Ordinary Differential Equation Second-Order

An ODE

$$y''+P(x)y'+Q(x)y = 0$$ (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

$${dy\over dx} = -z^2 {dy\over dz}$$ (2)

$${d^2y\over dx^2} = -z^2 {d\over dz}\left({-z^2 {dy\over dz}}\right) = -z^2\left({-2z {dy\over dz} - z^2 {d^2y\over dz^2}}\right)$$

$$= 2z^3 {dy\over dz} + z^4 {d^2y\over dz^2}.$$ (3)

Then (1) becomes

$$z^4 {d^2y\over dz^2} + [2z^3-z^2P(z)]{dy\over dz} + Q(z)y = 0.$$ (4)

Case (a): If

$$\alpha(z) \equiv {2z-P(z)\over z^2}$$ (5)

$$\beta(z) \equiv {Q(z)\over z^4}$$ (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

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

Define a function $z \equiv y(x)$,

$${dy\over dx} = {dz\over dx} {dy\over dz}$$ (8)

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

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

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

$$z \equiv B^{-1/2}\int [q(x)]^{1/2}\,dx.$$ (12)

Then

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

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

and

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

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

Equation (11) therefore becomes

$${d^2y\over dz^2} + {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2} {dy\over dz} + By = 0,$$ (16)

which has constant Coefficients provided that

$$A \equiv {q'(x)+2p(x)q(x)\over 2[q(x)]^{3/2}} B^{1/2} = {\rm [constant]}.$$ (17)

Eliminating constants, this gives

$$A'\equiv {q'(x)+2p(x)q(x)\over [q(x)]^{3/2}} = {\rm [constant]}.$$ (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

$${d^2y\over dz^2} + A {dy\over dz} + By = 0$$ (19)

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

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

$$y''+P(x)y'+Q(x)y = 0$$ (20)

can be transformed into standard form

$$z''+q(x)z=0$$ (21)

with the first-order term eliminated using the substitution

$$\ln y \equiv \ln z - {\textstyle{1\over 2}}\int P(x)\,dx.$$ (22)

Then

$${y'\over y}={z'\over z}-{\textstyle{1\over 2}}P(x)$$ (23)

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

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

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

$$= {z'^2\over z^2}-{z'\over z} P(x)+{\textstyle{1\over 4}}P^2(x)+{z''\over z}-{z'^2\over z^2}-{\textstyle{1\over 2}}P'(x),$$ (26)

so

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

Therefore,

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

where

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

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

$$y''+P(x)y'=0,$$ (30)

which can be solved by multiplying by

$$\mathop{\rm exp}\nolimits \left[{\int^x P(x')\,dx'}\right]$$ (31)

to obtain

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

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

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

$${dW\over W} = -P(x)\,dx,$$ (35)

where

$$W \equiv y_1y_2'-y_1'y_2$$ (36)

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

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

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

But

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

Combining (39) and (40) yields

$${d\over dx}\left({y_2\over y1}\right)= W(a) {\mathop{\rm exp}\nolimits [- \int^x_a P(x')\,dx']\over y_1^2}$$ (41)

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

$$y_2(x) = y_1(x)\int^x {\mathop{\rm exp}\nolimits \left[{- \int^{x'} P(x'')\,dx''}\right]\over [y_1(x')]^2}\,dx'.$$ (43)

If $P(x) = 0$, this simplifies to

$$y_2(x) = y_1(x) \int^x{dx'\over [y_1(x')]^2}.$$ (44)

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

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

let $v \equiv y'$, then

$${dv\over dx} = f(v,y) = {dv\over dy} {dy\over dx} = v {dv\over dy}.$$ (46)

So the first-order ODE

$$v {dv\over dy} = f(y,v),$$ (47)

if linear, can be solved for $v$ as a linear first-order ODE. Once the solution is known,

$${dy\over dx} = v(y)$$ (48)

$$\int {dy\over v(y)} = \int dx.$$ (49)

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

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

let $v \equiv y'$, then $v' = y''$, and the equation reduces to

$$v' = f(x,v),$$ (51)

which, if linear, can be solved for $v$ as a linear first-order ODE. Once the solution is known,

$$y = \int v(x)\,dx.$$ (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.