Use Laplace transform to solve $f'(t) = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t')$

Question

Consider the following differential equation for $f$ $$ \frac{df}{dt} = H(t - a) \int_{0}^{t-a} dt'\; g(t') f(t-t') $$ where we have some initial condition $f(0) = f_0$, the function $g(t)$ is known and $a>0$ is a parameter, and $H(t)$ is the Heaviside step function.

Assuming that $t-a>0$, we can ignore the Heaviside function that appears in the RHS.

I would like to solve the above using a Laplace transform, but the upper limit on the integral makes the RHS of the DE not exactly a convolution. I can write the RHS of the above as (assuming $t-a>0$) $$ H(t-a) \int_{0}^{t-a} dt'\; g(t') f(t-t') = \int_{0}^{t} ds\; g(t') H(t - a - t') f(t-t') = \int_{0}^{t} dt'\; \left[ H(t' - a) f(t) \right] g(t-t') $$ So the above can be written as a convolution of the function $g(t)$ and $e_{a}(t) := H(t - a) f(t)$. This means that taking the Laplace transform of my DE ends up in $$ s F(s) - f(0) = G(s) E_a(s) $$ where $\mathcal{L}\{ f \}(s) = F(s)$, and $\mathcal{L}\{ g \}(s) = G(s)$, and $\mathcal{L}\{ e_a \}(s) = E_a(s)$ (where $\mathcal{L}$ is the usual one-sided Laplace transform)

Using the above, I cannot isolate for $F(s)$ in my DE (since $E_a(s)$ does not depend easily on $F(s)$).

How do I solve this differential equation with the Laplace transform?

Answer 1

Upon a first look, one notices immediately that the derivative is zero in the region $t<a$ and is also continuous at $t=a$. We immediately conclude that $f$ assumes the piecewise constant form

$$f(t)=\begin{Bmatrix}f_0 &~,~t\leq a\\f_0+\int_{a}^th(x)dx&~,~t>a\end{Bmatrix}$$

One computes directly from this form that $\frac{df}{dt}=H(t-a)h(t)$, so $h(t)$ is to be identified with the convolution-like integral on the RHS of the ODE to be solved. Now we consider the ODE in the region $t>a$ and integrate it with a Laplace weight

$$\int_{a}^{\infty}\frac{df}{dt}e^{-st}dt=\int_{a}^{\infty}dt~ e^{-st}\int_{0}^{t-a}dt'g(t')f(t-t')$$ The left-hand side evaluates to $$LHS=s\int_{a}^{\infty}f(t)e^{-st}dt-f(a)e^{-sa}\equiv s F_a(s)-f_0e^{-sa}$$ In the right-hand side we change the order of integration carefully and after a shift in $t$ we obtain $$RHS=\int_{0}^{\infty}dt'g(t')\int_{t'+a}^{\infty}f(t-t')e^{-st}dt=\int_{0}^{\infty}g(t')e^{-st'}\int_{a}^{\infty}f(t)e^{-st}=G(s)F_a(s)$$ Equating the two and rearranging we obtain that $$F_a(s)=\frac{f_0 e^{-sa}}{s-G(s)}$$ and thus the Laplace transform of the function in the entire domain is given by $$F(s)=f_0\frac{1-e^{-sa}}{s}+\frac{f_0 e^{-sa}}{s-G(s)}$$ Of course this form could have been derived directly from the very first equation. Note that solving this equation by a Laplace transform may not pick up various solutions that may happen to not have a well-defined Laplace transform.

Finally, this reproduces the solution for the case where $g(t)=b=const.$, where the problem can be solved without using a Laplace transform:

$$f'(t)=b\int_{a}^t f(t')dt'\Rightarrow \int_a^t dt'f(t')=A(e^{\sqrt{b}t}+e^{\sqrt{b}(2a-t)})\Rightarrow f(t)=\frac{f_0}{2}\sinh((t-a)\sqrt{b})$$ whose Laplace transform reads (taking into account the constant part as well) $$F(s)=f_0\frac{1-e^{-sa}}{s}+f_0\frac{e^{-sa}}{s-\frac{b}{s}}$$ which directly confirms the result of the method presented here.

Answer 2

$$f'(t)=H(t-a)\int_{0}^{t-a} g(s) f(t-s)\mathrm{d}s =\int_{0}^\infty g(s)H\big((t-a)-s\big) f(t-s)\mathrm{d}s.$$

Note that for $t<a$, $f'(t)=0$ and hence $f(t)=f(0)$ for $0\leq t\leq a$. The Laplace transform of $H(t-a) f(t)$ is given by

\begin{align*} \mathcal{L}\big\{H(t-a) f(t)\big\}(s)&= % \int_{0}^\infty e^{-sx}H(x-a) f(x)\mathrm{d}s % =F(s)-\int_{0}^a e^{-sx}f(x)\mathrm{d}s\\ % &=F(s)+\frac1s e^{-sx}f(x)\biggm|_0^a-\frac1s\int_{0}^a e^{-sx}f'(x)\mathrm{d}s= % F(s)-\frac1s f(0)(1-e^{-as}). % \end{align*}

So,

$$F(s)=f(0)\frac{1-\frac1s(1-e^{-as})}{s-G(s)}.$$