On weakly singular equations and Frobenius' method

psie
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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?
 
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Both of these satisfy Definition 2; they just do so at multiple points. Note that in (b) you are asked for solutions in t &gt; 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.
 
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).##
 
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psie said:
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}<br /> q(t) &amp;= -\frac{2}{t(1-t)^2} \\<br /> &amp;= -\frac 2t (1 - t)^{-2} \\<br /> &amp; = -\frac2t \left(1 + (-2)(-t) +\frac{(-2)(-3)}{2!}(-t)^2 + \dots\right) \\<br /> &amp;= \frac{q_0}{t^2} + \frac{q_1}{t} +\dots \\<br /> \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 <br /> 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 &lt; u = 1 - t &lt; 1 is positive (and u^{\mu} is real and positive).
 
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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.
 
Having obtained one solution x_1, you can obtain a second linearly independent solution <br /> x_2(t) = x_1(t)\int_{t_0}^t \frac{W(s)}{x_1^2(s)}\,ds where W = x_1x_2&#039; - x_2x_1&#039; satisfies W&#039; = -pW. In this case p = 0 so you can take W = 1.
 
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##?
 
psie said:
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 <br /> 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 <br /> 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.
 
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