- #1
psie
- 315
- 40
- 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?
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?