22.13 problem 13

Internal problem ID [2376]
Internal file name [OUTPUT/2376_Tuesday_February_27_2024_08_36_27_AM_76446809/index.tex]

Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath. Boston. 1964
Section: Exercise 40, page 186
Problem number: 13.
ODE order: 2.
ODE degree: 1.

The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable", "second order series method. Taylor series method"

Maple gives the following as the ode type

[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

\[ \boxed {y^{\prime \prime }-\sin \left (y\right )=0} \] With initial conditions \begin {align*} \left [y \left (0\right ) = \frac {\pi }{4}, y^{\prime }\left (0\right ) = 0\right ] \end {align*}

With the expansion point for the power series method at \(x = \frac {\pi }{4}\).

The ode does not have its expansion point at \(x = 0\), therefore to simplify the computation of power series expansion, change of variable is made on the independent variable to shift the initial conditions and the expasion point back to zero. The new ode is then solved more easily since the expansion point is now at zero. The solution converted back to the original independent variable. Let \[ t = x -\frac {\pi }{4} \] The ode is converted to be in terms of the new independent variable \(t\). This results in \[ \frac {d^{2}}{d t^{2}}y \left (t \right ) = \sin \left (y \left (t \right )\right ) \] With its expansion point and initial conditions now at \(t = 0\). With initial conditions now becoming \begin {align*} y(0) &= \frac {\pi }{4}\\ y'(0) &= 0 \end {align*}

The transformed ODE is now solved. Solving ode using Taylor series method. This gives review on how the Taylor series method works for solving second order ode.

Let \[ y^{\prime \prime }=f\left ( x,y,y^{\prime }\right ) \] Assuming expansion is at \(x_{0}=0\) (we can always shift the actual expansion point to \(0\) by change of variables) and assuming \(f\left ( x,y,y^{\prime }\right ) \) is analytic at \(x_{0}\) which must be the case for an ordinary point. Let initial conditions be \(y\left ( x_{0}\right ) =y_{0}\) and \(y^{\prime }\left ( x_{0}\right ) =y_{0}^{\prime }\). Using Taylor series gives\begin {align*} y\left ( x\right ) & =y\left ( x_{0}\right ) +\left ( x-x_{0}\right ) y^{\prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{2}}{2}y^{\prime \prime }\left ( x_{0}\right ) +\frac {\left ( x-x_{0}\right ) ^{3}}{3!}y^{\prime \prime \prime }\left ( x_{0}\right ) +\cdots \\ & =y_{0}+xy_{0}^{\prime }+\frac {x^{2}}{2}\left . f\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\frac {x^{3}}{3!}\left . f^{\prime }\right \vert _{x_{0},y_{0},y_{0}^{\prime }}+\cdots \\ & =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . \frac {d^{n}f}{dx^{n}}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \end {align*}

But \begin {align} \frac {df}{dx} & =\frac {\partial f}{\partial x}\frac {dx}{dx}+\frac {\partial f}{\partial y}\frac {dy}{dx}+\frac {\partial f}{\partial y^{\prime }}\frac {dy^{\prime }}{dx}\tag {1}\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\\ \frac {d^{2}f}{dx^{2}} & =\frac {d}{dx}\left ( \frac {df}{dx}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {df}{dx}\right ) +\frac {\partial }{\partial y}\left ( \frac {df}{dx}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {df}{dx}\right ) f\tag {2}\\ \frac {d^{3}f}{dx^{3}} & =\frac {d}{dx}\left ( \frac {d^{2}f}{dx^{2}}\right ) \nonumber \\ & =\frac {\partial }{\partial x}\left ( \frac {d^{2}f}{dx^{2}}\right ) +\left ( \frac {\partial }{\partial y}\frac {d^{2}f}{dx^{2}}\right ) y^{\prime }+\frac {\partial }{\partial y^{\prime }}\left ( \frac {d^{2}f}{dx^{2}}\right ) f\tag {3}\\ & \vdots \nonumber \end {align}

And so on. Hence if we name \(F_{0}=f\left ( x,y,y^{\prime }\right ) \) then the above can be written as \begin {align} F_{0} & =f\left ( x,y,y^{\prime }\right ) \tag {4}\\ F_{1} & =\frac {df}{dx}\nonumber \\ & =\frac {dF_{0}}{dx}\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}y^{\prime \prime }\nonumber \\ & =\frac {\partial f}{\partial x}+\frac {\partial f}{\partial y}y^{\prime }+\frac {\partial f}{\partial y^{\prime }}f\tag {5}\\ & =\frac {\partial F_{0}}{\partial x}+\frac {\partial F_{0}}{\partial y}y^{\prime }+\frac {\partial F_{0}}{\partial y^{\prime }}F_{0}\nonumber \\ F_{2} & =\frac {d}{dx}\left ( \frac {d}{dx}f\right ) \nonumber \\ & =\frac {d}{dx}\left ( F_{1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{1}+\left ( \frac {\partial F_{1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{1}}{\partial y^{\prime }}\right ) F_{0}\nonumber \\ & \vdots \nonumber \\ F_{n} & =\frac {d}{dx}\left ( F_{n-1}\right ) \nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) y^{\prime \prime }\nonumber \\ & =\frac {\partial }{\partial x}F_{n-1}+\left ( \frac {\partial F_{n-1}}{\partial y}\right ) y^{\prime }+\left ( \frac {\partial F_{n-1}}{\partial y^{\prime }}\right ) F_{0} \tag {6} \end {align}

Therefore (6) can be used from now on along with \begin {equation} y\left ( x\right ) =y_{0}+xy_{0}^{\prime }+\sum _{n=0}^{\infty }\frac {x^{n+2}}{\left ( n+2\right ) !}\left . F_{n}\right \vert _{x_{0},y_{0},y_{0}^{\prime }} \tag {7} \end {equation} To find \(y\left ( x\right ) \) series solution around \(x=0\). Hence \begin {align*} F_0 &= \sin \left (y \left (t \right )\right )\\ F_1 &= \frac {d F_0}{dt} \\ &= \frac {\partial F_{0}}{\partial t}+ \frac {\partial F_{0}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{0}}{\partial \frac {d}{d t}y \left (t \right )} F_0 \\ &= \left (\frac {d}{d t}y \left (t \right )\right ) \cos \left (y \left (t \right )\right )\\ F_2 &= \frac {d F_1}{dt} \\ &= \frac {\partial F_{1}}{\partial t}+ \frac {\partial F_{1}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{1}}{\partial \frac {d}{d t}y \left (t \right )} F_1 \\ &= \sin \left (y \left (t \right )\right ) \left (\cos \left (y \left (t \right )\right )-\left (\frac {d}{d t}y \left (t \right )\right )^{2}\right )\\ F_3 &= \frac {d F_2}{dt} \\ &= \frac {\partial F_{2}}{\partial t}+ \frac {\partial F_{2}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{2}}{\partial \frac {d}{d t}y \left (t \right )} F_2 \\ &= \left (\frac {d}{d t}y \left (t \right )\right ) \left (4 \cos \left (y \left (t \right )\right )^{2}-3-\left (\frac {d}{d t}y \left (t \right )\right )^{2} \cos \left (y \left (t \right )\right )\right )\\ F_4 &= \frac {d F_3}{dt} \\ &= \frac {\partial F_{3}}{\partial t}+ \frac {\partial F_{3}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{3}}{\partial \frac {d}{d t}y \left (t \right )} F_3 \\ &= \sin \left (y \left (t \right )\right ) \left (\left (\frac {d}{d t}y \left (t \right )\right )^{4}-11 \left (\frac {d}{d t}y \left (t \right )\right )^{2} \cos \left (y \left (t \right )\right )+4 \cos \left (y \left (t \right )\right )^{2}-3\right )\\ F_5 &= \frac {d F_4}{dt} \\ &= \frac {\partial F_{4}}{\partial t}+ \frac {\partial F_{4}}{\partial y} \frac {d}{d t}y \left (t \right )+ \frac {\partial F_{4}}{\partial \frac {d}{d t}y \left (t \right )} F_4 \\ &= \left (\frac {d}{d t}y \left (t \right )\right )^{5} \cos \left (y \left (t \right )\right )+\left (-26 \cos \left (y \left (t \right )\right )^{2}+15\right ) \left (\frac {d}{d t}y \left (t \right )\right )^{3}+\left (34 \cos \left (y \left (t \right )\right )^{3}-33 \cos \left (y \left (t \right )\right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \end {align*}

And so on. Evaluating all the above at initial conditions \(t = 0\) and \(y \left (0\right ) = \frac {\pi }{4}\) and \(y^{\prime }\left (0\right ) = 0\) gives \begin {align*} F_0 &= \frac {\sqrt {2}}{2}\\ F_1 &= 0\\ F_2 &= {\frac {1}{2}}\\ F_3 &= 0\\ F_4 &= -\frac {\sqrt {2}}{2}\\ F_5 &= 0 \end {align*}

Substituting all the above in (7) and simplifying gives the solution as \[ y \left (t \right ) = \frac {\pi }{4}+\frac {\sqrt {2}\, t^{2}}{4}+\frac {t^{4}}{48}-\frac {\sqrt {2}\, t^{6}}{1440}+O\left (t^{7}\right ) \] \[ y \left (t \right ) = \frac {\pi }{4}+\frac {\sqrt {2}\, t^{2}}{4}+\frac {t^{4}}{48}-\frac {\sqrt {2}\, t^{6}}{1440}+O\left (t^{7}\right ) \] Unable to also solve using normal power series since not linear ode. Not currently supported. Replacing \(t\) in the above with the original independent variable \(xs\)using \(t = x -\frac {\pi }{4}\) results in \[ y = \frac {\pi }{4}+\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{2}}{4}+\frac {\left (x -\frac {\pi }{4}\right )^{4}}{48}-\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{6}}{1440}+O\left (\left (x -\frac {\pi }{4}\right )^{7}\right ) \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\pi }{4}+\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{2}}{4}+\frac {\left (x -\frac {\pi }{4}\right )^{4}}{48}-\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{6}}{1440}+O\left (\left (x -\frac {\pi }{4}\right )^{7}\right ) \\ \end{align*}

Verification of solutions

\[ y = \frac {\pi }{4}+\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{2}}{4}+\frac {\left (x -\frac {\pi }{4}\right )^{4}}{48}-\frac {\sqrt {2}\, \left (x -\frac {\pi }{4}\right )^{6}}{1440}+O\left (\left (x -\frac {\pi }{4}\right )^{7}\right ) \] Verified OK.

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying 2nd order Liouville 
trying 2nd order WeierstrassP 
trying 2nd order JacobiSN 
differential order: 2; trying a linearization to 3rd order 
trying 2nd order ODE linearizable_by_differentiation 
trying 2nd order, 2 integrating factors of the form mu(x,y) 
trying differential order: 2; missing variables 
`, `-> Computing symmetries using: way = 3 
`, `-> Computing symmetries using: way = exp_sym 
-> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-sin(_a) = 0, _b(_a)`   *** Sublevel 2 *** 
   Methods for first order ODEs: 
   --- Trying classification methods --- 
   trying a quadrature 
   trying 1st order linear 
   trying Bernoulli 
   <- Bernoulli successful 
<- differential order: 2; canonical coordinates successful 
<- differential order 2; missing variables successful`
 

✗ Solution by Maple

Order:=7; 
dsolve([diff(y(x),x$2)=sin(y(x)),y(0) = 1/4*Pi, D(y)(0) = 0],y(x),type='series',x=1/4*Pi);
 

\[ \text {No solution found} \]

✗ Solution by Mathematica

Time used: 0.0 (sec). Leaf size: 0

AsymptoticDSolveValue[{y''[x]==Sin[y[x]],{y[0]==Pi/4,y'[0]==0}},y[x],{x,1/4*Pi,6}]
 

Not solved