We need respond by creating index.html in current dir. Need inspect maybe empty. Need implement single-page web app self-contained. Need use canvas correct equations double pendulum, arms, colorful fading trail. Need likely just write file. Need maybe no external. Need ensure good app. Need include equations maybe in UI. Need maybe controls? User asked simulate chaotic double pendulum. We can implement.

Need create index.html in /tmp/ws. Need maybe use write. Need think through physics.

Double pendulum equations:
For equal masses? General.

theta1_ddot = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2)*2*(θ2dot^2 L2 + θ1dot^2 L1 cosΔ) - 2 sinΔ*m2*(θ2dot^2 L2 + θ1dot^2 L1 cosΔ)? Wait formula.

From MyPhysicsLab:
a = 2θ1'²L1 + θ2'²L2 cos(θ1-θ2)
b = θ2'²L2 sin(θ1-θ2)
c = 4θ1' + θ2' cos(θ1-θ2)
d = 2θ2' - θ1' cos(θ1-θ2)
θ1'' = -g(2 sinθ1 + a cos(θ1-θ2) + b sin(θ1-θ2)) / (L1 c + d sin(θ1-θ2))
θ2'' = 2 b (L1+L2) / (L1 c + d sin(θ1-θ2))
This for equal masses? Formula with masses equal. More general:
θ1dd = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2)*m2*(θ2dot^2*l2 + θ1dot^2*l1*cosΔ)) / (l1*(2m1+m2 - m2*cos(2Δ)))
θ2dd = (2 sinΔ*(θ1dot^2*l1*(m1+m2) + g*(m1+m2)*cosθ1 + θ2dot^2*l2*m2*cosΔ)) / (l2*(2m1+m2 - m2*cos(2Δ)))
Need check formula:
from Wikipedia:
theta1dd = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2)*m2*(θ2dot^2*l2 + θ1dot^2*l1*cos(θ1-θ2))) / (l1*(2m1+m2 - m2*cos(2θ1-2θ2)))
theta2dd = (2 sin(θ1-θ2) * (θ1dot^2*l1*(m1+m2) + g*(m1+m2)cosθ1 + θ2dot^2*l2*m2*cos(θ1-θ2))) / (l2*(2m1+m2 - m2*cos(2θ1-2θ2)))
Missing factors? Actually Wikipedia:
θ1dd = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2) m2 (θ2dot² l2 + θ1dot² l1 cos(θ1-θ2))) / (l1 (2m1+m2 - m2 cos(2θ1-2θ2)))
θ2dd = (2 sin(θ1-θ2) (θ1dot² l1 (m1+m2) + g (m1+m2) cosθ1 + θ2dot² l2 m2 cos(θ1-θ2))) / (l2 (2m1+m2 - m2 cos(2θ1-2θ2)))
Need include maybe factors? Actually formula has "2 sin(theta1-theta2)" and numerator includes "theta2dot^2 l2 + theta1dot^2 l1 cos(...)" no 2 inside? Let's verify with known code:
double numerator = -g*(2*m1 + m2)*sin(theta1);
numerator -= m2*g*sin(theta1 - 2*theta2);
numerator -= 2*sin(theta1-theta2)*m2;
numerator *= theta2_speed*theta2_speed*l2 + theta1_speed*theta1_speed*l1*cos(theta1-theta2);
double denom = l1*(2*m1 + m2 - m2*cos(2*theta1 - 2*theta2));
theta1_acc = numerator/denom;
numerator = 2*sin(theta1-theta2);
numerator *= theta1_speed*theta1_speed*l1*(m1+m2);
numerator += g*(m1+m2)*cos(theta1);
numerator += theta2_speed*theta2_speed*l2*m2*cos(theta1-theta2);
denom = l2*(2*m1 + m2 - m2*cos(2*theta1 - 2*theta2));
theta2_acc = numerator/denom;
This is another equivalent formula. Which is correct. The previous formula with sinθ1 and sinΔ maybe missing sin(θ1-2θ2) term? Let's derive. There are two equivalent forms? Let's check.

From Wikipedia:
\ddot θ1 = [ -g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2)m2(θ2dot² l2 + θ1dot² l1 cos(θ1-θ2)) ] / [l1(2m1+m2 - m2 cos(2θ1-2θ2))]
Wait m2 sin(θ1 - θ2) maybe should be m2 sin(θ1 - 2θ2)? Let's search memory:
Wikipedia:
\ddot θ_1 = \frac{-g(2m_1+m_2)\sin θ_1 - m_2\sin(θ_1-θ_2) - 2\sin(θ_1-θ_2)m_2(θ_2^2 l_2 + θ_1^2 l_1 \cos(θ_1-θ_2))}{l_1(2m_1+m_2 - m_2\cos(2θ_1 - 2θ_2))}
\ddot θ_2 = \frac{2\sin(θ_1-θ_2)(θ_1^2 l_1(m_1+m_2)+g(m_1+m_2)\cos θ_1+θ_2^2 l_2 m_2\cos(θ_1-θ_2))}{l_2(2m_1+m_2 - m_2\cos(2θ_1 - 2θ_2))}
I think yes sin(θ1-θ2), but common code has sin(θ1-2θ2). Are they equivalent? Let's test maybe not.

Let's derive or check known formula from Paul Bourke:
theta1acc = (-g(2m1+m2)sinθ1 - m2 sin(θ1-2θ2) - 2 sin(θ1-θ2)m2(θ2dot²l2 + θ1dot²l1cosΔ)) / (l1(2m1+m2 - m2cos(2Δ)))
theta2acc = (2sinΔ(θ1dot²l1(m1+m2)+g(m1+m2)cosθ1+θ2dot²l2m2cosΔ))/(l2(...)).
Yes Wikipedia maybe has sin(theta1 - 2theta2). Let me recall exactly: It is:
\ddot{\theta}_1 = \frac{-g(2m_1+m_2)\sin\theta_1 - m_2\sin(\theta_1-2\theta_2) - 2\sin(\theta_1-\theta_2)m_2(\dot{\theta}_2^2 l_2 + \dot{\theta}_1^2 l_1\cos(\theta_1-\theta_2))}{l_1(2m_1+m_2 - m_2\cos(2\theta_1-2\theta_2))}
\ddot{\theta}_2 = \frac{2\sin(\theta_1-\theta_2)(\dot{\theta}_1^2 l_1(m_1+m_2) + g(m_1+m_2)\cos\theta_1 + \dot{\theta}_2^2 l_2 m_2\cos(\theta_1-\theta_2))}{l_2(2m_1+m_2 - m_2\cos(2\theta_1-2\theta_2))}
Yes use sin(theta1 - 2theta2). Good.

Need implement simulation with RK4 maybe correct. Need canvas. Fading trail: Could draw transparent black overlay each frame instead of clearing, but if trail "colorful fading" with hue changes. Need draw full background with alpha each frame maybe. Use CSS. Need draw arms and tip trail. Trail of tip: maintain array of points with age/color; each frame update positions, push point, remove old. Draw trail segments with alpha decreasing and hue. Could draw line segments with strokeStyle hsla. Need fade over time.

Need single page. Include controls maybe reset, randomize, speed, pause, trail length. But user didn't ask controls; can include nice. Need no external. Need ensure file present. Need maybe include current directory. We can write.

Need maybe not use network.

Let's design index.html.

Features:
- Fullscreen canvas with dark background.
- Double pendulum with two rods, bobs, pivot, tip glow.
- Colorful fading trail of tip.
- Correct equations of motion, RK4 integrator, variable step maybe.
- UI overlay: title, controls (Pause/Reset/Randomize, speed, trail, gravity? maybe length? mass? maybe not).
- Info panel showing equations? Could include formula text maybe.

Need implement robust.

Technical:
Canvas resolution high DPI.

State:
const pendulum = {
l1: 170, l2: 170, m1: 1, m2: 1,
theta1: Math.PI/2 + 0.1, theta2: Math.PI/2 - 0.1,
omega1: 0, omega2: 0,
};
origin maybe canvas center x, y = height*0.20. Need scale arms to fit. If canvas small, use min(width,height)*0.24 maybe. But l values in pixels? Use scale. Use pixel lengths based on min dimension. For physics use l1/l2 maybe in pixels? g in pixels/s²? Could set l1=1, l2=1 and g=9.81 and scale for drawing. Simpler: lengths normalized 1.0; positions normalized from origin. Accelerations use l1/l2=1.0, g=9.81. Draw scale = min(w,h)*0.28. Then positions: x2 = originX + scale*l1*sinθ1; y2 = originY + scale*l1*cosθ1 (positive down). x3 = x2 + scale*l2*sinθ2; y3 = y2 + scale*l2*cosθ2. This makes bob at downward.
Need if canvas small.

Equation with l1/l2 = 1.0.
function accelerations(s):
const {theta1:t1, theta2:t2, omega1:w1, omega2:w2} = s;
const delta=t1-t2;
const denom = 2*m1+m2 - m2*Math.cos(2*delta);
const a1 = (-g*(2*m1+m2)*Math.sin(t1) - m2*Math.sin(t1-2*t2) - 2*Math.sin(delta)*m2*(w2*w2*l2 + w1*w1*l1*Math.cos(delta))) / (l1*denom);
const a2 = (2*Math.sin(delta)*(w1*w1*l1*(m1+m2) + g*(m1+m2)*Math.cos(t1) + w2*w2*l2*m2*Math.cos(delta))) / (l2*denom);
return [a1,a2];
Since l=1.

Integration:
Need stable. Use RK4 with substeps. requestAnimationFrame gives dt seconds. Cap dt maybe 0.033. speed slider maybe 0.2-3.0. If paused no update. Use fixed substep? Better:
let accumulator = 0;
const fixedDt = 1/120;
simTime += dt * speed;
while accumulator >= fixedDt: rk4(fixedDt); accumulator -= fixedDt.
Need avoid spiral of death. accumulator += Math.min(dt,0.05)*speed.
Use fixedDt = 1/240 maybe for accuracy, but performance okay. Use 1/120. Trail sample every frame.

Trail:
const trail = [];
function addTrailPoint(x,y,hue) { trail.push({x,y,hue,life:1}); if length > maxTrail remove shift; }
Each frame after update draw:
- Fade old life by dt maybe. Use trail.forEach(p => p.life -= dt * fadeRate); filter.
But if draw full background with alpha, old trail fades. But for colorful fading, array with alpha is fine. Need clearRect each frame and draw background.
Draw background: gradient maybe.
drawTrail:
for i=1; i<trail.length; i++:
p0=trail[i-1], p1=trail[i];
const a = Math.min(p0.life,p1.life);
ctx.strokeStyle = `hsla(${(p0.hue + p1.hue)/2}, 100%, 60%, ${a*0.85})`;
ctx.lineWidth = 1 + 3*a;
lineCap round.
Maybe draw many segments. If huge 2000, okay.

Need ensure canvas draws after resize.
Need maybe use devicePixelRatio.
function resizeCanvas() {
dpr = Math.min(window.devicePixelRatio || 1, 2);
canvas.width = Math.floor(innerWidth*dpr);
canvas.height = Math.floor(innerHeight*dpr);
ctx.setTransform(dpr,0,0,dpr,0,0);
}
Then drawing uses CSS pixels (innerWidth).
Need origin changes.

Need UI:
HTML:
<div id="hud">
<h1>Chaotic Double Pendulum</h1>
<p>Equal masses, RK4 integration, exact Lagrangian equations.</p>
<div class="row">
<button id="pause">Pause</button>
<button id="reset">Reset</button>
<button id="random">Randomize</button>
</div>
<label>Speed <input type=range id=speed min=0.1 max=3 step=0.05 value=1.2></label>
<label>Trail <input type=range id=trail min=100 max=3000 step=100 value=1600></label>
</div>
Maybe display equation.

Need avoid blocking.
CSS:
html, body { margin:0; height:100%; overflow:hidden; background:#05070d; color:#eaf6ff; font-family: system-ui; }
canvas { display:block; width:100vw; height:100vh; }
#hud { position:fixed; left:16px; top:16px; max-width:360px; padding:14px 16px; border:1px solid rgba...; background: rgba(4,8,18,.62); backdrop-filter: blur(12px); border-radius:16px; box-shadow; user-select:none; }
Maybe on small screen width.
Controls style.
#equation maybe small code.
Need no external fonts.

Need maybe include "Correct equations of motion" maybe in UI.
Maybe include hidden "tip" etc.

Need maybe initial conditions chaotic: θ1 = 120°, θ2 = -20°? Need ensure not too high. Use theta1 = Math.PI * 0.55; theta2 = Math.PI * 0.75? Since downward zero. Need interesting. Maybe theta1 = 1.95, theta2 = -1.85? If both large. Need not wrap. Use randomize:
theta1 = random(-Math.PI, Math.PI), theta2 = theta1 + random(-1.2,1.2), maybe near.
Need reset to initial.
Need if paused still draw.
Need on resize draw.

Need implement RK4:
function derivative(s) { return {theta1: s.omega1, theta2: s.omega2, omega1: a1, omega2: a2}; }
function addState(s, ds, k) { return {theta1:s.theta1+k.theta1*dt, theta2:s.theta2+k.theta2*dt, omega1:s.omega1+k.omega1*dt, omega2:s.omega2+k.omega2*dt};}
function rk4(dt) {
const d1 = derivative(state);
const s2 = add...
const d2...
state.theta1 += (dt/6)*(d1.theta1 + 2*d2.theta1 + 2*d3.theta1 + d4.theta1);
...
state.theta1 = normalize? We can keep not normalized to avoid discontinuity. But sin/cos okay. For display maybe modulo for text.
}
Need if state theta grows huge after long time; sin okay but maybe. Could wrap occasionally if abs > 1e6. But not necessary. Could implement wrapAngles() if abs > 1000.

Need if using fixed substeps and speed slider, maybe dt*dt.

Need draw:
function draw() {
const w=canvas.clientWidth; h=canvas.clientHeight;
ctx.clearRect(0,0,w,h);
const bg = ctx.createRadialGradient(w*0.5,h*0.25,0,w*0.5,h*0.25,Math.max(w,h)*0.75);
bg.addColorStop(0,'#111a33'); bg.addColorStop(0.45,'#070a14'); bg.addColorStop(1,'#02030a');
ctx.fillStyle=bg; ctx.fillRect(0,0,w,h);
draw faint grid maybe.

const scale = Math.min(w,h) * 0.27;
const ox = w*0.5; const oy = h*0.18; if h small maybe h*0.25.
const x1=ox, y1=oy;
const x2=x1 + scale*l1*Math.sin(theta1); y2=y1 + scale*l1*Math.cos(theta1);
const x3=x2 + scale*l2*Math.sin(theta2); y3=y2 + scale*l2*cos(theta2);
const hue = (performance.now()*0.035 + (theta1*30 + theta2*50)) % 360;
if (!paused) addTrailPoint(x3,y3,hue);
drawTrail();
// rods:
ctx.lineCap='round';
ctx.strokeStyle='rgba(230,245,255,.65)'; ctx.lineWidth=3; draw line x1 y1 x2 y2, x2 y2 x3 y3.
// shadows maybe.
draw pivot circles, bobs.
}
Need draw trail before arms.
Need draw bobs:
drawCircle(x,y,r,fill,glow)
function bob(x,y,r,fill,g) {
ctx.save(); ctx.shadowColor=fill; ctx.shadowBlur=20; ctx.fillStyle=fill; ctx.beginPath(); ctx.arc(x,y,r,0,TAU); ctx.fill(); ctx.restore();
ctx.strokeStyle='rgba(255,255,255,.45)'; ctx.lineWidth=1.5; ctx.stroke();
}
Need pivot:
ctx.fillStyle '#f5fbff'; circle.

Need trail point management:
let maxTrail=1600;
function pushTrail(x,y,hue) {
trail.push({x,y,hue,life:1});
if (trail.length > maxTrail) trail.splice(0, trail.length-maxTrail);
}
function updateTrail(dt) {
const fade = 0.35; // life per second
for p of trail p.life -= dt*fade;
while trail.length && trail[0].life <=0 trail.shift();
}
But if dt not exactly, okay.
Need if paused, trail not fade? Maybe no update if paused.
Need if trail slider changes, maxTrail set.

Need maybe if canvas not visible? okay.

Need maybe use ResizeObserver? window resize.

Potential issue with DPR transform and canvas CSS dimensions. If set canvas.width/height and style width/height 100vw. canvas.clientWidth/Height okay. But after ctx.setTransform(dpr,...), clearRect uses CSS coords with transform? Actually ctx.setTransform(dpr,0,0,dpr,0,0); ctx.clearRect(0,0,w,h) clears w*dpr by h*dpr due transform? clearRect is affected by transform, yes clears scaled area. Good. Or reset transform.

Need maybe draw background gradient after clear. Good.

Need maybe include "No external resources" not needed.

Need maybe "correct equations of motion" in JS comment.

Need maybe if user expects just artifact, final concise. Need write file. Need maybe use bash ls? Not necessary but maybe check cwd. Could write. Need maybe if file exists? We can overwrite. Use write.

Need ensure HTML complete. Let's craft.

Need maybe use exact file path /tmp/ws/index.html.

Potential issue: CSS backdrop-filter unsupported but okay.
Need maybe ensure "input range" labels.
Need maybe use buttons.

Need maybe include "energy" display? Not necessary. Could include "theta" display? maybe.
Need maybe include "Click/tap canvas to re-energize"? Could implement but no need.

Need maybe "simulate chaotic double pendulum on a canvas". We can include controls and explanation.

Let's write robust code.

Need maybe use `<meta name="viewport" content="width=device-width, initial-scale=1.0" />`
Need maybe `<canvas id="scene" aria-label="Chaotic double pendulum simulation"></canvas>`

JS:
(() => {
'use strict';
const canvas = document.getElementById('scene');
const ctx = canvas.getContext('2d');
const pauseBtn = ...
const speedInput = ...
...
const TAU = Math.PI*2;
const params = {l1:1,l2:1,m1:1,m2:1,g:9.81};
const initial = {theta1: Math.PI*0.58, theta2: -Math.PI*0.48, omega1: 0, omega2:0};
let state = {...initial};
let trail = [];
let maxTrail = Number(trailInput.value);
let paused = false;
let last = performance.now();
let accumulator = 0;
let hueClock = 0;

function resize() {
const dpr = Math.min(window.devicePixelRatio || 1, 2);
canvas.width = Math.floor(window.innerWidth * dpr);
canvas.height = Math.floor(window.innerHeight * dpr);
ctx.setTransform(dpr,0,0,dpr,0,0);
}
window.addEventListener('resize', resize); resize();

function accelerations(s) {
const { l1,l2,m1,m2,g } = params;
const t1=s.theta1, t2=s.theta2, w1=s.omega1, w2=s.omega2;
const delta = t1 - t2;
const c = Math.cos(delta), sDelta = Math.sin(delta);
const denom = l1 * (2*m1 + m2 - m2 * Math.cos(2*delta));
const a1 = (
-g*(2*m1+m2)*Math.sin(t1)
- m2*Math.sin(t1 - 2*t2)
- 2*sDelta*m2*(w2*w2*l2 + w1*w1*l1*c)
) / denom;
const a2 = (
2*sDelta*(w1*w1*l1*(m1+m2) + g*(m1+m2)*Math.cos(t1) + w2*w2*l2*m2*c)
) / (l2*(2*m1 + m2 - m2*Math.cos(2*delta)));
return {theta1:w1, theta2:w2, omega1:a1, omega2:a2};
}
Wait derivative returns theta dot. accelerations maybe returns all derivatives. Name derivative.

function derivative(s) { ... }

function integrate(dt) {
const d1=derivative(state);
const s2 = addScaled(state,d1,dt*0.5);
...
state.theta1 += dt/6*(...);
...
if (Math.abs(state.theta1)>100000 || Math.abs(state.theta2)>100000) wrap
}
Need maybe wrap after.
function wrapAngles() { state.theta1 %= TAU; state.theta2 %= TAU; }
Need preserve continuity? If wrap only huge, okay.

function animate(now) {
const rawDt = Math.min((now-last)/1000, 0.05);
last = now;
if (!paused) {
const speed = Number(speedInput.value);
accumulator += rawDt * speed;
const fixed = 1/180;
let steps = 0;
while (accumulator >= fixed && steps < 8) { integrate(fixed); accumulator -= fixed; steps++; }
if (steps === 8) accumulator = 0; // avoid spiral
updateTrail(rawDt*speed);
}
draw(now);
requestAnimationFrame(animate);
}
requestAnimationFrame(animate);

Need if speed slider value 0.1 and rawDt 0.05 -> accumulator 0.005, no steps until many frames. okay.
Maybe fixed 1/240 for accuracy. 8 steps cap at 8/240=0.033s; with speed 3 and rawDt .05 -> .15, steps cap 8 -> sim 0.033, not enough, but okay. Could use fixed=1/240 and cap steps 12. Or fixed=1/180. Use 1/240 and cap 16. Fine.
But if tab inactive dt capped 0.05, speed max 3 => 0.15, fixed .00417 => 36 steps. Cap 24 = 0.1 sec; okay. Use cap 24.

Need updateTrail fade. If speed high, fade should be real time not sim time. Use rawDt. fade = 0.28. If not paused, updateTrail(rawDt). Good.

Need drawTrail:
function drawTrail() {
if (trail.length < 2) return;
ctx.save();
ctx.globalCompositeOperation = 'lighter';
for (let i=1; i<trail.length; i++) {
const p0=trail[i-1], p1=trail[i];
const alpha = Math.max(0, Math.min(p0.life,p1.life));
if (alpha <= 0.01) continue;
const hue = (p0.hue + p1.hue)/2;
ctx.strokeStyle = `hsla(${hue}, 100%, ${55 + alpha*15}%, ${alpha*0.75})`;
ctx.lineWidth = 0.7 + 4.2*alpha;
ctx.beginPath(); ctx.moveTo(p0.x,p0.y); ctx.lineTo(p1.x,p1.y); ctx.stroke();
}
ctx.restore();
}
Need maybe if trail has many segments, lineWidth. Good.
Need maybe draw older segments first. Since trail array old to new. Fine.

Need drawPendulum:
function drawPendulum(now, points) { ... }
Need maybe draw background grid:
function drawBackground(w,h) {
const bg = ctx.createRadialGradient(w*0.5,h*0.2,0,w*0.5,h*0.2,Math.max(w,h)*0.8);
...
ctx.fillRect
// grid
ctx.save(); ctx.globalAlpha = 0.12; ctx.strokeStyle='rgba(120,180,255,0.15)'; ctx.lineWidth=1;
const grid = 48;
const ox = (performance.now()*0.01)%grid maybe moving? Could draw static.
for x = (w%grid); x<w; x+=grid...
ctx.restore();
}
Need no expensive.

Need maybe if paused, still animate draw for trail? If paused no update. If speed slider changed while paused? draw.
Need maybe if paused, requestAnimationFrame still draws maybe background. Good.

Need maybe "randomize" should clear trail, set state random:
function randomize() {
const t1 = random(-Math.PI, Math.PI);
const t2 = t1 + random(-1.4,1.4);
state.theta1=t1; state.theta2=t2; state.omega1=0; state.omega2=0; trail.length=0;
}
Need maybe "Reset" to initial. "Pause" toggles.
Need maybe "Nudge" on canvas click? Could implement:
canvas.addEventListener('pointerdown', (e)=> { if (!paused) { state.omega1 += random(-0.5,0.5); state.omega2 += random(-0.8,0.8); trail.length=0; }});
But not necessary. Could confuse. But nice.

Need maybe display stats:
<div class="stats"><span id="thetaReadout"></span> ...</div>
JS update every frame maybe.
function updateReadout() {
thetaReadout.textContent = `${deg(state.theta1).toFixed(1)}° / ${deg(state.theta2).toFixed(1)}°`;
}
Need maybe show "RK4 • exact equations".

Need maybe no external resources. Good.

Potential issue with exact formula for theta2 uses l1/l2. Good.

Let's verify formula with l1/l2=1.0. At rest t1=t2=0, denom=2, a1=0, a2=0. Good.
At t1=small, t2=0, w=0: a1 = -g(3)sin t1/(2) = -1.5g sin; a2 = 2*0? delta=t1, cos t1, numerator=2sin t1*g*2*cos t1 = 4g sin cos, denom=2 => 2g sin cos. Is that plausible? For small angles, linearized double pendulum equal masses/l=1: equations: 2θ1'' + θ2'' = -2g θ1? Wait Lagrange small: 2θ1'' + θ2'' = -2g θ1; θ1'' + θ2'' = -g θ2. For θ2=0, solve θ1'' =? 2a1+a2=-2gθ1, a1+a2=0 -> a2=-a1, so a1=-2gθ1. Our a1=-1.5g, a2=2gθ1, not correct. Maybe formula with sin(theta1-2theta2) vs sin(delta) matters? For t2=0, sin(t1-2t2)=sin t1, same. Denom? 2. So a1=-1.5g, a2=2g. Not matching linearization. Did I recall formula missing factor? Let's check. For equal masses and lengths, small-angle linearization should be:
2 θ1'' + θ2'' = -2 (g/l) θ1
θ1'' + θ2'' = - (g/l) θ2
For θ2=0: subtract: θ1'' = -2g θ1 + g θ2? Wait second: a1+a2 = -g θ2 = 0 => a2=-a1. First: 2a1+a2 = a1 = -2g θ1. So a1=-2g, a2=2g. Our a1=-1.5g. Hmm maybe linearization formula denominator for theta1 should l1*(2m1 + m2 - m2 cos2Δ) = 2, numerator -g(2m1+m2)sinθ1 - m2 sin(θ1-2θ2) = -3g sin. Gives -1.5. Is that wrong? Let's check Lagrange small equations maybe coefficient maybe 2θ1'' + θ2'' = -2g θ1; yes.

Let's derive exact formula maybe denominator? Wikipedia formula for theta1 denominator maybe l1(2m1+m2 - m2 cos(2θ1-2θ2)). At small t1,t2=0 denom=2. Numerator should -g(2m1+m2)sinθ1 - m2 sin(θ1 - 2θ2) - ... = -3g. So -1.5g. Contradiction? Maybe small linear equations for double pendulum with equal masses are:
2 θ1'' + θ2'' = -2g θ1
θ1'' + θ2'' = -g θ2
For θ2=0, a1=-2g, a2=2g. Yes.

Maybe exact formula's first gravitational term should -g(2m1+m2)sinθ1? That gives -3. Hmm maybe second term should -m2 sin(θ1 - 2θ2) = -sin; total -4? even worse. Wait if term is -m2 sin(θ1 - 2θ2) maybe -1. If term is -m2 sin(θ1-θ2) = 0. Then a1=-1.5g. If term is -m2 sin(θ1 - 2θ2) and denominator? no.

Let's rederive to ensure correct equations. There is known formula:
theta1dd = (-g (2 m1 + m2) sin theta1 - m2 sin(theta1 - theta2) - 2 sin(theta1 - theta2) m2 (theta2dot^2 l2 + theta1dot^2 l1 cos(theta1 - theta2))) / (l1 (2 m1 + m2 - m2 cos(2 theta1 - 2 theta2)))
theta2dd = (2 sin(theta1 - theta2) (theta1dot^2 l1 (m1+m2) + g (m1+m2) cos theta1 + theta2dot^2 l2 m2 cos(theta1 - theta2))) / (l2 (2m1+m2 - m2 cos(...)))
At rest small t2=0: numerator -3g sin - 0 = -3, a1=-1.5. Still.
Maybe small linear equations are:
2 θ1'' + θ2'' = -2g θ1
θ1'' + θ2'' = -g θ2
But maybe for double pendulum angles measured from vertical? yes. Let's derive with Lagrangian quickly to verify formula.
Positions: x1=l1 sinθ1, y1=-l1 cosθ1 (up?); x2=x1+l2 sinθ2, y2=y1-l2 cosθ2.
Kinetic: T=1/2 m1 l1² θ1'² + 1/2 m2[(l1² θ1'² + l2² θ2'² + 2l1l2 θ1'θ2' cos(θ1-θ2))].
Potential V= - (m1+m2) g l1 cosθ1 - m2 g l2 cosθ2.
L=T-V.
Equation θ1:
∂T/∂θ1'=(m1+m2)l1² θ1' + m2 l1l2 θ2' cosδ
d/dt = (m1+m2)l1² θ1'' + m2 l1l2[θ2'' cosδ - θ2' sinδ(θ1'-θ2')]
∂L/∂θ1 = -m2 l1l2 θ1'θ2' sinδ + (m1+m2)g l1 sinθ1? Wait V=-..., ∂V/∂θ1=(m1+m2)g l1 sinθ1, ∂L=-∂V = -(m1+m2)g l1 sinθ1. Plus T derivative -m2 l1l2 θ1'θ2' sinδ.
Equation: d/dt - ∂L =0:
(m1+m2)l1² θ1'' + m2l1l2 θ2'' cosδ - m2l1l2 θ2' sinδ(θ1'-θ2') + m2l1l2 θ1'θ2' sinδ + (m1+m2)g l1 sinθ1=0
Velocity term: -m2 l1l2 θ2'θ1' sinδ + m2l1l2 θ2'^2 sinδ + m2l1l2 θ1'θ2' sinδ = m2l1l2 θ2'^2 sinδ.
So (m1+m2)l1 θ1'' + m2l2 θ2'' cosδ + m2l2 θ2'^2 sinδ + (m1+m2)g sinθ1=0.
Equation θ2:
∂T/∂θ2'=m2l2² θ2' + m2l1l2 θ1' cosδ
d/dt=m2l2² θ2'' + m2l1l2[θ1'' cosδ - θ1' sinδ(θ1'-θ2')]
∂L/∂θ2 = m2l1l2 θ1'θ2' sinδ + m2g l2 sinθ2
Equation: m2l2² θ2'' + m2l1l2 θ1'' cosδ - m2l1l2 θ1'^2 sinδ + m2l1l2 θ1'θ2' sinδ - m2l1l2 θ1'θ2' sinδ - m2g l2 sinθ2=0
=> l2 θ2'' + l1 θ1'' cosδ - l1 θ1'^2 sinδ + g sinθ2=0.

For small δ, cos=1, sinδ≈δ; velocity terms second order. Linear:
(m1+m2)l1 θ1'' + m2l2 θ2'' + (m1+m2)g θ1=0
l2 θ2'' + l1 θ1'' + g θ2=0
Equal l=m=1:
2a1+a2=-2gθ1
a1+a2=-gθ2
Yes.

Solving exact equations formula should match. Let's solve general formula from these equations:
A = m1+m2
eq1: A l1 a1 + m2 l2 c a2 = -A g sin t1 - m2 l2 w2² sinδ
eq2: l1 c a1 + l2 a2 = -g sin t2 + l1 w1² sinδ
Solve:
D = A l1*l2 - m2 l1*l2 c² = l1 l2 (A - m2 c²) = l1l2(m1+m2 - m2 c²). Since c²=(1+cos2δ)/2, denominator l1l2(m1 + m2/2 - m2/2 cos2δ) = l1l2/2(2m1+m2 - m2cos2δ).
a1 numerator = b1*l2 - m2l2 c*b2 over l1l2(A-m2c²). b1=-Ag sin t1 - m2l2 w2² sinδ; b2=-g sin t2 + l1w1² sinδ.
a1 = [-Ag l2 sin t1 - m2l2² w2² sinδ - m2l2 c(-g sin t2 + l1w1² sinδ)] / [l1l2(A-m2c²)]
= [-A g sin t1 - m2 l2 w2² sinδ + m2 c g sin t2 - m2 l1 c w1² sinδ]/[l1(A-m2c²)].
Use identity? Need compare Wikipedia:
denom l1(2m1+m2 - m2 cos2δ)=2l1(A-m2c²).
Numerator maybe -g(2A?)? Let's transform:
For equal masses A=2. Denom 2l1(2-m2? wait m1=1,m2=1 A-m2c²=2-c²; 2(A-m2c²)=4-2c² = 2+ (1-cos2δ)? =3 - cos2δ? But Wikipedia denom 2m1+m2 - m2 cos2δ = 3 - cos2δ. Yes.
Our denom 2l1(2-c²)=2l1(1.5+0.5cos2δ)=3+cos2δ. Wait A=2, m2=1, A-m2c²=2-c² = 1.5+0.5cos2δ; times 2 =3+cos2δ. Wikipedia denom 3 - cos2δ. Something off due sign? Let's check determinant: A*l1 * l2 - (m2 l2 c)*(l1 c) = l1l2(A - m2 c²). A=2, c²=(1+cos2δ)/2 => A-c²=2 - .5 - .5cos2δ=1.5-.5cos2δ; times2 =3 - cos2δ. Yes I mis: 2(1.5 - .5cos)=3 - cos. Good.
At δ=0 denom=2l1. Good.

a1 formula with c sin t2. Wikipedia numerator for equal masses:
-g(3)sin t1 - sinδ - 2 sinδ(w2² + w1² c)
Denom l1(3 - cos2δ). At δ=0 denom=2l1, numerator=-3g sin, a=-1.5g. But our exact formula at δ=0 numerator=-2g sin + c g sin t2. For t2=0, -2g. Denom 2 => -g? Wait with equal masses, numerator = -2g sin t1 + c g sin t2 - velocity. At t2=0 denom=2, a1=-g. But linear solve gave -2g. Did I miss m1+m2=2 in b1; numerator -2g, denom? D = l1l2(A-m2c²)=2-1=1, a1=-2g. Wait denominator in formula l1(A-m2c²), not 2l1. At δ=0 A-m2=1, denom=l1. So a1=-2g. If multiply numerator and denom by 2, denom 2l1, numerator -4g. Wikipedia numerator -3g. So formula not matching. Maybe equations or determinant? Let's solve with A=m1+m2=2. Eq1: 2l1 a1 + l2 c a2 = -2g sin t1 - l2 w2² sinδ. Eq2: l1 c a1 + l2 a2 = -g sin t2 + l1 w1² sinδ.
D=2l1*l2 - l1l2 c² = l1l2(2-c²). At δ=0 D=l1l2. a1=(b1*l2 - l2 c*b2)/D = (-2g sin*l2 - 0)/l1l2 = -2g/l1. Good.
Wikipedia formula at δ=0 gives -3/2. So Wikipedia formula I recall maybe for m1=m2? It should match. Let's check known formula from Wikipedia maybe:
θ1dd = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2)m2(θ2dot²l2 + θ1dot²l1 cosΔ)) / (l1(2m1+m2 - m2 cos(2Δ)))
For equal masses at δ=0 numerator -3g, denom 2. Hmm. Maybe m1 and m2 in Wikipedia are not masses but something? No.
Let's check MyPhysicsLab formula for equal masses:
theta1_acc = (-g(2 sin theta1 + a cos delta + b sin delta))/(l1(4 theta1 + theta2 cos delta))? Not.

Let's use actual known code:
num = -g*(2*m1 + m2)*sin(theta1);
num -= m2*g*sin(theta1 - 2*theta2);
num -= 2*sin(theta1-theta2)*m2*(theta2_speed*theta2_speed*l2 + theta1_speed*theta1_speed*l1*cos(theta1-theta2));
den = l1*(2*m1 + m2 - m2*cos(2*theta1 - 2*theta2));
theta1_acc = num/den;
For equal rest t2=0: num=-3g, den=2 -> -1.5. Many demos use this. But Lagrangian says -2? Did I define theta2 relative to vertical? In double pendulum second angle maybe relative to first? Ah! In many formulas, θ2 is angle of second rod relative to first rod, not vertical. In Wikipedia, I think θ1 and θ2 are angles with vertical? Actually maybe θ2 is angle of second pendulum relative to first? Let's check. If θ2 relative to first, at rest both vertical θ2=0, small equations differ. My derivation uses θ2 absolute from vertical. Wikipedia formula maybe θ2 absolute? It has cos(θ1-θ2), so likely absolute. Hmm.

Let's verify with known MyPhysicsLab equal masses formula for absolute angles:
a = 2*theta1_v*theta1_v*len1 + theta2_v*theta2_v*len2*cos(theta1-theta2)
b = theta2_v*theta2_v*len2*sin(theta1-theta2)
c = 4*theta1_v + theta2_v*cos(theta1-theta2)
d = 2*theta2_v - theta1_v*cos(theta1-theta2)
theta1_acc = -g*(2*sin(theta1) + a*cos(theta1-theta2) + b*sin(theta1-theta2)) / (len1*(4 + d*sin(theta1-theta2)))
theta2_acc = 2*b*(len1+len2)/(len1*len2*(4 + d*sin(...)))
At rest small t2=0, w=0: a=0,b=0,c=0,d=0, theta1_acc = -2g sin /4 = -0.5g? That seems wrong? Wait MyPhysicsLab maybe θ2 angle relative to vertical? But formula includes 4 because equal masses? At small t1 only, a1=-0.5g, not.

Maybe I'm mixing variables. Let's test double pendulum with second mass free. If θ1 small, θ2=0, top bob and bottom bob both displaced same direction. Top bob experiences gravity torque from both masses: -2gθ1; lower rod constraint? Acceleration of top rod should maybe -? Let's simulate physical: both masses at x=lθ1, second at x=2lθ1. Gravity torque about pivot = -(m1 g l θ1 + m2 g 2l θ1) = -3gθ1. Moment of inertia about pivot = m1 l² + m2(2l)² = 5l². If θ2=θ1 and no relative acceleration? a1 maybe -3/5g, not -2. But constraints with θ2'' maybe. My linear equations solve a1=-2g, a2=2g, so relative acceleration huge. Is that plausible? If both rods initially vertical? Hmm.

Let's derive Lagrange carefully signs. Potential with y downward maybe. Use y = -l cosθ (up positive?); potential V = mgy with y upward? Let's set y upward = -l cosθ, V=mgy=-mgl cosθ. At θ=0 down, V=-mgl min. ∂V/∂θ=mgl sinθ. Lagrange d/dt∂T - ∂T/∂θ + ∂V =0. I used ∂L/∂θ=∂T/∂θ - ∂V. ∂T/∂θ for θ1 = -m2 l1l2 θ1'θ2' sinδ. ∂V=(m1+m2)g l1 sinθ1. Equation: d/dt∂Tdot - ∂Tθ + ∂V=0 => d/dt∂Tdot + m2... + (m1+m2)g... =0. I wrote d/dt - ∂L = d/dt - ∂T + ∂V =0. Same. Good.
For θ2: ∂T/∂θ = +m2 l1l2 θ1'θ2' sinδ; ∂V=m2gl2 sinθ2. Eq d/dt∂Tdot - m2... + m2g...=0. I wrote d/dt - ∂L = d/dt - ∂T + ∂V; yes.
Linear equations correct.

Let's solve for initial θ1=small θ2=0, ω=0. θ1''=-2gθ1, θ2''=2gθ1. That means lower rod angle accelerates same sign? θ2 positive (to right) while top also negative? If θ1 positive right, a1 negative, a2 positive. Lower rod swings further right, top left. Maybe.

Why known formula different? Let's derive explicit solution from equations to compare known formula. General formula from our linear solve:
a1 = [ -A g l2 sin t1 + m2 g l2 c sin t2 - m2 l2² w2² sinδ - m2 l1l2 c w1² sinδ ] / [l1l2(A - m2 c²)]
= [ -A g sin t1 + m2 g c sin t2 - m2(l2 w2² + l1 c w1²) sinδ ] / [l1(A - m2 c²)].
Use denominator multiply by 2: l1(2A - 2m2c²)=l1(2m1+2m2 - m2(1+cos2δ))=l1(2m1+m2 - m2 cos2δ). Numerator multiply by 2:
-2A g sin t1 + 2m2 g c sin t2 - 2m2(... ) sinδ.
Use identity 2c sin t2 - 2 sinδ? Let δ=t1-t2.
2c sin t2 - 2 sinδ = 2cosδ sin t2 - 2sin(t1-t2). Need transform to - sin(t1-2t2)? Let's see:
sin(t1-2t2)=sinδ cos t2 - cosδ sin t2? Since t1-2t2=δ - t2. sin(δ-t2)=sinδ cos t2 - cosδ sin t2. Thus 2c sin t2 - 2sinδ? Not same.
Known numerator -g(2m1+m2)sin t1 - m2g sin(t1-2t2). For m2=1, A=2 => -3g sin t1 - g(sinδ cos t2 - cosδ sin t2). Expand = -3g sin t1 - g(sin t1 cosδ cos t2 - cos t1 sinδ cos t2 - cosδ sin t2). Not equal to -4? Hmm.

Let's compare our transformed numerator for equal:
-4g sin t1 + 2g cosδ sin t2 - velocity.
Known: -3g sin t1 - g sin(t1-2t2) = -3g sin t1 - g[sin t1 cos2t2 - cos t1 sin2t2]? Not.

Maybe our derivation equations have sign of potential? If θ measured from upward vertical instead of downward, rest at π, not.

Let's check known formula from Wikipedia maybe actually:
θ1dd = (-g(2m1+m2)sinθ1 - m2 sin(θ1-θ2) - 2 sin(θ1-θ2)m2(θ2dot²l2 + θ1dot²l1 cosΔ)) / (l1(2m1+m2 - m2 cos(2Δ)))
At rest small θ1=θ2=0 okay. For θ2=0, a1=-1.5. Could be because θ2 is angle of second rod relative to first? If θ2_rel=0, absolute second angle=θ1+θ2_rel. Then small equations different. But formula has cos(θ1-θ2), if θ2_rel? Hmm.

Let's search memory more: Wikipedia formula:
\ddot{\theta}_1 = \frac{-g(2m_1+m_2)\sin\theta_1 - m_2\sin(\theta_1-2\theta_2) - 2\sin(\theta_1-\theta_2)m_2(\dot{\theta}_2^2l_2+\dot{\theta}_1^2l_1\cos(\theta_1-\theta_2))}{l_1(2m_1+m_2-m_2\cos(2\theta_1-2\theta_2))}
\ddot{\theta}_2 = \frac{2\sin(\theta_1-\theta_2)(\dot{\theta}_1^2l_1(m_1+m_2)+g(m_1+m_2)\cos\theta_1+\dot{\theta}_2^2l_2m_2\cos(\theta_1-\theta_2))}{l_2(2m_1+m_2-m_2\cos(2\theta_1-2\theta_2))}
This is widely used. It might be correct for θ2 measured from vertical? Let's test energy conservation maybe.

Let's derive exact equations from Lagrange solving fully with sympy mentally. Maybe I made error in ∂T/∂θ1 velocity term sign. Let's rederive carefully.

Coordinates with y downward? Let x1=l1 sinθ1, y1=-l1 cosθ1 (up positive? Actually y upward = -l cos). x2=x1+l2 sinθ2, y2=y1-l2 cosθ2.
Velocities:
v1²=l1² θ1'².
v2² = l1² θ1'² + l2² θ2'² + 2l1l2 θ1'θ2'(sinθ1 sinθ2 + cosθ1 cosθ2) = +2 cosδ. Good.
T = 1/2 m1 l1² w1² + 1/2 m2(... + 2l1l2w1w2 cosδ).
∂T/∂θ1 = 1/2 m2*2l1l2w1w2 * (-sinδ) = -m2l1l2w1w2 sinδ.
∂T/∂θ1dot = m1l1²w1 + m2(l1²w1 + l1l2w2 cosδ)
d/dt = (m1+m2)l1² a1 + m2l1l2(a2 cosδ - w2 sinδ (w1-w2)).
L=T-V, V=-(m1+m2)gl1 cosθ1 - m2gl2 cosθ2.
∂V/∂θ1=(m1+m2)gl1 sinθ1.
Euler: d/dt∂L/∂θ1dot - ∂L/∂θ1=0 = d/dt∂Tdot - (∂Tθ - ∂Vθ)=d/dt∂Tdot - ∂Tθ + ∂Vθ=0.
= (m1+m2)l1²a1 + m2l1l2a2 c - m2l1l2w2 sinδ(w1-w2) + m2l1l2w1w2 sinδ + (m1+m2)gl1 sinθ1 =0.
Velocity: -m2l1l2w2w1 sinδ + m2l1l2w2² sinδ + m2l1l2w1w2 sinδ = m2l1l2w2² sinδ. Correct.
Eq1: (m1+m2)l1²a1 + m2l1l2 c a2 + m2l1l2 w2² sinδ + (m1+m2)gl1 sinθ1=0.

θ2:
∂T/∂θ2 = +m2l1l2w1w2 sinδ.
∂T/∂θ2dot = m2l2²w2 + m2l1l2w1 c
d/dt=m2l2²a2 + m2l1l2(a1 c - w1 sinδ(w1-w2))
∂V/∂θ2=m2gl2 sinθ2.
Eq: d/dt - ∂Tθ + ∂V = m2l2²a2 + m2l1l2 a1 c - m2l1l2 w1² sinδ + m2l1l2 w1w2 sinδ - m2l1l2w1w2 sinδ + m2gl2 sinθ2=0.
Eq2: l2 a2 + l1 c a1 - l1 w1² sinδ + g sinθ2=0. Correct.

Solve exactly. Let A=m1+m2.
Eq1 divide l1: A l1 a1 + m2 l2 c a2 = -A g sin t1 - m2 l2 w2² sinδ.
Eq2: l1 c a1 + l2 a2 = -g sin t2 + l1 w1² sinδ.
Solve:
D = A l1*l2 - m2 l1l2 c² = l1l2(A - m2 c²).
a1 numerator = b1*l2 - m2l2 c*b2, b1 RHS eq1, b2 RHS eq2.
= l2[-A g sin t1 - m2l2w2² sinδ] - m2l2 c[-g sin t2 + l1w1² sinδ]
= -A g l2 sin t1 - m2l2²w2² sinδ + m2g l2 c sin t2 - m2l1l2 c w1² sinδ.
Divide l1l2(A-m2c²):
a1 = [-A g sin t1 - m2 l2 w2² sinδ + m2g c sin t2 - m2 l1 c w1² sinδ]/[l1(A-m2c²)].
For equal, t2=0, denom l1(2-1)=l1, numerator -2g sin, a1=-2g. Good.
a2 numerator = A l1*b2 - m2l1 c*b1 over D.
= A l1[-g sin t2 + l1w1² sinδ] - m2l1 c[-A g sin t1 - m2l2w2² sinδ]
= -A g l1 sin t2 + A l1²w1² sinδ + m2A g l1 c sin t1 + m2² l1l2 w2² c sinδ.
Divide l1l2(A-m2c²):
a2 = [-A g sin t2 + A l1 w1² sinδ + m2A g c sin t1 + m2² l2 w2² c sinδ]/[l2(A-m2c²)].
For equal t2=0,t1 small: numerator 2g c sin =2g, denom1 => 2g. Good.

Can we transform to known formula? Let's use identities. Denom multiply by 2: l1(2A - 2m2c²)=l1(2m1+2m2 - m2(1+cos2δ))=l1(2m1 + m2 - m2 cos2δ). Numerator multiply by 2:
N1 = -2A g sin t1 + 2m2g c sin t2 - 2m2(l2w2² + l1 c w1²) sinδ.
Need express gravitational part. Use 2c sin t2 = sin(t1) + sin(2t2-t1)? Since 2 cos(t1-t2) sin t2 = sin(t1) + sin(2t2 - t1). So -2A g sin t1 + 2m2g c sin t2 = -2(m1+m2)g sin t1 + m2g[sin t1 + sin(2t2-t1)] = -g(2m1+m2) sin t1 + m2g sin(2t2-t1) = -g(2m1+m2) sin t1 - m2g sin(t1-2t2). Ah known! Great. Velocity part: -2m2(l2w2² + l1 c w1²) sinδ. Known velocity term is -2 sinδ m2(θ2dot²l2 + θ1dot²l1 cosδ). Same. Great. So known formula is equivalent. At t2=0, N1=-3g sin - sin(t1)= -4g sin? Wait identity gave -g(3)sin + sin(-t1) = -4g sin. Denom 2 -> -2g. Good. I earlier thought known numerator -3g only, but with sin(t1-2t2) at t2=0 gives -sin t1, total -4. Good. So formula with sin(t1-2θ2) correct. The variant with sin(θ1-θ2) in gravitational term is wrong? Let's see if identity for 2c sin t2 - 2sin t1 = -? -2A +? Not.
Thus use sin(t1 - 2*t2). Good.

Need formula in code exactly known. Good.

Need maybe include "correct equations" maybe in comment:
const a1 = (
-g * (2*m1 + m2) * Math.sin(t1)
- m2 * g * Math.sin(t1 - 2*t2)
- 2 * Math.sin(delta) * m2 * (w2*w2*l2 + w1*w1*l1*Math.cos(delta))
) / (l1 * (2*m1 + m2 - m2 * Math.cos(2*delta)));
const a2 = (
2 * Math.sin(delta) * (w1*w1*l1*(m1+m2) + g*(m1+m2)*Math.cos(t1) + w2*w2*l2*m2*Math.cos(delta))
) / (l2 * (2*m1 + m2 - m2 * Math.cos(2*delta)));
Yes.

Need maybe if masses equal but formula general. Good.

Need maybe implement "correct equations of motion" in UI:
<pre>θ̈₁ = ...</pre> maybe too long. But user didn't ask explanation. Could include minimal.

Need maybe canvas should be "chaotic". With initial conditions θ1=2.0, θ2=-2.1 maybe. Need maybe not too high. Use 110° and -80°. It will move.

Need maybe "fading trail" should be of tip. We draw tip trail. Good.

Need maybe "arms" rods. Good.

Need maybe "colorful" trail. Use hue.

Need maybe "self-contained file named index.html in current directory". We'll write.

Let's draft index.html.

Need maybe use "accent-color" CSS for sliders.

Potential issue: `ctx.roundRect` not supported in older browsers? Use custom rounded rect? We can use CSS. Canvas maybe no roundRect needed. If use, could fail? Modern. Avoid.

Need maybe use `PointerEvent`? Not needed.

Need maybe `backdrop-filter` okay.

Let's craft full file.

Potential issue with resize: `canvas.clientWidth` after setting width/height maybe 0? CSS width 100vw height 100vh. Good.

Need maybe update `maxTrail` on input.

Need maybe if trail length changes smaller, trim.

Need maybe `trailInput.addEventListener('input', () => { maxTrail = +trailInput.value; if (trail.length > maxTrail) trail.splice(0, trail.length - maxTrail); });`
Need speed input label display.
HTML:
<label class="control">
<span>Speed <output id="speedValue">1.20×</output></span>
<input id="speed" type="range" min="0.10" max="3.00" step="0.05" value="1.20">
</label>
Trail.
Need maybe `button` IDs.
Need maybe `small` equation.

Need maybe "Tap canvas for a kick" optional.

Need maybe use `aria-live`? not necessary.

Need maybe body:
<div class="hud">
<div class="title">...
<div class="subtitle">Correct Lagrangian equations • RK4 integration</div>
<div class="buttons">...
<div class="controls">...
<div class="stats">...
</div>
<div class="hint">Click/tap canvas to kick the pendulum.</div>
</div>
<canvas id=scene></canvas>

Need maybe no external resources. Good.

Need maybe script at bottom.

Need maybe ensure `document.addEventListener('visibilitychange')` resets last to avoid dt jump:
document.addEventListener('visibilitychange', () => { last = performance.now(); accumulator = 0; });
Need if paused? okay.

Need maybe `canvas.addEventListener('pointerdown', e => { if (e.target !== canvas?) Actually canvas only. Kick:
const kick = 0.35 + Math.random()*0.65;
state.omega1 += (Math.random() - .5)*kick;
state.omega2 += (Math.random() - .5)*kick*1.6;
trail.length=0;
});
Need if user clicks UI not canvas because UI above. Fine.
Need maybe pointerdown on canvas when paused? "kick" maybe unpause? Could not.
Need maybe prevent if event target? no.

Need maybe draw trail after background. Good.

Need maybe if trail array empty and paused after reset, no trail. Draw pendulum.

Need maybe if trail fade and no update when paused. Good.

Need maybe `drawScene` uses `points` computed. Need maybe update readout.

Potential issue with `ctx.setTransform(dpr,0,0,dpr,0,0)` after resize; if not reset each draw, okay. But if canvas width/height reset, context state resets. Good.

Need maybe if CSS canvas width/height 100%, but `clientWidth` includes fractional? Use `window.innerWidth` for draw. Could use `const w = canvas.width / dpr; const h=canvas.height / dpr;` Need dpr variable. Define `let dpr=1;`. Then draw with w/h. Good.
In resize, `dpr = Math.min(window.devicePixelRatio || 1, 2); canvas.width = Math.floor(window.innerWidth*dpr); ...`.
In draw, `const w = canvas.width / dpr; const h = canvas.height / dpr;`.
Good.

Need maybe if dpr changes, resize.

Need maybe use `ctx.clearRect(0,0,w,h)` with transform. Good.

Need maybe if `ctx.globalCompositeOperation='lighter'` and then restore.

Need maybe draw energy? no.

Need maybe "correct equations" maybe formula in comments:
```
// Closed-form Lagrangian accelerations for a double pendulum:
// θ1̈ = ...
```
Good.

Need maybe no external.

Let's write.