MHB Solving Regular Singularities with Frobenius Expansion: Series Solution Question

[nacho-man]

Referring to the attached image.

I have found the solution first solution, it had regular singularities of $x_0=1$ annd $x_0=0$ so we can use frobenius expansion.

The indicial equation is r= 1 or 0, and the solution I found was for r=1.

****Is there only going to be one solution for this $y_1(x)$ for this question?

Also, is the ROC going to be 'at least one' because that is the closest regular singularity? My recurrence relation I determined was:

$a_{m+1} = \frac{a_0((2m-1)!)^2}{4^mm!(m-1)!}$

***How do I determine the ROC, and re-write this as a solution?

**Why is it that we only needed one initial condition, $y'(0)=1$, which I did not even use to find my first solution?

edit: I just read up, if$r_1-r_2=0$ then the second solution will be of the form $y_2 = y_1\ln\left({x}\right)$ and if you differentiate this you get $\frac{1}{X}$ and if you sub in 0 that is undefined. But why does this mean we only need one boundary condition? Also since it's undefined, doesn't that mean the boundary condition isn't being met?

Any help is appreciated!

 

[Euge]

Referring to the attached image.

I have found the solution first solution, it had regular singularities of $x_0=1$ annd $x_0=0$ so we can use frobenius expansion.

The indicial equation is r= 1 or 0, and the solution I found was for r=1.

****Is there only going to be one solution for this $y_1(x)$ for this question?

Also, is the ROC going to be 'at least one' because that is the closest regular singularity? My recurrence relation I determined was:

$a_{m+1} = \frac{a_0((2m-1)!)^2}{4^mm!(m-1)!}$

***How do I determine the ROC, and re-write this as a solution?

**Why is it that we only needed one initial condition, $y'(0)=1$, which I did not even use to find my first solution?

edit: I just read up, if$r_1-r_2=0$ then the second solution will be of the form $y_2 = y_1\ln\left({x}\right)$ and if you differentiate this you get $\frac{1}{X}$ and if you sub in 0 that is undefined. But why does this mean we only need one boundary condition? Also since it's undefined, doesn't that mean the boundary condition isn't being met?

Any help is appreciated!

Hi nacho,

The indicial equation has roots $r = 0$ or $r = 1$; it is not the roots themselves that are the indicial equation. There's an error in the formula for the sequence $a_m$. It should be

$$a_m = \frac{[(2m - 1)!]^2}{4^m (m + 1)! m!}a_0$$.

This gives

$$y_1(x) = a_0x\sum_{m = 1}^\infty \frac{[(2m - 1)!]^2}{4^m (m + 1)! m!}x^m = a_0 x\, {}_2F_1\left({\frac{1}{2}, \frac{1}{2} \atop 2}\, ; x\right)$$.

Finding a solution $y_2(x)$ linearly independent from $y_1(x)$ will take a lot of work. Since the indicial roots differ by 1 and the smallest indicial root is 0,

$$y_2(x) = \frac{\partial}{\partial r}(ry(r,x))|_{r = 0}$$,

where

$$y(r,x) = x^r \sum_{m = 0}^\infty a_m(r) x^m$$.

Once you find the expression for $y_2(x)$, then you can express the general solution as $y(x) = c_1 y_1(x) + c_2 y_2(x)$.

Prescribing a value to $y'(0)$ uniquely determines $y$. For setting $x = 0$ in the differential equation, you get $0y''(0) - y(0) = 0$, i.e., $y(0) = 0$. Since $y(0)$ and $y'(0)$ are known, the uniqueness theorem for linear ODE justifies uniqueness of $y(x)$.

To find the radius of convergence of the solution will be the smaller of the two radii of convergence of $y_1$ and $y_2$. Apply the ratio test to $y_1$ and $y_2$ to find those radii.

 

[nacho-man]

Thanks for responding Euge!
I have a few questions in regards to your response,
firstly:

Finding a solution $y_2(x)$ linearly independent from $y_1(x)$ will take a lot of work. Since the indicial roots differ by 1 and the smallest indicial root is 0,

$$y_2(x) = \frac{\partial}{\partial r}(ry(r,x))|_{r = 0}$$,

where

$$y(r,x) = x^r \sum_{m = 0}^\infty a_m(r) x^m$$.

I don't mean to question you, but are you absolutely sure this is necessary?
From my readings I've seen that in the instance where the difference between the indicial roots is 0, the second solution $y_2(x)$ is simply $\ln\left({x}\right)y_1(x)$

This would also be consistent with what we've done in the course so far, we have not touched on partial fractions in this particular course (even though a pre-req course for this one had covered them).

Secondly, I am not quite familiar with this notation

$$y_1(x) = a_0 x\, {}_2F_1\left({\frac{1}{2}, \frac{1}{2} \atop 2}\, ; x\right)$$.

is it customary to simplify the solution to this form?

Finally,

Prescribing a value to $y'(0)$ uniquely determines $y$. For setting $x = 0$ in the differential equation, you get $0y''(0) - y(0) = 0$, i.e., $y(0) = 0$. Since $y(0)$ and $y'(0)$ are known, the uniqueness theorem for linear ODE justifies uniqueness of $y(x)$.

Is this to answer the question why only one solution was required to fully specify the solution?

 

[Euge]

Thanks for responding Euge!
I have a few questions in regards to your response,
firstly:

I don't mean to question you, but are you absolutely sure this is necessary?
From my readings I've seen that in the instance where the difference between the indicial roots is 0, the second solution $y_2(x)$ is simply $\ln\left({x}\right)y_1(x)$

This would also be consistent with what we've done in the course so far, we have not touched on partial fractions in this particular course (even though a pre-req course for this one had covered them).

The difference between the roots is 1, not 0. As you have already determined, the indicial roots are not equal.

Secondly, I am not quite familiar with this notation

$$y_1(x) = a_0 x\, {}_2F_1\left({\frac{1}{2}, \frac{1}{2} \atop 2}\, ; x\right)$$.

is it customary to simplify the solution to this form?

Yes, but you don't need to write it. The ${}_2F_1$ symbol is known as Euler's hypergeometric function. It's defined as

$${}_2F_1\left({a, b \atop c}; x\right) = \sum_{n = 0}^\infty \frac{(a)_n (b)_n}{n!\, (c)_n}x^n$$,

where $(q)_n := q(q + 1) \cdots (q + n - 1)$ denotes the rising factorial. If you want, I can go through the details of how I got a ${}_2F_1(1/2, 1/2; 2; x)$, but it's not really necessary for the problem.

Finally, Is this to answer the question why only one solution was required to fully specify the solution?

You're talking about the paragraph concerning uniqueness of $y(x)$? That only answers the last part of the question. You don't need to find $y_1$ or $y_2$ to deduce uniqueness of solution. You just need to know the values of $y(0)$ and $y'(0)$, which I had explained.

 

[nacho-man]

The difference between the roots is 1, not 0. As you have already determined, the indicial roots are not equal.

My brain has betrayed me for the last time, thinking 1-0 = 0...! :P
Thanks haha

 

[chisigma]

Referring to the attached image.

I have found the solution first solution, it had regular singularities of $x_0=1$ annd $x_0=0$ so we can use frobenius expansion.

The indicial equation is r= 1 or 0, and the solution I found was for r=1.

****Is there only going to be one solution for this $y_1(x)$ for this question?

Also, is the ROC going to be 'at least one' because that is the closest regular singularity? My recurrence relation I determined was:

$a_{m+1} = \frac{a_0((2m-1)!)^2}{4^mm!(m-1)!}$

***How do I determine the ROC, and re-write this as a solution?

**Why is it that we only needed one initial condition, $y'(0)=1$, which I did not even use to find my first solution?

edit: I just read up, if$r_1-r_2=0$ then the second solution will be of the form $y_2 = y_1\ln\left({x}\right)$ and if you differentiate this you get $\frac{1}{X}$ and if you sub in 0 that is undefined. But why does this mean we only need one boundary condition? Also since it's undefined, doesn't that mean the boundary condition isn't being met?

Any help is appreciated!

It is required to find an analytical solution 'somewhere around' x=0 to the ODE ...

$\displaystyle y'' = \frac{y}{4\ x\ (1-x)},\ y' (0) = 1\ (1)$

We note first that the only initial condition is defined relative to the derivative at x = 0 and immediately understand why. We write so ...

$\displaystyle y = a_{0} + a_{1}\ x + a_{2}\ x^{2} + ... + a_{n}\ x^{n} + ... \implies y'' = 2\ a_{2} + 6\ a_{3}\ x + ... + n\ (n-1)\ x^{n-2} + ...\ (2)$

... and...

$\displaystyle \frac{1}{4\ x\ (1-x)} = \frac{1}{4}\ (\frac{1}{x} + 1 + x + ... + x^{n-1} + ...)\ (3)$

Now we seek the values ​​of the $a_{n}$ that satisfy (1) imposing...

$\displaystyle 2\ a_{2} + 6\ a_{3}\ x + ... + n\ (n-1)\ x^{n-2} + ... = \frac{a_{0}}{4\ x} + \frac{a_{0} + a_{1}}{4} + \frac{a_{0} + a_{1} + a_{2}}{4}\ x + ... + \frac{a_{0} + a_{1} + ... + a_{n-1}}{4}\ x^{n-2} + ...\ (4)$

A look at (4) makes us realize immediately that it must be $a_{0}=0$. The initial condition states that is $a_{1}=1$. The other n are found iteratively with the formula ...

$\displaystyle a_{n+1} = \frac{a_{1} + a_{2} + ... + a_{n}}{4\ n\ (n+1)},\ a_{1}=1\ (5)$

Some values are $a_{2} = \frac{1}{8}$, $a_{3} = \frac{3}{64}$, $a_{4} = \frac{25}{1024}$, ... A general more compact formula of the $a_{n}$ as solution of the difference equation (5) will be sought in the near posted ...

Kind regards

$\chi$ $\sigma$

 

[Euge]

It is required to find an analytical solution 'somewhere around' x=0 to the ODE ...

$\displaystyle y'' = \frac{y}{4\ x\ (1-x)},\ y' (0) = 1\ (1)$

We note first that the only initial condition is defined relative to the derivative at x = 0 and immediately understand why. We write so ...

$\displaystyle y = a_{0} + a_{1}\ x + a_{2}\ x^{2} + ... + a_{n}\ x^{n} + ... \implies y'' = 2\ a_{2} + 6\ a_{3}\ x + ... + n\ (n-1)\ x^{n-2} + ...\ (2)$

... and...

$\displaystyle \frac{1}{4\ x\ (1-x)} = \frac{1}{4}\ (\frac{1}{x} + 1 + x + ... + x^{n-1} + ...)\ (3)$

Hi chisigma,

Since there is a regular singularity at $x = 0$, $y$ cannot be assumed analytic near 0. Otherwise, y'' will also be analytic near 0. This cannot happen since $y'' = \frac{1}{4x(1 - x)}y$ has a pole at $x = 0$.

 

[chisigma]

Hi chisigma,

Since there is a regular singularity at $x = 0$, $y$ cannot be assumed analytic near 0. Otherwise, y'' will also be analytic near 0. This cannot happen since $y'' = \frac{1}{4x(1 - x)}y$ has a pole at $x = 0$.

Let's rewrite first the ODE with its initial condition ...

$\displaystyle y'' = \frac{y}{4\ x\ (1-x)},\ y'(0)=1\ (1)$

The fact that it is imposed at x = 0 the value of the derivative y' means that You should search for [if it exists ...] an analytical solution at x = 0. In the specific case the second term of (1) in x = 0 does not have a pole but a so called removable singularity... very improper term in my opinion that can only lead to confusion, as in this case (Headbang) ...

In fact in x = 0 is $\displaystyle y'' = \frac{a_{1}}{4} = 2\ a_{2} = \frac{1}{4}$ and (1) is fully satisfied...

Kind regards

$\chi$ $\sigma$

 

[Euge]

Let's rewrite first the ODE with its initial condition ...

$\displaystyle y'' = \frac{y}{4\ x\ (1-x)},\ y'(0)=1\ (1)$

The fact that it is imposed at x = 0 the value of the derivative y' means that You should search for [if it exists ...] an analytical solution at x = 0. In the specific case the second term of (1) in x = 0 does not have a pole but a so called removable singularity... very improper term in my opinion that can only lead to confusion, as in this case (Headbang) ...

In fact in x = 0 is $\displaystyle y'' = \frac{a_{1}}{4} = 2\ a_{2} = \frac{1}{4}$ and (1) is fully satisfied...

Kind regards

$\chi$ $\sigma$

Yes, I am aware of removable singularities, and you're right -- y'' has a removable singularity at 0. I see that, although you are using a power series method, it is non-Frobenius, which is not the usual when 0 is a singular point.

 

[chisigma]

It is required to find an analytical solution 'somewhere around' x=0 to the ODE ...

$\displaystyle y'' = \frac{y}{4\ x\ (1-x)},\ y' (0) = 1\ (1)$

We note first that the only initial condition is defined relative to the derivative at x = 0 and immediately understand why. We write so ...

$\displaystyle y = a_{0} + a_{1}\ x + a_{2}\ x^{2} + ... + a_{n}\ x^{n} + ... \implies y'' = 2\ a_{2} + 6\ a_{3}\ x + ... + n\ (n-1)\ x^{n-2} + ...\ (2)$

... and...

$\displaystyle \frac{1}{4\ x\ (1-x)} = \frac{1}{4}\ (\frac{1}{x} + 1 + x + ... + x^{n-1} + ...)\ (3)$

Now we seek the values ​​of the $a_{n}$ that satisfy (1) imposing...

$\displaystyle 2\ a_{2} + 6\ a_{3}\ x + ... + n\ (n-1)\ x^{n-2} + ... = \frac{a_{0}}{4\ x} + \frac{a_{0} + a_{1}}{4} + \frac{a_{0} + a_{1} + a_{2}}{4}\ x + ... + \frac{a_{0} + a_{1} + ... + a_{n-1}}{4}\ x^{n-2} + ...\ (4)$

A look at (4) makes us realize immediately that it must be $a_{0}=0$. The initial condition states that is $a_{1}=1$. The other n are found iteratively with the formula ...

$\displaystyle a_{n+1} = \frac{a_{1} + a_{2} + ... + a_{n}}{4\ n\ (n+1)},\ a_{1}=1\ (5)$

Some values are $a_{2} = \frac{1}{8}$, $a_{3} = \frac{3}{64}$, $a_{4} = \frac{25}{1024}$, ... A general more compact formula of the $a_{n}$ as solution of the difference equation (5) will be sought in the near posted ...

The difference equation that gives the $a_{n}$ can be written as...

$\displaystyle a_{n+1} = \frac{4\ n^{2} - 4\ n + 1}{4\ n^{2} + 4\ n}\ a_{n},\ a_{1}=1\ (1)$

... and with a little of patience one finds that for n>1 is...

$\displaystyle a_{n} = \frac{\{(2\ n - 3)! \}^{2}}{4^{n-1}\ n!\ (n-1)!}\ (2)$

... so that the solution to the ODE is...

$\displaystyle y(x) = x + \sum_{n=2}^{\infty} \frac{\{(2\ n - 3)! \}^{2}}{4^{n-1}\ n!\ (n-1)!}\ x^{n}\ (3)$

Kind regards

$\chi$ $\sigma$

 

[nacho-man]

Is it odd that I found the discussion above exciting?