Solving a pair of kinetic equations using the Laplace transform

Question

I have the following set of kinetic equations:

\begin{align} \frac{dx}{dt}&=r_1\delta(t-t_0)-(r_2+r_3)x(t)\\ \frac{dy}{dt}&=r_3x(t)-r_5y(t). \end{align}

How can I solve for $y(t-t_0)$ using the Laplace transform?

Answer 1

Step 1: Solving for $x$

First, I apply the Laplace transform to $\frac{dx}{dt}$. This gives me

$$ x'_L(s)=-x(0)+sx_L(s), $$

where $x'_L(s)$ is the Laplace transform of $\frac{dx}{dt}$ and $x_L(s)$ is the Laplace transform of $x(t)$.

I'm going to make the assumption that $t-t_0\geq0$ -- i.e., $t\geq t_0$. Since, by definition, $t$ cannot be less than $0$, if $t=0$ (as it does above), then $t_0=0$. I'm also going to assume that $x(0)=y(0)=0$. Hence, we have

$$ x'_L(s)=sx_L(s). $$

Substituting this into the equation for $\frac{dx}{dt}$ at the top, we have

\begin{align} sx_L(s)&=\mathcal{L}\{r_1\delta(t-t_0)-(r_2+r_3)x(t)\}\\ &=r_1\mathcal{L}\{\delta(t-t_0)\}-(r_2+r_3)x_L(s). \end{align}

In my table of commonly encountered Laplace transforms, I have

$$ \mathcal{L}\{\delta(t-c)\}=e^{-cs}. $$

Hence, we have

\begin{align} sx_L(s)&=r_1e^{-t_0s}-(r_2+r_3)x_L(s)\\ sx_L(s)+(r_2+r_3)x_L(s)&=r_1e^{-t_0s}\\ x_L(s)(s+r_2+r_3)&=r_1e^{-t_0s}\\ x_L(s)&=\frac{r_1e^{-t_0s}}{s+r_2+r_3}\\ x_L(s)&=r_1e^{-t_0s}f_L(s), \end{align}

where $f_L(s)=\frac{1}{s+r_2+r_3}$.

Now we take the inverse Laplace tranform to solve for $x$:

\begin{align} x(t)&=r_1\mathcal{L}^{-1}\{e^{-t_0s}f_L(s)\}. \end{align}

In my table, I have

\begin{align} \mathcal{L}\{H(t-c)f(t-c)\}&=e^{-cs}f_L(s)\\ \mathcal{L}\{e^{at}\}&=\frac{1}{s-a}, \end{align}

where $H$ is the Heaviside function and $s>a$.

Hence, we have

\begin{align} x(t)&=r_1H(t-t_0)e^{(-r_2-r_3)(t-t_0)}. \end{align}

Since $t-t_0\geq0$, we have

$$ x(t)=r_1e^{(-r_2-r_3)(t-t_0)}. $$

Step 2: Solving for $y$

First, I apply the Laplace transform to $\frac{dy}{dt}$. This gives me

\begin{align} y'_L(s)&=-y(0)+sy_L(s)\\ &=sy_L(s). \end{align}

Substituting this into the equation for $\frac{dy}{dt}$ at the top, we have

\begin{align} sy_L(s)&=\mathcal{L}\{r_3x(t)-r_5y(t)\}\\ &=r_3\mathcal{L}\{x(t)\}-r_5y_L(s)\\ &=r_3\mathcal{L}\{r_1e^{(-r_2-r_3)(t-t_0)}\}-r_5y_L(s)\\ &=r_3\bigg(r_1\mathcal{L}\{e^{(-r_2-r_3)t}\}\bigg)-r_5y_L(s)\\ sy_L(s)+r_5y_L(s)&=r_3r_1\mathcal{L}\{e^{(-r_2-r_3)t}\}\\ y_L(s)(s+r_5)&=r_3r_1\mathcal{L}\{e^{(-r_2-r_3)t}\}\\ y_L(s)&=\frac{r_3r_1\mathcal{L}\{e^{(-r_2-r_3)t}\}}{s+r_5}\\ y_L(s)&=\frac{r_3r_1\frac{1}{s+r_2+r_3}}{s+r_5}\\ y_L(s)&=r_3r_1\frac{1}{s+r_2+r_3}\frac{1}{s+r_5}.\\ \end{align}

In my table, I have

$$ \mathcal{L}\bigg\{\int_0^tf(t-\tau)g(\tau)d\tau\bigg\}=f_L(s)g_L(s). $$

With this, we can take the inverse Laplace transform to solve for $y$:

$$ y(t)=r_3r_1\mathcal{L}^{-1}\bigg\{\frac{1}{s+r_2+r_3}\frac{1}{s+r_5}\bigg\} $$

We know from above that if $f_L(s) = \frac{1}{s+r_2+r_3}$, then $f(t)=e^{(-r_2-r_3)t}$. Similarly, if $g_L(s) = \frac{1}{s+r_5}$, then $g(t)=e^{-r_5t}$.

Hence, we have

\begin{align} y(t)&=r_3r_1\int_0^t{e^{(-r_2-r_3)(t-\tau)}e^{-r_5\tau}d\tau}\\ &=r_3r_1e^{(-r_2-r_3)t}\int_0^t{e^{(r_2+r_3-r_5)\tau}d\tau}\\ &=r_3r_1e^{(-r_2-r_3)t}\bigg[\frac{e^{(r_2+r_3-r_5)\tau}}{r_2+r_3-r_5}\bigg]_0^t\\ &=r_3r_1e^{(-r_2-r_3)t}\bigg(\frac{e^{(r_2+r_3-r_5)t}}{r_2+r_3-r_5}-\frac{1}{r_2+r_3-r_5}\bigg)\\ &=r_3r_1\frac{e^{-r_5t}-e^{(-r_2-r_3)t}}{r_2+r_3-r_5}. \end{align}

Ta da!