how can I write a function of another function of two variables

How can I wtite

where are constant.

in such a way I can evaluate vectors in it, For example ,where

Lt=linespace(0,1,10) and Lt=linespace(0,10,10)

Answer 1

[Star Strider]

Create the ρ function and then call it in the μ function —

rho = @(t,z) something;

mu = @(t,z) c .* log(rho(t,z)*b ./ (1 - rho(t,z)*b)) + rho(t,z)*b./(1-rho(t,z)*b);

Example —

b = rand

b = 0.4246

c = rand

c = 0.0166

rho = @(t,z) t./z;
mu = @(t,z) c .* log(rho(t,z)*b ./ (1 - rho(t,z)*b)) + rho(t,z)*b./(1-rho(t,z)*b);

Li = linspace(0, 1, 10);
Lz = linspace(0, 10, 10);

Mu = mu(Li, Lz)

Mu = 1x10

NaN -0.0072 -0.0072 -0.0072 -0.0072 -0.0072 -0.0072 -0.0072 -0.0072 -0.0072

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.

[Maryam Albuhayr]

Thank you for your fast response.

but I am actually don't have a specific expression for , but I am using a numerical method and the value of it will be taken from the previous step.

[Star Strider]

My pleasure!

Can you be more specific about what actually is? My approach should work regardless. It is all just a matter of coding .

[Maryam Albuhayr]

yes,

I am actually applying FDM as follows,

and I have tried several ways to do it correctly but still it didn't work.

rho=zeros(N,M);

% BC

rho(1,:)=7000; %mol/m^3

rho(N,:)=0;

% % IC

%

rho(:,1)=7000+(7000/D)*z;

for i=2:N-1

for n=1:M-1

mu(rho(i,n))= 3 .* log(rho(i,n)*4 ./ (1 - rho(i,n)*4)) + rho(i,n)*4./(1-rho(i,n)*4);

rho(i,n+1)=Lambda*rho(i+1,n)+(1-2*Lambda)*rho(i,n)+Lambda*rho(i-1,n)+ mu(rho(i,n));

end

end

[Torsten]

Didn't you forget time stepsize dt and spatial stepsize dz somewhere in the update for rho ?

rho=zeros(N,M);
% BC
rho(1,:)=7000; %mol/m^3
rho(N,:)=0;
% % IC
% 
rho(:,1)=7000+(7000/D)*z;
for n=1:M-1
  for i=2:N-1
    mu = 3*log(rho(i,n)*4/(1 - rho(i,n)*4)) + rho(i,n)*4/(1-rho(i,n)*4);
    rho(i,n+1)=Lambda*rho(i+1,n)+(1-2*Lambda)*rho(i,n)+Lambda*rho(i-1,n) + mu;
  end 
end

[Maryam Albuhayr]

there are many details that I did not lit here

[Maryam Albuhayr]

of course I defineded them, here is what I wrote before

D=8;

T=1;

dz=0.2;

dt=0.02;

%-----------------------------------------------------

Lambda=dt/(dz^2);

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

N=D/dz+1; % space nodes

M=T/dt+1; % time nodes

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

z=zeros(N,1);

t=zeros(M,1);

for i=1:N

z(i)=(i-1)*dz;

end

for n=1:M

t(n)=(n-1)*dt;

end

[Torsten]

It's very unusual to give dt/dz^2 the name Lambda. Lambda is thermal conductivity in general.

So your code works now ? At least it should no longer give syntax errors.

[Maryam Albuhayr]

Unfortunately, not yet