Problem with derivative

[siavash]

Hi guys. i am totally new to this forum and i am would be very grateful if you answer my question.
Here is my Equation:

df(t)/dt = = f(t) + g(t)

Actually this equation investigates the changes in concentration of pollutant in soil (f(t)) which came from ground water (g(t))

Here is the original equation. note that Only Cw (concentration in water) and Cs (Concentration in soil) are time dependent and other parameters are constant. How can i solve this equation.

 

[Deleted member 4993]

Hi guys. i am totally new to this forum and i am would be very grateful if you answer my question.
Here is my Equation:

df(t)/dt = = f(t) + g(t)

Actually this equation investigates the changes in concentration of pollutant in soil (f(t)) which came from ground water (g(t))

Here is the original equation. note that Only Cw (concentration in water) and Cs (Concentration in soil) are time dependent and other parameters are constant. How can i solve this equation.

View attachment 4585

This is first order DE of the type:

y' + y * p(x) = h(x)

You will need to use an integrating factor of the form $\displaystyle \displaystyle{e^{\int p dx}}$

In your case

$\displaystyle p(t) = \lambda_N + \lambda_{1C} + \lambda_{1I}$

 

[HallsofIvy]

Since this is alinear equation- and especially since it is a linear equation with constant coefficients, there is a simpler approach to it. First, find the general solution to the "associated homogeneous equation". That is the equation we get by dropping "g(t)", the only part that has no "f(t)", the unknown function in it.

That gives $\displaystyle f'= -A f$ where "A" is your $\displaystyle \lambda_N+\lambda_{1C}+ \lambda_{1I}$.

We can write that as $\displaystyle \frac{df}{f}= -A dt$ and integrate both sides to get $\displaystyle ln(f)= -\int A dt$. Since A is a constant here, that is $\displaystyle ln(f)= -At+ C$ for C an arbitrary constant. Then, taking the exponential of both sides, $\displaystyle f(t)= e^{-At+ C}= C' e^{-At}$ where $\displaystyle C'= e^C$.

Now we look for a single solution to the entire equation using a method called "variation" of parameters. We look for a solution to the entire equation of the for $\displaystyle f(t)= u(t)e^{-At}$ for some function u(t) replacing the constant C'. (That is the "parameter" we are "varying".)

With $\displaystyle f(t)= u(t)e^{-At}$, we use the product rule to differentiate: $\displaystyle f'(t)= u'(t)e^{-At}- Au(t)e^{-At}$. Putting those into the differential equation, \(\displaystyle u'(t)e^{-At}- Au(t)e^{-At}= -Au(t)e^{-At}+ g(t). The two "\(\displaystyle -Aue^{-At}\)" cancel as a consequence of $\displaystyle e^{-At}$ being a solution to the associated homogeneous equation. That leaves $\displaystyle u'e^{-At}= g(t)$ so that $\displaystyle u'= g(t)e^{At}$ and $\displaystyle u(t)= \int g(t)e^{At}dt$.

Because, in your problem, $\displaystyle g(t)= \frac{V_{irr}}{d}C_w(t)$ that can be writtenas $\displaystyle u(t)= \frac{V_{irr}}{d}\int_0^t C_w(s)e^{As}ds+ D$. Notice that I have changed the "dummy variable" in the integral to s, to avoid confusion with the "real" variable, t, and have written the "indefinite" integral as a definite integral from 0 to t with the added constant, D, to allow for other possible starting values than "0".

Since $\displaystyle f(t)= u(t)e^{-At}$, we now have $\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-At}\right)\int_0^t C_w(s)e^{As}ds+ De^{-At}$ or,
since your $\displaystyle A= \lambda_N+\lambda_{1C}+ \lambda_{1I}$, $\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}\right)\int_0^t C_w(s)e^{(\lambda_N+\lambda_{1C}+ \lambda_{1I})s}ds+ De^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}$\)

 

[siavash]

Another Question

Since this is alinear equation- and especially since it is a linear equation with constant coefficients, there is a simpler approach to it. First, find the general solution to the "associated homogeneous equation". That is the equation we get by dropping "g(t)", the only part that has no "f(t)", the unknown function in it.

That gives $\displaystyle f'= -A f$ where "A" is your $\displaystyle \lambda_N+\lambda_{1C}+ \lambda_{1I}$.

We can write that as $\displaystyle \frac{df}{f}= -A dt$ and integrate both sides to get $\displaystyle ln(f)= -\int A dt$. Since A is a constant here, that is $\displaystyle ln(f)= -At+ C$ for C an arbitrary constant. Then, taking the exponential of both sides, $\displaystyle f(t)= e^{-At+ C}= C' e^{-At}$ where $\displaystyle C'= e^C$.

Now we look for a single solution to the entire equation using a method called "variation" of parameters. We look for a solution to the entire equation of the for $\displaystyle f(t)= u(t)e^{-At}$ for some function u(t) replacing the constant C'. (That is the "parameter" we are "varying".)

With $\displaystyle f(t)= u(t)e^{-At}$, we use the product rule to differentiate: $\displaystyle f'(t)= u'(t)e^{-At}- Au(t)e^{-At}$. Putting those into the differential equation, \(\displaystyle u'(t)e^{-At}- Au(t)e^{-At}= -Au(t)e^{-At}+ g(t). The two "\(\displaystyle -Aue^{-At}\)" cancel as a consequence of $\displaystyle e^{-At}$ being a solution to the associated homogeneous equation. That leaves $\displaystyle u'e^{-At}= g(t)$ so that $\displaystyle u'= g(t)e^{At}$ and $\displaystyle u(t)= \int g(t)e^{At}dt$.

Because, in your problem, $\displaystyle g(t)= \frac{V_{irr}}{d}C_w(t)$ that can be writtenas $\displaystyle u(t)= \frac{V_{irr}}{d}\int_0^t C_w(s)e^{As}ds+ D$. Notice that I have changed the "dummy variable" in the integral to s, to avoid confusion with the "real" variable, t, and have written the "indefinite" integral as a definite integral from 0 to t with the added constant, D, to allow for other possible starting values than "0".

Since $\displaystyle f(t)= u(t)e^{-At}$, we now have $\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-At}\right)\int_0^t C_w(s)e^{As}ds+ De^{-At}$ or,
since your $\displaystyle A= \lambda_N+\lambda_{1C}+ \lambda_{1I}$, $\displaystyle f(t)= \left(\frac{V_{irr}}{d}e^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}\right)\int_0^t C_w(s)e^{(\lambda_N+\lambda_{1C}+ \lambda_{1I})s}ds+ De^{-(\lambda_N+\lambda_{1C}+ \lambda_{1I})t}$\)

\(\displaystyle



Thanks a lot for your kind answer. My initial condition is that the Cs(0)=0 so i think the dummy variable should be zero. am i right?
The main function that define Cw is a combination of GREEN function and Error function which is extremely difficult to integrate. I want to know that in my case can i put the numerical value of Cw into the integral? I mean instead of inserting the complex equations of mass transport, i just put the number in the integral? (for example if Cw(40)=100, instead of integrating the hole equation i put 100 in the integral)
for more clarification i attached the CW equation:
\)

 

[HallsofIvy]

I am puzzled how you could know about or at least know the words "Green's Function" and "Error Function" but not know what a "dummy variable", which is used in Calculus, is. Perhaps you have just forgotten. No, the dummy variable is NOT 0. A dummy variable can't have a specific value. A "dummy variable" is a variable that takes on values that have meaning only inside a given calculation. For example, $\displaystyle \int_a^b 2xt dt= \left[xt^2\right]_a^b= (b^2- a^2)x$. "t" is a dummy variable because the result has NO "t" in it. t only has meaning inside the integral.