Perturbation Methods and Multiple Scales, differential equations lesson

With $t$ as time and $y' = dy/dt$ , the oscillator $y'' + (1 + \varepsilon)^2 y = 0$ differs from $y'' + y = 0$ only by a slightly higher frequency, so treating $\varepsilon$ as a small correction looks safe. After a time of about $\pi/\varepsilon$ , though, the two oscillations are half a cycle apart. A small parameter can also multiply the highest derivative of an equation, and then setting it to zero changes the kind of problem being solved. This lesson builds expansions in a small parameter, shows how they fail over long times and in thin layers, and repairs them with multiple scales and with inner and outer approximations, each valid on a stated interval.

Key terms

Regular perturbation expansion

Definition A small parameter is a positive number $\varepsilon$ in a problem, usually dimensionless, that is small compared with $1$ , and statements about it describe the limit $\varepsilon \to 0^+$ . A regular perturbation expansion seeks the solution in the form

y = y_0 + \varepsilon y_1 + \varepsilon^2 y_2 + \cdots,

where the functions $y_k$ do not depend on $\varepsilon$ . The expansion is substituted into the equation and the side conditions, everything is expanded in powers of $\varepsilon$ , and the coefficient of each power is set to zero. The $\varepsilon^0$ problem is the unperturbed problem with $\varepsilon = 0$ , and the problem for each later $y_k$ has a right side built from $y_0, \dots, y_{k-1}$ . The problem is a regular perturbation on a set $S$ of values of the independent variable when, for each $N$ , the truncated sum $y_0 + \varepsilon y_1 + \cdots + \varepsilon^N y_N$ differs from the solution by at most $C\varepsilon^{N+1}$ at every point of $S$ and for every small $\varepsilon$ , with a constant $C$ that may depend on $N$ but not on the point or on $\varepsilon$ .

In plain words Solve the easy problem with $\varepsilon = 0$ first, then correct it in layers of size $\varepsilon$ , $\varepsilon^2$ and so on. Each correction solves a linear problem built from the earlier terms, and the expansion can be trusted only where each layer stays much smaller than the one before it.

Example Let $t \ge 0$ be the independent variable and $y' = dy/dt$ . The oscillator problem

\begin{gathered} y'' + (1 + \varepsilon)^2 y = 0, \\[0.4em] y(0) = 1, \qquad y'(0) = 0 \end{gathered}

has the exact solution $y = \cos((1 + \varepsilon)t)$ . Since $(1 + \varepsilon)^2 = 1 + 2\varepsilon + \varepsilon^2$ , the powers $\varepsilon^0$ and $\varepsilon^1$ give

\begin{gathered} y_0'' + y_0 = 0, \\[0.4em] y_1'' + y_1 = -2y_0, \end{gathered}

with $y_0(0) = 1$ , $y_0'(0) = 0$ and zero initial data for $y_1$ . So $y_0 = \cos t$ , and $y_1$ solves $y_1'' + y_1 = -2\cos t$ , which gives $y_1 = -t\sin t$ . Indeed $y_1'' = -2\cos t + t\sin t$ , so $y_1'' + y_1 = -2\cos t$ , and $y_1(0) = y_1'(0) = 0$ . The two-term expansion is $\cos t - \varepsilon t \sin t$ .

Secular term

Definition A secular term is a term of a perturbation expansion that grows without bound relative to the leading term as the independent variable $t \to \infty$ , typically a power of $t$ multiplying an oscillation of the leading solution, as in $t\sin t$ . It appears when the forcing in a correction equation contains a solution of the unperturbed homogeneous equation. This is the resonance case of undetermined coefficients, which needs an extra factor of $t$ . Once $\varepsilon t$ is no longer small, the term $\varepsilon y_1$ is as large as $y_0$ , the ordering of the expansion breaks down, and the expansion is not uniformly valid for large $t$ . A secular term is a defect of the approximation. It does not show that the true solution grows.

In plain words The correction keeps growing because it tries to describe a slow drift, here a shift of phase, by adding larger and larger amounts of a fixed shape. Over short times this works, and over times of order $1/\varepsilon$ it gives nonsense.

Example In the oscillator the forcing $-2\cos t$ solves $y'' + y = 0$ , so the correction $y_1 = -t\sin t$ carries the factor $t$ . The exact solution never exceeds $1$ in absolute value, while the two-term expansion reaches values of size about $\varepsilon t$ . With $\varepsilon = 0.1$ and $t = 10.5\pi \approx 32.99$ , the exact value is $\cos(1.1t) \approx 0.156$ , but $\cos t - \varepsilon t \sin t \approx -3.30$ . The trouble comes from expanding the phase, since

\begin{aligned} &\cos(t + \varepsilon t) \\[0.4em] &\quad = \cos t - \varepsilon t \sin t + \cdots \end{aligned}

is accurate only while $\varepsilon t$ is small.

Method of multiple scales

Definition The method of multiple scales introduces a slow time $T = \varepsilon t$ and seeks the solution as $y(t) = Y(t, \varepsilon t)$ , where

\begin{aligned} Y(t, T) &= Y_0(t, T) \\[0.4em] &\quad + \varepsilon Y_1(t, T) + \cdots \end{aligned}

and $t$ and $T$ are treated as independent variables. By the chain rule along $T = \varepsilon t$ ,

\begin{gathered} \frac{d}{dt} = \partial_t + \varepsilon \partial_T, \\[0.4em] \frac{d^2}{dt^2} = \partial_t^2 + 2\varepsilon \partial_t \partial_T + \varepsilon^2 \partial_T^2. \end{gathered}

Collecting powers of $\varepsilon$ gives an equation in $t$ for each $Y_k$ , whose coefficients are still free functions of $T$ . These functions are chosen so that the forcing of the next correction contains no term that resonates with the unperturbed operator. The resulting solvability conditions are differential equations in $T$ , and they remove the secular terms. The approximation is usually intended for times up to order $1/\varepsilon$ , but that claim needs its own error estimate, especially for nonlinear problems.

In plain words Let the solution oscillate on the fast clock $t$ while its amplitude and phase drift on the slow clock $\varepsilon t$ . Instead of letting a correction grow, build the slow drift into the leading term from the start, and use the requirement that the next correction stay bounded to find how the drift goes.

Example For the oscillator, the $\varepsilon^0$ equation $\partial_t^2 Y_0 + Y_0 = 0$ gives $Y_0 = A(T)\cos t + B(T)\sin t$ . The $\varepsilon^1$ equation is $\partial_t^2 Y_1 + Y_1 = -2\partial_t\partial_T Y_0 - 2Y_0$ , and working out the right side gives

\begin{aligned} &\partial_t^2 Y_1 + Y_1 \\[0.4em] &\quad = -2(A + B')\cos t \\[0.4em] &\qquad + 2(A' - B)\sin t, \end{aligned}

where primes on $A$ and $B$ mean $d/dT$ . Both forcing terms resonate, so both coefficients must vanish, which gives $A' = B$ and $B' = -A$ . At leading order the data $y(0) = 1$ and $y'(0) = 0$ give $A(0) = 1$ and $B(0) = 0$ , so $A = \cos T$ , $B = -\sin T$ and

\begin{aligned} Y_0 &= \cos T \cos t \\[0.4em] &\quad - \sin T \sin t \\[0.4em] &= \cos(t + T). \end{aligned}

With $T = \varepsilon t$ this is $\cos((1 + \varepsilon)t)$ , the exact solution. The slow phase $\varepsilon t$ now sits inside the cosine instead of being expanded into the growing term $-\varepsilon t\sin t$ . That the leading term is exact is special to this linear example.

Singular perturbation and boundary layer

Definition A perturbation problem is singular on a domain when no regular expansion approximates its solution uniformly there. A common cause in boundary-value problems is a small parameter multiplying the highest derivative. Setting $\varepsilon = 0$ then gives a reduced equation of lower order, which in general cannot satisfy all the boundary conditions. A solution of the reduced equation that keeps some of the conditions is an outer solution. Near a boundary where a condition was dropped, the solution has a boundary layer, a region whose width tends to $0$ with $\varepsilon$ and across which the solution changes by an amount that does not tend to $0$ . The layer is examined with a stretched inner variable, such as $\xi = x/\varepsilon$ for a layer of width of order $\varepsilon$ at $x = 0$ , in which the layer has width of order $1$ . By this definition the oscillator above is a regular perturbation on every bounded time interval but singular on $[0, \infty)$ .

In plain words Dropping the small term removes the derivative that lets the solution bend sharply, so the reduced equation cannot reach every boundary value. The true solution meets the missed condition by changing quickly in a thin layer, and only a magnified variable shows the shape of that change.

Example On $0 \le x \le 1$ , with $x$ the independent variable and $y' = dy/dx$ , consider

\begin{gathered} \varepsilon y'' + y' = 0, \\[0.4em] y(0) = 0, \qquad y(1) = 1. \end{gathered}

Setting $\varepsilon = 0$ leaves $y' = 0$ , whose solutions are constants, and no constant meets both conditions. The exact solution is

y = \frac{1 - e^{-x/\varepsilon}}{1 - e^{-1/\varepsilon}},

which is close to $1$ except near $x = 0$ , where it drops to $0$ within a few multiples of $\varepsilon$ . With $\varepsilon = \tfrac{1}{10}$ it already equals about $0.632$ at $x = 0.1$ and $0.950$ at $x = 0.3$ . In the inner variable $\xi = x/\varepsilon$ , with $Y(\xi) = y(\varepsilon\xi)$ , each $x$ derivative brings a factor $1/\varepsilon$ , and after multiplying by $\varepsilon$ the equation becomes

\frac{d^2Y}{d\xi^2} + \frac{dY}{d\xi} = 0.

Its solutions are $C_1 + C_2 e^{-\xi}$ , and the one with $Y(0) = 0$ that approaches the outer value $1$ as $\xi \to \infty$ is $Y = 1 - e^{-\xi}$ . The layer sits at $x = 0$ because $e^{-\xi}$ decays into the interval. Near $x = 1$ the variable $\eta = (1 - x)/\varepsilon$ gives $d^2Y/d\eta^2 - dY/d\eta = 0$ , whose nonconstant solutions grow like $e^{\eta}$ , so no layer can form there.

Composite approximation

Definition Let $y_{\text{out}}$ be an outer approximation, valid away from a layer at $x = 0$ , and let $Y_{\text{in}}$ be an inner approximation in the stretched variable $\xi$ , valid inside the layer. Leading-order matching requires the inner solution far out in the layer and the outer solution near the layer to have the same limit,

\begin{aligned} \lim_{\xi \to \infty} Y_{\text{in}}(\xi) &= \lim_{x \to 0^+} y_{\text{out}}(x) \\[0.4em] &= c. \end{aligned}

The composite approximation adds the two and subtracts the common part, which would otherwise be counted twice,

\begin{aligned} y_{\text{comp}}(x) &= y_{\text{out}}(x) \\[0.4em] &\quad + Y_{\text{in}}(x/\varepsilon) - c. \end{aligned}

Inside the layer it is close to the inner approximation, and outside the layer it is close to the outer one. It is meant to be accurate on the whole interval at once, and that claim is checked by an error estimate.

In plain words Each approximation is right in its own region and wrong in the other. Adding them counts the value they share in the overlap twice, so that shared value is subtracted once.

Example For $\varepsilon y'' + y' = 0$ above, the outer solution is $y_{\text{out}} = 1$ , the inner solution is $Y_{\text{in}} = 1 - e^{-\xi}$ , and both tend to $c = 1$ , so

\begin{aligned} y_{\text{comp}} &= 1 + (1 - e^{-x/\varepsilon}) - 1 \\[0.4em] &= 1 - e^{-x/\varepsilon}. \end{aligned}

Write $a = e^{-1/\varepsilon}$ , so that $0 < a < 1$ . The exact solution is $y = y_{\text{comp}}/(1 - a)$ , so

y - y_{\text{comp}} = \frac{a\,y_{\text{comp}}}{1 - a}.

Since $y_{\text{comp}}$ increases from $0$ to $1 - a$ on $[0, 1]$ , the largest error is exactly $a$ , attained at $x = 1$ . For $\varepsilon = \tfrac{1}{10}$ that is $e^{-10} \approx 4.54 \times 10^{-5}$ . The composite function satisfies the differential equation exactly and meets $y(0) = 0$ , so its whole error comes from missing the value $y(1) = 1$ by $a$ . A zero residual in the equation does not by itself show that a boundary-value problem is solved.

Interval of validity

Definition The interval of validity of an approximation is a set of values of the independent variable on which its error is controlled, either by a stated bound or by a bound that tends to $0$ as $\varepsilon \to 0^+$ , uniformly at every point of the set. An approximation has no meaningful accuracy apart from such a set. The set may change with $\varepsilon$ , as in $0 \le t \le t_1/\varepsilon$ , or it may leave out a layer, as in $\delta \le x \le 1$ with a fixed $\delta > 0$ . Here the word refers to the accuracy of an approximation, not to the interval on which a solution exists.

In plain words An approximation comes with a place where it is good. Before trusting it, ask how its error depends on $t$ or $x$ and on $\varepsilon$ , and whether the problem reaches times or positions where that error is large. Numerical work needs the same care, because a step size that jumps over a thin layer can hide the layer completely.

Example For the oscillator, the error of the two-term regular expansion,

\begin{aligned} R(t) &= \cos((1 + \varepsilon)t) \\[0.4em] &\quad - (\cos t - \varepsilon t \sin t), \end{aligned}

satisfies $\lvert R(t) \rvert \le \tfrac{1}{2}(\varepsilon t)^2$ . This follows from Taylor’s theorem in the phase shift $\varepsilon t$ , because the second derivative of $\cos(t + p)$ with respect to $p$ never exceeds $1$ in absolute value. So the expansion is valid with error $O(\varepsilon^2)$ on every fixed interval $0 \le t \le t_1$ , and with $\varepsilon = 0.01$ its error stays below $0.01$ for $0 \le t \le 14.1$ . On $0 \le t \le 1/\varepsilon = 100$ the bound gives nothing small, and the actual error there reaches about $0.48$ . In the boundary layer problem, the outer solution $y = 1$ has error at most $e^{-\delta/\varepsilon}/(1 - e^{-1/\varepsilon})$ on $\delta \le x \le 1$ but error $1$ at $x = 0$ , while the composite approximation is valid on all of $[0, 1]$ .

Check yourself

Let $t \ge 0$ be the independent variable and $y' = dy/dt$ . Find the first two terms $y_0$ and $y_1$ of the regular expansion of $y' = -y + \varepsilon y^2$ with $y(0) = 1$ . Does a secular term appear?

Show the answer

At order $\varepsilon^0$ , $y_0' = -y_0$ with $y_0(0) = 1$ , so $y_0 = e^{-t}$ . At order $\varepsilon^1$ , $y_1' = -y_1 + y_0^2$ with $y_1(0) = 0$ , that is $y_1' + y_1 = e^{-2t}$ . With the integrating factor $e^t$ this reads $(e^t y_1)' = e^{-t}$ , so $e^t y_1 = 1 - e^{-t}$ and

y_1 = e^{-t} - e^{-2t}.

No secular term appears. The correction satisfies $0 \le y_1 \le \tfrac{1}{4}$ for all $t \ge 0$ , with the maximum at $t = \ln 2$ , and $y_1/y_0 = 1 - e^{-t}$ stays below $1$ , because the forcing $e^{-2t}$ is not the mode $e^{-t}$ of the unperturbed equation. For $0 < \varepsilon < 1$ the exact solution is $y = 1/(\varepsilon + (1 - \varepsilon)e^t)$ , and with $w = 1 - e^{-t}$ the error of the two-term expansion is

y - y_0 - \varepsilon y_1 = \frac{\varepsilon^2 e^{-t} w^2}{1 - \varepsilon w}.

Since $0 \le w < 1$ , this is at most $\varepsilon^2/(1 - \varepsilon)$ for every $t \ge 0$ , so the expansion is uniformly valid on $[0, \infty)$ .

On $0 \le x \le 1$ , with $y' = dy/dx$ , consider $\varepsilon y'' - y' = 0$ with $y(0) = 0$ and $y(1) = 1$ . A classmate copies the example of this lesson, keeps the outer solution $y = 1$ and puts the layer at $x = 0$ . Show that this fails, and find the correct leading-order composite approximation and its largest error.

Show the answer

With $\xi = x/\varepsilon$ the equation becomes $d^2Y/d\xi^2 - dY/d\xi = 0$ , whose solutions are $C_1 + C_2 e^{\xi}$ . The condition $Y(0) = 0$ forces $C_1 = -C_2$ , and a solution that approaches $1$ as $\xi \to \infty$ would need $C_2 = 0$ , which leaves $Y = 0$ . So no layer can form at $x = 0$ , and the outer solution must keep the left condition, $y_{\text{out}} = 0$ . Near $x = 1$ the variable $\eta = (1 - x)/\varepsilon$ gives $d^2Y/d\eta^2 + dY/d\eta = 0$ . The solution with $Y = 1$ at $\eta = 0$ that tends to the outer value $0$ is $Y = e^{-\eta}$ . The common value is $c = 0$ , so

y_{\text{comp}} = e^{-(1 - x)/\varepsilon}.

The exact solution is

y = \frac{e^{(x - 1)/\varepsilon} - e^{-1/\varepsilon}}{1 - e^{-1/\varepsilon}},

and with $a = e^{-1/\varepsilon}$ the error is $y - y_{\text{comp}} = -a(1 - y_{\text{comp}})/(1 - a)$ . Its absolute value is largest at $x = 0$ , where it equals $a = e^{-1/\varepsilon}$ . The sign of the first-derivative term decides the side of the layer, because the inner solution must decay as it moves into the interval.

Apply multiple scales to $y'' + (1 + \varepsilon)^2 y = 0$ with $y(0) = 0$ and $y'(0) = 1$ , where $y' = dy/dt$ . Compare the leading term with the exact solution $\sin((1 + \varepsilon)t)/(1 + \varepsilon)$ for all $t \ge 0$ , and with the two-term regular expansion.

Show the answer

The solvability conditions come from the equation, not from the data, so they are again $A' = B$ and $B' = -A$ . Now the data give $A(0) = 0$ and $B(0) = 1$ at leading order, because $Y_0(0, 0) = A(0)$ and $\partial_t Y_0(0, 0) = B(0)$ . Hence $A = \sin T$ and $B = \cos T$ , and

\begin{aligned} Y_0 &= \sin T \cos t \\[0.4em] &\quad + \cos T \sin t \\[0.4em] &= \sin(t + T). \end{aligned}

With $T = \varepsilon t$ this is $\sin((1 + \varepsilon)t)$ . Since $1 - 1/(1 + \varepsilon) = \varepsilon/(1 + \varepsilon)$ , its difference from the exact solution $y$ is

Y_0 - y = \frac{\varepsilon \sin((1 + \varepsilon)t)}{1 + \varepsilon},

so the error is at most $\varepsilon/(1 + \varepsilon) < \varepsilon$ for every $t \ge 0$ . The phase is right for all time, and only the amplitude is off by a relative amount of order $\varepsilon$ . The regular expansion has $y_0 = \sin t$ and $y_1'' + y_1 = -2\sin t$ with zero data, so $y_1 = t\cos t - \sin t$ . Its term $t\cos t$ is secular, and the two-term regular expansion is useful only while $\varepsilon t$ is small.

Perturbation Methods and Multiple Scales

Key terms

Regular perturbation expansion

Secular term

Method of multiple scales

Singular perturbation and boundary layer

Composite approximation

Interval of validity

Check yourself

How this lesson connects

Builds on