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Ordinary Differential Equation Second-Order

An ODE

\begin{displaymath}
y''+P(x)y'+Q(x)y = 0
\end{displaymath} (1)

has singularities for finite $x=x_0$ under the following conditions: (a) If either $P(x)$ or $Q(x)$ diverges as $x\to
x_0$, but $(x-x_0)P(x)$ and $(x-x_0)^2Q(x)$ remain finite as $x\to
x_0$, then $x_0$ is called a regular or nonessential singular point. (b) If $P(x)$ diverges faster than $(x-x_0)^{-1}$ so that $(x-x_0)P(x)\to \infty$ as $x\to
x_0$, or $Q(x)$ diverges faster than $(x-x_0)^{-2}$ so that $(x-x_0)^2Q(x) \to \infty$ as $x\to
x_0$, then $x_0$ is called an irregular or essential singularity.


Singularities of equation (1) at infinity are investigated by making the substitution $x \equiv z^{-1}$, so $dx =
-z^{-2} \,dz$, giving

\begin{displaymath}
{dy\over dx} = -z^2 {dy\over dz}
\end{displaymath} (2)


$\displaystyle {d^2y\over dx^2}$ $\textstyle =$ $\displaystyle -z^2 {d\over dz}\left({-z^2 {dy\over dz}}\right)
= -z^2\left({-2z {dy\over dz} - z^2 {d^2y\over dz^2}}\right)$  
  $\textstyle =$ $\displaystyle 2z^3 {dy\over dz} + z^4 {d^2y\over dz^2}.$ (3)

Then (1) becomes
\begin{displaymath}
z^4 {d^2y\over dz^2} + [2z^3-z^2P(z)]{dy\over dz} + Q(z)y = 0.
\end{displaymath} (4)

Case (a): If
$\displaystyle \alpha(z)$ $\textstyle \equiv$ $\displaystyle {2z-P(z)\over z^2}$ (5)
$\displaystyle \beta(z)$ $\textstyle \equiv$ $\displaystyle {Q(z)\over z^4}$ (6)

remain finite at $x = \pm \infty$ ($y=0$), then the point is ordinary. Case (b): If either $\alpha(z)$ diverges no more rapidly than $1/z$ or $\beta(z)$ diverges no more rapidly than $1/z^2$, then the point is a regular singular point. Case (c): Otherwise, the point is an irregular singular point.


Morse and Feshbach (1953, pp. 667-674) give the canonical forms and solutions for second-order ODEs classified by types of singular points.


For special classes of second-order linear ordinary differential equations, variable Coefficients can be transformed into constant Coefficients. Given a second-order linear ODE with variable Coefficients

\begin{displaymath}
{d^2y\over dx^2} + p(x){dy\over dx} + q(x)y = 0.
\end{displaymath} (7)

Define a function $z \equiv y(x)$,
\begin{displaymath}
{dy\over dx} = {dz\over dx} {dy\over dz}
\end{displaymath} (8)


\begin{displaymath}
{d^2y\over dx^2} = \left({dz\over dx}\right)^2 {d^2y\over dz^2} + {d^2z\over dx^2} {dy\over dz}
\end{displaymath} (9)


\begin{displaymath}
\left({dz\over dx}\right)^2 {d^2y\over dz^2} + \left[{{d^2z\over dx^2} + p(x){dz\over dx}}\right]{dy\over dz} + q(x)y = 0
\end{displaymath} (10)


\begin{displaymath}
{d^2y\over dz^2} + \left[{{d^2z\over dx^2} + p(x){dz\over dx...
...^2}\right]y \equiv {d^2y\over dz^2} + A {dy\over dz} + By = 0.
\end{displaymath} (11)

This will have constant Coefficients if $A$ and $B$ are not functions of $x$. But we are free to set $B$ to an arbitrary Positive constant for $q(x)\geq 0$ by defining $z$ as
\begin{displaymath}
z \equiv B^{-1/2}\int [q(x)]^{1/2}\,dx.
\end{displaymath} (12)

Then
\begin{displaymath}
{dz\over dx} = B^{-1/2}[q(x)]^{1/2}
\end{displaymath} (13)


\begin{displaymath}
{d^2z\over dx^2} = {\textstyle{1\over 2}}B^{-1/2}[q(x)]^{-1/2}q'(x),
\end{displaymath} (14)

and
$\displaystyle A$ $\textstyle =$ $\displaystyle {{\textstyle{1\over 2}}B^{-1/2}[q(x)]^{-1/2}q'(x)+B^{-1/2}p(x)[q(x)]^{1/2}\over B^{-1}q(x)}$  
  $\textstyle =$ $\displaystyle {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2}.$ (15)

Equation (11) therefore becomes
\begin{displaymath}
{d^2y\over dz^2} + {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2} {dy\over dz} + By = 0,
\end{displaymath} (16)

which has constant Coefficients provided that
\begin{displaymath}
A \equiv {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2} = {\rm [constant]}.
\end{displaymath} (17)

Eliminating constants, this gives
\begin{displaymath}
A'\equiv {q'(x)+2p(x)q(x)\over [q(x)]^{3/2}} = {\rm [constant]}.
\end{displaymath} (18)

So for an ordinary differential equation in which $A'$ is a constant, the solution is given by solving the second-order linear ODE with constant Coefficients
\begin{displaymath}
{d^2y\over dz^2} + A {dy\over dz} + By = 0
\end{displaymath} (19)

for $z$, where $z$ is defined as above.


A linear second-order homogeneous differential equation of the general form

\begin{displaymath}
y''+P(x)y'+Q(x)y = 0
\end{displaymath} (20)

can be transformed into standard form
\begin{displaymath}
z''+q(x)z=0
\end{displaymath} (21)

with the first-order term eliminated using the substitution
\begin{displaymath}
\ln y \equiv \ln z - {\textstyle{1\over 2}}\int P(x)\,dx.
\end{displaymath} (22)

Then
\begin{displaymath}
{y'\over y}={z'\over z}-{\textstyle{1\over 2}}P(x)
\end{displaymath} (23)


\begin{displaymath}
{yy''-y'^2\over y^2}={zz''-z'^2\over z^2}-{\textstyle{1\over 2}}P'(x)
\end{displaymath} (24)


\begin{displaymath}
{y''\over y}-\left({y'\over y}\right)^2={z''\over z}-{z'^2\over z}-{z'^2\over z^2}-{\textstyle{1\over 2}}P'(x)
\end{displaymath} (25)


$\displaystyle {y''\over y}=\left[{{z'\over z}-{\textstyle{1\over 2}}P(x)}\right]^2+{z''\over z}-{z'^2\over z^2}-{\textstyle{1\over 2}}P'(x)$      
  $\textstyle =$ $\displaystyle {z'^2\over z^2}-{z'\over z} P(x)+{\textstyle{1\over 4}}P^2(x)+{z''\over z}-{z'^2\over z^2}-{\textstyle{1\over 2}}P'(x),$ (26)

so


$\displaystyle {y''\over y}+P(x){y'\over y}+Q(x)$ $\textstyle =$ $\displaystyle -{z'\over z}P(x)+{\textstyle{1\over 4}}P^2(x)+{z''\over z}-{\text...
...{1\over 2}}P'(x)+P(x)\left[{{z'\over z}-{\textstyle{1\over 2}}P(x)}\right]+Q(x)$ (27)

Therefore,


\begin{displaymath}
z''+[Q(x)-{\textstyle{1\over 2}}P'(x)-{\textstyle{1\over 4}}P^2(x)]z \equiv z''(x)+q(x)z = 0,
\end{displaymath} (28)

where
\begin{displaymath}
q(x) \equiv Q(x)-{\textstyle{1\over 2}}P'(x)-{\textstyle{1\over 4}}P^2(x).
\end{displaymath} (29)


If $Q(x)= 0$, then the differential equation becomes

\begin{displaymath}
y''+P(x)y'=0,
\end{displaymath} (30)

which can be solved by multiplying by
\begin{displaymath}
\mathop{\rm exp}\nolimits \left[{\int^x P(x')\,dx'}\right]
\end{displaymath} (31)

to obtain
\begin{displaymath}
0 = {d\over dx}\left\{{\mathop{\rm exp}\nolimits \left[{\int^x P(x')\,dx'}\right]{dy\over dx}}\right\}
\end{displaymath} (32)


\begin{displaymath}
c_1 = \mathop{\rm exp}\nolimits \left[{\int^x P(x')\,dx'}\right]{dy\over dx}
\end{displaymath} (33)


\begin{displaymath}
y = c_1\int^x {dx\over \mathop{\rm exp}\nolimits \left[{\int^x P(x')\,dx'}\right]}+c_2.
\end{displaymath} (34)


If one solution ($y_1$) to a second-order ODE is known, the other ($y_2$) may be found using the Reduction of Order method. From the Abel's Identity

\begin{displaymath}
{dW\over W} = -P(x)\,dx,
\end{displaymath} (35)

where
\begin{displaymath}
W \equiv y_1y_2'-y_1'y_2
\end{displaymath} (36)


\begin{displaymath}
\int_a^x {dW\over W} = \int^x_a P(x')\,dx'
\end{displaymath} (37)


\begin{displaymath}
\ln\left[{W(x)\over W(a)}\right]= \int^x_a P(x')\,dx'
\end{displaymath} (38)


\begin{displaymath}
W(x) = W(a)\mathop{\rm exp}\nolimits \left[{- \int^x_a P(x')\,dx'}\right].
\end{displaymath} (39)

But
\begin{displaymath}
W \equiv y_1y_2'-y_1'y_2 = y_1^2{d\over dx}\left({y_2\over y_1}\right).
\end{displaymath} (40)

Combining (39) and (40) yields
\begin{displaymath}
{d\over dx}\left({y_2\over y1}\right)= W(a) {\mathop{\rm exp}\nolimits [- \int^x_a P(x')\,dx']\over y_1^2}
\end{displaymath} (41)


\begin{displaymath}
y_2(x) = y_1(x)W(a)\int_b^x {\mathop{\rm exp}\nolimits [- \int^{x'}_a P(x'')\,dx'']\over [y_1(x')]^2}\,dx'.
\end{displaymath} (42)

Disregarding $W(a)$, since it is simply a multiplicative constant, and the constants $a$ and $b$, which will contribute a solution which is not linearly independent of $y_1$,
\begin{displaymath}
y_2(x) = y_1(x)\int^x {\mathop{\rm exp}\nolimits \left[{- \int^{x'} P(x'')\,dx''}\right]\over [y_1(x')]^2}\,dx'.
\end{displaymath} (43)

If $P(x) = 0$, this simplifies to
\begin{displaymath}
y_2(x) = y_1(x) \int^x{dx'\over [y_1(x')]^2}.
\end{displaymath} (44)


For a nonhomogeneous second-order ODE in which the $x$ term does not appear in the function $f(x,y,y')$,

\begin{displaymath}
{d^2y\over dx^2} = f(y,y'),
\end{displaymath} (45)

let $v \equiv y'$, then
\begin{displaymath}
{dv\over dx} = f(v,y) = {dv\over dy} {dy\over dx} = v {dv\over dy}.
\end{displaymath} (46)

So the first-order ODE
\begin{displaymath}
v {dv\over dy} = f(y,v),
\end{displaymath} (47)

if linear, can be solved for $v$ as a linear first-order ODE. Once the solution is known,
\begin{displaymath}
{dy\over dx} = v(y)
\end{displaymath} (48)


\begin{displaymath}
\int {dy\over v(y)} = \int dx.
\end{displaymath} (49)


On the other hand, if $y$ is missing from $f(x,y,y')$,

\begin{displaymath}
{d^2y\over dx^2} = f(x,y'),
\end{displaymath} (50)

let $v \equiv y'$, then $v' = y''$, and the equation reduces to
\begin{displaymath}
v' = f(x,v),
\end{displaymath} (51)

which, if linear, can be solved for $v$ as a linear first-order ODE. Once the solution is known,
\begin{displaymath}
y = \int v(x)\,dx.
\end{displaymath} (52)

See also Abel's Identity, Adjoint Operator


References

Arfken, G. ``A Second Solution.'' §8.6 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 467-480, 1985.

Boyce, W. E. and DiPrima, R. C. Elementary Differential Equations and Boundary Value Problems, 4th ed. New York: Wiley, 1986.

Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 667-674, 1953.



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© 1996-9 Eric W. Weisstein
1999-05-26