); ry = zeros (Np,Np); r = zeros (Np,Np );
f ( 1 :Nb) = u ( 1 :Nb ) ; % Di r i c h l e t Boundary c ond i t i ons
for i =1:Np
    for j =1:Np
        rx ( i , j ) = x ( i ) -x ( j ) ;
        ry ( i , j ) = y ( i ) - y ( j ) ;
        r ( i , j ) = sqrt ( rx ( i , j )^2 + ry ( i , j )^2 ) ;
    end
end
index = 1 ;
for shape =1.0: -0.01: 0.1;
    H( 1 :Nb , : ) = mq( r ( 1 :Nb , : ) , shape ) ; % enf or c e boudary c ond i t i ons
    Hxx = mqDerivatives( r (Nb+1:Np , : ) , rx (Nb+1:Np , : ) , shape , 2 ) ; % enf or c e PDE
    Hyy = mqDerivatives( r (Nb+1:Np , : ) , ry (Nb+1:Np , : ) , shape , 2 ) ;
    H(Nb+1:Np , : ) = Hxx + Hyy ; % e v a l u a t i on matrix
    kappa ( index ) = cond(H) ;
    warning off
    lambda = H\ f ; % expansion c o e f f i c i e n t s v i a Gaussian e l imi na t i on
    warning on
    B = mq( r , shape ) ; % system matrix
    uh = B*lambda ; % approximate s o l u t i o n
    e r ( index ) = norm(u?uh , i n f ) ;
    sh ( index ) = shape ;
    index = index + 1 ;
end
semilogy( sh , er , 'b' ) , xlabel ’ shape parameter ’ , ylabel ’max e r r o r ’
figure , semilogy( sh , kappa ) , xlabel ’ shape parameter ’ , ylabel ’ \kappa (H) ’    

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Walter Roberson 2022-3-9

编辑：Walter Roberson 2022-3-9

在 MATLAB Online 中打开

clc
Nb = 22; % number of boundary point s
Np = 60; % t o t a l number of c ent e r , i n t e r i o r and boundary
centers = dlmread('centersUnitCircle60.txt' ,'_');
x = centers(:,1);y = centers(:,2);
u = 65./(65+(x-0.2).^2+(y+0.1).^2 );% e x ac t s o l u t i o n
f = 130./(65+(x-0.2).^2+(y+0.1).^2 ).^3.*(2.*x-0.4).^2 - ...
260./(65+(x-0.2).^2+(y+0.1).^2).^2+...
130./(65+(x-0.2).^2+(y+0.1).^2).^3.*(2.*y+0.2).^2;
H = zeros (Np,Np ); rx = zeros (Np,Np ); ry = zeros (Np,Np); r = zeros (Np,Np );
f ( 1 :Nb) = u ( 1 :Nb ) ; % Di r i c h l e t Boundary c ond i t i ons
for i =1:Np
for j =1:Np
rx ( i , j ) = x ( i ) -x ( j ) ;
ry ( i , j ) = y ( i ) - y ( j ) ;
r ( i , j ) = sqrt ( rx ( i , j )^2 + ry ( i , j )^2 ) ;
end
end
index = 1 ;
for shape =1.0: -0.01: 0.1;
H( 1 :Nb , : ) = mq( r ( 1 :Nb , : ) , shape ) ; % enf or c e boudary c ond i t i ons
Hxx = mqDerivatives( r (Nb+1:Np , : ) , rx (Nb+1:Np , : ) , shape , 2 ) ; % enf or c e PDE
Hyy = mqDerivatives( r (Nb+1:Np , : ) , ry (Nb+1:Np , : ) , shape , 2 ) ;
H(Nb+1:Np , : ) = Hxx + Hyy ; % e v a l u a t i on matrix
kappa ( index ) = cond(H) ;
warning off
lambda = H\f ; % expansion c o e f f i c i e n t s v i a Gaussian e l imi na t i on
warning on
B = mq( r , shape ) ; % system matrix
uh = B*lambda ; % approximate s o l u t i o n
er( index ) = norm(u - uh , inf ) ;
sh( index ) = shape ;
index = index + 1 ;
end
semilogy( sh , er , 'b' ) , xlabel 'shape parameter' , ylabel 'max error'
figure , semilogy( sh , kappa ) , xlabel 'shape parameter', ylabel '\kappa (H)'   

Walter Roberson 2022-3-9

编辑：Walter Roberson 2022-3-9

在 MATLAB Online 中打开

shape = 2.5;
N = 30;
Np = N^2;
[X,Y] = meshgrid(linspace(0,1,N),linspace(0,1,N));
x = reshape(X,N^2,1);
y = reshape(Y,N^2,1);
f = -2*(2*y.^3-3*y.^2+1)+6*(1-x.^2).*(2*y-1);
u = (1-x.^2).*(2*y.^3-3*y.^2+1); % exact solution
H =  zeros(Np,Np);
rx = zeros(Np,Np);
ry = zeros(Np,Np);
r = zeros(Np,Np);
dirichletBCs = find(x==0 | x==1); % identify boundry and interior centers
neumannBCs =   find ((y==0 | y==1)& ~(x==0 | x==1));
interior = find(x~=0 & x ~ = 1 & y~ = 0 & y~ = 1 );
f(dirichletBCs)=u(dirichletBCs);
f(neumannBCs)= 0;
for i =1:Np
for j =1:Np
rx( i , j ) = x( i )-x ( j ) ;
ry( i , j ) = y( i )-y ( j ) ;
r( i , j ) = sqrt( rx( i , j )^2 + ry( i , j )^2 );
end
end
H(dirichletBCs,:)=mq(r(dirichletBCs,:),shape); % e v a l u a t i on matrix
H(neumannBCs,:)= mqDerivatives(r(neumannBCs,:),ry(neumannBCs,:),shape,1);
Hxx = mqDerivatives(r(interior,:),rx(interior,:),shape,2);
Hyy = mqDerivatives(r(interior,:),ry(interior,:),shape,2);
H(interior,:)=Hxx+Hyy;
kappa=cond(H);
alpha=H\f;
B = mq( r , shape ) ; % system matrix
uh = B*alpha;
error = norm(uh-u,inf);
t = delaunay(x,y);trisurf(t,x,y,abs(uh-u));
xlabel  'x' , ylabel  'y' colormap('Summer')

Akhtar Jan 2022-3-10

编辑：Akhtar Jan 2022-3-10

在 MATLAB Online 中打开

%its about Solving PDEs by Asymmetric MQ Collocation  the above code is already given in the book which %i have 
%pasted below i havn't much experience in MATALAB thats why i shared it here...
%Neumann Boundary Conditions  
%The function f is set to f(x, y) = -2(2y3 - 3y2 + 1) + 6(1 - x2)(2y - 1) and Dirichlet boundary conditions
%of u(0, y) = 2y2 - 3y2 + 1 and u(1, y) = 0 are applied as well as Neumann boundary conditions of
%∂u/∂y = 0, along y = 0 and y = 1.
%A shape parameter of ε = 2.5 results in an evaluation matrix to be inverted that has a very high condition %number 
%κ(H) ≈ 4.6e19.   

Jan 2022-3-10

@Walter Roberson: Although the OP does not want to do it, I want to say: Thank you for spending your time for fixing the code.

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Answer 1

Image Analyst 2022-3-9

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1667274-how-can-i-run-this-code-need-help#answer_913989

The function is called 'linspace', not 'linespace'.

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Akhtar Jan 2022-3-10

%its about Solving PDEs by Asymmetric MQ Collocation the above code is already given in the book which %i have

%pasted below i havn't much experience in MATALAB thats why i shared it here...

%Neumann Boundary Conditions

%The function f is set to f(x, y) = -2(2y3 - 3y2 + 1) + 6(1 - x2)(2y - 1) and Dirichlet boundary conditions

%of u(0, y) = 2y2 - 3y2 + 1 and u(1, y) = 0 are applied as well as Neumann boundary conditions of

%∂u/∂y = 0, along y = 0 and y = 1.

%A shape parameter of ε = 2.5 results in an evaluation matrix to be inverted that has a very high condition %number

%κ(H) ≈ 4.6e19.

Jan 2022-3-10

编辑：Jan 2022-3-10

@Akhtar Jan: Your code contains to many spaces. Your description contains a lot of % characters. It is a good idea to spend the time to post clean messages, if you want to motivate others to help you.

You have been told already, that statements as "but not.....run" are not useful. Post the error message you see, or even better: read the message and try to fix the specific problem.

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