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February 11th, 2016, 07:53 AM   #1
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Please help me with definite integral

Hello everyone,

my question looks like this

dS/dt = A/(B + t) - CS/(D+S)

1) A, B, C, D are all constants;
2) Integrate S from S (t=0) to S (t=t) and from t=0 to t=t;
3) S (t=0) >0 and S (t=t) >0;

-------------------------------------
I tried with similar techniques as the below example. However, I don't think it is correct.
Here is how I tired:

dS = (A/(B+t)) dt - (CS/(D+S)) dt

Integrate from t=0 to t=t for only right hand side:

dS = A*ln((B+t)/B) - (CS *t)/(D+S)

Then,

(CS *t)/(D+S) + dS = A*ln((B+t)/B) (integrate just left hand side?? but the 1st term does not have dS??)

------------------

I have an example of this question but simpler:

dS/dt = -ES/(F+S)

1) E, F are all constants;
2) Integrate S from S (t=0) to S (t=t) and from t=0 to t=t;
3) S (t=0) >0 and S (t=t) >0;

(F+S)/S dS = -E dt

Integrate both side:

(F ln(abs(S))+S) [from S(t=0) to S(t=t)] = -E*t [from t=0 to t=t]

therefore,

F ln(S(t=t)/S(t=0))+(S(t=t)-S(t=0)) = -E*t
-------------------------------

Thank everyone in advance!!

Last edited by Johnny223; February 11th, 2016 at 08:06 AM.
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February 11th, 2016, 08:55 AM   #2
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How did this problem arise? Is it from a book?
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February 11th, 2016, 12:47 PM   #3
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Quote:
Originally Posted by Johnny223 View Post
Integrate from t=0 to t=t for only right hand side:

dS = A*ln((B+t)/B) - (CS *t)/(D+S)
I can't think of any circumstance or method that would do this. It's like squaring just one part of an equation. It isn't a valid technique.

Mathematica won't solve the equation. It's apparently a really really nasty integral.

-Dan
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February 11th, 2016, 04:49 PM   #4
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Quote:
Originally Posted by skipjack View Post
How did this problem arise? Is it from a book?
Thank you for your question.

When I was trying to describe my situation, I found several mistakes in my previous question. However, I am still not able to solve the definite integral after corrections.

My area of study is wastewater treatment. In my equation, there are 2 core concepts which we need to agree with.

1) Mass Balance in wastewater treatment

https://books.google.ca/books?id=GJe...ewater&f=false

2) Monod equation for microbial kinetics

https://en.wikipedia.org/wiki/Monod_equation

-------------------------------------
In the book mentioned above, equation 2.17 is the fundamental equation to solve my problem:

d(C*V)/dt = Q*C0 - Q*C + rp*V - rc*V (eq. 2.17) Pg. 32

C*(dV/dt) + V*(dC/dt) = Q*C0 - Q*C + rp*V - rc*V (eq. 2.18 )

my equation

C*(dV/dt) + V*(dC/dt) = Qin*Cin - Qout*C + r*V (my equation)

where,

C = target compound concentration in the reactor (mass/volume)
Cin = influent target compound concentration (mass/volume)
V = volume of reactor (volume)
Qin = influent flow rate (volume/time)
Qout = effluent flow rate (volume/time)
r = Monod equation (mass/volume-time)
----------------------

In my situation, there are 2 major differences from the book:

1) V = function(time) = V0 + Qin*t (where, V0 = Volume of Reactor @ t=0)
2) Qout = 0
------------------------

Substitute the following to my equation:

1) C to become S
2) V = V0 + Qin*t
3) dV/dt = Qin
4) Qout = 0

therefore,

S*Qin + (V0 + Qin*t)*(dS/dt) = Qin*Sin + r*(V0 + Qin*t)

Divide Qin (Qin>0) in both sides

S + (V0/Qin + t)*(dS/dt) = Sin + r*(V0/Qin + t)

Let
constantA = Sin
constantB = V0/Qin

Therefore,

S + (constantB + t)*(dS/dt) = constantA + r*(constantB + t)

(constantB + t)*(dS/dt) = constantA - S + r*(constantB + t)

Divide (constantB + t) in both sides (constantB >0)

dS/dt = (constantA - S)/(constantB + t) + r

Let
r = -(constantC * S)/(constantD + S)

dS/dt = (constantA - S)/(constantB + t) - (constantC * S)/(constantD + S)

Omit "constant" character

dS/dt = (A-S)/(B+t) - (C*S)/(D+S)

-----------------------

Now the new equation still looks similar as the previous one. I still do not have any idea how to tackle this equation.

However, thank everyone for your interest!!!
Thanks from greg1313 and topsquark

Last edited by Johnny223; February 11th, 2016 at 04:56 PM.
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February 12th, 2016, 11:50 AM   #5
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So, is there any chance that we can solve this integral?

dS/dt = (A-S)/(B+t) - (C*S)/(D+S)

from S(t=0) to S(t=t), and
from t=0 to t=t ??

A, B, C, D are constants
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February 12th, 2016, 12:19 PM   #6
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As Mathematica fails to solve this, it's certainly not easy in general. However, various special cases can be solved, some of them leading to a linear solution, some to a more complicated solution. As a first step, using simple substitutions to simplify the equation would be a good idea. It might be possible to determine all the special cases that can be solved.
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February 17th, 2016, 09:27 AM   #7
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Quote:
Originally Posted by Johnny223 View Post
Hello everyone,

I have an example of this question but simpler:

dS/dt = -ES/(F+S)

1) E, F are all constants;
2) Integrate S from S (t=0) to S (t=t) and from t=0 to t=t;
3) S (t=0) >0 and S (t=t) >0;

(F+S)/S dS = -E dt

Integrate both side:

(F ln(abs(S))+S) [from S(t=0) to S(t=t)] = -E*t [from t=0 to t=t]

therefore,

F ln(S(t=t)/S(t=0))+(S(t=t)-S(t=0)) = -E*t
-------------------------------

Thank everyone in advance!!
I did come up with an analytical solution of integrating this type of equation.

I will use the simpler example for illustration.
-------------------------------
Given S(t=0) = S0

At t=0, S(t=0)=S0

dS/dt (t=0, S=S(t=0)) = -E(S(t=0))/(F + S(t=0)) = some number

-------------------------

Let t = t + increment

-------------------------

Therefore

At t= increment

S(t=increment) = S(t=0) + (increment)*(dS/dt (t=0, S=S(t=0))

dS/dt (t=increment, S=S(t=increment)) = -E(S(t=increment))/(F + S(t=increment)) = some number2

----------------------------

At t = increment*2

S(t=increment*2) = S(t=increment) + (increment)*(dS/dt (t=increment, S=S(t=increment))

dS/dt (t=increment*2, S=S(t=increment*2)) = -E(S(t=increment*2))/(F + S(t=increment*2)) = some number3

----------------------------

As increment --> 0, then I can approximate S(t=t+increment) to be the integral form of dS/dt = -ES/(F+S)

I confirm my answer with the known integrated form F ln(S(t=t)/S(t=0))+(S(t=t)-S(t=0)) = -E*t

----------------------------

As this is a realistic question, I do have expectations of the magnitude of my increment and maximum t.

Therefore, I can solve the harder equation by this method.

Do you guys think if this is reasonable?

Last edited by Johnny223; February 17th, 2016 at 09:31 AM.
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