[{"id":246989,"date":"2026-04-25T11:16:12","date_gmt":"2026-04-25T16:16:12","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=246989"},"modified":"2026-04-25T11:18:14","modified_gmt":"2026-04-25T16:18:14","slug":"exact-solution-nonlinear-pendulum","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/04\/25\/exact-solution-nonlinear-pendulum\/","title":{"rendered":"Closed-form solution to the nonlinear pendulum equation"},"content":{"rendered":"

The previous post<\/a> looks at the nonlinear pendulum equation and what difference it makes to the solutions if you linearize the equation.<\/p>\n

If the initial displacement is small enough, you can simply replace sin \u03b8 with \u03b8. If the initial displacement is larger, you can improve the accuracy quite a bit by solving the linearized equation and then adjusting the period.<\/p>\n

You can also find an exact solution, but not in terms of elementary functions; you have to use Jacobi elliptic functions. These are functions somewhat analogous to trig functions, though it’s not helpful to try to pin down the analogies. For example, the Jacobi function sn is like the sine function in some ways but very different in others, depending on the range of arguments.<\/p>\n

We start with the differential equation<\/p>\n

\u03b8\u2033(t<\/em>) + c<\/em>\u00b2 sin( \u03b8(t<\/em>) ) = 0<\/p>\n

where c<\/em>\u00b2 = g<\/em>\/L<\/em>, i.e. the gravitational constant divided by pendulum length, and initial conditions \u03b8(0) = \u03b80<\/sub> and \u03b8\u2032(0) = 0. We assume \u2212\u03c0 < \u03b80<\/sub> < \u03c0.<\/p>\n

Then the solution is<\/p>\n

\u03b8(t<\/em>) = 2 arcsin( a<\/em> cd(c<\/em>t<\/em> | m<\/em> ) )<\/p>\n

where a<\/em> = sin(\u03b80<\/sub>\/2), m<\/em> =\u00a0a<\/em>\u00b2, and cd is one of the 12 Jacobi elliptic functions. Note that cd, like all the Jacobi functions, has an argument and a parameter. In the equation above the argument is\u00a0ct<\/em> and the parameter is\u00a0m<\/em>.<\/p>\n

The last plot in the previous post was misleading, showing roughly equal parts genuine difference and error from solving the differential equation numerically. Here’s the code that was used to solve the nonlinear equation.<\/p>\n

from scipy.special import ellipj, ellipk\r\nfrom numpy import sin, cos, pi, linspace, arcsin\r\nfrom scipy.integrate import solve_ivp\r\n\r\ndef exact_period(\u03b8):\r\n    return 2*ellipk(sin(\u03b8\/2)**2)\/pi\r\n\r\ndef nonlinear_ode(t, z):\r\n    x, y = z\r\n    return [y, -sin(x)]    \r\n\r\ntheta0 = pi\/3\r\nb = 2*pi*exact_period(theta0)\r\nt = linspace(0, 2*b, 2000)\r\n\r\nsol = solve_ivp(nonlinear_ode, [0, 2*b], [theta0, 0], t_eval=t)\r\n<\/pre>\n
The solution is contained in sol.y[0]<\/code>.<\/p>\n
Let’s compare the numerical solution to the exact solution.<\/p>\n
def f(t, c, theta0):\r\n    a = sin(theta0\/2)\r\n    m = a**2\r\n    sn, cn, dn, ph = ellipj(c*t, m)\r\n    return 2*arcsin(a*cn\/dn)\r\n<\/pre>\n
There are a couple things to note about the code. First,SciPy doesn’t implement the cd function, but it can be computed as cn\/dn. Second, the function ellipj<\/code> returns four functions at once because it takes about as much time to calculate all four as it does to compute one of them.<\/p>\n
Here is a plot of the error in solving the differential equation.<\/p>\n
<\/p>\n
And here is the difference between the exact solution to the nonlinear pendulum equation and the stretched solution to the linear equation.<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
The previous post looks at the nonlinear pendulum equation and what difference it makes to the solutions if you linearize the equation. If the initial displacement is small enough, you can simply replace sin \u03b8 with \u03b8. If the initial displacement is larger, you can improve the accuracy quite a bit by solving the linearized […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[9],"tags":[47,129],"class_list":["post-246989","post","type-post","status-publish","format-standard","hentry","category-math","tag-differential-equations","tag-special-functions"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/246989","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/comments?post=246989"}],"version-history":[{"count":0,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/246989\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=246989"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=246989"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=246989"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":246988,"date":"2026-04-25T08:12:39","date_gmt":"2026-04-25T13:12:39","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=246988"},"modified":"2026-04-25T08:12:39","modified_gmt":"2026-04-25T13:12:39","slug":"nth-derivative-of-a-quotient","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/04\/25\/nth-derivative-of-a-quotient\/","title":{"rendered":"nth derivative of a quotient"},"content":{"rendered":"
There’s a nice formula for the\u00a0n<\/em>th derivative of a product. It looks a lot like the binomial theorem.<\/p>\n
\"(gh)^{(n)}<\/p>\n
There is also a formula for the\u00a0n<\/em>th derivative of a quotient, but it’s more complicated and less known.<\/p>\n
We start by writing the quotient rule in an unusual way.<\/p>\n
\"\\left(\\frac{g}{h}\\right)^{(1)}<\/p>\n
Applying the quotient rule twice gives the following.<\/p>\n
\"\\left(\\frac{g}{h}\\right)^{(2)}<\/p>\n
And here’s the general rule in all its glory.<\/p>\n
\"\\left(\\frac{g}{h}\\right)^{(n)}<\/p>\n
 <\/p>\n
Source: V. F. Ivanoff. The n<\/em>th Derivative of a Fractional Function. The American Mathematical Monthly, Vol. 55, No. 8 (Oct., 1948), p. 491<\/p>\n","protected":false},"excerpt":{"rendered":"
There’s a nice formula for the\u00a0nth derivative of a product. It looks a lot like the binomial theorem. There is also a formula for the\u00a0nth derivative of a quotient, but it’s more complicated and less known. We start by writing the quotient rule in an unusual way. Applying the quotient rule twice gives the following. […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[1],"tags":[],"class_list":["post-246988","post","type-post","status-publish","format-standard","hentry","category-uncategorized"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/246988","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/comments?post=246988"}],"version-history":[{"count":0,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/posts\/246988\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/media?parent=246988"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/categories?post=246988"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.johndcook.com\/blog\/wp-json\/wp\/v2\/tags?post=246988"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}},{"id":246986,"date":"2026-04-24T18:53:29","date_gmt":"2026-04-24T23:53:29","guid":{"rendered":"https:\/\/www.johndcook.com\/blog\/?p=246986"},"modified":"2026-04-25T11:19:01","modified_gmt":"2026-04-25T16:19:01","slug":"nonlinear-pendulum","status":"publish","type":"post","link":"https:\/\/www.johndcook.com\/blog\/2026\/04\/24\/nonlinear-pendulum\/","title":{"rendered":"How nonlinearity affects a pendulum"},"content":{"rendered":"
The equation of motion for a pendulum is the differential equation<\/p>\n
\"\\theta''<\/p>\n
where\u00a0g<\/em> is the acceleration due to gravity and \u2113 is the length of the pendulum. When this is presented in an introductory physics class, the instructor will immediately say something like “we’re only interested in the case where \u03b8 is small, so we can rewrite the equation as<\/p>\n
\"\\theta''<\/p>\n
Questions<\/h2>\n
This raises a lot of questions, or at least it should.<\/p>\n
    \n
  1. Why not leave sin \u03b8 alone?<\/li>\n
  2. What justifies replacing sin \u03b8 with just \u03b8?<\/li>\n
  3. How small does \u03b8 have to be for this to be OK?<\/li>\n
  4. How do the solutions to the exact and approximate equations differ?<\/li>\n<\/ol>\n
    First, sine is a nonlinear function, making the differential equation nonlinear. The nonlinear pendulum equation cannot be solved using mathematics that students in an introductory physics class have seen. There is a closed-form solution<\/a>, but only if you extend “closed-form” to mean more than the elementary functions a student would see in a calculus class.<\/p>\n
    Second, the approximation is justified because sin \u03b8 \u2248 \u03b8 when \u03b8 is small. That’s true, but it’s kinda subtle. Here’s a post<\/a> unpacking that.<\/p>\n
    The third question doesn’t have a simple answer, though simple answers are often given. An instructor could make up an answer on the spot and say “less than 10 degrees” or something like that. A more thorough answer requires answering the fourth question.<\/p>\n
    I address how nonlinear affects the solutions in a post<\/a> a couple years ago. This post will expand a bit on that post.<\/p>\n
    Longer period<\/h2>\n
    The primary difference between the nonlinear and linear pendulum equations is that the solutions to the nonlinear equation have longer periods. The solution to the linear equation is a cosine. Solving the equation determines the frequency, amplitude, and phase shift of the cosine, but qualitatively it’s just a cosine. The solution to the corresponding nonlinear equation, with sin \u03b8 rather than \u03b8, is not exactly a cosine, but it looks a lot like a cosine, only the period is a little longer [1].<\/p>\n
    OK, the nonlinear pendulum has a longer period, but how much longer? The period is increased by a factor\u00a0f<\/em>(\u03b80<\/sub>) where \u03b80<\/sub>\u00a0is the initial displacement.<\/p>\n
    You can find the exact answer in my earlier post<\/a>. The exact answer depends on a special function called the \u201ccomplete elliptic integral of the first kind,\u201d but to a good approximation<\/p>\n
    \"f(\\theta)<\/p>\n
    The earlier post compares this approximation to the exact function.<\/p>\n
    Linear solution with adjusted period<\/h2>\n
    Since the nonlinear pendulum equation is roughly the same as the linear equation with a longer period, you can approximate the solution to the nonlinear equation by solving the linear equation but increasing the period. How good is that approximation?<\/p>\n
    Let’s do an example with \u03b80<\/sub> = 60\u00b0 = \u03c0\/3 radians. Then sin \u03b80<\/sub> = 0.866 but \u03b80<\/sub>, in radians, is 1.047, so we definitely can’t say sin \u03b80<\/sub> is approximately \u03b80<\/sub>. To make things simple, let’s set \u2113 =\u00a0g<\/em>. Also, assume the pendulum starts from rest, i.e. \u03b8'(0) = 0.<\/p>\n
    Here’s a plot of the solutions to the nonlinear and linear equations.<\/p>\n
    <\/p>\n
    Obviously the solution to the nonlinear equation has a longer period. In fact it’s 7.32% longer. (The approximation above would have estimated 7.46%.)<\/p>\n
    Here’s a plot comparing the solution of the nonlinear equation and the solution to the linear equations with period stretched by 7.32%.<\/p>\n
    <\/p>\n
    The solutions differ by less than the width of the plotting line, so it’s too small to see. But we can see there’s a difference when we subtract the two solutions.<\/p>\n
    Here’s a plot of the solutions to the nonlinear and linear equations.<\/p>\n
    <\/p>\n
    Update<\/b>: The plot above is misleading. Part of what it shows is numerical error from solving the pendulum equation. When we redo the plot using the exact solution the error is about half as large. And the error is periodic, as we’d expect. See this post<\/a> for more on the exact solution using Jacobi functions and the differential equation solver that was used to make the original plot.<\/p>\n
    <\/p>\n
    Related posts<\/h2>\n