I am given the general wave equation: $$\sum_{n=0}^{\infty}a_n \frac{\partial^n F(x,t)}{\partial x^n}=\sum_{l=1}^{\infty}b_l\frac{\partial^l F(x,t)}{\partial t^l}$$
I am supposed to show that $F(x,t)=Ae^{i(kx-\omega t)}$ solves this equation and then determine the type of equation that results when plugging in the function $F(x,t)$.
What I have got so far:
$$\sum_{n=0}^{\infty}a_n \frac{\partial^n Ae^{i(kx-\omega t)}}{\partial x^n}=\sum_{l=1}^{\infty}b_l\frac{\partial^l Ae^{i(kx-\omega t)}}{\partial t^l} \\ \iff \sum_{n=0}^{\infty}a_n(ik)^nF(x,t)=\sum_{l=1}^{\infty}b_l (-i\omega )^lF(x,t)$$
There are three problems/questions that arise:
- Why does the right sum start at $l=1$ and not at $l=0$ like the left sum?
- In order for this equation to hold, $$a_n(ik)^n \stackrel{!}{=} b_l(-i \omega)^n$$ but I would have to shift the index of the second sum right?
- Woul it be correct to say that the resulting equation is a system of linear equations?
Thanks for your help!
Consider you have the term with $l=0$: $b_0F(x,t)$, but with $a_0F(x,t)$, you can redefine $a'_0=a_0-b_0$ and recover the sums with different start index.
$\displaystyle\sum_{n=0}^{\infty}a_n(ik)^nF(x,t)=\sum_{l=1}^{\infty}b_l (-i\omega )^lF(x,t)\implies\sum_{n=0}^{\infty}a_n(ik)^n=\sum_{l=0}^{\infty}b_l (-i\omega )^l$
we cannont assume that the sums are always equal term by term. Consider, e.g. $f(x)=\displaystyle\sum_{n=0}^{\infty}a_nx^n=e^x$ and $g(x)=\displaystyle\sum_{n=0}^{\infty}b_nx^n=e^{x^2}$ (starting at $n=0$, but we can redefine $a_0$!!)
Let's assume that the series are equal term by term, $\dfrac{(ik)^{2n}}{(2n)!}=\dfrac{(-i\omega)^{2n}}{n!};\dfrac{(ik)^{2n+1}}{(2n+1)!}=0\;\forall n$ That leads to $k=\omega=0$ as the only solution when clearly, $e^{ik}=e^{(-i\omega)^2}$ or $ik=-\omega^2+2\pi i m$
So, the equation to solve is $f(ik)=g(-i\omega)$ or $k=-if^{-1}(g(i\omega))$ to plug it into the solution...