how to replace variable by another variable

Hn=-2*P/(Mu*exp(n^2*pi^2*T/Mu))

I want to replace T by t and want to do calculation

[Venkat Siddarth]

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]

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

Answer 1

[Walter Roberson]

if you are using the symbolic toolbox then subs()

[vikas singh]

if i want to give some numeric value to new variable then how to give

[Walter Roberson]

syms L Mu n P T t
Pi = sym(pi);
Hn=-2*P/(Mu*exp(n^2*Pi^2*T/Mu))

Hn = 

Hn_new = subs(Hn, T, t/L)

Hn_new = 

subs(Hn_new, t, (1:5).')

ans = 

[vikas singh]

thanks sir

[vikas singh]

it is a great help for me.

[vikas singh]

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.

[Walter Roberson]

subs(Hn_new, t, (5:5:1000).')

Answer 2

[vikas singh]

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)

[Walter Roberson]

I think you are getting some underflow, exp(-number) terms that evaluate in floating point to zero even though mathematically they are not zero.

[vikas singh]

suppose I have a variable z and it has 100 values for each time variable t (suppose t= 0,1,5, 10,15). suppose one variable cell V{} contains 100 value for different time, then how to minus z and V{} values

[vikas singh]

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

[Walter Roberson]

r.*f

Answer 3

[vikas singh]

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.

[Walter Roberson]

for n=1:M 
    for i=1:N
        if x(i)>=250 && x(i)<=350
            r(i)=w2;
        else
            r(i)=w1;
        end
        
    end

r is a vector of length N after that for i loop.

    if t(n)<=182.5
        f(n)=sin(2*pi*t(n)/365).^2;
    else
        f(n)=0;
    end 

f is a vector of length n after that if test -- it is being expanded as do more interations of for n

    f2=r.*f

r is 1 x N vector. f is a 1 x n vector. They can only be .* together if the changing variable n == N or n == 1 (in which case 1 x N .* 1 x 1 would be scalar expansion.)

[vikas singh]

thanks

[vikas singh]

I want to plot between x and -(z(n+1,i)).^0.5. how to get 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=(k*delt*H)/epc;
w1=1*86400*10^-8;
w2=2*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 
        
        if t(n)<=182.5
            f(n)=sin(2*pi*t(n)/365).^2;
            
        else
            f(n)=0;
        end 
        Q(n,i)=(2*r(i)*f(n)*mu)/aa;
    end 
end


for n=1:M
    z(n,1)=0;
    z(n,N)=0;
end
for i=1:N
    z(1,i)=0;
end
for n=1:M-1
    for i=2:N-1
        z(n+1,i)=z(n,i)+d*(z(n,i+1)-2*z(n,i)+z(n,i-1))+Q(n,i)*dt;
        
        
    end
end

[Walter Roberson]

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?

[vikas singh]

tell me for both the conditons as well