Solving differential equation that includes an (extra) unknown function











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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?










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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 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 at 1:17












  • Noted. And I do see your point. Thank you for your input.
    – Christopher
    Nov 19 at 23:47















up vote
0
down vote

favorite












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?










share|cite|improve this question






















  • 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 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 at 1:17












  • Noted. And I do see your point. Thank you for your input.
    – Christopher
    Nov 19 at 23:47













up vote
0
down vote

favorite









up vote
0
down vote

favorite











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?










share|cite|improve this question













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






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asked Nov 19 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 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 at 1:17












  • Noted. And I do see your point. Thank you for your input.
    – Christopher
    Nov 19 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 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 at 1:17












  • Noted. And I do see your point. Thank you for your input.
    – Christopher
    Nov 19 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 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 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 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 at 1:17














Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 at 23:47




Noted. And I do see your point. Thank you for your input.
– Christopher
Nov 19 at 23:47










2 Answers
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0
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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.






share|cite|improve this answer





















  • Thank you, Robert, Riemann sums were a a good suggestion!
    – Christopher
    Nov 19 at 23:41


















up vote
0
down vote



accepted










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.






share|cite|improve this answer





















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    2 Answers
    2






    active

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    2 Answers
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    up vote
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    down vote













    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.






    share|cite|improve this answer





















    • Thank you, Robert, Riemann sums were a a good suggestion!
      – Christopher
      Nov 19 at 23:41















    up vote
    0
    down vote













    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.






    share|cite|improve this answer





















    • Thank you, Robert, Riemann sums were a a good suggestion!
      – Christopher
      Nov 19 at 23:41













    up vote
    0
    down vote










    up vote
    0
    down vote









    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 19 at 1:44









    Robert Israel

    314k23206453




    314k23206453












    • Thank you, Robert, Riemann sums were a a good suggestion!
      – Christopher
      Nov 19 at 23:41


















    • Thank you, Robert, Riemann sums were a a good suggestion!
      – Christopher
      Nov 19 at 23:41
















    Thank you, Robert, Riemann sums were a a good suggestion!
    – Christopher
    Nov 19 at 23:41




    Thank you, Robert, Riemann sums were a a good suggestion!
    – Christopher
    Nov 19 at 23:41










    up vote
    0
    down vote



    accepted










    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.






    share|cite|improve this answer

























      up vote
      0
      down vote



      accepted










      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.






      share|cite|improve this answer























        up vote
        0
        down vote



        accepted







        up vote
        0
        down vote



        accepted






        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 19 at 23:45









        Christopher

        11




        11






























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