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Function $f_n(t)$ is defined for all positive integer $n$ and is continuously differential in $t$ for $t\in[0,T]$.

We have a recursive differential equation that we would like to solve.

$$\frac{\partial f_n(t)}{\partial t}-\alpha\left(f_n(t)-f_{n-1}(t)\right)+\alpha\beta=0,$$ with the terminal condition $f_n(T)=0$.

Now we know that for $n=1$, we have $f_1(t)=\beta-\beta e^{-\alpha(T-t)}$. So we would like to find $f_n(t)$ for $n\ge 2$.

The first thought I had was solving using the integrating factor method. Doing that, we get this solution $$f_n(t)=\int_{t}^{T}\left(\alpha\beta + \alpha f_{n-1}(s)\right)e^{-\alpha\left(s - t\right)}ds.$$ This is nice but for higher $n$, solving the integral is very messy.

Is there another way to find a closed-form expression for $f_n(t)$?

I got a hint on using the Laplace Transform, the closed-form expression of $f_n(t)$ is something like $k + (a+bx+cx^2+…) e^{-\alpha x}$ where $x=T-t$.

So using this hint, I started with $n=2$ but I don't get this form. I can compute $f_n(t)$ for each $n$ but not a closed-form for any $n$.

Could someone help please?

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  • $\begingroup$ Have you tried performing Laplace transform to the equation without plugging specific $n$ in? $\endgroup$ Commented Jul 17, 2022 at 13:37
  • $\begingroup$ Thank you for the comment. I thought about that but I wasn't sure how to handle $f_{n-1}(t)$. I'm very new to all this! $\endgroup$ Commented Jul 17, 2022 at 13:45
  • $\begingroup$ Just define $F_n(s)$ to be the Laplace transform of $f_n(t)$, so you will obtain an algebraic recurrence relationship for $F_n(s)$. $\endgroup$ Commented Jul 17, 2022 at 13:47

1 Answer 1

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I see a way using a "characteristic function approach".

I mean by that, the use an auxiliary variable, say $x$, generating a formal series obeying a "big" differential equation gathering all the "small" differential equations. Let us be more precise.

Let us multiply the $n$-th differential equation by $x^n$ and sum up all of them in this way:

$$\sum_{n=1}^{\infty}\frac{\partial f_n(t)}{\partial t}x^n-\alpha \sum_{n=1}^{\infty}f_n(t)x^n+\alpha \sum_{n=1}^{\infty} f_{n-1}(t)x^n+\alpha\beta \sum_{n=1}^{\infty}x^n=0 \tag{1}$$

Let us use the following notation

$$\Phi(t,x):=\sum_{n=1}^{\infty}f_n(t)x^n\tag{2}$$

Using (2), (1) becomes:

$$\frac{\partial \Phi(t,x)}{\partial t}-\alpha \Phi(t,x)+\alpha x(\Phi(t,x)\color{red}{-f_1(t)})+\alpha \beta\frac{x}{1-x}=0 \tag{3}$$

(please note the handling (in red) of the first term). Otherwise said:

$$\frac{\partial \Phi(t,x)}{\partial t}+\alpha(x-1)\Phi(t,x)=\alpha x f_1(t)-\alpha \beta\frac{x}{1-x} \tag{4}$$

with (as indicated)

$$f_1(t)=\beta-\beta e^{-\alpha(T-t)}\tag{5}$$

giving:

$$\frac{\partial \Phi(t,x)}{\partial t}+\alpha(x-1)\Phi(t,x)=\alpha \beta\frac{x^2}{1-x}-\alpha \beta x e^{-\alpha(T-t)} \tag{6}$$

(6) is an ordinary first order differential equation in variable $t$ (with $x$ considered as a parameter, i.e., a constant).

This constant coefficients differential equation can be solved using Laplace Transform. Here is the how, denoting by $P$ the Laplace Transform of $\Phi$, with customary variable $s$ in the Laplace domain:

$$s P(s,x)-\alpha(x-1) P(s,x)=\alpha \beta \left(\dfrac{s^3}{1-s^3}-xe^{-\alpha T}\dfrac{1}{s-\alpha}\right)$$

$$P(s,x)=\dfrac{\alpha \beta}{s-\alpha(x-1)} \left(\dfrac{s^3}{1-s^3}-xe^{-\alpha T}\dfrac{1}{s-\alpha}\right)$$

I haven't done the computations of the inverse Laplace Transform of $P$, in order to obtain a closed form expression for $\Phi$ but as they involve rational fractions, they are standard (ask Wolfram Alpha for example).

Having obtained $\Phi$, it will remain to expand it as a formal series in $x^n$ and "harvest" its coefficients which are the $f_n(t)$.

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  • $\begingroup$ I have added some further computations. Any comment ? $\endgroup$ Commented Jul 17, 2022 at 16:31
  • $\begingroup$ Thank you. I am trying to do all the calculations step-by-step and I'm very slow at it! $\endgroup$ Commented Jul 18, 2022 at 9:48
  • $\begingroup$ Is your issue connected to Picard iterative method for solving differential equations ? $\endgroup$ Commented Jul 18, 2022 at 14:28
  • $\begingroup$ This is actually the first time I'm hearing that! I don't see how we can get that $k, a, b, c,\ldots$ coefficients. $\endgroup$ Commented Jul 19, 2022 at 20:51

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