I am a physics student modeling a quantum mechanical system for my undergraduate research. I am currently solving for the time-dependent coefficients $c_1(t), c_2(t), c_3(t)$ on a super position of three tensor product states and have derived a system of coupled ODEs. I ran this system by my research professor and he confirmed that my system and its derivation are okay. So, I would now like to solve it but have been having some difficulty in doing so. The system is as follows:
- $$\dot{c}_1= \alpha c_1 + \beta c_2 + \beta c_3$$
- $$\dot{c}_2= (\beta+\gamma)c_1 + \alpha c_2$$
- $$ \dot{c}_3 = \beta c_1 +(\alpha + \delta) c_3 $$
where $\alpha, \beta \in \mathbb{C}$ and $\gamma,\delta \in \mathbb{R}$ are constants.
I first applied the ansatz $c_i(t) = v_i e^{\lambda t}$ where $i \in \{1,2,3\}$. In this approach, $v_i$ is the component of an eigenvector and $\lambda$ an eigenvalue. However, the resulting characteristic polynomial turns out to be a cubic and somewhat complicated.
My professor was of the opinion that a Laplace transform is the better approach. So, I have tried that instead, but I am having trouble bringing the first result into an inverse-Laplace-friendly form.
For the Laplace transform, I use the initial conditions $c_1(0)=1, c_2(0)=0, c_3(0)=0$ and thus obtain
- $$ s C_1(s)-1= \alpha C_1(s) + \beta C_2(s)+\beta C_3(s)$$
- $$ s C_2(s) = (\beta + \gamma) C_1(s) + \alpha C_2(s) $$
- $$ s C_3(s) = \beta C_1(s) + (\alpha + \delta) C_3(s) $$
Solving for everything in terms of $C_1(s)$, I obtain
$$ C_2(s) = \frac{\beta + \gamma}{s-\alpha}C_1(s) $$
and
$$C_3(s) = \frac{\beta}{s-\alpha-\delta} C_1(s)$$
Applying these to (4), I then get
$$ C_1(s) \bigg[s-\alpha - \frac{\beta (\beta + \gamma)}{s-\alpha} - \frac{\beta^2}{s-\alpha-\delta} \bigg] = 1$$
I am trying not to "muddle" the factor on $C_1(s)$ because I recall that we want to keep fractions and expressions containing $s$ as neatly parsed out as possible so that taking the inverse Laplace transform is easier. However, it seems like I must take the inverse of the factor on $C_1(s)$ such that I end up with
$$C_1(s) = \bigg[s-\alpha - \frac{\beta (\beta + \gamma)}{s-\alpha} - \frac{\beta^2}{s-\alpha-\delta} \bigg] ^{-1}$$
I have tried checking my Laplace tables for anything helpful from this point, but it seems that I must "condense and flip" such that I have
$$C_1(s) = \frac{(s-\alpha)(s-\alpha-\delta)}{(s-\alpha)^2(s-\alpha-\delta)-\beta (\beta + \gamma)(s-\alpha-\delta) - \beta^2 (s-\alpha)}$$
I would like to somehow parse the fraction apart via partial fraction decomposition, but I cannot see a way to helpfully factor the denominator. Did I perhaps miss a step I could've taken prior to this? I have tried to see if I can somehow take the inverse Laplace of the initial form
$$ C_1(s) \bigg[s-\alpha - \frac{\beta (\beta + \gamma)}{s-\alpha} - \frac{\beta^2}{s-\alpha-\delta} \bigg] = 1$$
because the coefficient on $C_1(s)$ is already in such a tidy form, but none of the tables I can find indicate that I could take the inverse Laplace of the product $C_1(s) \bigg[...\bigg]$. Have I made a mistake somewhere, or is Laplace not the best approach here?
Thank you for any help you can provide.