Coupled functions $f,g$ satisfy $Lf(\lambda) = \frac{\beta}{\beta + \lambda} Lg(\lambda)$, solve for $f$

Question

Let $\beta, \delta \geq 0$, $\beta + \delta > 0$. Let $f(t), g(t) : \mathbb{R}_{\geq 0} \rightarrow [0,1]$ satisfying the following equations: $$f(t) = \int_{0}^t \beta e^{-\beta (t-s)} g(s) ds$$ $$g(t) = e^{-\delta t} + \int_{0}^{t} \delta e^{-\delta(t-s)}f(s) ds$$ From this I was able to derive the following identity: $$Lf(\lambda) = \frac{\beta}{\beta + \lambda} Lg(\lambda),\text{ } \lambda > 0$$ where $L$ denotes the Laplace transform, that is $Lf(\lambda) = \int_{0}^\infty f(t)e^{-\lambda t}dt$. Now I would like to show that $$f(t) = \frac{\beta}{\beta + \delta} \left(1-e^{-(\beta + \delta)t} \right)$$ I was unable to show this. I have tried playing around with the expressions, for example plugging in the definition of $g$ into the Laplace identity equation, but this lead nowhere. I am not very familiar with Laplace transforms, is there maybe some property I could use? Maybe apply an inverse Laplace transform? What can I try here?

Answer 1

Let's first compute $Lg(\lambda)$: \begin{align} Lg(\lambda)&=\int_0^{\infty}e^{-\lambda t}g(t)\,dt \\ &=\int_0^{\infty}e^{-\lambda t}\left(e^{-\delta t} + \int_{0}^{t} \delta e^{-\delta(t-s)}f(s)\,ds \right)dt \\ &=\frac{1}{\lambda+\delta}+\delta\int_0^{\infty}\int_0^te^{-(\lambda+\delta)t+\delta s}f(s)\,dsdt. \tag{1} \end{align} Interchanging the order of integration in the double integral yields \begin{align} Lg(\lambda)&=\frac{1}{\lambda+\delta}+\delta\int_0^{\infty}\int_s^{\infty} e^{-(\lambda+\delta)t+\delta s}f(s)\,dtds \\ &=\frac{1}{\lambda+\delta}+\delta\int_0^{\infty}e^{\delta s}f(s)\frac{e^{-(\lambda+\delta)s}}{\lambda+\delta}\,ds \\ &=\frac{1}{\lambda+\delta}+\frac{\delta}{\lambda+\delta}\int_0^{\infty} e^{-\lambda s}f(s)\,ds \\ &=\frac{1}{\lambda+\delta}\left[1+\delta\, Lf(\lambda)\right]. \tag{2} \end{align} Plugging $(2)$ into the identity $$ Lf(\lambda) = \frac{\beta}{\beta + \lambda} Lg(\lambda) \tag{3} $$ and solving for $Lf(\lambda)$ gives us $$ Lf(\lambda)=\frac{\beta}{\lambda^2+(\beta+\delta)\lambda}=\frac{\beta}{\beta+\delta}\left(\frac{1}{\lambda}-\frac{1}{\lambda+\beta+\delta}\right). \tag{4} $$ Applying the inverse Laplace transform to both sides of $(4)$ we finally obtain $$ f(t) = \frac{\beta}{\beta + \delta} \left(1-e^{-(\beta + \delta)t} \right). \tag{5} $$

Answer 2

(Partial answer)

If I were you, I would use Leibniz integral rule instead of Laplace transform. Indeed, one gets by differentiating $f$ and $g$ : $$ \begin{cases} f'(t) = \beta g(t) - \beta f(t) \\ g'(t) = \delta f(t) - \delta g(t) \end{cases} $$ This pair of differential equations may be recasted in a matrix form as $$ \frac{\mathrm{d}}{\mathrm{d}t} \begin{pmatrix} f \\ g \end{pmatrix} = \begin{pmatrix} -\beta & \beta \\ \delta & -\delta \end{pmatrix} \begin{pmatrix} f \\ g \end{pmatrix} $$ and solved by $\Psi(t) = e^{At}\,\Psi_0$, where $\Psi(t) := (f(t),g(t))$ are the unknowns with the initial conditions $\Psi_0 = (f(0),g(0)) = (0,1)$ and $A$ is the matrix above. I let you compute its exponential and the final result.