Internal problem ID [1795]
Internal file name [OUTPUT/1796_Sunday_June_05_2022_02_31_41_AM_47261104/index.tex]
Book: Differential equations and their applications, 3rd ed., M. Braun
Section: Section 2.8.2, Regular singular points, the method of Frobenius. Page 214
Problem number: 3.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\sin \left (t \right ) y^{\prime \prime }+\cos \left (t \right ) y^{\prime }+\frac {y}{t}=0} \] With the expansion point for the power series method at \(t = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \sin \left (t \right ) y^{\prime \prime }+\cos \left (t \right ) y^{\prime }+\frac {y}{t} = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(t) y^{\prime } + q(t) y &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {\cos \left (t \right )}{\sin \left (t \right )}\\ q(t) &= \frac {1}{\sin \left (t \right ) t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([\pi Z, 0, \pi Z]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \cos \left (t \right ) y^{\prime } t +\sin \left (t \right ) y^{\prime \prime } t +y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \cos \left (t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right ) t +\sin \left (t \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) t +\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Expanding \(\sin \left (t \right ) t\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} \sin \left (t \right ) t &= t^{2}-\frac {1}{6} t^{4}+\frac {1}{120} t^{6} + \dots \\ &= t^{2}-\frac {1}{6} t^{4}+\frac {1}{120} t^{6} \end {align*}
Expanding \(\cos \left (t \right ) t\) as Taylor series around \(t=0\) and keeping only the first \(6\) terms gives \begin {align*} \cos \left (t \right ) t &= t -\frac {1}{2} t^{3}+\frac {1}{24} t^{5}-\frac {1}{720} t^{7} + \dots \\ &= t -\frac {1}{2} t^{3}+\frac {1}{24} t^{5}-\frac {1}{720} t^{7} \end {align*}
Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +6} a_{n} \left (n +r \right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right )}{24}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +2} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right ) \left (n +r -1\right )}{120} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) \left (n -5+r \right ) t^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )}{6}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r}}{6}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +6} a_{n} \left (n +r \right )}{720}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) t^{n +r}}{720}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {t^{n +r +4} a_{n} \left (n +r \right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) t^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {t^{n +r +2} a_{n} \left (n +r \right )}{2}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) t^{n +r}}{2}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) \left (n -5+r \right ) t^{n +r}}{120}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) t^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-\frac {a_{n -6} \left (n -6+r \right ) t^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {a_{n -4} \left (n -4+r \right ) t^{n +r}}{24}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-\frac {a_{n -2} \left (n +r -2\right ) t^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+t^{n +r} a_{n} \left (n +r \right )+a_{n} t^{n +r} = 0 \] When \(n = 0\) the above becomes \[ t^{r} a_{0} r \left (-1+r \right )+t^{r} a_{0} r +a_{0} t^{r} = 0 \] Or \[ \left (t^{r} r \left (-1+r \right )+t^{r} r +t^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+1\right ) t^{r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2}+1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= i\\ r_2 &= -i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+1\right ) t^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([i, -i]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +i}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -i} \end {align*}
\(y_{1}\left (t \right )\) is found first. Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r \left (2+r \right )}{6 r^{2}+24 r +30} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = 0 \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {r \left (7 r^{3}+56 r^{2}+137 r +100\right )}{360 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right )} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = 0 \] For \(6\le n\) the recursive equation is \begin{equation} \tag{3} \frac {a_{n -4} \left (n -4+r \right ) \left (n -5+r \right )}{120}-\frac {a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )}{6}+a_{n} \left (n +r \right ) \left (n +r -1\right )-\frac {a_{n -6} \left (n -6+r \right )}{720}+\frac {a_{n -4} \left (n -4+r \right )}{24}-\frac {a_{n -2} \left (n +r -2\right )}{2}+a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {6 n^{2} a_{n -4}-120 n^{2} a_{n -2}+12 n r a_{n -4}-240 n r a_{n -2}+6 r^{2} a_{n -4}-120 r^{2} a_{n -2}-n a_{n -6}-24 n a_{n -4}+240 n a_{n -2}-r a_{n -6}-24 r a_{n -4}+240 r a_{n -2}+6 a_{n -6}}{720 \left (n^{2}+2 n r +r^{2}+1\right )}\tag {4} \] Which for the root \(r = i\) becomes \[ a_{n} = \frac {\left (-6 a_{n -4}+120 a_{n -2}\right ) n^{2}+\left (\left (24-12 i\right ) a_{n -4}+\left (-240+240 i\right ) a_{n -2}+a_{n -6}\right ) n +\left (6+24 i\right ) a_{n -4}+\left (-120-240 i\right ) a_{n -2}+\left (-6+i\right ) a_{n -6}}{720 n \left (2 i+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(\frac {r \left (2+r \right )}{6 r^{2}+24 r +30}\) | \(\frac {1}{48}+\frac {i}{16}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (7 r^{3}+56 r^{2}+137 r +100\right )}{360 \left (r^{2}+4 r +5\right ) \left (r^{2}+8 r +17\right )}\) | \(\frac {1}{57600}+\frac {217 i}{57600}\) |
| \(a_{5}\) | \(0\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \begin{align*} y_{1}\left (t \right )&= t^{i} \left (a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \right ) \\ &= t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (t \right )\) is found by taking the complex conjugate of \(y_{1}\left (t \right )\) which gives \[ y_{2}\left (t \right )= t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \] Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) + c_{2} t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right )+c_{2} t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right )+c_{2} t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right )+c_{2} t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+O\left (t^{6}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5 trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying with_periodic_functions in the coefficients --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 5`[0, y]
✓ Solution by Maple
Time used: 0.171 (sec). Leaf size: 45
Order:=6; dsolve(sin(t)*diff(y(t),t$2)+cos(t)*diff(y(t),t)+1/t*y(t)=0,y(t),type='series',t=0);
\[ y \left (t \right ) = c_{1} t^{-i} \left (1+\left (\frac {1}{48}-\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}-\frac {217 i}{57600}\right ) t^{4}+\operatorname {O}\left (t^{6}\right )\right )+c_{2} t^{i} \left (1+\left (\frac {1}{48}+\frac {i}{16}\right ) t^{2}+\left (\frac {1}{57600}+\frac {217 i}{57600}\right ) t^{4}+\operatorname {O}\left (t^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.036 (sec). Leaf size: 70
AsymptoticDSolveValue[Sin[t]*y''[t]+Cos[t]*y'[t]+1/t*y[t]==0,y[t],{t,0,5}]
\[ y(t)\to \left (\frac {1}{19200}+\frac {i}{57600}\right ) c_1 t^i \left ((22+65 i) t^4+(720+960 i) t^2+(17280-5760 i)\right )-\left (\frac {1}{57600}+\frac {i}{19200}\right ) c_2 t^{-i} \left ((65+22 i) t^4+(960+720 i) t^2-(5760-17280 i)\right ) \]