Mixture of 2 salts + Dirac

%I'm doing the Mixture of 2 salts equation and I would to like to add a
%dirac with the purpose in the real life to add just once 2lb of salt in this
%tank.
%So basically, I have a constat flow going in and out to this tank, and I
%want to add 2lb of salt in this tank just one time and I need do it with
%the dirac delta function. 

syms A(t) y
%f = diff(A) == Kin * A
%dsolve(f)

% k*A = the solution concentration 
% A: amount of salt  k: amount of water

Kin = 3;
Ain = 2;
Kout = 3;
Ain = Kin*Ain;
Aout = Kout * (A/300);

x=0:15;
idx = y == Inf;
y(idx) = 1;  

% After 5 minutes 2 lb the salt will be added just once.
f1 = diff(A) == (Ain - Aout) + (2 * dirac(x-5))

f1(t) = 

S = dsolve(f1)

Warning: Number of equations greater than number of indeterminates. Trying heuristics to reduce to square system.

Error using mupadengine/feval_internal
Unable to reduce to square system because the number of equations differs from the number of indeterminates.

Error in dsolve>mupadDsolve (line 334)
T = feval_internal(symengine,'symobj::dsolve',sys,x,options);

Error in dsolve (line 203)
sol = mupadDsolve(args, options);

S1 = dsolve(S,A(0)==50,t)

plot(x,S,'o-')
title 'Mixture of 2 salts'
xlabel ('Time (min)')
ylabel ('Amount of Salt (lb)')

Answer 1

[Paul]

Hi Adrielli,

Why is Ain defined and then redefined? Is that correct?

What is the role of the x and y variables in this problem?

f1(t) is a 1 x 16 array becasue x has 16 elements and is used in the defintion of f1(t). Why is x used here at all, insofar as t is the independendent variable.

Mabye the code should be:

syms A(t) y
%f = diff(A) == Kin * A
%dsolve(f)
% k*A = the solution concentration 
% A: amount of salt  k: amount of water
Kin = 3;
Ain = 2;
Kout = 3;
Ain = Kin*Ain;
Aout = Kout * (A/300);
%x=0:15;
%idx = y == Inf;
%y(idx) = 1;  
% After 5 minutes 2 lb the salt will be added just once.
f1 = diff(A) == (Ain - Aout) + (2 * dirac(t-5))  % change x to t

f1(t) = 

S = dsolve(f1)

S = 

[Adrielli]

Answer your questions:

Why is Ain defined and then redefined? Is that correct?

It was a fool mistake giving the name to the variable. My to intend was the amount of water+salt pumped in the tank was Ain=6. I just should rename one of these variables.

What is the role of the x and y variables in this problem?

I would like my X was the time, so when I reach the 5 minutes my dirac delta function should add the amount of salt I set it up.

I'm using the Y to setting my dirac funcion.

Honestly, I'm learning how to use the matlab, so maybe I just make a huge mess.

f1(t) is a 1 x 16 array because x has 16 elements and is used in the definition of f1(t). Why is x used here at all, insofar as t is the independent variable?

About this, I was trying to plot it in some way, trying to do it work. So probably, it is just a mess I did in a frustated tryied.

Let me try explain it again...

I had this mixture of 2 salts problem that I already solved (I'll add the image problem and the code at matlab).

What I'm trying to do now is: I want to add just one time 2lb of salt in this tank, and I need to do it using the dirac delta funtion.

syms A(t)
Kin = 2;
Ain1 = 3;
Kout = 3;
Ain = Kin*Ain1;
Aout = Kout * (A/300);
f1 = diff(A) == Ain - Aout
dsolve(f1);
dsolve(f1,A(0)==50,t)

[Paul]

The code you've posted matches the solution in the book, though it's using A(t) as the unknown function instead of x(t)

syms A(t)
Kin = 2;
Ain1 = 3;
Kout = 3;
Ain = Kin*Ain1;
Aout = Kout * (A/300);
f1 = diff(A) == Ain - Aout

f1(t) = 

dsolve(f1);
Asol = dsolve(f1,A(0)==50)  % deleted the third input argument

Asol = 

Same solution as is in the text.

Now, if we want to modify Ain, which has units of lb/min if I'm reading the problem correctly, s.t. the A(t) has step change of 2 lb at t = 5 min, then we can do as you propose, which is to modify Ain with the properly scaled delta function:

f2 = lhs(f1) == rhs(f1) + 2 * dirac(t - 5)

f2(t) = 

Asol1 = dsolve(f2,A(0)==50,t)

Asol1 = 

Plot both solutions over a ling time scale. They look very siimilar, which shiouldn't be surprising.

figure;
fplot([Asol Asol1],[0 1000])

But we see the 2 lb step change if we zoom in

figure
fplot([Asol Asol1],[4 6])

Another way to solve this problem would be to solve the first ode from t = 0 t = 5. Then take the end state A(5), add 2 lb, and use that as the boundary condition to sovle same ode from t = 5 forwards. The total solution for A(t) would be the piecewise function formed from the two solutions.

[Adrielli]

Thank you. Undoubtedly you managed to clarify all my doubts and I could understand how to use matlab to solve this problem.