Riccati Equations with a Known Particular Solution, differential equations lesson

Throughout this lesson $t$ is the independent variable and $y' = dy/dt$ . The equation $y' = y^2 - 1$ has two solutions that can be read off at once, the constants $1$ and $-1$ , because each makes the right side zero. The solution with $y(0) = 0$ , or one that starts above $1$ , is not visible in the same way. This lesson shows how one known solution of such a quadratic equation reduces the search for every other solution to a linear first-order equation. It also covers the one solution the reduction leaves out, the known solution itself, and the times at which other solutions become infinite.

Key terms

Riccati equation

Definition A Riccati equation is a first-order equation of the form

y' = a(t)y^2 + b(t)y + c(t),

where $a$ , $b$ and $c$ are continuous on an open interval $I$ and $a$ is not the zero function on $I$ . When $a$ is the zero function the equation is linear, and when $c$ is the zero function it is a Bernoulli equation with exponent $2$ . The right side $f(t, y) = a(t)y^2 + b(t)y + c(t)$ and its partial derivative $\partial f/\partial y = 2a(t)y + b(t)$ are continuous for every $t$ in $I$ and every real $y$ . So each initial condition $y(t_0) = y_0$ with $t_0$ in $I$ has a solution on some open interval containing $t_0$ , and two solutions with the same initial value agree wherever both are defined. The interval of a solution can be much shorter than $I$ .

In plain words The rate of change depends on the current state through a quadratic whose coefficients may vary with time. The square makes the equation nonlinear, and it can drive a solution to infinity in finite time even when the coefficients are continuous everywhere.

Example The equation $y' = y^2 - 1$ is a Riccati equation with $a = 1$ , $b = 0$ and $c = -1$ on the whole real line. The equation $y' = y^2 - 2/t^2$ is one on $I = (0, \infty)$ , or on $(-\infty, 0)$ , but not on an interval containing $0$ . The equation $y' = y^2 + 1$ shows how short the interval of a solution can be. Its solutions are $y = \tan(t - k)$ with a constant $k$ , each defined only for $k - \pi/2 < t < k + \pi/2$ , although its coefficients are continuous for every $t$ .

Known particular solution

Definition A particular solution of the Riccati equation is a differentiable function $y_p$ on an interval $J \subseteq I$ with

y_p' = a\,y_p^2 + b\,y_p + c

at every point of $J$ . The reduction in this lesson starts from one such function. It does not produce one, so $y_p$ has to come from inspection, from the structure of a model or from another calculation, and it counts as known only after substituting it into the equation confirms it. When $a$ , $b$ and $c$ are constants, each real root $r$ of $ar^2 + br + c = 0$ gives the constant solution $y_p = r$ .

In plain words You need one solution in hand before the method can start. Guessing it is fine, but a guess is only a candidate until you substitute it.

Example For $y' = y^2 - 1$ the roots of $r^2 - 1 = 0$ give the known solutions $y_p = 1$ and $y_p = -1$ . For $y' = y^2 - 2/t^2$ on $(0, \infty)$ , try $y = k/t$ with a constant $k$ . Then $y' = -k/t^2$ and $y^2 - 2/t^2 = (k^2 - 2)/t^2$ , so the trial function works exactly when $-k = k^2 - 2$ , that is, when $(k - 1)(k + 2) = 0$ . Both $y_p = 1/t$ and $y_p = -2/t$ are known particular solutions on $(0, \infty)$ .

Reciprocal substitution

Definition Let $y_p$ be a known particular solution on $J$ . The reciprocal substitution writes the unknown as

y = y_p + \frac{1}{u},

where $u$ is a new differentiable unknown with no zero on the interval considered. Then

y' = y_p' - \frac{u'}{u^2},

and expanding the square gives

\begin{aligned} &ay^2 + by + c \\[0.4em] &= ay_p^2 + by_p + c \\[0.4em] &\quad + \frac{2ay_p + b}{u} + \frac{a}{u^2}. \end{aligned}

The terms that involve only $y_p$ add up to $y_p'$ , because $y_p$ solves the equation, so they cancel against $y_p'$ on the left. What remains is

-\frac{u'}{u^2} = \frac{2ay_p + b}{u} + \frac{a}{u^2},

and multiplying by $-u^2$ gives

u' + \left(2ay_p + b\right)u = -a.

Equivalently, the gap $v = y - y_p$ satisfies the Bernoulli equation $v' - (2ay_p + b)v = av^2$ with exponent $2$ , and $u = 1/v$ is the usual substitution for it from Bernoulli Equations.

In plain words Describe every other solution by its gap from the known one, and write that gap as $1/u$ . The known terms cancel, and after clearing denominators the square no longer involves the unknown.

Example For $y' = y^2 - 1$ with $y_p = 1$ , the substitution $y = 1 + 1/u$ gives $y' = -u'/u^2$ and

y^2 - 1 = \frac{2}{u} + \frac{1}{u^2}.

Setting the two sides equal and multiplying by $-u^2$ gives $u' = -2u - 1$ , that is, $u' + 2u = -1$ .

Linear equation for the correction

Definition Let $y_p$ be a particular solution on an interval $J \subseteq I$ . The correction $u$ in $y = y_p + 1/u$ satisfies the linear first-order equation

u' + p(t)\,u = q(t),

with $p = 2ay_p + b$ and $q = -a$ , both continuous on $J$ . The two equations correspond exactly. If $u$ solves this linear equation on an interval $K \subseteq J$ and has no zero there, then $y_p + 1/u$ solves the Riccati equation on $K$ . If $y$ solves the Riccati equation on such an interval $K$ and $y \ne y_p$ at one point of $K$ , then by uniqueness $y \ne y_p$ at every point of $K$ , and $u = 1/(y - y_p)$ solves the linear equation on $K$ . With the integrating factor $\mu = e^{\int p\,dt}$ the linear equation can be solved on all of $J$ , and each initial value $u(t_0) = u_0$ with $t_0$ in $J$ selects exactly one solution there.

In plain words After the substitution, the correction obeys the same kind of equation as in Linear First-Order Equations and Integrating Factors, so an integrating factor finds every $u$ . Each $u$ that stays away from zero gives back a solution of the original equation, and every solution other than $y_p$ comes from such a $u$ .

Example Solve $y' = y^2 - 1$ with $y(0) = 0$ , using $y_p = 1$ . The correction satisfies $u' + 2u = -1$ . With the integrating factor $e^{2t}$ this reads $(e^{2t}u)' = -e^{2t}$ , so $e^{2t}u = -\tfrac{1}{2}e^{2t} + C$ and

u = Ce^{-2t} - \frac{1}{2}.

The initial value converts through $y(0) = 1 + 1/u(0)$ , so $0 = 1 + 1/u(0)$ and $u(0) = -1$ . Then $C - \tfrac{1}{2} = -1$ gives $C = -\tfrac{1}{2}$ , so $u = -\tfrac{1}{2}(1 + e^{-2t})$ , which is negative for every real $t$ . Hence

y = 1 - \frac{2}{1 + e^{-2t}},

which equals $-\tanh t$ and is defined for all real $t$ . To check it, write $w = e^{-2t}$ , so that $w' = -2w$ and $y = 1 - 2/(1 + w)$ . Then

\begin{aligned} y' &= \frac{2w'}{(1 + w)^2} \\[0.4em] &= -\frac{4w}{(1 + w)^2}, \end{aligned}

and the right side of the equation gives the same expression,

\begin{aligned} y^2 - 1 &= (y - 1)(y + 1) \\[0.4em] &= -\frac{2}{1 + w}\cdot\frac{2w}{1 + w}. \end{aligned}

The initial value also checks, since $y(0) = 1 - 2/2 = 0$ .

Omitted particular branch

Definition The known solution $y_p$ never has the form $y_p + 1/u$ , because $1/u$ is never zero, so the substitution omits it. By uniqueness, no other solution takes the value $y_p(t)$ at a time $t$ where both are defined. So a solution on an interval inside the interval $J$ of $y_p$ is either $y_p$ itself or $y_p + 1/u$ for a solution $u$ of the linear equation, taken on an interval where $u$ has no zero. The omitted branch has to be listed separately. Other constant solutions may still appear in the family, and which branch is omitted depends on the choice of $y_p$ .

In plain words The substitution measures every solution by its gap from the known one. The known solution has gap zero, which $1/u$ can never equal, so keep it as a separate solution rather than expect the formula to contain it.

Example For $y' = y^2 - 1$ with $y_p = 1$ , the family

y = 1 + \frac{1}{Ce^{-2t} - \tfrac{1}{2}}

contains $y = -1$ for $C = 0$ but never $y = 1$ . For each fixed $t$ the family approaches $1$ as $\lvert C \rvert \to \infty$ , a limit that no finite $C$ reaches. So the initial value $y(0) = 1$ cannot be converted into a value of $u(0)$ , since $1 = 1 + 1/u(0)$ has no solution. Yet the problem has the solution $y = 1$ , and by uniqueness no other.

Poles of solutions

Definition Let $u$ solve the linear equation for the correction on $J$ , and let $K$ be a largest open interval in $J$ on which $u$ has no zero, so that $y = y_p + 1/u$ solves the Riccati equation on $K$ . At an endpoint $t^*$ of $K$ that lies in $J$ the correction vanishes, $u(t^*) = 0$ , and $\lvert y(t) \rvert \to \infty$ as $t$ approaches $t^*$ from inside $K$ . The solution cannot be continued past $t^*$ , and $t^*$ is called a pole of the solution. If $a(t^*) \ne 0$ , the linear equation gives $u'(t^*) = -a(t^*) \ne 0$ , so the zero is simple and

\lim_{t \to t^*} (t - t^*)\,y(t) = -\frac{1}{a(t^*)},

with the limit taken from inside $K$ . Because $u$ exists on all of $J$ , a solution other than $y_p$ can stop existing at a point of $J$ only where $u = 0$ . On the other side of $t^*$ the same formula gives a different solution.

In plain words A pole is a time at which the correction $u$ reaches zero, so the gap $1/u$ between the solution and the known one becomes infinite. Its location is set by the initial value as well as by the coefficients, so different solutions of one equation can have poles at different times.

Example For $y' = y^2 - 1$ with $y(0) = 2$ , the conversion $2 = 1 + 1/u(0)$ gives $u(0) = 1$ , so $C - \tfrac{1}{2} = 1$ and $C = \tfrac{3}{2}$ . Then

\begin{gathered} u = \frac{3e^{-2t} - 1}{2}, \\[0.4em] y = 1 + \frac{2}{3e^{-2t} - 1}. \end{gathered}

The correction vanishes when $e^{-2t} = \tfrac{1}{3}$ , that is, at $t^* = \tfrac{1}{2}\ln 3 \approx 0.549$ , and it is positive for $t < t^*$ . So this solution exists on $(-\infty, \tfrac{1}{2}\ln 3)$ . It rises from values near $1$ for very negative $t$ and tends to $+\infty$ as $t$ approaches $t^*$ from the left, with $(t - t^*)\,y(t) \to -1$ because $a = 1$ . For $t > t^*$ the same formula gives a different solution, which comes up from $-\infty$ and tends to $-1$ . More generally, an initial value $y(0) = y_0 \ne 1$ gives

C = \frac{y_0 + 1}{2(y_0 - 1)},

so a pole occurs exactly when $C > 0$ , that is, when $\lvert y_0 \rvert > 1$ . It lies after $t = 0$ for $y_0 > 1$ and before it for $y_0 < -1$ . Initial values in $[-1, 1]$ give solutions defined for all real $t$ .

Check yourself

For $y' = y^2 - 1$ , use the other constant solution $y_p = -1$ . Find the linear equation for the correction and the family of solutions it produces. Which constant solution does this family omit, and which does it contain?

Show the answer

Here $2ay_p + b = -2$ and $-a = -1$ , so the correction satisfies $u' - 2u = -1$ . The integrating factor $e^{-2t}$ gives $(e^{-2t}u)' = -e^{-2t}$ , so $e^{-2t}u = \tfrac{1}{2}e^{-2t} + C$ and $u = \tfrac{1}{2} + Ce^{2t}$ . The family is

y = -1 + \frac{2}{1 + 2Ce^{2t}}.

Now $y = -1$ is the omitted branch, while $C = 0$ gives $y = -1 + 2 = 1$ . Which branch is left out depends on the choice of $y_p$ , not on the equation. As a check, $y(0) = 0$ gives $u(0) = 1$ and $C = \tfrac{1}{2}$ , and the function $-1 + 2/(1 + e^{2t})$ equals $-\tanh t$ , the solution found with $y_p = 1$ .

Check that $y_p = 1/t$ solves $y' = y^2 - 2/t^2$ on $(0, \infty)$ . Then find and solve the linear equation for $u$ in $y = 1/t + 1/u$ , and find the constant that gives the other power solution $-2/t$ .

Show the answer

The derivative of $1/t$ is $-1/t^2$ , and $(1/t)^2 - 2/t^2 = -1/t^2$ , so $y_p$ is a solution. With $a = 1$ and $b = 0$ the coefficient is $2ay_p + b = 2/t$ , so

u' + \frac{2}{t}\,u = -1.

On $(0, \infty)$ the integrating factor is $e^{2\ln t} = t^2$ , so $(t^2u)' = -t^2$ and $t^2u = C - t^3/3$ . Hence

\begin{gathered} u = \frac{C}{t^2} - \frac{t}{3}, \\[0.4em] y = \frac{1}{t} + \frac{3t^2}{3C - t^3}. \end{gathered}

The constant $C = 0$ gives $1/t - 3/t = -2/t$ . The branch $y = 1/t$ itself is the omitted one.

Solve $y' = y^2 - 2/t^2$ with $y(1) = 2$ , using question 2, and find the interval of existence. What happens at each end of that interval?

Show the answer

From $2 = 1 + 1/u(1)$ we get $u(1) = 1$ , so $C - \tfrac{1}{3} = 1$ and $C = \tfrac{4}{3}$ . Then

\begin{gathered} u = \frac{4 - t^3}{3t^2}, \\[0.4em] y = \frac{1}{t} + \frac{3t^2}{4 - t^3}. \end{gathered}

Substituting back confirms it, since $y'$ and $y^2 - 2/t^2$ both equal

-\frac{1}{t^2} + \frac{24t + 3t^4}{(4 - t^3)^2}.

The correction is positive for $0 < t < \sqrt[3]{4}$ and vanishes at $t^* = \sqrt[3]{4} \approx 1.587$ , so the interval of existence is $(0, \sqrt[3]{4})$ . The left end comes from the equation itself, because the coefficient $-2/t^2$ is not defined at $t = 0$ , and there $t\,y(t) \to 1$ as $t \to 0^+$ . The right end is a pole, where $y \to +\infty$ with $(t - t^*)\,y(t) \to -1$ . Other starting values $y(1) > 1$ move the pole, but the left end stays at $0$ .

Riccati Equations with a Known Particular Solution

Key terms

Riccati equation

Known particular solution

Reciprocal substitution

Linear equation for the correction

Omitted particular branch

Poles of solutions

Check yourself

How this lesson connects

Builds on