= Solution
Take a <smooth> variation $u+\varepsilon v$ with $v=0$ on the <boundary>. Differentiating under the <integral> gives the <first variation>
$$
\delta I[u](v)=\iint_R(F_uv+F_{u_\xi}v_\xi+F_{u_\zeta}v_\zeta)\,d\xi\,d\zeta.
$$
<Integration by parts> removes the <derivatives> of $v$, and the <boundary> contribution vanishes. The <fundamental lemma of the calculus of variations> therefore gives
$$
\boxed{F_u-\partial_\xi F_{u_\xi}-\partial_\zeta F_{u_\zeta}=0.}
$$
For a regular <constraint> $J[u]=c$, the <Lagrange multiplier> rule says that an extremal satisfies $\delta I=\lambda\delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a <linear functional> on the admissible variations. For the quadratic <constraint> below this holds at every nonzero admissible $u$; a zero <constraint> leaves only $u=0$.
In <polar coordinates> the constrained integrand is
$$
\mathcal L=r u_r^2+\frac1r u_\theta^2-\lambda r u^2.
$$
Its Euler–Lagrange equation is $-2\lambda r u-\partial_r(2ru_r)-\partial_\theta(2u_\theta/r)=0$, or
$$
\boxed{u_{rr}+\frac1r u_r+\frac1{r^2}u_{\theta\theta}+\lambda u=0,\qquad u(a,\theta)=0.}
$$
The <field> is single-valued and <periodic> in $\theta$. <Substitution> $u=f(r)\cos(n\theta)$ with $n\in\{0,1,2,\ldots\}$ gives
$$
\boxed{f''+\frac1r f'+\left(\lambda-\frac{n^2}{r^2}\right)f=0,\qquad f(a)=0.}
$$
The <centre> is part of the disk, so one must also select the regular radial solution: for $n\geq1$, smoothness gives $f(r)=O(r^n)$ and the regular <derivative> behaviour; for $n=0$, $f$ is <bounded> and $f'(0)=0$. A singular solution at zero does not define an admissible <smooth> <field> on the disk.
For fixed $n$, let $c_n=\int_0^{2\pi}\cos^2(n\theta)\,d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\geq1$. The angular <derivative> term has <integral> $n^2c_n$, including zero when $n=0$. Consequently
$$
\widehat I[f]=c_n\int_0^a\left(rf'^2+\frac{n^2}{r}f^2\right)\,dr,\qquad
\widehat J[f]=c_n\int_0^a rf^2\,dr.
$$
The one-dimensional Euler–Lagrange equation for $\widehat I-\lambda\widehat J$ is
$$
-(rf')'+\frac{n^2}{r}f=\lambda rf,
$$
exactly the same radial equation. The common factor $c_n$ cancels.
For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result is
$$
(\lambda_1-\lambda_2)\int_0^a rf_1f_2\,dr
=\left[r(f_1f_2'-f_1'f_2)\right]_0^a=0.
$$
At $a$, both functions vanish; at zero, regularity makes the <boundary> term tend to zero. Since the multipliers are distinct,
$$
\boxed{\int_0^a rf_1f_2\,dr=0.}
$$
This is weighted Sturm–Liouville <orthogonality> with weight $r$.
Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yield
$$
\int_0^a\left(rf'^2+\frac{n^2}{r}f^2\right)\,dr=\lambda\int_0^a rf^2\,dr.
$$
Multiplication by $c_n$ proves
$$
\boxed{\widehat I[f_i]-\lambda_i\widehat J[f_i]=0.}
$$
For any nonzero admissible solution the <energy> on the left before subtraction is positive, so its multiplier is the positive <Rayleigh quotient> $\lambda=\widehat I/\widehat J$.
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