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Varying the scalar and integrating by parts givesbecause the boundary term vanishes by the support assumption. The fundamental lemma of the calculus of variations therefore gives the Klein-Gordon equation

For a variation $y+\epsilon \eta$,Integration by parts givesFor fixed endpoints, $\eta (0)=\eta (x_0)=0$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationA solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditionsThus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the natural boundary conditions for a free endpoint.

Forthe two equations areSet $u=y+z$ and $v=y-z$. ThenThe conditions $y(0)=z(0)=0$ giveand hence the most general solution is

Free conditions at $x_0$ are $y^{\prime }(x_0)=z^{\prime }(x_0)=0$. Adding and subtracting them givesThus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely whenFor those values they form the one-parameter familywhere $C$ is arbitrary. This is the free-endpoint normal mode of a coupled variational functional.

For compactly supported smooth variations $(\varphi ,\psi )$, differentiating at $t=0$ givesThis is the weak formulation. Taking $\psi =0$ or $\varphi =0$ separates its two equations. When $X,Y$ are smooth, integration by parts transfers each derivative from the test function; the fundamental lemma of the calculus of variations then recovers exactly the two pointwise equations in part (a).

At an area critical point the assumed first-variation identity holds for every smooth $\phi$ compactly supported where ${\kappa }_g\ne 0$:The fundamental lemma of the calculus of variations gives
$({\kappa }_g^{-1}{)}^{\prime }=0$ on each component of $\{{\kappa }_g\ne 0\}$. Thus ${\kappa }_g$ is constant there. A nonzero constant component cannot have a boundary point where continuity makes ${\kappa }_g=0$. Since the closed curve is connected, either ${\kappa }_g\equiv 0$ or it is one nonzero constant everywhere. Therefore every area-maximizing curve has constant geodesic curvature.

Write $q_i=a^{ij}{D}_{j}u$. If $u$ is a classical solution, multiply $-{\partial }_i{q}_i=f$ by a smooth function $v$ and apply the divergence theorem. The Neumann boundary condition $q_i{N}_i=0$ removes the boundary term and givesThe density of smooth functions in a Sobolev space and boundedness of the coefficients extend this identity to every $v\in H^1(U)$, so $u$ is a weak solution.

Conversely, take $v$ to be a test function compactly supported in $U$. The weak formulation saysThe fundamental lemma of the calculus of variations gives the equation in $U$. Under the regularity implicit in the stated notion of a classical solution, it holds pointwise. Applying integration by parts again with arbitrary $v\in H^1(U)$ leaveswhere $T$ is the trace operator. Traces of smooth functions can be chosen arbitrarily on the boundary, so the boundary fundamental lemma of the calculus of variations gives $(a^{ij}{\partial }_{j}u)N_i=0$. Thus $u$ is a classical solution.

For $u_0\in L^{\infty }(\mathbb{R})$, call $u\in L^{\infty }({\mathbb{R}}_{+}\times \mathbb{R})$ a weak solution when, for every $\phi \in C_c^1([0,\infty )\times \mathbb{R})$,This follows by multiplying the equation by a test function and applying integration by parts in time and space; $(F\phi {)}_x$ includes the term $F_x\phi$ because the original transport operator is not in divergence form.

If $u$ is $C^1$, test functions supported away from $t=0$ show that $u_t+F{u}_x=g$ as a distributional identity, hence pointwise. Integrating this pointwise equation by parts in the weak identity leavesArbitrary boundary test functions and the fundamental lemma of the calculus of variations give $u(0,x)=u_0(x)$. Thus a $C^1$ weak solution is the unique classical solution from part a.

First choose a test function $v\in C_c^{\infty }(U)$. The weak formulation and integration by parts giveThe fundamental lemma of the calculus of variations implies $-\Delta u=f$ pointwise because $u\in C^2(\overline{U})$ and $f$ is continuous. The Dirichlet boundary condition on ${\Gamma }_1$ already follows from $u\in V$ and continuity of $u$.

For arbitrary $v\in V$, Green's first identity and the interior equation now reduce the weak identity toThe traces of smooth members of $V$ can be chosen freely on compact subsets of ${\Gamma }_2$. Another application of the fundamental lemma of the calculus of variations, now on the boundary, gives ${\partial }_{\nu }u=0$ pointwise on ${\Gamma }_2$. Hence $u$ is a classical solution of the complete mixed boundary value problem.

For every compactly supported $C^1$ test function $\phi$ on $[0,\infty )\times \mathbb{R}$, define a weak solution by the identityThe extra $\phi$ appears because ${\partial }_x(x\phi )=x{\phi }_x+\phi$. This identity is obtained from the linear transport equation by integration by parts in time and space.

Conversely, if $u$ and $u_0$ have the stated $C^1$ regularity, choosing test functions supported away from $t=0$ shows in the distributional sense that $u_t+x{u}_x=0$. Continuity makes the equation pointwise. Integrating that pointwise equation by parts in the displayed identity leavesfor all boundary test functions. The fundamental lemma of the calculus of variations gives $u(0,x)=u_0(x)$, so $u$ is a classical solution.

For an arbitrary smooth variation $u_t=u+t\phi$, differentiation under the integral givesSince $M$ is compact without boundary, integration by parts turns this intoThe fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationor equivalently $-{\Delta }_{g}u=f$ for the Laplace-Beltrami operator.