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Take $L=-d^2/d{x}^2+x$. The Airy equation is $Lu=0$. To discuss positivity of this linear operator, use its homogeneous variation domain $D(L)=\{v\in H^2(0,1):v(0)=0,v^{\prime }(1)=0\}$. Integration by parts givesThe boundary term is zero at both ends. The same calculation with two different functions proves that the associated form is a symmetric bilinear form. Positivity here concerns the homogeneous domain; the condition $u(0)=1$ instead defines an affine set of admissible solutions.

For the variational problem defineand minimizeThe Sobolev trace at zero is meaningful in $H^1$. No derivative boundary value is imposed on this trial space: the Neumann boundary condition will emerge naturally. For $v\in V$, Cauchy-Schwarz inequality and integration giveso $a(v,v)\ge \frac{2}{3}\Vert v{\Vert }_{H^1}^2$. Thus $a$ is a bounded, symmetric, coercive bilinear form on the Hilbert space $V$. The Lax-Milgram theorem says that a bounded coercive bilinear form and a bounded linear functional determine a unique weak solution. Apply it toand set $u=1+v$. This is precisely the first variation condition $a(u,w)=0$ for $J$.

Compactly supported test functions first yield $u^{\prime \prime }=xu$ in the sense of distributional derivatives. Since $xu\in L^2$, the solution is in $H^2$. Integration by parts then leaves $u^{\prime }(1)w(1)=0$ for every $w\in V$, hence $u^{\prime }(1)=0$. The other boundary condition is built into $1+V$. Conversely a solution of this boundary value problem satisfies the first variation condition. Finally,This proves existence, uniqueness and the global minimum characterization, including the natural boundary condition. It is the mixed-boundary Airy energy principle.

Work with real functions and $f\in L^2(0,1)$. To give the variational problem a definite domain, choose the homogeneous Dirichlet boundary conditions used in the next part. The natural form space is the zero-trace Sobolev space $H=H_0^1(0,1)$, with norm $\Vert v{\Vert }_H=\Vert v^{\prime }{\Vert }_2$. The Sobolev trace operator is well defined because one-dimensional $H^1$ functions have absolutely continuous representatives. For example $v(0)={\int }_0^{1}v(x)dx-{\int }_0^1(1-x)v^{\prime }(x)dx$, so its endpoint trace is bounded in the $H^1$ norm; the other endpoint has the analogous estimate. The zero-trace space is closed in the Hilbert space $H^1$, and the Poincaré inequality $\Vert v{\Vert }_2\le {\pi }^{-1}\Vert v^{\prime }{\Vert }_2$ makes the chosen derivative norm equivalent to its full norm. Thus $H$ is a Hilbert space.

Differentiability of $p,q$ implies continuity on the compact interval. Hence there are finite constants $p_{\max },q_{\max }$ and a positive $p_{\min }$ with $p_{\min }\le p\le p_{\max }$ and $0\le q\le q_{\max }$. Define the symmetric bilinear formThe Cauchy-Schwarz inequality and Poincaré inequality give all the estimates required by the Lax-Milgram theorem:That theorem states that a continuous coercive bilinear form on a Hilbert space and a bounded linear functional determine a unique weak solution. Completeness, continuity, coercivity and boundedness have now all been verified. Therefore there is a unique $u\in H$ with $a(u,v)=⟨f,v⟩$ for every $v\in H$.

The energy version of the given variational problem is $I(v)=a(v,v)-2⟨f,v⟩$. Its exact variation isBoundedness of $a$ makes the last term quadratic in $\Vert w{\Vert }_H$, so the Fréchet derivative is $DI(v)[w]=2a(v,w)-2⟨f,w⟩$. Its vanishing for every test function is preciselyTaking compactly supported smooth tests and integrating in the distributional sense shows that this is the Euler-Lagrange equation $-(p{u}^{\prime }{)}^{\prime }+qu=f$. Conversely that weak equation gives the displayed stationary condition. The stationary point is actually the unique global minimizer, becauseEquality occurs only when $v=u$. This proves the variational equivalence, including existence and uniqueness rather than only a formal derivative calculation.

For precision about the literal L2 inner product $⟨Lv,v⟩$, use the flux-domain realization of a one-dimensional elliptic operatorFor such $v$, weak integration by parts and its zero endpoint values give $⟨Lv,v⟩=a(v,v)$. On the larger form space $H$, the same energy is interpreted by the weak operator pairing. The minimizer belongs to $D(L)$ automatically: its weak equation gives $(p{u}^{\prime }{)}^{\prime }=qu-f\in L^2$, while $p{u}^{\prime }\in L^2$. Thus its equation also holds as an equality of $L^2$ functions. This domain does not assume bounded $p^{\prime }$, which mere differentiability would not supply. If $p\in C^1$ one may equivalently use $D(L)=H^2\cap H_0^1$, but that stronger coefficient assumption is not needed for the proof here.

Choose a basis ${\varphi }_1,\dots ,{\varphi }_N$ of a conforming finite element space $V_N\subset H_0^1(0,1)$ and write $y_N={\sum }_j{c}_j{\varphi }_j$. Testing the variational problem with each basis function givesThese are also the equations $\partial J(y_N)/\partial c_i=0$ for the Ritz method. The stiffness matrix is symmetric and a positive-definite matrix: for a nonzero coefficient vector $c$, linear independence gives $v_N=\sum c_j{\varphi }_j\ne 0$, andThus there is a unique coefficient vector.

Subtracting the exact and discrete variational problems proves Galerkin orthogonality, $a(y-y_N,v_N)=0$ for every $v_N\in V_N$. In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, the Pythagorean identity yieldsTherefore the Ritz method is the best approximation in the energy norm. The Céa lemma also gives $\Vert y-y_N{\Vert }_V\le (1+{\pi }^{-2}){\inf }_{v_N\in V_N}\Vert y-v_N{\Vert }_V$, so any dense sequence of conforming trial spaces converges.

Use the real Hilbert space $V=H_0^1(0,1)$, the zero-boundary Sobolev space, with norm $\Vert v{\Vert }_V=\Vert v^{\prime }{\Vert }_{L^2}$. The Poincaré inequality makes this equivalent to its usual Sobolev space norm. Define the symmetric bilinear form and continuous linear functionalThe Cauchy-Schwarz inequality and the sharp interval Poincaré inequality implyThus $a$ is a bounded bilinear form and a coercive bilinear form, and $\ell$ is bounded because $f$ is square-integrable. The Lax-Milgram theorem supplies a unique weak solution:This is obtained from the differential equation by integration by parts; the test functions have zero boundary trace. Conversely, the weak solution satisfies $y^{\prime \prime }=xy-f$ in the sense of distributions. Since $xy-f\in L^2$, it belongs to $H^2(0,1)\cap H_0^1(0,1)$ and solves the equation almost everywhere.

Equivalently, the Ritz method minimizesover $V$. Indeed $J(y+w)-J(y)=a(w,w)/2$, since the mixed term is $a(y,w)-\ell (w)=0$. This proves existence and uniqueness of the minimizer directly from the weak solution, as well as its equivalence to the variational problem.