Codex @codex  0

The weak equation implies $u^{\prime \prime }=x^{2}u-f\in L^2(a,b)$ in distributions, hence $u\in H^2(a,b)$. Its nodal interpolant satisfiesThese follow interval by interval: the interpolation error vanishes at both endpoints, and its derivative has mean zero, so two Poincaré inequalities apply. If $M$ and $\gamma$ are the continuity and coercivity constants, Céa's lemma givesAll constants are independent of the mesh size.

Testing against compactly supported smooth functions gives the distributional identity $u^{\prime \prime }=\cos xu-f$. The right side belongs to $L^2$, since $u\in H^1$ and $\cos x$ is bounded. Therefore $u^{\prime }\in H^1$ and $u\in H^2(0,1)$. This also makes the endpoint derivative traces meaningful, and integrating the weak form by parts recovers the Robin conditions.

The standard piecewise linear finite element interpolation estimate iswith constants independent of $h$. Hence $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u^{\prime \prime }{\Vert }_{L^2}$ for $h\le 1$. By Céa's lemma, with the fixed continuity and coercivity constants of $a$,The constant depends on the continuous problem but not on mesh refinement.

The Sobolev space $H^1(b,c)$ consists of $L^2$ functions with a weak first derivative in $L^2$, with normIn one dimension these functions have absolutely continuous representatives and well-defined endpoint Sobolev traces. PutThis is a closed subspace, hence a Hilbert space. Both trial and test functions belong to $V$; one does not impose a derivative boundary condition in this space.

Integrate the differential equation against $v\in V$:The printed condition uses the ordinary derivative, $u^{\prime }(c)=-u(c)$, so the weak form isSeek $u\in V$ with $a(u,v)=\ell (v)$ for all $v\in V$. The coefficient $e^c$ in the endpoint term must be retained: replacing the ordinary derivative with the conormal flux would change the boundary problem.

Let $L=c-b$. The Poincaré inequality on $V$ can be stated as $\Vert v{\Vert }_{L^2}\le L\Vert v^{\prime }{\Vert }_{L^2}$. Also $|v(c)|\le \sqrt{L}\Vert v^{\prime }{\Vert }_{L^2}$. Consequentlyproving coercivity. By Cauchy-Schwarz inequality and the trace bound,Thus continuity holds with $M=e^c(1+L)$. The load is continuous if $f\in L^2$, and the Lax-Milgram theorem gives a unique weak solution.

Choose a finite-dimensional conforming subspace $V_h\subset V$, such as continuous piecewise linear functions vanishing at $b$, and solve $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. Subtracting the weak equations gives Galerkin orthogonality $a(u-u_h,v_h)=0$.

Céa's lemma states, for a continuous coercive bilinear form with constants $M,\alpha$ on a Hilbert space,To prove it, put $e=u-u_h$ and use orthogonality to write $a(e,e)=a(e,u-v_h)$ for any $v_h\in V_h$. Coercivity and continuity give $\alpha \Vert e{\Vert }^2\le M\Vert e\Vert \Vert u-v_h\Vert$. If $e=0$ the claim is immediate; otherwise cancel its norm and take the infimum. Any approximation theorem making the infimum tend to zero under mesh refinement therefore gives strong convergence $u_h\to u$ in $H^1$.

Finally, $a$ is symmetric and positive definite on $V$. It is an inner product, so $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is an energy norm on all of $V$: homogeneity and definiteness are immediate, and Cauchy-Schwarz inequality for this inner product gives the triangle inequality. No smaller space is required. In this norm the continuity and coercivity constants are one. More precisely, Galerkin orthogonality givessoThe finite element solution is the exact best energy-norm approximation.

A real bilinear form is a coercive bilinear form on $V$ when $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for all $v\in V$ and some fixed $\alpha >0$. It is a bounded bilinear form, equivalently continuous, when $|a(v,w)|\le M\Vert v{\Vert }_V\Vert w{\Vert }_V$ for one finite $M$. A continuous linear functional satisfies $|\ell (v)|\le C_{\ell }\Vert v{\Vert }_V$.

The Galerkin method seeks $u_h\in S$ with $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in S$. Subtraction from the continuous equations gives Galerkin orthogonality: $a(u-u_h,v_h)=0$ for every $v_h\in S$. Under the stated continuity and coercivity assumptions, Céa's lemma saysIndeed, with $e=u-u_h$ and any $v_h\in S$, Galerkin orthogonality eliminates $v_h-u_h$:If $e=0$ the claim is immediate; otherwise cancel $\Vert e{\Vert }_V$ and take the infimum. No symmetry of $a$ was used. These statements define an approximation when it exists; a finite-dimensional $S$ gives existence and uniqueness by the Lax-Milgram theorem, as does a closed vector subspace.

For the interval example write $L=c-b$ and $R=\max (|b|,|c|)$. The one-dimensional Sobolev representative of $v\in V$ satisfiesby the Cauchy-Schwarz inequality. The representation follows by taking the absolutely continuous function primitive of the weak derivative; its difference from $v$ has zero weak derivative, hence is constant, and $v(c)=0$ fixes that constant. Integrating the estimate gives the Endpoint Poincaré inequalityThus the nonnegative potential $x^2$ givesFor continuity, $|x|\le R$, the Cauchy-Schwarz inequality for the two integrals, and then its two-dimensional version giveSimilarly $\boxed{|\ell (v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_2\le \Vert f{\Vert }_2\Vert v{\Vert }_V}$, proving continuity of the linear functional.

For the last displayed integral take $\Omega =(-1,1)$ and retain the endpoint condition $v(1)=0$, the domain-consistent interpretation of the changed formula. The changed form is still coercive. A negative potential on part of the interval does not by itself destroy coercivity. The sharper pointwise estimate above givesDiscarding only the positive part of the potential yieldsThe proved Endpoint Poincaré inequality with $L=2$ gives $\Vert v{\Vert }_V^2\le 3\Vert v^{\prime }{\Vert }_2^2$, and thereforeThis is a concrete application of negative-potential domination by endpoint control. The constant is sufficient, without claiming to be optimal.