Identify an unrecognized function or variable
古いコメントを表示
clear all; close all;
syms T U t w
a=0.3; M=1; h0=1; h1=1; k0=1; k1=1; p=0.5; d=0.5; Z=10;
%t=0:1:10; %w=0.75;
H=@(s) (p*Z-d-exp(-s))*k1;
K=@(r,g) p*Z*k1-(d+exp(-r))*k1-2*exp(a*r).*g;
for n=1:10
y=int(H(U),0,10);
yy= h0 + ((1-a)./M)*H(t).*h1 + (a./M)*y;
h1=yy;
end
for n=1:10
z=int(K(U,T),0,10);
zz= k0 + ((1-w)./M).*K(t,yy).*k1 + (w./M)*z;
k1=zz;
end
zzfcn = matlabFunction(zz)
t=0:0.1:10;
w = 0.1:0.1:0.75;
[T,W] = ndgrid(t,w);
figure
surf(T, W, zzfcn(T,W))
xlabel('t')
ylabel('\omega')
zlabel('k(t)')
set(gca, 'ZScale','log')
colormap(turbo)
colorbar
回答 (1 件)
Note that the function zzfcn has three input variables: @(T,t,w)
Here is the corrected code:
clear all; close all;
syms T U t w
a=0.3; M=1; h0=1; h1=1; k0=1; k1=1; p=0.5; d=0.5; Z=10;
%t=0:1:10; %w=0.75;
H=@(s) (p*Z-d-exp(-s))*k1;
K=@(r,g) p*Z*k1-(d+exp(-r))*k1-2*exp(a*r).*g;
for n=1:10
y=int(H(U),0,10);
yy= h0 + ((1-a)./M)*H(t).*h1 + (a./M)*y;
h1=yy;
end
for n=1:10
z=int(K(U,T),0,10);
zz= k0 + ((1-w)./M).*K(t,yy).*k1 + (w./M)*z;
k1=zz;
end
zzfcn = matlabFunction(zz)
t=0:0.1:10;
w = 0.1:0.1:0.75;
[T,W] = ndgrid(t,w);
figure
surf(T, W, zzfcn(T, T,W))
xlabel('t')
ylabel('\omega')
zlabel('k(t)')
set(gca, 'ZScale','log')
colormap(turbo)
colorbar
11 件のコメント
a=linspace(0.1,0.1,0.95)
w=0.75;
a is empty, w has only one element. You can't use "plot3" in this case.
a=linspace(0.1,0.1,0.95)
What values for "a" do you want to set here ? At the moment, you set no values.
Mathew
2025 年 3 月 10 日
Walter Roberson
2025 年 3 月 10 日
Maybe
a = 0.1:0.1:0.95 ;
??
However, this will end at 0.9 not at 0.95
Walter Roberson
2025 年 3 月 10 日
for i=1:length(t)
y=int(H(U),0,10);
yy= h0 + ((1-w)./M).*H(t).*h1 + (w./M).*y;
h1=yy;
end
t is a vector.
yy involves H(t) where t is the entire t vector. So yy returns a vector the same size as the row vector t.
for i=1:length(t)
z=int(K(yy,t),0,10);
zz= k0 + ((1-w)./M).*K(yy,t).*k1 + (w./M).*z;
k1=zz;
end
yy is a vector, and t is a vector, and these are the same size. K(yy,t) is working with vectors.
K=@(g,s) p*Z*k1-(d+exp(-s))*k1-2*exp(a.*s).*g;
Working with two vectors the same size is okay in K, with a vector the same size being returned.
So the int(K(yy,t),0,10) is a vector. The loop is not changing yy or t during the iterations, so z will be the same each iteration.
Then zz involves computing with the vector K(yy,t) together with the vector z, so zz is going to end up being a vector the same size as t.
Except...
K involves computation with the vector a and the vector s and those vectors are not the same size, so you would generate an error in the computation exp(a.*s). Unless, that is, you deliberately made a the same size as t by using something such as
a = linspace(0.1, 0.95, numel(t));
Still, the loops over length(t) that use the entire vector t are suspicious. It would make more sense if you were to be doing something such as
for i=1:length(t)
y=int(H(U),0,10);
yy= h0 + ((1-w)./M).*H(t(i)).*h1 + (w./M).*y;
h1=yy;
end
and
for i=1:length(t)
z=int(K(yy,t(i)),0,10);
zz= k0 + ((1-w)./M).*K(yy,t(i)).*k1 + (w./M).*z;
k1=zz;
end
a=linspace(0.1,0.95,10);
?
It's important that t and a are both row vectors of equal size (here:10). Otherwise your code will not work.
Mathew
2025 年 3 月 14 日
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