Can an integral that is a variable of itself be solved analytically?

[kmarinas86]

Almost two months ago I posted the following question:

Under the assumption that the voltage is $V_f\left(1-e^{-\frac{t}{RC}}\right)$, where $V_f$ is the final voltage, how would I determine the relationship between current $I$ and time $t$?

$$I = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI}{L} \,dt \,$$

$L$ the magnetic inductance, $R$ the resistance, and $C$ the capacitance, are constants.

How would I plot current $I$ as a function of time $t$? (The only variables here are $I$ and $t$.) Let's assume initial conditions of $I=0$ and $t=0$. My problem here is that the variable I am trying to calculate is a variable inside the integral that is used in deriving the variable itself! How are such problems handled? Any help is appreciated!

I probably wasn't specific enough in my question to really get the answer I wanted. So I now ask, "Can an integral that is a variable of itself be solved analytically?"

 

[Simon Bridge]

If:
$$I=\int f(I)dt$$
(where I is a function of t)

Then:
$$\frac{dI}{dt}=f(I)$$

... will have the same solution for I.

Which is 1st order non-linear.
So you will have analytic solutions under the same conditions.

(note: differentiating both sides was suggested in your original post.)

 

[kmarinas86]

If:
$$I=\int f(I)dt$$
(where I is a function of t)

Then:
$$\frac{dI}{dt}=f(I)$$

... will have the same solution for I.

Which is 1st order non-linear.
So you will have analytic solutions under the same conditions.

(note: differentiating both sides was suggested in your original post.)

I guess what I want is an algebraic solution. Let's try this now:

Under the assumption that the voltage is $V_f\left(1-e^{-\frac{t}{RC}}\right)$, where $V_f$ is the final voltage, how would I determine the relationship between current $I$ and time $t$?

$$I = \int_0^T \frac{V_f\left(1-e^{-\frac{t}{RC}}\right) - RI}{L} \,dt \,$$

$L$ the magnetic inductance, $R$ the resistance, and $C$ the capacitance, are constants.

How would I plot current $I$ as a function of time $t$? (The only variables here are $I$ and $t$.) Let's assume initial conditions of $I=0$ and $t=0$. My problem here is that the variable I am trying to calculate is a variable inside the integral that is used in deriving the variable itself! How are such problems handled? Any help is appreciated!

$$\frac{dI}{dt} = \frac{V_f \left(1-e^{-\frac{t}{RC}}\right) - RI}{L}$$

$$L\frac{dI}{dt} = V_f\left(1-e^{-\frac{t}{RC}}\right) - RI$$
$$RI = V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}$$
$$I = \frac{V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}}{R}$$

Now the problem is, "What is $\frac{dI}{dt}$?" I already had defined it in terms of $I$. Now I am simply back where I started. I just swapped the terms, and I still don't see a closed-form solution.

 

[Mute]

If:
$$I=\int f(I)dt$$
(where I is a function of t)

Then:
$$\frac{dI}{dt}=f(I)$$

Your notation could be confusing. Without the limits someone might think you are differentiating with respect to the dummy integration variable, which would not be good. You should write

If

$$I(t) = \int_0^t dt' f(I(t'),t'),$$

then

$$\frac{dI(t)}{dt} = f(I(t),t).$$

(I added an extra argument for t on its own, since it depends on t separately too).

I guess what I want is an algebraic solution. Let's try this now:

$$\frac{dI}{dt} = \frac{V_f \left(1-e^{-\frac{t}{RC}}\right) - RI}{L}$$

$$L\frac{dI}{dt} = V_f\left(1-e^{-\frac{t}{RC}}\right) - RI$$
$$RI = V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}$$
$$I = \frac{V_f\left(1-e^{-\frac{t}{RC}}\right)-L\frac{dI}{dt}}{R}$$

Now the problem is, "What is $\frac{dI}{dt}$?" I already had defined it in terms of $I$. Now I am simply back where I started. I just swapped the terms, and I still don't see a closed-form solution.

Do you know what a differential equation is? Do you know how to solve one? Your equation is relatively simple and can be solved with an integrating factor. See here.

 

[jackmell]

I suggest you look at easier ones first. See, "A First Course in Integral Equations" by Abdul-Majid Wazwas. No, I'm not kidding. That is his name. Don't make fun. Start with this one:

$$I(t)=\int_0^t k\left(1-e^{-x/c}\right) I(x)dx$$

Now differentiate:

$$\frac{dI}{dt}=k\left(1-e^{-t/c}\right)I(t)$$

separate variables:

$$\frac{dI}{I}=k\left(1-e^{-t/c}\right) dt$$

integrate:

$$\int_{I_0}^I \frac{dI}{I}=k\int_{t_0}^{t} \left(1-e^{-t/c}\right)dt$$

$$\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t$$

Do all that and I get:

$$I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

 

[Simon Bridge]

I guess what I want is an algebraic solution

The differential equation does yield an algebraic solution. It's the first step.
Especially as in your case f(I)=RI/L

@Mute: Yes I noticed that and was concerned - now you've pointed it out I can relax :) OP seems not to understand the differential equation.

Semi walk-through:

OPs integral is of the form

$$y(x)=\int f(x).dx - \int ay(x).dx$$
... and only the second term is causing trouble.
Focussing on the problematic term - differentiate both sides:

$$\frac{dy}{dx}=ay$$

...rearrange and integrate both sides:

$$\int \frac{dy}{y}=a\int dx$$

which yields:

$$\ln{|y|}=ax+c$$
... where c is the constant of integration - the actual problem has definite integrals so apply limits. Anyway - need to make y the subject so we take the natural exponent of both sides.

$$y=Ce^{ax}$$
... where C=e^c

... so what did I miss?

Of course, it may be more like:

$$I=\frac{R}{L}\int_0^T i(t)dt$$
... which is to say the integral is expected to turn out a single number, rather than:

$$I=\frac{R}{L}\int_0^t i(t^\prime)dt^\prime$$
... but I don't want to make it too easy :)
[between this and #5, it should be easy.]

 

[kmarinas86]

I suggest you look at easier ones first. See, "A First Course in Integral Equations" by Abdul-Majid Wazwas. No, I'm not kidding. That is his name. Don't make fun. Start with this one:

$$I(t)=\int_0^t k\left(1-e^{-x/c}\right) I(x)dx$$

Now differentiate:

$$\frac{dI}{dt}=k\left(1-e^{-t/c}\right)I(t)$$

separate variables:

$$\frac{dI}{I}=k\left(1-e^{-t/c}\right) dt$$

integrate:

$$\int_{I_0}^I \frac{dI}{I}=k\int_{t_0}^{t} \left(1-e^{-t/c}\right)dt$$

$$\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t$$

Do all that and I get:

$$I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

It appears that one can go from:

$$\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t$$

to

$$\ln(I)=\ln(I_0)+k\left(t+ce^{-t/c}\right)_{t_0}^t$$

to

$$I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

Now I see why that is (Answer: Logarithm of the product gives us the sum of the logarithm of the factors). I just didn't know how to recognize starting with the sum of the logarithms first (i.e. Sum of the logarithm of the factors gives the logarithm of the product).

If I start with:

$$I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

And then work backwards, then I get:

$$I(t)-I_0=\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$ [hindsight 20/20: WTF?]
$$ln(I(t)-I_0)=k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]$$
$$ln(I(t)-I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t$$

Yet you had:

$$\ln(I)-\ln(I_0)=k\left(t+ce^{-t/c}\right)_{t_0}^t$$

Which is clearly not the same thing.

 

[jackmell]

If I start with:

$$I(t)=I_0\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

And then work backwards, then I get:

$$I(t)-I_0=\exp\left\{k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})]\right\}$$

That's not correct. You'd have to take log of both sides first as in:

$$\log(I(t))=\log(I_0)+k[(t+ce^{-t/c})-(t_0+ce^{-t_0/c})$$

 

[kmarinas86]

The differential equation does yield an algebraic solution. It's the first step.
Especially as in your case f(I)=RI/L

@Mute: Yes I noticed that and was concerned - now you've pointed it out I can relax :) OP seems not to understand the differential equation.

Semi walk-through:

OPs integral is of the form

$$y(x)=\int f(x).dx - \int ay(x).dx$$
... and only the second term is causing trouble.
Focussing on the problematic term - differentiate both sides:

$$\frac{dy}{dx}=ay$$

...rearrange and integrate both sides:

$$\int \frac{dy}{y}=a\int dx$$

which yields:

$$\ln{|y|}=ax+c$$
... where c is the constant of integration - the actual problem has definite integrals so apply limits. Anyway - need to make y the subject so we take the natural exponent of both sides.

$$y=Ce^{ax}$$
... where C=e^c

... so what did I miss?

Of course, it may be more like:

$$I=\frac{R}{L}\int_0^T i(t)dt$$
... which is to say the integral is expected to turn out a single number, rather than:

$$I=\frac{R}{L}\int_0^t i(t^\prime)dt^\prime$$
... but I don't want to make it too easy :)
[between this and #5, it should be easy.]

http://en.wikipedia.org/wiki/Integrating_factor

The integrating factor cannot allow $Q(x)$ to be zero for all $x$, otherwise the quantity:

$$y = \frac{\int Q(x) M(x)\, dx}{M(x)}$$

...would also be zero. This would not help me.

So I have to set $$Q(x)$$ to the value of the first term. Therefore, given an ordinary differential equation of the form:

$$y'+P(x)y = Q(x)$$

$$I' + \frac{R}{L}I = \frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)$$
$$y' = I'$$
$$P = \frac{R}{L}$$
$$y = I$$
$$Q = \frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)$$

The integrating factor:
$$M(x)=e^{\int P(x)\,dx}$$

...is therefore:

$$M(x)=e^{t\frac{R}{L}}$$

Therefore, $y$ equals:

$$y = \frac{\int \left(\frac{V}{L}\left(1-e^{-t\frac{1}{RC}}\right)\right) \left(e^{t\frac{R}{L}}\right)\, dt}{e^{t\frac{R}{L}}}$$

To solve integral of the numerator, I used:

http://www.quickmath.com/webMathema...1=(V/L)*(1-e^(-t/a))*(e^(t/b))&v2=t&v3=0&v4=T

http://www.quickmath.com/msolver//graphs/2011-12-05/3e/ed/6d/3eed6d27b0b2dd008c1be88cce8245fc-3.png?t=1323052868

If my math is correct, then the solution is:

$$y=\frac{\frac{V}{R}\left(\frac{(((L/R)^2 - LC)e^{T/(RC)} + LC)e^{\frac{TR}{L}-\frac{T}{RC}} - \left(\frac{L}{R}\right)^2}{\frac{L}{R} - RC}\right)}{e^{\frac{TR}{L}}}$$

However, when $T$ becomes modestly large, my MS Excel spreadsheet doesn't seem to be able to handle it. So I can't verify the correctness of this using MS Excel.

At least I can plot it on the QuickMath site. It is also in the expected shape:

http://www.quickmath.com/webMathema...=T&v8=y&v9=0&v10=20&v11=0&v12=1&v16=light-red

http://www.quickmath.com/msolver//graphs/2011-12-05/e5/38/eb/e538ebf6aa9c832b718fb73d770c1245-1.png?t=1323057486

http://www.quickmath.com/msolver//graphs/2011-12-05/8d/97/d2/8d97d20d75f4ac3ace6dfedcd6be1a34-1.png?t=1323057564

L/R=10
RC=1
V=100
L=1000
R=100
V/R=1

 

[Simon Bridge]

That looks over-complicated.
However, you have seen that your original question has been answered.

In the following I will use lower case for the time varying current i(t) and upper case I to denote fixed values - I find that easier to read. I'll repeat the first part of what you did for clarification. As always, check my working[1]: this is supposed to be illustrative, not correct.

In standard form, the DE is:

$$L\frac{d}{dt}i + Ri = V_f \left ( 1- e^{-t/RC} \right )$$

So you need an integrating factor of $e^{Rt/L}$:

$$i(t)=\frac{V_f}{L}e^{-Rt/L}\int e^{Rt/L}(1-e^{-t/RC})dt$$
... expand the integrand:


$$i(t)=\frac{V_f}{L}e^{-Rt/L}\left ( i_1 - i_2\right )$$
... where:
$$\begin{align} i_1 & =\int e^{Rt/L}dt\\ i_2 & =\int \exp{\left [(\frac{R}{L}-\frac{1}{RC})t \right ]}dt \end{align}$$

... using $a\int e^{at}dt = e^{at}$, divide through by L, gives:

$$i= V_f\left [ \frac{1}{R}\exp{\left [ ( \frac{R}{L} )t \right ] } - \frac{RC}{R^2C-L}\exp{\left [(\frac{R}{L}-\frac{1}{RC})t \right ]} \right ] e^{-Rt/L} +c$$
... where c is the constant of integration, determined from initial condition (i.e. $i(0)=I_0$)
$$I_0 = \frac{V_f}{R}- \frac{RCV_f}{R^2C-L} + c$$

If I expand the brackets and then collect like terms I get:

$$i= V_f\left [ \frac{1}{R} - \frac{RC}{R^2C-L}\exp{\left [-(\frac{1}{RC})t \right ]} \right ] +c$$

Which should give an idea if how it behaves.
When $t \gg RC$ the exponential term vanishes, leaving:

$$i(t \gg RC) \rightarrow \frac{V_f}{R} + c =I_f$$

Putting I₀=0 and (given figures) RC=1, V=100, L=1000, R=100; the equation becomes:

$$i(t)= \frac{1}{9}\left ( e^{-t} - 1 \right )$$

The current starts at 0 and decays exponentially to -(1/9) units with a mean-time of 1s.

----------------------
[1] I am not going to guarantee any of these calculations are 100% correct or correctly performed. Finding my mistakes is left as an exercise for the student xD

 

[kmarinas86]

To solve integral of the numerator, I used:

http://www.quickmath.com/webMathema...1=(V/L)*(1-e^(-t/a))*(e^(t/b))&v2=t&v3=0&v4=T



If my math is correct, then the solution is:

$$y=\frac{\frac{V}{R}\left(\frac{(((L/R)^2 - LC)e^{T/(RC)} + LC)e^{\frac{TR}{L}-\frac{T}{RC}} - \left(\frac{L}{R}\right)^2}{\frac{L}{R} - RC}\right)}{e^{\frac{TR}{L}}}$$

However, when $T$ becomes modestly large, my MS Excel spreadsheet doesn't seem to be able to handle it. So I can't verify the correctness of this using MS Excel.

At least I can plot it on the QuickMath site. It is also in the expected shape:

http://www.quickmath.com/webMathema...=T&v8=y&v9=0&v10=20&v11=0&v12=1&v16=light-red

http://www.quickmath.com/msolver//graphs/2011-12-05/e5/38/eb/e538ebf6aa9c832b718fb73d770c1245-1.png?t=1323057486

http://www.quickmath.com/msolver//graphs/2011-12-05/8d/97/d2/8d97d20d75f4ac3ace6dfedcd6be1a34-1.png?t=1323057564

L/R=10
RC=1
V=100
L=1000
R=100
V/R=1

Whooops! A bit of typo there. It's actually supposed to be:

$$y=\frac{\frac{V}{L}\left(\frac{(((L/R)^2 - LC)e^{T/(RC)} + LC)e^{\frac{TR}{L}-\frac{T}{RC}} - \left(\frac{L}{R}\right)^2}{\frac{L}{R} - RC}\right)}{e^{\frac{TR}{L}}}$$

The graphs I showed reflect this equation (not the typo version).

By the way: The graph at http://www.quickmath.com/webMathema...=T&v8=y&v9=0&v10=20&v11=0&v12=1&v16=light-red only works if you clear the second and third equations, $y=sin(x)$ and $x^2+y^2=1$ as they are not needed in the problem, otherwise one gets the message "When defining an equation to graph please use variables T and y only. You have: x in `y = sin(x)`". After those two are deleted, leaving behind the line where

y=(((10^2-10)*%e^(T)+10)*%e^(T/10-T)/(10-1)-10^2/(10-1))*100/1000/(%e^(T/10))

is written, then the graph can be successfully plotted by clicking on the plot button.

To avoid that issue, it is simpler just to click this link http://www.quickmath.com/webMathema...1))*100/1000/(%e^(x/10))&v2=0&v3=20&v4=0&v5=1

Equation

http://www.quickmath.com/msolver//graphs/2011-12-05/3f/e4/4b/3fe44bf4b326bd8d36e1ac63b00b661e-1.png?t=1323087170

Result

http://www.quickmath.com/msolver//graphs/2011-12-05/3f/e4/4b/3fe44bf4b326bd8d36e1ac63b00b661e-2.png?t=1323087170