Solving a DE with a Series Index Shifts & Recurrence Relations

[Benny]

Ok I'm giving these another go. I found the following DE from a reduction of order problem and figured that it would be an alright question if I turned it into one requiring a series solution. However I'm stuck. I think it's just a matter of index shifts to get an appropriate recurrence relation between the coefficients. Here is the equation and what I've done.

$$2x^2 y'' + 3xy' - y = 0$$

$$y = \sum\limits_{n = 0}^\infty {c_n x^n } \Rightarrow y' = \sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } \Rightarrow y'' = \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} }$$

The reason why I haven't just straight away shifted the starting indices for y' and y'' is that the coefficients of those derivatives in the DE, are not constant. So I think that initially leaving the derivatives with different starting values makes it easier to decide on index shifts later on. Is that the right way to start?

$$2x^2 y'' = 2x^2 \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right)x^{n - 2} } = \sum\limits_{n = 2}^\infty {2c_n n\left( {n - 1} \right)x^n }$$

$$3xy' = 3x\sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } = \sum\limits_{n = 1}^\infty {3c_n nx^n }$$

I know that I need to get a 'common power' for x in each of the summations eg. x^n, x^(n+1), x^(n+2) etc. I can't think of a way to do this. Does anyone have any suggestions?

 

[HallsofIvy]

Ok I'm giving these another go. I found the following DE from a reduction of order problem and figured that it would be an alright question if I turned it into one requiring a series solution. However I'm stuck. I think it's just a matter of index shifts to get an appropriate recurrence relation between the coefficients. Here is the equation and what I've done.
$$2x^2 y'' + 3xy' - y = 0$$
$$y = \sum\limits_{n = 0}^\infty {c_n x^n } \Rightarrow y' = \sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } \Rightarrow y'' = \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} }$$
The reason why I haven't just straight away shifted the starting indices for y' and y'' is that the coefficients of those derivatives in the DE, are not constant. So I think that initially leaving the derivatives with different starting values makes it easier to decide on index shifts later on. Is that the right way to start?
$$2x^2 y'' = 2x^2 \sum\limits_{n = 2}^\infty {n\left( {n - 1} \right)x^{n - 2} } = \sum\limits_{n = 2}^\infty {2c_n n\left( {n - 1} \right)x^n }$$
$$3xy' = 3x\sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } = \sum\limits_{n = 1}^\infty {3c_n nx^n }$$
I know that I need to get a 'common power' for x in each of the summations eg. x^n, x^(n+1), x^(n+2) etc. I can't think of a way to do this. Does anyone have any suggestions?

You can't think of a way to do that? Isn't it obvious that you already have "a common power"- n? If it's the fact that each sum starts with a different value, you could do the cases n= 0, n= 1, n> 1 separately.
Actually, this is an "equipotential" equation. also known as an "Euler-type equation", (the coefficient of each derivative is a power of x- the power is the same as the order of the derivative): you don't need to use a power series method.
Either: Let y= x^r, substitute into the equation to get a "characteristic equation" just as you do with e^rx for equations with constant coefficients or
Make the substitution u= ln x to convert to an equation in u with constant coefficients.

 

[saltydog]

Benny, your equation has a singular point at x=0 and therefore if you wished to solve it via power series, you would need to use:

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

where c is a root of the indicial equation. Depending if the difference of the roots is nonintegral, 0 or integer, logarithms may enter into the solutions. Check out "solutions near regular singular points" and the indicial equation for these types of DEs.

 

[Benny]

I don't know what I was thinking. Yeah, I already had a common power. The starting index being different for the summations shouldn't be a problem. I could always just do something starting each summation at a common value of n by taking out the first few terms from the relevant summations. This will allow me to set some coefficients equal to zero and from the recurrence this should mean that a whole lot of other coefficients are also zero. Something like that.

Saltydog - From the reduction of order question, a given solution is y = 1/x. So that is probably part of the problem with my series 'solution' attempt.

Now instead of just trying to make up a question (there's not much point in taking the reduction of order questions I have because the equations are all Cauchy - Euler equations), I'll take an actual series question. The examples associated with this question only use the standard trial series for y so the following shouldn't require any modifications to be made to y. Here it is.

$$x^2 y'' + x^2 y' + xy = 0,y\left( 0 \right) = 1,y'\left( 0 \right) = 1$$

$$y = \sum\limits_{n = 0}^\infty {c_n x^n } \Rightarrow y' = \sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } \Rightarrow y'' = \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} }$$

Substitute into the equation.

$$x^2 \sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)x^{n - 2} } + x\sum\limits_{n = 1}^\infty {c_n nx^{n - 1} } + x^2 \sum\limits_{n = 0}^\infty {c_n x^n } = 0$$

$$\sum\limits_{n = 2}^\infty {c_n n\left( {n - 1} \right)} x^n + \sum\limits_{n = 1}^\infty {c_n nx^n } + \sum\limits_{n = 0}^\infty {c_n x^{n + 2} } = 0$$

$$\sum\limits_{n = 0}^\infty {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right)x^{n + 2} } + \sum\limits_{n = 0}^\infty {c_n x^{n + 2} } + \sum\limits_{n = 1}^\infty {c_n nx^n } = 0$$

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + c_n } \right]} x^{n + 2} + \sum\limits_{n = - 1}^\infty {c_{n + 2} \left( {n + 2} \right)x^{n + 2} } = 0$$

Extract the n = -1 term to get:

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)\left( {n + 1} \right) + \left( {n + 3} \right)c_n } \right]} x^{n + 1} + c_1 x = 0$$

Edit: Erased my query about the IC contradicting the equation.

Edit 2: I wrote this problem down on paper and basically got down to the following.

$$\sum\limits_{n = 0}^\infty {\left[ {c_{n + 2} \left( {n + 2} \right)^2 + c_n } \right]} x^{n + 2} + c_1 x = 0$$

This implies c_1 = 0 which is consistent with the ICs. This leaves me with $$c_{n + 2} \left( {n + 2} \right)^2 + c_n = 0$$.

$$c_{n + 2} = - \frac{{c_n }}{{\left( {n + 2} \right)^2 }}$$

$$n = 1:c_3 = - c_1 \times stuff = 0$$ since c_1 = 0.

$$\Rightarrow c_{2n + 1} = 0,\forall n \in N$$

Concentrating on the even coefficients:

$$c_{2n} = - \frac{{c_{2n - 2} }}{{\left( {2n} \right)^2 }}$$

Iterating a few times I find that $$c_{2n} = \frac{{\left( { - 1} \right)^n c_0 }}{{\left( {2^n n!} \right)^2 }}$$.

Using c_0 = 1 I get:

$$y\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n }}{{\left( {2^n n!} \right)^2 }}x^{2n} }$$

That's my best attempt. I'm really being bogged down by this final topic.

 

[HallsofIvy]

Benny, this is again an equation that is singular at x= 0 (the coefficient of y" is 0 at x= 0) so a simple power series may not be enough. As salty dog said, you should use "Frobenius' method": try a series of the form
$$y(x)=\sum_{n=0}^\infty a_n x^{n+c}$$
Because that "c" can be chosen in many different ways, assume that a₀ is not 0. Put that into your differential equation, collect like terms and set the coefficient of the lowest power term equal to 0 (setting n=0). That will typically give you a formula in c times a₀. Since a₀ is not 0 you can solve for c (and, typically, since this is a second order equation, that "indicial equation" will be quadratic and will give two different solutions for c).

 

[Benny]

Ok I'll try that method out. It's just that from the examples that were given in the section containing this problem, only a standard power series was required. But I'll try saltydog's suggestion anyway. Thanks for the help.

 

[saltydog]

Benny, you sure your initial conditions are correct? I suspect y(0) can't be 0. This is why:

When I use the trial solution:

$$\sum_0^{\infty} a_nx^{n+c}$$

and substitute into the DE and shifted indexes so that all the sums have the same power of x, I obtain:

**** edited to show c's ****

$$\sum_0 a_n(n+c)(n+c-1)x^{n+c}+\sum_1 a_{n-1}(n-1+c)x^{n+c}+\sum_1 a_{n-1}x^{n+c}$$

The indicial equation comes from the n=0 term when this method is used so therefore it's:

$$c(c-1)=0$$

The roots are therefore 0 and 1. The difference is integral so there is a likelyhood a logarithm will enter into the solution.

Letting c=0:

$$\sum_0^\infty a_n n(n-1)x^n+\sum_1^\infty a_{n-1}(n+1-1)x^n$$

Sure enough, when I solve for the coefficients, I obtain:

$$a_0=0$$

$$a_1=\text{arbitrary}$$

$$a_n=-\frac{a_{n-1}}{n-1};\quad n\ge 2$$

That's only 1 arbitrary constant so at most I can obtain only one solution unless I use another method (such as the logarithmic method).

Anyway, when I form the product for n ge 2:

$$a_2a_3a_4\cdots a_n=\frac{(-1)a_1}{1}\frac{(-1)a_2}{2}\frac{-1)a_3}{3}\cdots\frac{(-1)a_{n-1}}{n-1}$$

and cancel, I'm left with:

$$a_n=\frac{(-1)^na_1}{(n-1)!}$$

Thus, one solution is:

$$y(x)=\sum_1^\infty \frac{(-1)^na_n}{(n-1)!} x^n$$

Using the magic of Mathematica:

[tex]y(x)=\sum_1^\infty \frac{(-1)^na_n}{(n-1)! }x^n[/tex]

yd1[x_]=D[y[x],x];

yd2[x_]=D[y[x],{x,2};

Simplify[x^2 yd2[x]+x^2 yd1[x]+x y[x]

Mathematica returns 0 so I have some confidence this series solution is correct.

Note that y(0) can only be 0 and the logarithm solution will not admit x=0.