Order Reduction in Nonlinear Equations, differential equations lesson

The equation $y'' = (y')^2$ , where $y' = dy/dx$ , is nonlinear, so the theory of linear equations from the previous lessons does not apply to it. Still, the unknown $y$ never appears by itself, only through its derivatives, so the equation is really a first-order equation for the velocity $y'$ . This lesson shows how to lower the order when either $y$ or $x$ is missing from the equation, how to rebuild $y$ with both initial values, and where the new coordinate stops being valid.

Key terms

Nonlinear order reduction

Definition Consider a second-order equation $y'' = F(x, y, y')$ for an unknown function $y$ of $x$ , with $y' = dy/dx$ . Nonlinear order reduction replaces it by first-order equations when one variable is absent from $F$ .

If $y$ is absent, so the equation reads $y'' = F(x, y')$ , the velocity $v = y'$ is treated as a function of $x$ .
If $x$ is absent, so the equation reads $y'' = G(y, y')$ , the velocity is treated as a function of $y$ on an interval where $y' \ne 0$ .

In both cases one solves a first-order equation for $v$ and then the first-order equation $y' = v$ for $y$ . The two steps supply two constants, fixed by $y(x_0)$ and $y'(x_0)$ . The method works whether or not the equation is linear, and the integrals it leads to need not have closed forms.

In plain words Before trying anything else, look at which letters appear. A missing letter means the equation is really first order in the velocity, and one more integration recovers the position.

Example In $y'' = 1 + (y')^2$ neither $x$ nor $y$ appears, so either version applies. With $v$ as a function of $x$ the equation reads $v' = 1 + v^2$ , and since $1 + v^2$ is never zero, separation gives $\arctan v = x + C_1$ , so $v = \tan(x + C_1)$ . One more integration gives

y = -\ln\bigl(\cos(x + C_1)\bigr) + C_2

on the interval $-\pi/2 < x + C_1 < \pi/2$ , of length $\pi$ . Differentiating confirms $y' = \tan(x + C_1)$ and $y'' = \sec^2(x + C_1) = 1 + (y')^2$ . By contrast, $y'' = xy$ contains both $x$ and $y$ , so neither substitution lowers its order.

Velocity substitution

Definition The velocity substitution sets $v = y'$ and writes $y''$ in terms of $v$ . If $v$ is regarded as a function of $x$ , then $y'' = dv/dx$ . If $y' \ne 0$ on an interval $J$ , then $y$ is strictly monotone on $J$ and has a differentiable inverse $x = x(y)$ , so $v$ can be regarded as a function of $y$ , and the chain rule gives

\begin{aligned} y'' &= \frac{dv}{dx} = \frac{dv}{dy}\,\frac{dy}{dx} \\[0.4em] &= v\,\frac{dv}{dy}. \end{aligned}

The first form keeps $x$ as the independent variable. The second makes $y$ the independent variable and is valid only where $y' \ne 0$ .

In plain words The velocity can be recorded against $x$ or against the position $y$ . Against $x$ the acceleration is simply the derivative of $v$ . Against $y$ the chain rule brings in the factor $v$ , and the record makes sense only while $y$ moves in one direction, so that each position is passed once.

Example For $y = e^{2x}$ the velocity is $v = 2e^{2x} = 2y$ , so as a function of $y$ it is $v(y) = 2y$ with $dv/dy = 2$ . Then $v\,dv/dy = 4y = 4e^{2x}$ , which is indeed $y''$ . For $y = x^2$ the second form breaks down near $x = 0$ . The velocity $v = 2x$ equals $2\sqrt{y}$ for $x > 0$ and $-2\sqrt{y}$ for $x < 0$ , so no single function of $y$ describes it on an interval around the turning point $x = 0$ , where $y' = 0$ .

Missing dependent variable

Definition An equation $y'' = F(x, y')$ in which $y$ does not appear is reduced by $v(x) = y'(x)$ to the first-order problem

\begin{gathered} v' = F(x, v), \\[0.4em] v(x_0) = y'(x_0). \end{gathered}

If $v$ solves this problem on an interval $I$ containing $x_0$ , then

y(x) = y(x_0) + \int_{x_0}^{x} v(s)\,ds

solves the original problem on $I$ , so $y$ lives on the same interval as $v$ . The velocity datum belongs to the first-order problem and the height datum to the final integration. If $F(x, 0) = 0$ for all $x$ , then $v = 0$ is a solution of the reduced equation, and it gives the constant solutions $y = c$ , which a division by $v$ can lose.

In plain words With $y$ missing, the equation is a first-order equation for the slope. Solve for the slope with the slope datum, then integrate the slope starting from the height datum.

Example Solve $y'' = (y')^2$ with $y(0) = 2$ and $y'(0) = 1$ . The velocity satisfies $v' = v^2$ with $v(0) = 1$ . The constant $v = 0$ also solves $v' = v^2$ , and since solutions of this equation are unique, the solution with $v(0) = 1$ never reaches $0$ , so dividing by $v^2$ is safe. Separation gives $-1/v = x + C$ , and $v(0) = 1$ gives $C = -1$ , so

v = \frac{1}{1 - x}

on $x < 1$ . Integrating from the height datum gives

y = 2 - \ln(1 - x)

on $x < 1$ . Its derivatives are $y' = (1 - x)^{-1}$ and $y'' = (1 - x)^{-2} = (y')^2$ , with $y(0) = 2$ and $y'(0) = 1$ . The maximal interval is $(-\infty, 1)$ , and both $y$ and $y'$ tend to $\infty$ as $x \to 1^-$ . The constant solutions $y = c$ from the branch $v = 0$ solve the equation too, but their slope $0$ does not match the data.

Missing independent variable

Definition An equation $y'' = G(y, y')$ in which $x$ does not appear is called autonomous. On an interval where $y' \ne 0$ , the velocity substitution with $v$ as a function of $y$ reduces it to the first-order problem

\begin{gathered} v\,\frac{dv}{dy} = G(y, v), \\[0.4em] v(y_0) = y_1, \end{gathered}

where $y_0 = y(x_0)$ and $y_1 = y'(x_0) \ne 0$ . After solving for $v(y)$ , the position follows from the separable equation $y' = v(y)$ with $y(x_0) = y_0$ . A constant solution $y = c$ has $v = 0$ and is not described by any function $v(y)$ . The constant solutions are exactly the constants $c$ with $G(c, 0) = 0$ , and they should be found from the original equation before dividing by $v$ .

In plain words With $x$ missing, the acceleration depends only on where you are and how fast you move, so the velocity can be recorded as a function of position. That record satisfies a first-order equation, and one more integration says where you are at each $x$ .

Example Solve $y'' = 2y^3$ with $y(0) = 1$ and $y'(0) = 1$ . The only constant solution is $y = 0$ , since $2c^3 = 0$ forces $c = 0$ , and it does not match the data. The reduced equation is $v\,dv/dy = 2y^3$ , and integrating in $y$ gives

\frac{v^2}{2} = \frac{y^4}{2} + C.

The data $y = 1$ , $v = 1$ give $C = 0$ , so $v^2 = y^4$ , and the positive initial slope selects $v = y^2$ . Then $y' = y^2$ with $y(0) = 1$ separates into $-1/y = x - 1$ , so

y = \frac{1}{1 - x}, \qquad x < 1.

This function has $y' = (1 - x)^{-2} = y^2$ and $y'' = 2(1 - x)^{-3} = 2y^3$ , with $y(0) = 1$ and $y'(0) = 1$ . It is positive and increasing on all of $(-\infty, 1)$ , so $y' \ne 0$ along the whole solution and the coordinate $y$ stays valid. As $x \to 1^-$ the solution blows up.

First integral

Definition A first integral of an autonomous equation $y'' = G(y, y')$ is a continuously differentiable function $H(y, v)$ , not constant on any open set, such that $H\bigl(y(x), y'(x)\bigr)$ is constant on the interval of every solution. For $y'' = f(y)$ with $f$ continuous and $\Phi$ an antiderivative of $f$ , the function

H(y, v) = \frac{v^2}{2} - \Phi(y)

is a first integral, because along every solution

\begin{aligned} \frac{dH}{dx} &= y'y'' - f(y)\,y' \\[0.4em] &= y'\bigl(y'' - f(y)\bigr) = 0. \end{aligned}

Each solution therefore lies on a level set $H = E$ , with $E$ fixed by the initial data, even across points where $y' = 0$ . Solving the level set for the velocity gives $v = \pm\sqrt{2\bigl(E + \Phi(y)\bigr)}$ . The sign is part of the data, and one level set can carry several different solutions as well as equilibria.

In plain words A first integral is a conserved quantity, like the energy of a frictionless particle. It replaces the second-order equation by a first-order relation between position and velocity, and only the sign of the velocity is left to choose.

Example For $y'' = 2y^3$ take $\Phi(y) = y^4/2$ . Doubling $H$ gives the conserved quantity $(y')^2 - y^4$ . The data $y(0) = 1$ , $y'(0) = 1$ give the value $0$ , and so do the data $y(0) = 1$ , $y'(0) = -1$ . Both solutions lie on the level set $v^2 = y^4$ , which in the $(y, v)$ plane consists of the curves $v = y^2$ and $v = -y^2$ , meeting at the equilibrium $(0, 0)$ . The first data follow $v = y^2$ and give $y = 1/(1 - x)$ on $(-\infty, 1)$ . The second data follow $v = -y^2$ , so $y' = -y^2$ and

y = \frac{1}{1 + x}, \qquad x > -1,

which has $y' = -(1 + x)^{-2}$ and $y'' = 2(1 + x)^{-3} = 2y^3$ , with the right initial values. One level set holds two different solutions, defined on two different intervals.

Local coordinate restriction

Definition Using $y$ as the independent coordinate is valid only on an interval of $x$ where $y' \ne 0$ . Since $y'$ is continuous, such an interval exists around every point $x_0$ with $y'(x_0) \ne 0$ , and $y$ is strictly monotone on it. At a turning point, where $y' = 0$ and $y'' \ne 0$ , the velocity changes sign and the description $v(y)$ switches from one branch to the other. There the reduced equation $y' = v(y)$ can have several solutions, including constant ones that do not solve the original equation, so the motion is continued with the original second-order equation. That equation has unique solutions when $G$ is continuously differentiable.

In plain words The position can label the points of a solution only while it keeps moving in one direction. When the motion stops and turns back, start a new chart, and let the original equation say which way the motion goes.

Example Take $y'' = y - 2y^3$ with $y(0) = 1$ and $y'(0) = 0$ , which starts at a turning point. Here $\Phi(y) = y^2/2 - y^4/2$ , the doubled first integral $(y')^2 - y^2 + y^4$ equals $0$ for these data, and so $v^2 = y^2(1 - y^2)$ . The original equation gives $y''(0) = 1 - 2 = -1 < 0$ , so $x = 0$ is a local maximum of $y$ . The velocity is negative just to the right of $0$ and positive just to the left, which selects the branches

\begin{gathered} v = -y\sqrt{1 - y^2} \;\text{ for } x > 0, \\[0.4em] v = y\sqrt{1 - y^2} \;\text{ for } x < 0. \end{gathered}

Separating either branch leads to the even function

y = \operatorname{sech} x = \frac{2}{e^{x} + e^{-x}},

which satisfies the equation and the data and is therefore the solution on all of $\mathbb{R}$ . It decreases for $x > 0$ and increases for $x < 0$ , so each branch describes one half of the motion.

The reduced equation $y' = -y\sqrt{1 - y^2}$ with $y(0) = 1$ is also solved by the constant $y = 1$ , because its right side vanishes at $y = 1$ . That constant is not a solution of the original equation, which demands $y'' = 1 - 2 = -1$ wherever $y = 1$ . Only the second-order equation decides what happens at the turning point.

Check yourself

Solve $y'' = 2x(y')^2$ with $y(0) = 1$ and $y'(0) = -1$ , and give the interval on which the solution exists.

Show the answer

The variable $y$ is missing, so set $v = y'$ . Then $v' = 2xv^2$ with $v(0) = -1$ . The constant $v = 0$ does not match the data, and by uniqueness the solution never reaches it, so separation is safe and gives $-1/v = x^2 + C$ . The datum $v(0) = -1$ gives $C = 1$ , so $v = -1/(1 + x^2)$ , which is defined for every $x$ . Integrating from $y(0) = 1$ gives

y = 1 - \arctan x.

Indeed $y' = -1/(1 + x^2)$ and $y'' = 2x/(1 + x^2)^2 = 2x(y')^2$ . The solution exists on all of $\mathbb{R}$ .

A classmate solves $y'' = (y')^2$ with $y(0) = 2$ and $y'(0) = 0$ by separating $v' = v^2$ into $-1/v = x + C$ , and then cannot find $C$ . What went wrong, and what is the solution?

Show the answer

Separation divides by $v^2$ and so assumes $v \ne 0$ . The data give $v(0) = 0$ , and the constant $v = 0$ solves $v' = v^2$ . Since $v^2$ is continuously differentiable, it is the only solution with $v(0) = 0$ . So $y' = 0$ and $y = 2$ for all $x$ . This constant solves the original equation, since $y'' = 0 = (y')^2$ , and it is one of the constant solutions that the division lost.

Reduce $y'' = y + (y')^2$ to a first-order equation for $w = (y')^2$ as a function of $y$ , solve it, and find the first integral for the data $y(0) = 0$ , $y'(0) = 1$ . Explain why $y$ can serve as a coordinate along the whole solution.

Show the answer

The variable $x$ is missing. With $v = y'$ as a function of $y$ , the identity $v\,dv/dy = \frac{1}{2}\,dw/dy$ turns the equation into $\frac{1}{2}\,dw/dy = y + w$ , a linear equation,

\frac{dw}{dy} - 2w = 2y.

The integrating factor $e^{-2y}$ gives $\frac{d}{dy}\bigl(e^{-2y}w\bigr) = 2ye^{-2y}$ and then $e^{-2y}w = -ye^{-2y} - \frac{1}{2}e^{-2y} + C$ , so $w = Ce^{2y} - y - \frac{1}{2}$ . At $y = 0$ we need $w = 1$ , so $C = \frac{3}{2}$ and

(y')^2 = \tfrac{3}{2}e^{2y} - y - \tfrac{1}{2}.

Multiplying by $e^{-2y}$ shows that the first integral is $H = e^{-2y}\bigl((y')^2 + y + \frac{1}{2}\bigr)$ , with the value $\frac{3}{2}$ for these data. Its derivative along any solution is $e^{-2y}y'\bigl(2y'' - 2(y')^2 - 2y\bigr) = 0$ , so the relation holds on the whole interval of the solution, not only where $y' \ne 0$ . The right side $g(y) = \frac{3}{2}e^{2y} - y - \frac{1}{2}$ of the relation has $g'(y) = 3e^{2y} - 1$ , which vanishes only at $y = -\frac{1}{2}\ln 3$ , where $g$ takes its minimum value $\frac{1}{2}\ln 3 > 0$ . So $y'$ is never zero, and since $y'(0) = 1 > 0$ the solution satisfies $y' = \sqrt{g(y)}$ throughout. It is strictly increasing, so $y$ is a valid coordinate along all of it. The last step $x = \int_0^{y} ds/\sqrt{g(s)}$ can be left as an integral or evaluated numerically.

Order Reduction in Nonlinear Equations

Key terms

Nonlinear order reduction

Velocity substitution

Missing dependent variable

Missing independent variable

First integral

Local coordinate restriction

Check yourself

How this lesson connects

Builds on