Exercises — Euler-Lagrange equation — derivation

WHAT: treat the three slots $x$, $y$, $y^{\prime }$ as independent labels and differentiate slot by slot. WHY: the partial $\partial /\partial y$ means "freeze $x$ and $y^{\prime }$, they are just constants for this step".

• $\frac{\partial L}{\partial y}$: only the $y^2$ term contains $y$, so $\frac{\partial L}{\partial y}=2y$.
• $\frac{\partial L}{\partial y^{\prime }}$: only the $x{y}^{\prime 2}$ term contains $y^{\prime }$; $x$ is frozen, so $\frac{\partial L}{\partial y^{\prime }}=2x{y}^{\prime }$. Answer: $\boxed{\frac{\partial L}{\partial y}=2y,\frac{\partial L}{\partial y^{\prime }}=2x{y}^{\prime }}$

WHAT: hunt for a standalone $x$ that is not hidden inside $y$ or $y^{\prime }$.

• (a) $\sqrt{1+y^{\prime 2}}$ — no lone $x$. ✅ Beltrami applies.
• (b) $x{y}^{\prime 2}$ — the factor $x$ sits there explicitly. ❌ Beltrami does not apply.
• (c) $\sqrt{(1+y^{\prime 2})/y}$ — only $y$ and $y^{\prime }$ appear. ✅ Beltrami applies. WHY it matters: Beltrami is the statement "if the rules don't change as you move along $x$, then a certain combination is conserved". A visible $x$ means the rules do change with position, so nothing is conserved.

WHAT / WHY: $\frac{\partial L}{\partial y}=0$ (no $y$ appears), so E–L collapses to $\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}=0$. If the derivative of a thing is zero, the thing is constant.

• $\frac{\partial L}{\partial y^{\prime }}=\frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=c$.
• Squaring and solving — mind the branches. Squaring gives $y^{\prime 2}=c^2(1+y^{\prime 2})$, hence $y^{\prime 2}=\frac{c^2}{1-c^2}$. Taking the root introduces a $\pm$: $y^{\prime }=\pm \frac{c}{\sqrt{1-c^2}}$. WHY the sign is not a problem here: the right side is a fixed constant (call the whole thing $m$, including its sign), because $c$ is a constant. Whichever branch we are on, $y^{\prime }$ never changes value along the curve — squaring cannot create a curve whose slope flips, since $y^{\prime }$ is pinned to one constant. So $y^{\prime }=m=\text{const}$, and the sign of $m$ is fixed later purely by the endpoints (here it comes out positive).
• WHAT IT LOOKS LIKE (Fig. s01): a constant slope means the extremal is a straight ramp. In the figure the orange line $y=2x$ is that extremal; a teal dashed competitor $y+\epsilon \eta$ bulges away and is visibly longer, and the plum dots are the two fixed endpoints the wiggle $\eta$ must not disturb ($\eta =0$ at both ends).
• So $y=mx+b$. Endpoints: $y(0)=0\Rightarrow b=0$; $y(1)=2\Rightarrow m=2$ (positive branch selected by the data). Answer: $\boxed{y=2x}$.

WHAT: apply E–L directly. $\frac{\partial L}{\partial y}=2y$, $\frac{\partial L}{\partial y^{\prime }}=2y^{\prime }$.

• E–L: $2y-\frac{d}{dx}(2y^{\prime })=0\Rightarrow y^{\prime \prime }-y=0$.
• WHY the solution is built from $e^x$ and $e^{-x}$ (grounded, not asserted): the equation $y^{\prime \prime }=y$ says "the second derivative gives back the original function". Ask: which functions are their own second derivative? The exponential $e^{kx}$ has $\frac{d^2}{d{x}^2}{e}^{kx}=k^2{e}^{kx}$, so $y^{\prime \prime }=y$ forces $k^2=1$, i.e. $k=+1$ or $k=-1$. That is why exactly two building blocks appear, $e^x$ (a curve that keeps steepening as it rises) and $e^{-x}$ (a curve that decays), and every solution is a blend $y=P{e}^x+Q{e}^{-x}$.
• WHY rewrite as $\sinh ,\cosh$: the two natural symmetric blends are $\cosh x=\frac{1}{2}(e^x+e^{-x})$ (even, value $1$ at $x=0$) and $\sinh x=\frac{1}{2}(e^x-e^{-x})$ (odd, value $0$ at $x=0$). Since our left endpoint demands $y(0)=0$, the $\cosh$ piece (which is $1$ at $0$) must be switched off — so $\sinh$ is the tailor-made basis here.
• WHAT IT LOOKS LIKE (Fig. s02): the figure plots $e^x$ (teal) and $e^{-x}$ (plum), and their difference $\sinh x$ (orange) — notice the orange curve threads through the origin ($y(0)=0$) exactly as the boundary condition requires, while its gentle upward bow is the extremal shape.
• General solution $y=A\sinh x+B\cosh x$. $y(0)=0\Rightarrow B=0$. $y(1)=1\Rightarrow A=\frac{1}{\sinh 1}$. Answer: $\boxed{y=\frac{\sinh x}{\sinh 1}}$.

WHAT: $\partial L/\partial x=0$, so Beltrami gives $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C$.

• $\frac{\partial L}{\partial y^{\prime }}=\sqrt{y}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}$.
• Beltrami: $\sqrt{y}\sqrt{1+y^{\prime 2}}-\sqrt{y}\frac{y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\sqrt{y}\cdot \frac{(1+y^{\prime 2})-y^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\frac{\sqrt{y}}{\sqrt{1+y^{\prime 2}}}=C$.
• WHY the middle step: common denominator, and $(1+y^{\prime 2})-y^{\prime 2}=1$ — the $y^{\prime 2}$ terms annihilate, which is the whole point of Beltrami. Answer (first-order ODE): $\boxed{\frac{y}{1+y^{\prime 2}}=C^2}$.

WHAT: Beltrami produces a valid first integral of E–L whenever $\partial L/\partial x=0$. So consistency is guaranteed by the derivation of Beltrami itself — we re-trace it to be sure.

• Beltrami was derived from $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}$.
• Here $\partial L/\partial x=0$, so $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=C$ (constant). This is an integrated form of E–L; differentiating it once returns $y^{\prime }\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)=0$.
• WHY consistent — and the edge case $y^{\prime }=0$ made explicit. The product $y^{\prime }\cdot (\text{E–L expression})=0$ gives two ways to satisfy it: either the E–L bracket vanishes, or $y^{\prime }=0$ (a horizontal point/segment). At an isolated turning point where $y^{\prime }=0$ momentarily, the equation is satisfied trivially by the $y^{\prime }=0$ factor, so at that single instant Beltrami tells you nothing new about the E–L bracket — but E–L is a continuous condition, and everywhere the curve is genuinely sliding ($y^{\prime }\ne 0$) the bracket is forced to zero; by continuity the bracket stays zero through the isolated $y^{\prime }=0$ point too. The only genuinely degenerate case is a curve that is horizontal on a whole interval ($y^{\prime }\equiv 0$ there): then Beltrami degenerates to $L=C$ and carries no E–L information on that stretch, and you must fall back on the full E–L equation directly. For the brachistochrone the curve is strictly descending except at its start, so this degenerate stretch never occurs.
• Now do the algebra for this $L$ explicitly so nothing is hidden. Write $L=\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{y}}$.
  • Step 1 — the momentum term: $\frac{\partial L}{\partial y^{\prime }}=\frac{1}{\sqrt{y}}\cdot \frac{y^{\prime }}{\sqrt{1+y^{\prime 2}}}=\frac{y^{\prime }}{\sqrt{y}\sqrt{1+y^{\prime 2}}}$. WHY: chain rule on $\sqrt{1+y^{\prime 2}}$, with $1/\sqrt{y}$ a frozen constant for this partial.
  • Step 2 — form $L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}$: $\frac{\sqrt{1+y^{\prime 2}}}{\sqrt{y}}-\frac{y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{(1+y^{\prime 2})-y^{\prime 2}}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=\frac{1}{\sqrt{y}\sqrt{1+y^{\prime 2}}}.$ WHY: common denominator $\sqrt{y}\sqrt{1+y^{\prime 2}}$, then $(1+y^{\prime 2})-y^{\prime 2}=1$ — the same $y^{\prime 2}$-cancellation that makes Beltrami useful.
  • Step 3 — set equal to the Beltrami constant $C$ and square: $\frac{1}{\sqrt{y}\sqrt{1+y^{\prime 2}}}=C\Rightarrow y(1+y^{\prime 2})=\frac{1}{C^2}=\text{const}$. Answer: the first integral $y(1+y^{\prime 2})=\text{const}$ is a legitimate consequence of E–L; no contradiction.

WHAT: free right endpoint means the boundary term ${\left[\frac{\partial L}{\partial y^{\prime }}\eta \right]}_0^1$ from Integration by Parts does not vanish automatically at $x=1$; it forces a natural boundary condition.

• Interior E–L: $\frac{\partial L}{\partial y}=1$, $\frac{\partial L}{\partial y^{\prime }}=y^{\prime }$. So $1-\frac{d}{dx}(y^{\prime })=0\Rightarrow y^{\prime \prime }=1$.
• Integrate: $y=\frac{1}{2}{x}^2+Ax+B$. Fixed left end $y(0)=0\Rightarrow B=0$.
• Natural BC at $x=1$: ${\frac{\partial L}{\partial y^{\prime }}|}_{x=1}=y^{\prime }(1)=0$. WHY: with $\eta (1)$ free, the only way the boundary term dies is $\partial L/\partial y^{\prime }=0$ there.
• $y^{\prime }=x+A$, so $y^{\prime }(1)=1+A=0\Rightarrow A=-1$. Answer: $\boxed{y=\frac{1}{2}{x}^2-x}$, with $y(1)=-\frac{1}{2}$.

WHAT: with several unknown functions $q_i(t)$, there is one E–L equation per function — you wiggle each independently.

• For $q_1$: $\frac{\partial L}{\partial q_1}=-q_1$, $\frac{\partial L}{\partial {\dot{q}}_1}={\dot{q}}_1$. E–L: $-q_1-\frac{d}{dt}{\dot{q}}_1=0\Rightarrow {\ddot{q}}_1=-q_1$.
• By identical algebra: ${\ddot{q}}_2=-q_2$.
• WHY / WHAT IT LOOKS LIKE: $\ddot{q}=-q$ is simple harmonic motion; each coordinate oscillates like $\cos t,\sin t$. Two independent oscillators. Answer: $\boxed{{\ddot{q}}_1=-q_1,{\ddot{q}}_2=-q_2}$ — two uncoupled simple-harmonic oscillators.

WHAT: compute the total derivative of the candidate conserved quantity and show it is zero.

• Total derivative of $L$: $\frac{dL}{dx}=\frac{\partial L}{\partial x}+\frac{\partial L}{\partial y}{y}^{\prime }+\frac{\partial L}{\partial y^{\prime }}{y}^{\prime \prime }$.
• Now compute $\frac{d}{dx}\left(y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=y^{\prime \prime }\frac{\partial L}{\partial y^{\prime }}+y^{\prime }\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}$.
• Subtract: $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}+y^{\prime }\underset{=0\text{by E–L}}{\underbrace{\left(\frac{\partial L}{\partial y}-\frac{d}{dx}\frac{\partial L}{\partial y^{\prime }}\right)}}$.
• WHY the bracket dies: that bracket is the Euler–Lagrange expression, zero on any extremal.
• So $\frac{d}{dx}\left(L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}\right)=\frac{\partial L}{\partial x}$. If $\partial L/\partial x=0$, the left side is a constant. Answer: $\boxed{L-y^{\prime }\frac{\partial L}{\partial y^{\prime }}=\text{const}}$ (see Beltrami Identity).

WHAT: $L=y\sqrt{1+y^{\prime 2}}$, no explicit $x$ → Beltrami.

• $\frac{\partial L}{\partial y^{\prime }}=\frac{y{y}^{\prime }}{\sqrt{1+y^{\prime 2}}}$.
• Beltrami: $y\sqrt{1+y^{\prime 2}}-\frac{y{y}^{\prime 2}}{\sqrt{1+y^{\prime 2}}}=\frac{y}{\sqrt{1+y^{\prime 2}}}=a$. WHY: the $y^{\prime 2}$ terms cancel exactly, as in L2.3. Where does $a$ come from? It is simply the Beltrami constant — the fixed number the left side equals. In L2.3 we called it $C$; here we rename that same constant $a$ because, as we will see two lines down, $a$ turns out to be a length (the neck radius), and calling it $a$ makes that geometric meaning obvious. So $a$ is defined as $a:=\frac{y}{\sqrt{1+y^{\prime 2}}}$, the conserved Beltrami value along the curve.
• Rearrange: $\sqrt{1+y^{\prime 2}}=\frac{y}{a}\Rightarrow y^{\prime 2}=\frac{y^2}{a^2}-1\Rightarrow y^{\prime }=\pm \sqrt{\frac{y^2}{a^2}-1}$.
• Integrate — the key step, shown in full. Separate variables: $\frac{dy}{\sqrt{y^2/a^2-1}}=\pm dx$. Multiply top and bottom by $a$: $\frac{ady}{\sqrt{y^2-a^2}}=\pm dx$. WHY this integral: the shape $\frac{1}{\sqrt{y^2-a^2}}$ is the tell-tale signature of the inverse hyperbolic cosine, because $\frac{d}{dy}\mathrm{arcosh}(y/a)=\frac{1}{\sqrt{y^2-a^2}}$. Integrating both sides: $a\mathrm{arcosh}(y/a)=\pm (x-b)$, where $b$ is the constant of integration. Solving for $y$: $\mathrm{arcosh}(y/a)=\pm \frac{x-b}{a}$, and since $\cosh$ is even the $\pm$ is absorbed, giving $\boxed{y=a\cosh \left(\frac{x-b}{a}\right)}$.
• Which branch? Both are legitimate — and we need both. The $+$ branch describes the film rising away from the throat (increasing $y$), the $-$ branch describes it falling toward the throat. WHY it does not change the answer: $y=a\cosh \left(\frac{x-b}{a}\right)$ has $y^{\prime }=\sinh \left(\frac{x-b}{a}\right)$, which is negative left of the throat $x=b$ (the $-$ branch) and positive right of it (the $+$ branch). A single symmetric catenary automatically stitches both branches together, meeting at the throat where $y^{\prime }=0$. So picking "the positive branch" is really just choosing to describe the right half.
• WHAT IT LOOKS LIKE (Fig. s03): the orange curve is that catenary; the plum dot at the bottom is the throat where $y^{\prime }=0$ and $y=a$ (here $a=1$); the two teal vertical bars are the rings the film must reach at $x=\pm 1$. Revolving this profile about the horizontal axis sweeps out the catenoid soap film.
• With $a=1,b=0$: $y=\cosh x$. Check endpoints: $y(\pm 1)=\cosh 1\approx 1.5431$. ✓ Answer: $\boxed{y=a\cosh \left(\frac{x-b}{a}\right)}$; symmetric case $y=\cosh x$.

WHAT: two symmetries → two conserved quantities.

• Time-translation ($\partial L/\partial t=0$): Beltrami-in-$t$ gives $E={\dot{q}}_1\frac{\partial L}{\partial {\dot{q}}_1}+{\dot{q}}_2\frac{\partial L}{\partial {\dot{q}}_2}-L=\frac{1}{2}({\dot{q}}_1^2+{\dot{q}}_2^2)+\frac{1}{2}(q_1^2+q_2^2)$ = total energy, conserved.
• Rotational symmetry in the $(q_1,q_2)$ plane: Noether's Theorem gives the conserved angular momentum $\ell =q_1{\dot{q}}_2-q_2{\dot{q}}_1$.
• WHY: every continuous symmetry of the action yields one conserved quantity — that is precisely Noether's theorem, of which Beltrami (for time symmetry) is the special case. Answer: energy $E=\frac{1}{2}({\dot{q}}_1^2+{\dot{q}}_2^2+q_1^2+q_2^2)$ and angular momentum $\ell =q_1{\dot{q}}_2-q_2{\dot{q}}_1$.

Beltrami applies when :: $L$ has no explicit $x$ ($\partial L/\partial x=0$). A free endpoint replaces a fixed height with :: the natural boundary condition $\partial L/\partial y^{\prime }=0$ there. $n$ unknown functions give how many E–L equations :: exactly $n$, one per function. The catenoid neck radius equals :: the Beltrami constant $a$, where $y^{\prime }=0$.