Does the Solution Lie in the Ring of Exponential Sums?

[mathmari]

Hey!

We consider the differential equation $Ly=f$ in the ring of exponential sums $\mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$ so we have that $f=\sum_{i=0}^n C_i e^{\lambda_i x}$.

If we apply the superposition principle we have to solve differential equations of the form $Ly=e^{bx}$.

If $b$ is a root of the characteristic equation of the homogeneous equation of multiplicity $M$, then the solution is of the form $Cx^Me^{b x}\notin \mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$.

If $b$ is not a root of the characteristic equation of the homogeneous equation, then the solution is of the form $Ce^{b x}\in \mathbb{C}[e^{\lambda x} \mid \lambda \in \mathbb{C}]$. Therefore, the differential equation has a solution in the ring if $b$ is not a root of the characteristic equation, right?

This is equivalent to $L(e^{b x}) \neq 0$, right?

Does this stand also for the original differential equation? So is it $$L\left (\sum_{i=0}^n C_i e^{\lambda_i x}\right ) \neq 0$$ ? (Wondering)

 

[jvicens]

Hello! Yes, you are correct. If $b$ is not a root of the characteristic equation, then the solution will be in the ring of exponential sums. This is because the solution will be of the form $Ce^{bx}$, which can be written as a sum of exponential terms in the form $e^{\lambda x}$.

And yes, this also holds for the original differential equation. The superposition principle allows us to break down the original equation into simpler equations, and as long as $L(e^{bx}) \neq 0$, the solution will be in the ring of exponential sums. So for the original equation $Ly=f$, we have $L\left (\sum_{i=0}^n C_i e^{\lambda_i x}\right ) \neq 0$.

I hope this helps clarify things for you! Let me know if you have any other questions.