Mixing
Solving differential equation that includes an (extra) unknown function
I am trying to find a solution for $C_a(t)$ in the below differential equation:
$$
V cdot frac {dC_a(t)} {dt}=R(t)-Q_v cdot C_a(t) \
$$
$
text{Where} \
qquad V text{ is the volume of the test chamber } [m^3] \
qquad C_a text{ is the concentration in test chamber } Big[frac{kg}{m^3} Big] \
qquad R text{ is the non-constant emission rate from test sample } Big[frac{kg}{m^2 cdot s} Big] \
qquad Q_v text{ is the constant volume flow rate of ventilation air passing through the test chamber } Big[frac{m^3}{s} Big] \
$
The non-constant emission rate is the "unknown function" mentioned in the question title. Stating that the function that governs emission rate is unknown is somewhat inaccurate as I have a vector containing values for emission for every examined time t.
Besides knowing values for $R$ at given times $t$, we know that $Ca(0) = 0$ and that $R(0) = 0$.
I am in the process of writing a small Matlab script, and therefore I am hoping to find a solution on a form that allows implementation and evaluation at given points in time (corresponding to the values in known vector $R$).
I have already given it my best, and I have arrived at the below expression (that I sincerely hope is correct). However, this expression considers $R(x)$ as a continuous function and will not be implemented easily.
$$
C_a(t) = exp(-Q_v t V^{-1}) int_{0}^{t}
exp(-Q_v x V^{-1}) R(x) V^{^-1} dx, quad for C_a(0)=0
$$
Can anybody help me find a simpler (explicit and fully analytical) solution for $C_a(t,R(t))$ that is easy to implement?
differential-equations matlab
asked Nov 19 '18 at 1:09
Christopher
11
add a comment |
I am trying to find a solution for $C_a(t)$ in the below differential equation:
$$
V cdot frac {dC_a(t)} {dt}=R(t)-Q_v cdot C_a(t) \
$$
$
text{Where} \
qquad V text{ is the volume of the test chamber } [m^3] \
qquad C_a text{ is the concentration in test chamber } Big[frac{kg}{m^3} Big] \
qquad R text{ is the non-constant emission rate from test sample } Big[frac{kg}{m^2 cdot s} Big] \
qquad Q_v text{ is the constant volume flow rate of ventilation air passing through the test chamber } Big[frac{m^3}{s} Big] \
$
The non-constant emission rate is the "unknown function" mentioned in the question title. Stating that the function that governs emission rate is unknown is somewhat inaccurate as I have a vector containing values for emission for every examined time t.
Besides knowing values for $R$ at given times $t$, we know that $Ca(0) = 0$ and that $R(0) = 0$.
I am in the process of writing a small Matlab script, and therefore I am hoping to find a solution on a form that allows implementation and evaluation at given points in time (corresponding to the values in known vector $R$).
I have already given it my best, and I have arrived at the below expression (that I sincerely hope is correct). However, this expression considers $R(x)$ as a continuous function and will not be implemented easily.
$$
C_a(t) = exp(-Q_v t V^{-1}) int_{0}^{t}
exp(-Q_v x V^{-1}) R(x) V^{^-1} dx, quad for C_a(0)=0
$$
Can anybody help me find a simpler (explicit and fully analytical) solution for $C_a(t,R(t))$ that is easy to implement?
differential-equations matlab
asked Nov 19 '18 at 1:09
Christopher
11
If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47
add a comment |
I am trying to find a solution for $C_a(t)$ in the below differential equation:
$$
V cdot frac {dC_a(t)} {dt}=R(t)-Q_v cdot C_a(t) \
$$
$
text{Where} \
qquad V text{ is the volume of the test chamber } [m^3] \
qquad C_a text{ is the concentration in test chamber } Big[frac{kg}{m^3} Big] \
qquad R text{ is the non-constant emission rate from test sample } Big[frac{kg}{m^2 cdot s} Big] \
qquad Q_v text{ is the constant volume flow rate of ventilation air passing through the test chamber } Big[frac{m^3}{s} Big] \
$
The non-constant emission rate is the "unknown function" mentioned in the question title. Stating that the function that governs emission rate is unknown is somewhat inaccurate as I have a vector containing values for emission for every examined time t.
Besides knowing values for $R$ at given times $t$, we know that $Ca(0) = 0$ and that $R(0) = 0$.
I am in the process of writing a small Matlab script, and therefore I am hoping to find a solution on a form that allows implementation and evaluation at given points in time (corresponding to the values in known vector $R$).
I have already given it my best, and I have arrived at the below expression (that I sincerely hope is correct). However, this expression considers $R(x)$ as a continuous function and will not be implemented easily.
$$
C_a(t) = exp(-Q_v t V^{-1}) int_{0}^{t}
exp(-Q_v x V^{-1}) R(x) V^{^-1} dx, quad for C_a(0)=0
$$
Can anybody help me find a simpler (explicit and fully analytical) solution for $C_a(t,R(t))$ that is easy to implement?
differential-equations matlab
asked Nov 19 '18 at 1:09
Christopher
11
I am trying to find a solution for $C_a(t)$ in the below differential equation:
$$
V cdot frac {dC_a(t)} {dt}=R(t)-Q_v cdot C_a(t) \
$$
$
text{Where} \
qquad V text{ is the volume of the test chamber } [m^3] \
qquad C_a text{ is the concentration in test chamber } Big[frac{kg}{m^3} Big] \
qquad R text{ is the non-constant emission rate from test sample } Big[frac{kg}{m^2 cdot s} Big] \
qquad Q_v text{ is the constant volume flow rate of ventilation air passing through the test chamber } Big[frac{m^3}{s} Big] \
$
The non-constant emission rate is the "unknown function" mentioned in the question title. Stating that the function that governs emission rate is unknown is somewhat inaccurate as I have a vector containing values for emission for every examined time t.
Besides knowing values for $R$ at given times $t$, we know that $Ca(0) = 0$ and that $R(0) = 0$.
I am in the process of writing a small Matlab script, and therefore I am hoping to find a solution on a form that allows implementation and evaluation at given points in time (corresponding to the values in known vector $R$).
I have already given it my best, and I have arrived at the below expression (that I sincerely hope is correct). However, this expression considers $R(x)$ as a continuous function and will not be implemented easily.
$$
C_a(t) = exp(-Q_v t V^{-1}) int_{0}^{t}
exp(-Q_v x V^{-1}) R(x) V^{^-1} dx, quad for C_a(0)=0
$$
Can anybody help me find a simpler (explicit and fully analytical) solution for $C_a(t,R(t))$ that is easy to implement?
differential-equations matlab
differential-equations matlab
asked Nov 19 '18 at 1:09
Christopher
11
asked Nov 19 '18 at 1:09
Christopher
11
asked Nov 19 '18 at 1:09
Christopher
11
asked Nov 19 '18 at 1:09
Christopher
11
asked Nov 19 '18 at 1:09
Christopher
11
11
If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47
add a comment |
If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47
If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47
add a comment |
2 Answers
2
active
oldest
votes
The solution with initial value $C_a(0) = C_0$ is
$$C_a(t) = exp(-Q_v t/V) left( C_0 + int_0^t exp(Q_v s/V) frac{R(s)}{V}; ds right) $$
You can approximate the integral with Riemann sums.
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
add a comment |
Turns out that Matlab does have a function that allows for solving ODEs with time dependent terms.
Thank you all very much for your time and help.
answered Nov 19 '18 at 23:45
Christopher
11
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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The solution with initial value $C_a(0) = C_0$ is
$$C_a(t) = exp(-Q_v t/V) left( C_0 + int_0^t exp(Q_v s/V) frac{R(s)}{V}; ds right) $$
You can approximate the integral with Riemann sums.
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
add a comment |
The solution with initial value $C_a(0) = C_0$ is
$$C_a(t) = exp(-Q_v t/V) left( C_0 + int_0^t exp(Q_v s/V) frac{R(s)}{V}; ds right) $$
You can approximate the integral with Riemann sums.
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
add a comment |
The solution with initial value $C_a(0) = C_0$ is
$$C_a(t) = exp(-Q_v t/V) left( C_0 + int_0^t exp(Q_v s/V) frac{R(s)}{V}; ds right) $$
You can approximate the integral with Riemann sums.
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
The solution with initial value $C_a(0) = C_0$ is
$$C_a(t) = exp(-Q_v t/V) left( C_0 + int_0^t exp(Q_v s/V) frac{R(s)}{V}; ds right) $$
You can approximate the integral with Riemann sums.
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
answered Nov 19 '18 at 1:44
Robert Israel
318k23208457
318k23208457
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
add a comment |
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
Thank you, Robert, Riemann sums were a a good suggestion!
– Christopher
Nov 19 '18 at 23:41
add a comment |
Turns out that Matlab does have a function that allows for solving ODEs with time dependent terms.
Thank you all very much for your time and help.
answered Nov 19 '18 at 23:45
Christopher
11
add a comment |
Turns out that Matlab does have a function that allows for solving ODEs with time dependent terms.
Thank you all very much for your time and help.
answered Nov 19 '18 at 23:45
Christopher
11
add a comment |
Turns out that Matlab does have a function that allows for solving ODEs with time dependent terms.
Thank you all very much for your time and help.
answered Nov 19 '18 at 23:45
Christopher
11
Turns out that Matlab does have a function that allows for solving ODEs with time dependent terms.
Thank you all very much for your time and help.
answered Nov 19 '18 at 23:45
Christopher
11
answered Nov 19 '18 at 23:45
Christopher
11
answered Nov 19 '18 at 23:45
Christopher
11
answered Nov 19 '18 at 23:45
Christopher
11
11
add a comment |
add a comment |
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If $R$ is only given at discrete times then you will probably need to write your own finite difference code, or else use some suitable interpolant of the data in the analytical solution.
– Ian
Nov 19 '18 at 1:16
By the way, in general when asking a question, it's best to strip down your problem as much as you possibly can without omitting information that is needed to answer the question. In this situation, your problem can be stripped down to $y'(t)=ay(t)+f(t)$ where $a$ is a constant and $f(t)$ is a function which is "really" continuous but is nevertheless only known only at discrete times.
– Ian
Nov 19 '18 at 1:17
Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 '18 at 23:47