how to replace variable by another variable

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vikas singh
vikas singh 2023-2-8
评论： vikas singh 2023-4-29
Hn=-2*P/(Mu*exp(n^2*pi^2*T/Mu))
I want to replace T by t and want to do calculation
  2 个评论
Venkat Siddarth
Venkat Siddarth 2023-2-8
Hi @vikas singh,
Can you elaborate the query,on whats the result you are getting when you replace "T" with "t" in the equation itself i.e
Hn=-2*P/(Mu*exp(n^2*pi^2*t/Mu))?
vikas singh
vikas singh 2023-2-10
actually I have to replace T by t/L and again i have to give different value to t for calculation. and some for loop is also involve in the equation for n

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回答（3 个）

Walter Roberson
Walter Roberson 2023-2-8
if you are using the symbolic toolbox then subs()
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vikas singh
vikas singh 2023-2-13
can you please tell me if I want to give value of t as 5, 10, 15, 20,25,.......,1000 then how to give here in the code. thanks in advance.

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vikas singh
vikas singh 2023-2-15
when I'm using the following code to evaluate z value . I'm getting the Q value and all the value after Q is zero. I m not getting where I'm wrong.
clc;
clear all;
close all;
l=10000;
x=0:100:l;
t=[0;1;5;10;15;20;25];
H=30;
Mu=2.8e-1;
D=31;
k=0.01132;
k1=l/H;
for n=1:100
a=2*H;
b=(1-(-1)^n*exp(-k*l))/((n*pi+(k^2*l^2)/(n*pi)));
c=1/(n*pi);
Cn(n)=a*(b-c);
end
s=0;
for n=1:100
d= sin(n*pi*x/l);
q= exp((-n^2*pi^2*t)/(Mu^2*l));
aa=Cn(n).*d.*q;
s=s+aa;
end
syms X T
for n=1:100
H0=@(X,T)Cn(n).*sin(n*pi*X).*exp(-n^2*pi^2*T/Mu)+H*X/(D-H)+H/(D-H);
y=diff(H0*diff(H0,X),X);
Q=int(y*sin(n*pi*X),X,0,1);
P=int(Q*exp(n^2*pi^2*T/Mu),T,0,T);
Hn=-2*P/(Mu*exp(n^2*pi^2*T/Mu));
H1=Hn.*sin(n*pi*X);
for i=1:size(t,1)
for j=1:length(x)
H=(subs(H1,{X,T},{x(j)/l,t(i,1)/(Mu*l)}));
ZZ(j,1)=double(H);
end
XX(i,1)=sum(ZZ);
clear ZZ
end
s1=((D-H)^2).*XX/l;
end
j=(H*x)/l;
z=s-j-s1
plot(x,z)
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vikas singh
vikas singh 2023-4-25
I want to multiply r and f value in the following code. how to do it.
clc;
clear all;
close all;
L=600;
T=365;
delt=0.025;
k=8.64;
aa=delt*k*(1+delt);
epc=0.3;
H=50;
mu=epc/(k*delt*H);
w1=86400*1*10^-8;
w2=8*86400*10^-8;
dx=1;
dt=1;
d=(mu*dt)/(dx^2);
N=L/dx +1;
M=T/dt +1;
for i=1:N
x(i)=0+(i-1)*dx;
end
for n=1:M
t(n)=0+(n-1)*dt;
end
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
end
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
end

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vikas singh
vikas singh 2023-4-25
编辑：Walter Roberson 2023-4-25
Ran in:
I'm writing the following but it is giving me zero and showing matrix dimension is not agree
clc;
clear all;
close all;
L=600;
T=365;
delt=0.025;
k=8.64;
aa=delt*k*(1+delt);
epc=0.3;
H=50;
mu=epc/(k*delt*H);
w1=86400*1*10^-8;
w2=8*86400*10^-8;
dx=1;
dt=1;
d=(mu*dt)/(dx^2);
N=L/dx +1;
M=T/dt +1;
for i=1:N
x(i)=0+(i-1)*dx;
end
for n=1:M
t(n)=0+(n-1)*dt;
end
for n=1:M
for i=1:N
if x(i)>=250 && x(i)<=350
r(i)=w2;
else
r(i)=w1;
end
end
if t(n)<=182.5
f(n)=sin(2*pi*t(n)/365).^2;
else
f(n)=0;
end
f2=r.*f
end
f2 = 1×601
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Arrays have incompatible sizes for this operation.
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Walter Roberson
Walter Roberson 2023-4-28
I want to plot between x and -(z(n+1,i)).^0.5
Do you mean that you have an independent variable 1:N on the x axis, and you want to treat the variable x and that particular expression as dependent variables to be drawn and you want to fill the area between the two lines?
Or is your x variable to be treated as the x axis and you want to draw -(z(n+1,i)).^0.5 even though that appears to be independent of x?

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