I am attempting to resolve this first order nonlinear ODE $$y'(t) = a\left(e^{-k_1t}-e^{-k_2t}\right) - b\,y(t) - c\,y^2(t)$$ where $a$, $b$, $c$, $k_1$ and $k_2$ are positive constants and $0\leq t<\infty$, with initial condition $y(0) = 0$.

Many moons ago, I typed the equation into Mathematica and I recall the solution being a ratio of differences of Bessel functions of different kinds. Vaguely from memory, it looked something like $$y(t) = \frac{c_1K(\cdot)-c_2Y(\cdot)}{c_3J(\cdot)-c_4B(\cdot)}$$ where all capital letters are some form of Bessel functions, though I cannot remember the actual solution and I can't find where I wrote it down :(

For some reason, Mathematica no longer will solve this equation. So now I am attempting to solve the equation myself but have run into some roadblocks.

My Attempt

A priori the equation is a Riccati equation. So making the substitution $$y=\frac{u'(t)}{c\,u(t)}$$ gives the following second order linear DE with variable coefficients $$u''(t) + b\,u'(t) - ac\left(e^{-k_1t}-e^{-k_2t}\right)u(t)=0$$

My Mathematica does not like solving this one either. Power series isn't the best approach due to the last term in the equation. I can solve it using the Wronskian if I can find just one fundamental solution. Though I am not clever enough to find a fundamental solution to get the Wronskian method off the ground. However computation of the Wronskian is simple, $$W_{u_1,u_2}(t)=Ce^{-bt}$$ where $C\neq0$ is arbitrary.

I then turned to Laplace transforms which looked promising at first $$s^2U(s)-su(0)-u'(0)+b(sU(s)-u(0))-ac\left(U(s+k_1)-U(s+k_2)\right)=0$$ where $U(s)=\mathcal{L}_t\{u\}(s)$, but the shifts in the last two $U$ terms makes the solution mysterious to me.

Any suggestions on how to continue, or any alternative approaches to obtain the solution will be greatly appreciated.

  • $\begingroup$ I'm not sure in what I say. But maybe you should first try the first technique for the homogeneous equation than substitute the appropriate constants with function of $t$. $\endgroup$ – kolobokish Nov 27 '17 at 2:10

It may not be the best method, but one can at the very least write down a series solution to the original problem.

Let $$y(t) = \sum_{n=0}^{\infty} A_n t^n $$

Then $y'(t) = \sum_{n=0}^{\infty} (n+1)A_{n+1}t^n$, $y^2(t) = \sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} A_k A_{n-k}\right) t^n$. So we arrive at

$$\sum_{n=0}^{\infty} (n+1)A_{n+1}t^n - a\sum_{n=0}^{\infty}\frac{(-1)^n(k_1^n - k_2^n)}{n!}t^n + b\sum_{n=0}^{\infty}A_nt^n+c\sum_{n=0}^{\infty} \left( \sum_{k=0}^{n} A_k A_{n-k}\right) t^n = 0$$

after substituting into the original problem. Gathering things up, and noting that $a_0 = 0$ from the initial condition, we have the following recurrence relation for the coefficients:

$$\begin{cases} (n+1)A_{n+1} -a(-1)^n \frac{k_1^n-k_2^n}{n!} + bA_n + c \sum_{k=0}^{n} A_k A_{n-k} = 0 \\ A_0 = 0 \end{cases}$$

A couple of coefficients I found:

$A_1 = 0$

$A_2 = -\frac{1}{2}a(k_1 - k_2)$

$A_3 = \frac{1}{6}a(k_1 - k_2)(b + (k_1 + k_2))$

$A_4 = - \frac{1}{24}a(k_1 - k_2)(b(b+(k_1 + k_2))+k_1^2 + k_1 k_2 + k_2^2)$

EDIT: I wanted to make a quick mention for computing these coefficients efficiently since you're using Mathematica: use "memoization" to cut down on computation time, it allows you to compare the series solution to NDSolve pretty easily.

  • $\begingroup$ This is helpful. I didn't try the power series because I didn't know how to handle the $y^2$ term. I'll see what I can come up with using your recursive formula. $\endgroup$ – MasterYoda Nov 27 '17 at 4:13
  • $\begingroup$ Ah, yes, definitely don't forget about cauchy products when thinking about multiplying series together, it can be very helpful when writing it out. It also avoids the headache of a double sum(sort of) by only have the second sum dealing with the coefficients of $t^n$. There seems to be some sort of pattern with the coefficients, but I'm not sure it's going to be easily discerned. $\endgroup$ – DaveNine Nov 27 '17 at 4:31
  • $\begingroup$ Further, it is worth noting that you can show that the trivial solution is the only solution for when $a = 0$ or $k_1 = k_2$. $\endgroup$ – DaveNine Nov 27 '17 at 5:08
  • $\begingroup$ Thank you for your help. I will remember the Cauchy product for next time. Using your recursion relation, I came up with the following formula for the coefficients $A_n$ for $n\geq2$: $$A_n=\frac{(-1)^{n-1}a}{n!}\sum^{n-1}_{i=1}b^{n-1-i}\left(k_1^i-k_2^i\right)$$ Your answer will be marked as accepted. $\endgroup$ – MasterYoda Nov 27 '17 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.