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When solving an ODE via a power series at an ordinary (nonsingular) point, the initial guess is $y = \sum_{n = 0}^\infty a_n x^n$.

When solving an Euler ODE, the second order equation $x^2 y'' + pxy' + qy = 0$ where $p$ and $q$ are constant coefficients for example, the initial guess is $y = x^s$.

When solving an ODE via The Method Of Frobenius at a regular singular point, the initial guess is said to be the product of the two prior guesses, $y = x^s \sum_{n = 0}^\infty a_n x^n$.

My question is whether The Method Of Frobenius is a generalization of each of the two prior solution techniques, and if not, what its relationship is to those techniques. To break this down:

  1. Does applying The Method Of Frobenius to an ODE at an ordinary point always produce $s = 0$ with algebraic multiplicity equal to the order of the ODE?

  2. Does applying The Method Of Frobenius to an ODE and finding that $s = 0$ with algebraic multiplicity equal to the order of the ODE entail that the point approximated around was ordinary?

  3. Does applying The Method Of Frobenius to an Euler ODE always produce a power series factor $\sum_{n = 0}^\infty a_n x^n$ in the guess identically equal to $1$?

  4. Does applying The Method Of Frobenius to an ODE and finding that the power series factor $\sum_{n = 0}^\infty a_n x^n$ in the guess is identically equal to $1$ entail that the ODE was an Euler ODE?

If the answers are not all "yes," then The Method Of Frobenius is not a generalization of each of the other solution techniques. In this case, what is the correct, high-level way to think about how these solution techniques relate, aside from symbolic similarity?

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  • $\begingroup$ Is the following link useful for your answer ? math.stackexchange.com/questions/326758/… $\endgroup$
    – nmasanta
    Commented Feb 19, 2021 at 8:38
  • $\begingroup$ @nmasanta I had read this before I posted the question. SasQ's answer is interesting, and may indeed contain the key takeaway, but I was hoping my question would provide some insight from a different angle. $\endgroup$
    – user10478
    Commented Feb 20, 2021 at 21:07
  • $\begingroup$ Let's see @user10478 $\endgroup$
    – nmasanta
    Commented Feb 21, 2021 at 3:16

1 Answer 1

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I'm going to first remark that the respective assumptions $x^s\sum a_nx^n$ and $\sum a_nx^n$ are equivalent, and that $x^s$ is a relevant special case of either. I'll also remark that if we accept the notion of a generalization to have some structural relevance as opposed to simply looking like it is more general, then the series assumption and the Frob. assumption are the same. For example, I could say $1+x^s$ is more general than $1+x$, and in a naïve way I would be correct by seeing that $s=1$ produces $1+x$--but the keen eye will note that by changing $x\to x^s$, we get the same form.

To 3.: Clearly for the Euler ODE, if we assume $y$ is some polynomial of degree $n$, then the first term $x^2y''$ remains degree $n$, and the term $pxy'$ remains degree $n$, and the term $qy$ remains degree $n$. Since our $x$ is not a root and is independent, one must solve for the coefficients such that each new coefficient at the $n$th term is $0$. In other words, assuming $y=a_0+a_1x+\cdots + a_nx^n$ produces $x^2y'' +pxy'+qy = A_nx^n+A_{n-1}x^{n-1}+\cdots + A_0=0$ and so each $A_k=0$. More exactly, $$x^2y'' + pxy' +qy = \sum [n^2+(p-1)n+q]a_nx^n$$ and so $$n^2+(p-1)n+q=0$$ Therefore $n$ only takes on two values for the sum to be valid. This is the same if I had assumed $x^s$ in the original equation to obtain $$s^2+(p-1)s+q=0$$ which obviously produces $$y=C_1x^{s_1}+C_2x^{s_2}.$$

And clearly, it is not true that in the guess $y=x^s\sum a_nx^n$ will have the term $\sum a_nx^n=1$, but rather some combination of indices telling you essentially the same thing.

To 4.: Your statement in (4) is clearly false unless you call all ODEs of the form $$x^ny^{(n)}+a_{n-1}x^{n-1}y^{(n-1)}+\cdots =0$$ Euler ODEs. However, it's rather arbitrarily false since what is called the general Euler-Cauchy ODE is the second-degree case $n=2$. Your statement (4) is, respectfully, not a requirement for the guess $y=x^s\sum a_nx^n$ to be a more general kind of guess. It's actually true that the guess $y=\sum a_nx^n$ suffices for an Euler ODE, because you'll find the conditions necessary for the index.

Lastly, the assumption $y=uv$ or knowing that $u'v+v'u = (uv)'$ is probably one of the most important in differential equations, alongside using those assumptions to bend the equation to your will, into the suitable forms you choose or which make the problem more obvious for you with respect to your solution goals. And this is why the Frob. method works its magic in certain relevant cases.

I hope this may have answered your questions.

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