On weakly singular equations and Frobenius' method

[psie]

Homework Statement

Find a basis for the solution space of (a) $t(t-1)^2x''-2x=0, 0<t<1$ and (b) $4(t^2+t^3)x''-4tx'+(3+t)x=0, t>0$.

Relevant Equations

This exercise appears in a section on the Frobenius' method. Some results related to this method are given below.

Definition 2. The differential equation $$x''(z)+p(z)x'(z)+q(z)x(x)=0\tag1$$ is called weakly singular at the origin if $p(z)$ has at most a simple pole and $q(z)$ at most a double pole there, in other words, if $$p(z)=\frac{p_0}{z}+p_1+p_2z+\ldots,\quad q(z)=\frac{q_0}{z^2}+\frac{q_1}{z}+q_2+q_3z\ldots .$$

Theorem 6. A basis for the solution space of $(1)$ is $$z^\mu a(z),\quad z^\nu b(z),$$ or $$z^\mu a(z),\quad (z^\mu \log(z))a(z)+z^\nu b(z).$$ Here $a(z)$ and $b(z)$ are analytic functions in a neighborhood of the origin.

$\mu$ is a root of the so-called indicial equation $I(\lambda)=\lambda(\lambda-1)+p_0\lambda+q_0$. $\nu$ can also be a root of the indicial equation, or we may have $\mu=\nu$

My attempt so far is trying to characterize both equations according to Definition 2, as well as identifying $p_0$ and $q_0$, so I can obtain the indicial equation. Then I could probably solve this myself.

My strategy is to rewrite the equation on the form given in Definition 2, so for (a) $$x''-\frac{2}{t(t-1)^2}x=0,$$ and (b) $$x''-\frac{t}{t^2+t^3}x'+\frac{3+t}{t^2+t^3}x=0\iff x''-\frac{1}{t(1+t)}x'+\frac{3+t}{t^2(1+t)}x=0.$$ I struggle with the fact that both these equations do not seem to satisfy Definition 2. In particular, there seems to be a pole both at the origin and at $z=\pm1$. I have not encountered a problem like this before and I'm a little puzzled on how to continue. Any ideas?

 

[pasmith]

Both of these satisfy Definition 2; they just do so at multiple points. Note that in (b) you are asked for solutions in $t > 0$, so the behaviour near the other singular point at -1 is of no consequence.

In (a), you can either obtain a basis whose behaviour near the origin is known, or a basis whose behaviour near 1 is known. Doing both and expressing them in terms of each other is probably beyond the scope of the question.

 

[psie]

Take for example (a). We can use partial fraction decomposition to get $$x''-\frac{2}{t(t-1)^2}x=0\iff x''-\left(\frac2{(t-1)^2}-\frac2{t-1}+\frac2{t}\right)x=0.$$ I'm confused about how to identify $q_0$ here, since $q(z)$ is not of the form $q(z)=\frac{q_0}{z^2}+\frac{q_1}{z}+q_2+q_3z\ldots$.

Moreover, should my attempt/ansatz be modified so that the power series is centered at $t=1$, i.e. $t^\mu a(t)=t^\mu\sum_{k=0}^\infty a_k(t-1)^k$? This seems to cause trouble with the fraction $\frac2{t}$ in the equation...

Since I have the solution to (a), I'll just post it here: $x(t)=A\frac{t}{1-t}+B\left(t+1+\frac{2t}{1-t}\log(t)\right).$

 

[pasmith]

Take for example (a). We can use partial fraction decomposition to get $$x''-\frac{2}{t(t-1)^2}x=0\iff x''-\left(\frac2{(t-1)^2}-\frac2{t-1}+\frac2{t}\right)x=0.$$ I'm confused about how to identify $q_0$ here, since $q(z)$ is not of the form $q(z)=\frac{q_0}{z^2}+\frac{q_1}{z}+q_2+q_3z\ldots$.

$$\begin{split} q(t) &= -\frac{2}{t(1-t)^2} \\ &= -\frac 2t (1 - t)^{-2} \\ & = -\frac2t \left(1 + (-2)(-t) +\frac{(-2)(-3)}{2!}(-t)^2 + \dots\right) \\ &= \frac{q_0}{t^2} + \frac{q_1}{t} +\dots \\ \end{split}$$ so that $q_0 = 0$ and $q_1 = -4$ (EDIT: -2 is correct).

Moreover, should my attempt/ansatz be modified so that the power series is centered at $t=1$, i.e. $t^\mu a(t)=t^\mu\sum_{k=0}^\infty a_k(t-1)^k$? This seems to cause trouble with the fraction $\frac2{t}$ in the equation...

You want $$x(t) = (1 - t)^{\mu}a(t) = (1 - t)^{\mu} \sum_{k=0}^\infty a_k(1-t)^k.$$ This follows from substituting $t = 1 - u$ so that $0 < u = 1 - t < 1$ is positive (and $u^{\mu}$ is real and positive).

 

[psie]

Ok, so far I've been able to find one solution to (a). Identifying $q_0=0$ and $p_0=0$ in (a), the indicial equation reads $$\lambda(\lambda-1)=0,$$ with roots $\mu=1$ and $\nu=0$. I'll try the Ansatz $$\sum_{k=0}^{\infty}a_k t^{k}.$$ Expanding $\frac{2}{(t-1)^2}$, the ODE reads $$\sum a_k k(k-1)t^{k-2} = \sum 2(k+1)t^k \sum a_kt^{k-1},$$ where the index runs from $k\geq0$. Here we observe that $a_0=0$, since the RHS would otherwise contain a term with $1/t$, whereas the LHS does not. So, the ODE reads $$\sum a_{k+2}(k+2)(k+1)t^k = \sum 2(k+1)t^k \sum a_{k+1}t^k,$$where sums again go from $k\geq0$. Put $c_k = 2(k+1)$ and $d_k = a_{k+1}$. From products of power series, we know that the coefficients in the RHS will be $\sum_{j=0}^k d_j c_{k-j} = \sum_{j=0}^k a_{j+1}\cdot 2(k-j+1)$. Thus we obtain the recurrence relation $$a_{k+2}(k+2)(k+1) = \sum_{j=0}^k a_{j+1}\cdot 2(k-j+1)\iff a_{k+2} = \sum_{j=0} \frac{2(k-j+1)}{(k+2)(k+1)}a_{j+1}.$$ $a_1$ seems to be quite arbitrary, so put $a_1=C\in\mathbb C$, then it follows by induction that $a_k=a_1$ for $k\geq 1$. We thus obtain the solution $$x_0(t)=C\frac{t}{1-t}.$$

The other solution will either be of the form $t b(t)$ or $\log(t)a(t)+t b(t)$, where $a(t)=x_0(t)$ and $b(t)$ is analytic, according to the theorem above. How do I know which one to guess? I'm looking for the path of least work.

 

[pasmith]

Having obtained one solution $x_1$, you can obtain a second linearly independent solution $$x_2(t) = x_1(t)\int_{t_0}^t \frac{W(s)}{x_1^2(s)}\,ds$$ where $W = x_1x_2' - x_2x_1'$ satisfies $$W' = -pW.$$ In this case $p = 0$ so you can take $W = 1$.

 

[psie]

Thank you. I have found a second solution to (a) and indeed, the general solution agrees with the solution I posted earlier.

I'm trying to convince myself that (b) also satisfies Definition 2 (so I can find $p_0$ and $q_0$). Recall $$x''-\frac{t}{t^2+t^3}x'+\frac{3+t}{4(t^2+t^3)}x=0\iff x''-\frac{1}{t(1+t)}x'+\frac{3+t}{4t^2(1+t)}x=0.$$ I guess to see that it is of the form as given in definition 2, we need to expand $1/(1+t)$, but can we do that when the series for $1/(1+t)$ only converges for $|t|<1$ and we are looking for solutions in $t>0$?

 

[pasmith]

I guess to see that it is of the form as given in definition 2, we need to expand $1/(1+t)$, but can we do that when the series for $1/(1+t)$ only converges for $|t|<1$ and we are looking for solutions in $t>0$?

Yes. You should not expect the power series for $p$ and $q$ about the origin to have radius of convergence greater than 1, due to the singularity at $-1$. The resulting series solution may not have radius of convergence greater than 1 for the same reason; however it may be possible to continue it analytically to some open region of $\mathbb{C}$ which contains $[1, \infty)$.

Here, however, it is not necessary to do a series expansion: you can see from $$q(t) = \frac{3 + t}{4t^2(1 + t)} = \frac{1}{4t^2}\left( 1 + \frac{2}{1 + t} \right)$$ that $t^2q(t)$ is analytic at the origin. Hence $$t^2 q(t) = q_0 + q_1 t + \dots \Rightarrow q(t) = q_0t^{-2} + q_1t^{-1} + \dots$$ so that evaluating $\lim_{t \to 0} t^2q(t)$ will give $q_0 = 3/4$. In the same way you can see that $tp(t)$ is analytic at the origin, so that $p_0 = \lim_{t \to 0} tp(t) = -1$.