Series Solution for O.D.E (Frobenius method?)

In summary: However, in order to do this, you need to find a point at which the power series converges. This is where the problem becomes strange, as you need to find a point at which the cosine function has a zero at x=0. Once you find this point, you can then write out the power series for x=0.
  • #1
JacobHempel
3
0

Homework Statement


(For Physics 306: Theoretical methods of Physics... Text book: Mathematical Tools for Physics (really good!)...
Assumption: 300 level class = considerably junior level class)

-- Find a series solution about x=0 for y''+ysec(x) = 0, at least to a few terms.

(Ans: a[itex]_{0}[/itex][1-(1/2)x[itex]^{2}[/itex]+0x[itex]^{4}[/itex]+1/720x[itex]^{6}[/itex]+...]+a[itex]_{1}[/itex][x-(1/6)x[itex]^{3}[/itex]-(1/60)x[itex]^{5}[/itex]+...]


Homework Equations



Frobenius Solution: Supposing there is a singularity point at x=x0, a possible solution is given by the power series, that is

Ʃ(from k=0 to inf) a[itex]_{k}[/itex]x[itex]^{k+s}[/itex]


The Attempt at a Solution



There is a singularity point not at x=0, but at x= npi/2. The problem specifies, though, to try finding a power series solution about x=0.

Let x(t) = Ʃ a[itex]_{k}[/itex]x[itex]^{k+s}[/itex]

Using this, plug it into the equation to obtain

Ʃ a[itex]_{k}[/itex]x[itex]^{k+s-2}[/itex] (k+s)(k+s-1) +Ʃ a[itex]_{k}[/itex]x[itex]^{k+s}[/itex] = 0

after that I matched the terms up by pushing the first summation up by two indicese. To do this, I wrote out two terms, then shifted the k values by +2.

a[itex]_{0}[/itex](s)(s-1)x[itex]^{s-2}[/itex]+a[itex]_{1}[/itex](1+s)(s)x[itex]^{s-1}[/itex]+Ʃ a[itex]_{k+2}[/itex]x[itex]^{k+s}[/itex] (k+s+2)(k+s+1) +Ʃ a[itex]_{k}[/itex]x[itex]^{k+s}[/itex] = 0

This is where I stopped. I'm unsure as to what to do next. You match up the terms in a regular Forbenius method, but this time it has a cos(x) in the second term, which is what makes this problem strange to me.

I think the next step is to find s, then just write out a few terms of both the series.

Also sorry about the mess regarding the equations. I got rid of all the sigma indices, so when you see a sigma, it is always from k=0 to k=∞.
 
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  • #2
Expand the secant function into a power series.
 
  • #3
Since there's no singularity at x=0, you can use a plain old power series and avoid trying to find s from an indicial equation.
 

What is the Frobenius method?

The Frobenius method, also known as the power series method, is a technique used to find a series solution for a second-order ordinary differential equation (ODE) with a regular singular point. It involves expressing the unknown function as a power series and substituting it into the ODE to determine the coefficients of the series.

When is the Frobenius method used?

The Frobenius method is used when solving second-order ODEs with a regular singular point, which is a point where the coefficient of the highest derivative term becomes zero. This method is also useful when the ODE cannot be solved by other techniques such as separation of variables or substitution.

What is a regular singular point?

A regular singular point is a point in the domain of a differential equation where the coefficient of the highest derivative term becomes zero. This can cause the ODE to become singular, meaning its solutions are no longer unique. The Frobenius method is used to find a series solution at this point.

How does the Frobenius method work?

The Frobenius method involves expressing the unknown function as a power series and substituting it into the ODE. This results in an infinite series of equations, where each term is multiplied by a different power of the independent variable. By equating the coefficients of each power, a recurrence relation is obtained, which can then be used to find the coefficients of the power series and hence the series solution of the ODE.

When is the Frobenius method not applicable?

The Frobenius method is not applicable when the ODE has an irregular singular point, which is a point where the coefficient of the highest derivative term is non-zero but still causes the ODE to become singular. In this case, other techniques such as the Laplace transform or numerical methods may be used to find a solution.

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