What is the next step in solving this Frobenius method problem?

[TheFerruccio]

I am having an issue understanding the Frobenius method. I plan on posting multiple questions in this thread as I go. I searched these forums and saw some other threads pointing to it, but the questions were slightly different. I am able to get through the problem to a certain extent, but after a point, I don't know where to go on.

Example 1:x^{2}y''-3xy'+4y = 0

This can be put into the format:

y''-\frac{3x}{x^{2}}y'+\frac{4}{x^{2}}y

Which is consistent with the form that can be solved using the Frobenius method.

I assume the solution y(x) and its derivatives to be in this form, multiplying by the power x^{r}:

y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}
y'(x)=\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1}
y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r-2}

I plug these into the original equation, getting:

\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r}-3\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r}+4\sum_{n=0}^{\infty}a_{n}x^{n+r} = 0

I find that the lowest power in this equation is x^{r}. Plugging n=0 into the equation, and pulling out x^{r}, I get:

(r-1)ra_{0} - 3ra_{0} + 4a_{0} = 0
a_{0}(r^{2}-r-3r+4) = 0
a_{0}(r^{2}-4r+4) = 0

Assuming a_{0} \neq 0, making the solution non-trivial, I get the roots for r, which are r_{1},r_{2} = 2

I use one of the roots, r=2, and plug in (here's my attempt to get the TeX to work again)

\sum_{n=0}^{\infty}(n+1)(n+2)a_{n}x^{n+2}-3\sum_{n=0}^{\infty}(n+2)a_{n}x^{n+2}+4\sum_{n=0}^{\infty}a_{n}x^{n+2}

I do not know where to go from this point. I cannot get a_{n+1} or a_{n+2} in terms of a_{n+1} or a_{0}. Every single result results in some constant, multiplied by a_{n} equals 0. So, at this point. I am lost.

NOTE: I don't want the answer. I just want some clues on how I should continue.

 

[ross_tang]

In fact, what you have show is that, x^2 is one of the solution. Since for all other terms other than a_0, they have to be zero in order for the expression to equal to zero.

 

[TheFerruccio]

I'm a tad confused, because what I'm ending up with is a kind of useless term.

The x^{2} terms in the last equation reduce to 0a_{0} = 0 which doesn't tell me anything!

For the higher order terms, I do end up with some coefficient * a_{n}, which shows that the higher order terms have to equal 0.

I'm still left with not enough information, though. I'm confused as to how x^{2} term can be a solution.

 

[ross_tang]

Please substitute y = x^2 to your differential equation, and you will find that it is indeed one of the solution.

y = x^2
y' = 2x
y'' = 2

Therefore:

x^2 y'' -3x y' + 4y = 2 x^2 - 6 x^2 + 4x^2 = 0

 

[HallsofIvy]

I am having an issue understanding the Frobenius method. I plan on posting multiple questions in this thread as I go. I searched these forums and saw some other threads pointing to it, but the questions were slightly different. I am able to get through the problem to a certain extent, but after a point, I don't know where to go on.

Example 1:


x^{2}y''-3xy'+4y = 0

This can be put into the format:

y''-\frac{3x}{x^{2}}y'+\frac{4}{x^{2}}y

Which is consistent with the form that can be solved using the Frobenius method.

I assume the solution y(x) and its derivatives to be in this form, multiplying by the power x^{r}:

y(x)=\sum_{n=0}^{\infty}a_{n}x^{n+r}
y'(x)=\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r-1}
y''(x)=\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r-2}

I plug these into the original equation, getting:

\sum_{n=0}^{\infty}(n+r-1)(n+r)a_{n}x^{n+r}-3\sum_{n=0}^{\infty}(n+r)a_{n}x^{n+r}+4\sum_{n=0}^{\infty}a_{n}x^{n+r} = 0

I find that the lowest power in this equation is x^{r}. Plugging n=0 into the equation, and pulling out x^{r}, I get:

(r-1)ra_{0} - 3ra_{0} + 4a_{0} = 0
a_{0}(r^{2}-r-3r+4) = 0
a_{0}(r^{2}-4r+4) = 0

Assuming a_{0} \neq 0, making the solution non-trivial, I get the roots for r, which are r_{1},r_{2} = 2

I use one of the roots, r=2, and plug in (here's my attempt to get the TeX to work again)

\sum_{n=0}^{\infty}(n+1)(n+2)a_{n}x^{n+2}-3\sum_{n=0}^{\infty}(n+2)a_{n}x^{n+2}+4\sum_{n=0}^{\infty}a_{n}x^{n+2}

This can be written as a single sum:
\sum_{n=0}^\infty [(n+1)(n+2)- 3(n+2)+ 4)a_n= 0
so that you must have either a_n= 0 or (n+1)(n+2)- 3(n+2)+ 4= n^2+ 3n+ 2-3n- 6+ 4= n^2= 0
That is, a_n= 0 for all n> 0. Since your solution is of the form "Cx^{n+ 2}" that means that your solution is Cx^2.

And ross_tang has verified that, indeed, y= Cx^2 is a solution.

I do not know where to go from this point. I cannot get a_{n+1} or a_{n+2} in terms of a_{n+1} or a_{0}. Every single result results in some constant, multiplied by a_{n} equals 0. So, at this point. I am lost.

NOTE: I don't want the answer. I just want some clues on how I should continue.

 

[TheFerruccio]

This can be written as a single sum:
\sum_{n=0}^\infty [(n+1)(n+2)- 3(n+2)+ 4)a_n= 0
so that you must have either a_n= 0 or (n+1)(n+2)- 3(n+2)+ 4= n^2+ 3n+ 2-3n- 6+ 4= n^2= 0
That is, a_n= 0 for all n> 0. Since your solution is of the form "Cx^{n+ 2}" that means that your solution is Cx^2.

And ross_tang has verified that, indeed, y= Cx^2 is a solution.

Yes, I can see that it is verified that x^{2} is a solution, just by plugging in, I'm still having trouble seeing how that answer was arrived at. Let me think a bit.

The ODE reduces to the form of \sum_{n=0}^{\infty}n^{2}a_{n}x^{n+2} So either a_{n} = 0 or n^{2} = 0 for all n. It's been assumed that a_{n}\neq0 so n^{2}=0 So n=0 for the statement to be true. So, one solution becomes the form:

\sum_{n=0}^{\infty}a_{n}x^{n+2} Thus, the first solution is a_{0}x^{2}?

I think I realized where my confusion came from. Of course the equation in the form of summations would reduce to 0, because that's not one of the solutions! That's the ODE. I was mixing up the solution with the ODE. I just put it in summation form to solve for n, then plugged that n back into y(x).

Thanks for the help. If what I typed is correct, then I understand this problem so far. I will work on finding the second solution.

 

[LawlQuals]

I am not sure I understand what you are saying with the whole n^2 = 0 business. Maybe a closer inspection of all posts on this thread would help me see. Put another way (or, perhaps the same way),

You found a value for r = 2 under the proviso that a_0 \neq 0. Your "recursive" relation (which relates the coefficients of one term to the coefficients of others), reduces to (as you have shown, with r = 2):

(const)*a_n = 0

so that all a_n terms are indeed zero, except the one that you enforced to not be so (a_0 as you had stated). You can piece together the Frobenius solution then by noting you were seeking a solution of the form:

y(x) = \sum_{n = 0}^{\infty} a_n x^{n+r}

but r = 2,

y(x) = \sum_{n = 0}^{\infty} a_n x^{n+2} = y(x) = x^2 \sum_{n = 0}^{\infty} a_n x^{n} = x^2 \left( a_0 x^0 + a_1 x^1 + a_2 x^2 + \ldots\right)

you have just shown that a_0 \neq 0, while a_n = 0 ,\, n\neq 1. this means all terms vanish except a_0, so we have the above reducing to:

y(x) = a_0x^2

a_0 may have been the first coefficient, but in a Frobenius solution, the first term n = 0 corresponds to a power of x^{n+r} = x^{r} = x^2.

 

[TheFerruccio]

I am not sure I understand what you are saying with the whole n^2 = 0 business. Maybe a closer inspection of all posts on this thread would help me see. Put another way (or, perhaps the same way),

You found a value for r = 2 under the proviso that a_0 \neq 0. Your "recursive" relation (which relates the coefficients of one term to the coefficients of others), reduces to (as you have shown, with r = 2):

(const)*a_n = 0

so that all a_n terms are indeed zero, except the one that you enforced to not be so (a_0 as you had stated). You can piece together the Frobenius solution then by noting you were seeking a solution of the form:

y(x) = \sum_{n = 0}^{\infty} a_n x^{n+r}

but r = 2,

y(x) = \sum_{n = 0}^{\infty} a_n x^{n+2} = y(x) = x^2 \sum_{n = 0}^{\infty} a_n x^{n} = x^2 \left( a_0 x^0 + a_1 x^1 + a_2 x^2 + \ldots\right)

you have just shown that a_0 \neq 0, while a_n = 0 ,\, n\neq 1. this means all terms vanish except a_0, so we have the above reducing to:

y(x) = a_0x^2

a_0 may have been the first coefficient, but in a Frobenius solution, the first term n = 0 corresponds to a power of x^{n+r} = x^{r} = x^2.

Thanks for the help. I understand the process for doing this DE with its indicial equation having repeated roots of r now. This time, it is a very specific question.

I have another question, regarding an ODE with two solutions. I understand what a basis of solutions is, and why you need at least two coefficients in a second order ODE to form the solution space. However, the following confuses me.

Given an ODE:

x^2y''-4xy'+(x^2+6)y = 0
y_1(x) = a_0x^2sin(x)
y_2(x) = A_0x^2cos(x)+A_1x^2sin(x)

Why does the A_1x^2sin(x) term get removed from y_2? It's been said that it is y_1, so removing it simplifies the solution. However, how is A_1x^2sin(x) the same as a_0x^2sin(x) given that a_0 is not A_1? The book doesn't explain this part, nor do the solutions.

I understand that, if A_1 = a_0, then removing y_1 makes sense.

 

[LawlQuals]

Just to be clear, the fact that this differential equation happened to have a one-term solution has nothing to do the fact that equation held repeated roots in the parameter r. This was a coincidence. We are not even usually so fortunate to be able to express a Frobenius solution in closed form, which does not have anything to do with whether or not the roots are repeated. Did you end up finding the second solution to the first differential equation by the way? Seeking solutions by way of Frobenius' method is a delicate process in that how you go about finding those solutions strongly depends on how the two roots (if a second order ODE) r are related (repeated roots, distinct, differ by an integer, perhaps I am preaching to the choir).

For your followup question, where and when does A_1x^2\sin x get removed? It sounds like your confusion possibly is coming from the phrasing used in the textbook (which may have translated into the question you have posed, as I cannot follow exactly what the question really is.). What I suspect though is that your text provides the ODE. It then lists two representative solutions, y_1(x) and y_2(x):

y_1(x) = a_0 x^2\sin x
y_2(x) = A_0x^2\cos x + A_1 x^2\sin x

Put any of these into the ODE, and you may verify they are indeed solutions. It sounds like your text was trying to construct the most general form of a solution to an ODE from two known solutions y_1 and y_2. One characteristic of a linear differential equation is that a linear combination of any solutions is also a solution. So, that you may write a general solution as:

y(x) = y_1(x) + y_2(x) = a_0x^2\sin x + A_0x^2\cos x + A_1x^2\sin x = (a_0 + A_1)x^2\sin x + A_0\cos x = A_2 x^2\sin x + A_0x^2 \cos x

where A_2 = a_0 + A_1 = const, it really does not matter what the constant actually is, we are seeking a general solution. The most general form is of the above form, there is no need to represent

A_2x^2 \sin x as \alpha x^2\sin x + \beta x^2\sin x any more than there is reason to write \gamma x as \kappa x + \epsilon x + \delta x + \ldots. These are redundant, you may consolidate them all into one constant.

I imagine that is what your textbook is getting at, but I could be off. In which case, a more exact excerpt from your text would help clarify the context.

 

[TheFerruccio]

Yes, I managed to answer the remainder of the previous problem in the first part of the thread.

I guess I am still having trouble understanding what, exactly, these solutions mean for a differential equation. It looks to me like you combined y_1(x) and y_2(x) to equal something that looks exactly like another y_2(x), especially given that it doesn't matter what the constants are. Why is y_1(x) even necessary to list as a solution, anymore? Adding it to y_2(x) seems to just result in y_2(x).

What do you mean by linear combination?

I'm sorry, but I'm left with even more questions than before. Now, I'm completely lost as to the meaning of these solutions, when I previously thought I had some grasp. All I know is that it can be verified that they are solutions, and I have verified them. I just don't understand what constitutes a basis of solutions.

I was told that a basis of solutions is much like having basis vectors in space, sufficient to describe every point in that space by multiplying those vectors by some constant. I'm having trouble seeing how this carries over to a basis of solutions.

 

[LawlQuals]

No problem about being confused, I should be clearer.

I do not have your text in front of me so I cannot gauge the actual context the author is presenting without speculation, but what I was suggesting is that the text's motive was to list two candidate solutions y_1 and y_2, it then proceeds to establish a basis in solution space (exactly as you understand it, both solutions constitute a basis to describe every solution in solution space constrained by any boundary conditions) by showing that the most general form of the solution is:

y(x) = b_0x^2\sin x + b_1 x^2\cos x (This is a linear combination, a sum of two solutions multiplied each by arbitrary constants).

where b_0 and b_1 are arbitrary constants that may be uniquely determined given boundary conditions. I suspect what the author wants to do is to label the two solutions as y_1 and y_2, then we could write the most general solution to the differential equation as

y(x) = y_1(x) + y_2(x)

Proceeding, the author suggests two solutions that do happen to solve the differential equation, but the question is: are they distinct solutions, and further what is the general solution? Inputting the suggestions by the author into the above form:

y(x) = \underbrace{a_0x^2\sin x}_{y_1} + \underbrace{A_0x^2\cos x + A_1x^2\sin x}_{y_2}
= \underbrace{(a_0 + A_1)}_{A_2}x^2\sin x + A_0x^2\cos x
\Rightarrow y(x) = A_2x^2\sin x + A_0 x^2\cos x

is the most general representation of the above, since a_0 and A_1 can be combined. Nothing is being removed, the solution is preserved. All that was found was that y_1 + y_2 as written, was not the most general form of the solution since we had a repeated term (x^2\sin x). Now, we may relabel the solutions to be compact, where we write one independent solution in y_1 and the other as y_2,

y_1(x) = a_0 x^2\sin x (does not have to be a_0, it can be anything, it is only a constant, I am just sticking to the notation given)
y_2(x) = A_0 x^2\cos x

Such that,

y(x) = y_1(x) + y_2(x) is the most general form (no repeated terms, each solution is independent). That was all they were doing I believe. Alternatively, you could just relabel y_2 as having two solutions that are independent and discard y_1, same thing. The bottom line is just that you need two independent solutions, the extra term in y_2 had a little "extra" on it that was unneeded, since just the other term itself also satisfies the differential equation. Two solutions, independent of each other, need only be written as just above, there is no reason to have an extra term in there for y_2

I think the confusion you have is just that the author seems to be doing something that is possible backwards to your thinking. The thought experiment was to seek the general solution a posteriori, rather than working from the beginning towards the end. Given two solutions, construct the general solution (rather than, given an ODE, find the general solution. In which case, you would have found those two solutions anyway). In doing it the way the author did, given these two solutions, you can combine two of the terms as described above because they are, for all intents and purposes, the same.

 

[TheFerruccio]

Thanks for the information. I think this is beginning to make more sense. I will need to do some more practice to really nail down the concept of independent solutions, and what terms can be eliminated, but your help has given me lots of direction in understanding this.

 

[LawlQuals]

Glad to hear it and happy to help.