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 [itex]t > 0[/itex], 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##.

[tex]\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}[/tex] so that [itex]q_0 = 0[/itex] and [itex]q_1 = -4[/itex] (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 [tex]
x(t) = (1 - t)^{\mu}a(t) = (1 - t)^{\mu} \sum_{k=0}^\infty a_k(1-t)^k.[/tex] This follows from substituting [itex]t = 1 - u[/itex] so that [itex]0 < u = 1 - t < 1[/itex] is positive (and [itex]u^{\mu}[/itex] 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 [itex]x_1[/itex], you can obtain a second linearly independent solution [tex]
x_2(t) = x_1(t)\int_{t_0}^t \frac{W(s)}{x_1^2(s)}\,ds[/tex] where [itex]W = x_1x_2' - x_2x_1'[/itex] satisfies [tex]W' = -pW.[/tex] In this case [itex]p = 0[/itex] so you can take [itex]W = 1[/itex].

 

[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 [itex]p[/itex] and [itex]q[/itex] about the origin to have radius of convergence greater than 1, due to the singularity at [itex]-1[/itex]. 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 [itex]\mathbb{C}[/itex] which contains [itex][1, \infty)[/itex].

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