# Solving for unknown functions

I am not a mathematician, so excuse if my question is silly or badly stated. I have the following problem. I have 2 conditions on two unknown continuously differentiable functions:

$$A(t)=\frac{1}{B(t)}+C \\ B(t)=D-A(t)-\int_0^t A(\tau) d\tau.$$

C and D are constants. I also know $A(0)$ and $B(0)$. I am looking for a way to get the value of $A(t)$ and $B(t)$ for small $t>0$. So far I have a numerical solution, but that involves a lot of interpolation and I don't think it is very good.

I was wondering if there is some way to get an analytic solution for this problem. I don't expect you to solve the problem for me, I'm willing to learn and I'd be very grateful if you could point me towards possible readings.

Thanks in advance.

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## 2 Answers

Differentiate both equations, then mess with the variables, to get $$(B^2 - 1)B' + CB^2 + B = 0.$$

This does not appear to have a solution in elementary functions, although Wolfram Alpha is telling me that for $k$ an arbitrary constant, one has $$k - x = \frac{B(x)}{C} - \log(B(x)) + \frac{C^2 - 1}{C}\log(CB(x) + 1),$$ which you can easily check (if it's true! - I haven't checked yet).

From this you can approximate $B$ (and hence $A$) as well as you can like.

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It might be easier to just substitute the value of$A(t)$ from the first formula into the second formula, and then differentiate to get the DE. – copper.hat Jul 28 '12 at 15:03

UPDATED: Let's write everything in function of $E(t):=A(t)-C$ :

$$B(t)=\frac 1{E(t)}$$ $$\frac 1{E(t)}+E(t)=D-C-\int_0^t E(t)+C\ dt$$

after derivation the second equation becomes : $$E'(t)\left(1-\frac 1{E(t)^2}\right)=-E(t)-C$$

$$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{(E(t)+C)E(t)^2}=-1$$

$$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{C^2}\left(\frac {1}{E(t)+C}-\frac {E(t)-C}{E(t)^2}\right)=-1$$ $$\frac {E'(t)}{E(t)+C}-\frac {E'(t)}{C^2}\left(\frac {1}{E(t)+C}-\frac 1{E(t)}+\frac C{E(t)^2}\right)=-1$$

Integrating this we get : $$\log(E(t)+C)-\frac 1{C^2}\log(E(t)+C)+\frac 1{C^2}\log(E(t))+ \frac 1{CE(t)}=-t+F$$

that we may rewrite as (changing sign) : $$t-F=\frac {1-C^2}{C^2}\log(E(t)+C)-\frac 1{C^2}\log(E(t))- \frac 1{CE(t)}$$

with $F$ a constant and $E(t)=A(t)-C$

This is the same result as countinghaus' except that I expressed $t$ in function of $E(t)=A(t)-C$ while his result was in function of $B(t)=\dfrac 1{E(t)}$. Let's rewrite our result with $B(t)$ : $$t-F=\frac {1-C^2}{C^2}\log\left(\frac 1{B(t)}+C\right)+\frac 1{C^2}\log(B(t))- \frac {B(t)}C$$ $$\boxed{\displaystyle t=- \frac {B(t)}C-\frac {C^2-1}{C^2}\log(1+C B(t))+\log(B(t))+F}$$

Obtaining $A(t)$ (or $B(t)$) in function of $t$ in 'closed form' is probably not possible except for specific values of ($C$) :

• $C=1$ : solutions $B(t)=-1$ and $\displaystyle B(t)=-W\left(-e^{F-t}\right)$
• $C=-1$ : solutions $B(t)=1$ and $\displaystyle B(t)=W\left(e^{F-t}\right)$
• $C=0$ : solutions $\displaystyle B(t)=\pm\frac {\sqrt{W\left(-e^{F-2t}\right)}}i$
(with $W$ the Lambert W function and $F$ a constant ; you may search other solutions using Alpha by changing the ($C=1$ in the example) constant before $x^2$ in the query and get the plot of the result)
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