Solution verification of ODE using Frobenius' method

  • #1
psie
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Homework Statement
Determine for ##t>0## all solutions ##x(t)## of the equation ##2t^2x''+3tx'-(1+t)x=0##.
Relevant Equations
The Frobenius method (see e.g. Wikipedia).
I have no problems with solving this exercise, but my solution disagrees slightly with that given in the answers in the back of the book, and I do not know who's correct.

First, we rewrite the equation as $$x''+\frac{3}{2t}x'-\frac{(1+t)}{2t^2}x=0.\tag1$$ We recognize that this is so-called weakly singular, i.e. the coefficients in front of ##x'## has at most a simple pole and that in front of ##x## at most a double pole. The first term in the Laurent expansion of the coefficients are ##\frac{3}{2}## and ##-\frac{1}{2}## respectively. Thus the indicial polynomial reads $$\mu(\mu-1)+\frac{3}{2}\mu-\frac{1}{2},$$ which has roots ##-1## and ##\frac{1}{2}##. Since these do not differ by an integer, our two linearly independent solutions will be of the form $$y(t)=\sum_{k=0}^\infty a_k t^{k-1}\quad\text{and}\quad z(t)=\sum_{k=0}^\infty b_k t^{k+\frac{1}{2}}.$$ I will only solve for the coefficients ##b_k##, since I get the answer as in the book for the coefficients ##a_k##. We have that the first and second derivative of ##z(t)## are $$z'(t)=\sum_{k=0}^\infty \left(k+\frac{1}{2}\right) b_k t^{k-\frac{1}{2}}\quad\text{and}\quad z''(t)=\sum_{k=0}^\infty \left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)b_k t^{k-\frac{3}{2}}.$$ Plugging these into the equation ##(1)##, we get \begin{align}
\sum_{k=0}^\infty \left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)b_k t^{k-\frac{3}{2}}&+\sum_{k=0}^\infty \frac{3}{2}\left(k+\frac{1}{2}\right)b_k t^{k-\frac{3}{2}} -\sum_{k=0}^\infty \frac{1}{2} b_k t^{k-\frac{3}{2}}\nonumber \\ &- \sum_{k=0}^\infty \frac{1}{2} b_k t^{k-\frac{1}{2}}=0 \nonumber
\end{align}
We can bring the powers of ##t^{k-\frac{3}{2}}## under a single sum, so the equation reads
$$\sum_{k=0}^\infty \left(\left(k+\frac{1}{2}\right)\left(k-\frac{1}{2}\right)+\frac{3}{2}\left(k+\frac{1}{2}\right)-\frac{1}{2}\right)b_k t^{k-\frac{3}{2}}-\sum_{k=0}^\infty \frac{1}{2} b_k t^{k-\frac{1}{2}}=0.$$ Here we can see that the first term in the first sum vanishes, so it really starts from ##k=1##, i.e. $$\sum_{k=1}^\infty \Big(\ldots\Big)b_k t^{k-\frac{3}{2}}-\sum_{k=0}^\infty \frac{1}{2} b_k t^{k-\frac{1}{2}}=0.$$ Re-indexing the first sum so that it starts from ##k=0## again, we get $$\sum_{k=0}^\infty \left(\left(k+\frac{3}{2}\right)\left(k+\frac{1}{2}\right)+\frac{3}{2}\left(k+\frac{3}{2}\right)-\frac{1}{2}\right)b_{k+1} t^{k-\frac{1}{2}}-\sum_{k=0}^\infty \frac{1}{2} b_k t^{k-\frac{1}{2}}=0.$$ Here we can again write everything under a single sum and since that sum equals zero, the coefficients vanish. You can read off the recurrence relation from the previous equation, it is \begin{align}
&\qquad&\frac{1}{2}b_k&=\left(\left(k+\frac{3}{2}\right)\left(k+\frac{1}{2}\right)+\frac{3}{2}\left(k+\frac{3}{2}\right)-\frac{1}{2}\right)b_{k+1} \nonumber\\
&\iff&b_k&=(k+1)(2k+5)b_{k+1} \nonumber
\end{align}
We see that ##b_0## is quite arbitrary, so put ##b_0=C##, then $$b_1=\frac{C}{5},\quad b_2=\frac{C}{2\cdot5\cdot 7},\quad b_3=\frac{C}{2\cdot 3\cdot5\cdot 7\cdot 9},\quad \ldots.$$ We reach the conclusion that $$b_k=\frac{C}{k!(2k+3)!!},\quad k=1,2,\ldots.$$So the solution reads $$Ct^{\frac{1}{2}}\left(1+\sum_{k=1}^\infty\frac{t^k}{k!(2k+3)!!}\right).$$ The solution given in the book is $$Ct^{\frac{1}{2}}\left(\sum_{k=0}^\infty\frac{t^k}{k!(2k+3)!!}\right).$$
This is not the same as my solution. Any ideas?
 
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  • #2
psie said:
so put ##b_0=C##, then $$b_1=\frac{C}{5},\quad b_2=\frac{C}{2\cdot5\cdot 7},\quad b_3=\frac{C}{2\cdot 3\cdot5\cdot 7\cdot 9},\quad \ldots.$$ We reach the conclusion that $$b_k=\frac{C}{k!(2k+3)!!},\quad k=1,2,\ldots.$$
This is wrong. The ##(2k + 5)## term starts at 5 when ##k=0## so it gives rise to a factor ##1/(5\cdot 7 \cdot \ldots \cdot(2k + 3)) \neq 1/(3\cdot 5\cdot \ldots \cdot (2k + 3)) = 1/(2k +3)!!##
 
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  • #3
You're right, I was wrong there. To resolve this, I am tempted to just write ##b_0=C/3##, yet this doesn't feel right for some reason. The solution given in the book confuses me.
 
  • #4
psie said:
You're right, I was wrong there. To resolve this, I am tempted to just write ##b_0=C/3##, yet this doesn't feel right for some reason. The solution given in the book confuses me.
Yes, you solve it by letting ##b_0 = C/3## … and then you get the expression from the book since ##0!(2\cdot 0 +3)!! = 3##.
 
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FAQ: Solution verification of ODE using Frobenius' method

What is Frobenius' method for solving ODEs?

Frobenius' method is a technique used to find solutions to linear ordinary differential equations (ODEs) near a singular point. This method involves expressing the solution as a power series with coefficients that are determined recursively. It is particularly useful for handling equations where the standard power series method fails due to the presence of a singularity.

When can Frobenius' method be applied?

Frobenius' method can be applied to linear ODEs with a regular singular point. A regular singular point is a point where the ODE has singular coefficients, but the singularity is not too severe. Specifically, the coefficients of the ODE can have poles of order up to one at the singular point.

How do you determine the indicial equation in Frobenius' method?

The indicial equation is derived from the lowest power of the series solution when substituted into the ODE. By substituting the series solution into the ODE and equating the lowest power term to zero, we obtain the indicial equation. This equation is a polynomial whose roots determine the possible values of the exponent in the series solution.

What are the steps involved in applying Frobenius' method?

The steps involved in applying Frobenius' method are: (1) Identify the regular singular point of the ODE. (2) Assume a series solution of the form \( y = x^r \sum_{n=0}^{\infty} a_n x^n \). (3) Substitute this series into the ODE. (4) Determine the indicial equation by setting the coefficient of the lowest power of \( x \) to zero and solve for \( r \). (5) Use the roots of the indicial equation to find the recurrence relations for the coefficients \( a_n \). (6) Solve the recurrence relations to obtain the series solution.

What happens if the roots of the indicial equation differ by an integer?

If the roots of the indicial equation differ by an integer, the Frobenius method may yield only one linearly independent solution directly. In such cases, the second solution can often be found using a generalized Frobenius method or by constructing a solution that involves a logarithmic term. This second solution is typically more complex and requires additional techniques to derive.

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