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In conformal gauge, ${\gamma }_{\mu \nu }=e^{2\omega }\mathrm{diag}(-1,1)$. The Weyl invariance of the Polyakov action removes $\omega$, leavingIts complete variation isFix the string embedding map on the initial and final time slices. The interior Euler-Lagrange field equation is the wave equationAt each spatial endpoint the remaining variation vanishes under either Dirichlet boundary conditions, $\delta X^m=0$, or Neumann boundary conditions, $X^{m\prime }=0$, independently for each target-space component. Thus a fixed endpoint or a free endpoint makes the boundary contribution vanish. Fixing a spatial point throughout time imposes Dirichlet boundary conditions on its spatial coordinates, while the time coordinate may retain Neumann boundary conditions.

The independent metric tensor equation must also be retained after gauge fixing. Its Virasoro constraints areTogether with the wave equation, these ensure equivalence to the Nambu–Goto action; varying only the already fixed metric would lose these Virasoro constraints.

Let $T$ be the $M\times M$ tridiagonal matrix with diagonal two and adjacent entries minus one, incorporating zero Dirichlet boundary conditions. The actual PDF scheme isAn orthonormal basis of eigenvectors of $T$ has components $\sqrt{2/(M+1)}\sin (jm\pi /(M+1))$, and substitution givesThey form an orthonormal basis by the spectral theorem for real symmetric matrices. Hence $I+\mu T$ is invertible for every $\mu >0$, and the amplification eigenvalues are $g_j=(1+\mu {\lambda }_j{)}^{-1}\in (0,1)$. For the discrete L2 norm $\Vert U{\Vert }_h^2=h{\sum }_m|U_m{|}^2$,This bound has constant one independent of time step and mesh, which is the required stability of a numerical method. Therefore every positive Courant number is stable. If a forcing or local-defect sequence is added, the discrete variation-of-constants formula gives a bound by the initial norm plus the sum of those perturbation norms; this also justifies the global error conclusion in part (a).

For a sufficiently regular solution with zero Dirichlet boundary conditions, differentiate its squared L2 norm. Since $u_t=-i(\Delta u-Vu)$,Integration by parts and the zero boundary trace give $\int \overline{u}\Delta u=-\int |\nabla u{|}^2$. The potential contribution is real because $V$ is real. The quantity in parentheses is consequently real, and its product with $-i$ has zero real part. HenceThis is the energy method for a skew-adjoint evolution. For general L2 space data, the same result follows from the strongly continuous unitary group generated by a self-adjoint realization of $\Delta -V$. A sufficient assumption is a bounded real potential with the Dirichlet Laplacian; arbitrary real potentials require appropriate domain and self-adjointness assumptions. The differentiated proof presumes the regularity needed for the displayed integrals, rather than asserting it for every pointwise real function $V$.

The starting point of a conforming finite element method is a weak formulation in a Hilbert space $X$. For example, a homogeneous Dirichlet boundary condition on a bounded Lipschitz domain gives $X=H_0^1(\Omega )$. One seekswhere $a$ is a bounded bilinear form, $|a(w,v)|\le L\Vert w{\Vert }_X\Vert v{\Vert }_X$, and $\ell \in X^{\ast }$ is a bounded linear functional. A conforming finite element space $X_h\subset X$ is finite dimensional and is typically built from functions that are polynomial on each element of a finite element mesh. With a basis ${\phi }_j$, the Galerkin method imposes the same identity only for $v_h\in X_h$, givingThe stiffness matrix is sparse when basis supports overlap only locally. Local element integrals assemble its entries; the mass matrix represents an $L^2$ term. This separates the choice of trial functions from the variational principle determining their coefficients.

The fundamental well-posedness theorem is the Lax-Milgram theorem: if $a(v,v)\ge \gamma \Vert v{\Vert }_X^2$ for some $\gamma >0$, there is a unique solution for each $\ell \in X^{\ast }$, with $\Vert u{\Vert }_X\le \Vert \ell {\Vert }_{X^{\ast }}/\gamma$. The same theorem on $X_h$ gives a unique discrete solution. The constants must be independent of the mesh for uniform stability of a numerical method. Boundary conditions are built into the trial space when they are essential constraints; flux conditions arise through boundary terms in integration by parts and are natural constraints in the weak formulation.

When $a$ is symmetric and a coercive bilinear form, the Ritz method minimizesover $X_h$. Its first variation is exactly the Galerkin method, so Ritz minimization and Galerkin testing coincide for symmetric coercive problems. The energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ turns the approximation into an orthogonal projection. Indeed, subtracting the exact and discrete weak equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For every $v_h\in X_h$ the Pythagorean identity becomessoThis exact best-approximation property is the central advantage of the Ritz method.

Symmetry is unnecessary for the Galerkin method. For any bounded, coercive form, the Céa lemma givesTo prove it, put $e=u-u_h$. Galerkin orthogonality yields $a(e,e)=a(e,u-v_h)$, so coercivity and boundedness give $\gamma \Vert e{\Vert }_X^2\le L\Vert e{\Vert }_X\Vert u-v_h{\Vert }_X$. Divide when $e\ne 0$ and take the infimum. Thus approximation capability plus uniform coercivity gives convergence. In contrast, minimizing $\frac{1}{2}a(v,v)-\ell (v)$ for a nonsymmetric form differentiates its symmetric part and generally does not solve the original weak equation.

A concrete symmetric example is the one-dimensional reaction-diffusion problem from Question 5. Continuous piecewise-linear hat functions give the diffusion stiffness matrix $S$ with diagonal $2/h$ and neighbouring entries $-1/h$, and the mass matrix $M$ with diagonal $2h/3$ and neighbouring entries $h/6$. Therefore $K=S+M$ and $F_i=i{h}^2$. Its positive-definite matrix is the coordinate form of the continuous energy. The exact smooth solution $u(x)=x-\sinh (x)/\sinh 1$ makes this an explicit test of the method, not just an abstract existence result.

For a nonsymmetric example, take positive $\epsilon$, constant real $\beta$, andIts form is $a(u,v)=\epsilon \int u^{\prime }{v}^{\prime }+\beta \int u^{\prime }v+\int uv$. Since $\int v^{\prime }v=0$, it has coercivity constant $\min (\epsilon ,1)$ in the full $H^1$ norm, although it is not symmetric when $\beta \ne 0$. The Galerkin method remains well posed. On the uniform hat basis, $K=\epsilon S+\beta C+M$, where $C_{i,i+1}=1/2$, $C_{i,i-1}=-1/2$ and $C_{ii}=0$. The skew-symmetric matrix $C$ cancels from $c^{T}Kc$, proving nonsingularity. A Ritz energy would omit precisely this advection contribution. Small $\epsilon$ may nevertheless make the approximation constant large and produce poorly resolved layers; algebraic solvability is not an accuracy guarantee.

Quantitative convergence uses a finite element interpolation estimate. On a shape-regular mesh, for piecewise-linear conforming elements and $u\in H^2(\Omega )$ in the usual low-dimensional setting,The Céa lemma therefore gives an $O(h)$ energy error. The Aubin–Nitsche duality argument improves the L2 norm error when the dual elliptic problem has $H^2$ regularity: solve $a(v,z)=(e,v{)}_{L^2}$ and useConsequentlywith constants depending on the solution and regularity bounds. The second estimate is conditional on dual regularity; corners or insufficient data regularity can reduce the rate.

The same Galerkin method handles evolution equations. For the homogeneous heat equation, the semidiscrete finite element heat equation is $M\dot{c}+Sc=0$. Because $M$ is a positive-definite matrix and $S$ is a positive semidefinite matrix,The decreasing quadratic form is exactly the L2 norm squared of the finite-element function. This separates spatial stability of a numerical method from the subsequent time integrator, just as in the Schrödinger equation example. Conformity, coercivity and approximation estimates together explain convergence.

For an axisymmetric separated solution of the Laplace equation whose angular factor is the Legendre polynomial $P_n(\cos \theta )$, the radial ordinary differential equation has the two powers $r^n$ and $r^{-n-1}$. Thusfor constants $A,B$.

Write $x=\cos \theta$. Since $3\cos (2\theta )=6x^2-3=4P_2(x)-P_0(x)$, only the spherical harmonics of degrees zero and two occur. The degree-zero radial coefficient satisfying its two Dirichlet boundary conditions is $1-2/r$. For degree two, write $C{r}^2+D{r}^{-3}$; thenso $C=-4/31$ and $D=128/31$. By linearity and uniqueness of this boundary value problem,Direct substitution at $r=1,2$ verifies both boundary values.

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

Write $h=\Delta x$ and set $u_0=u_{M+1}=0$. In the discrete L2 norm $\Vert u{\Vert }_h^2=h{\sum }_{m=1}^M|u_m{|}^2$, the energy method givesTo obtain this identity, shift the indices in the centered second-difference sum. The centered first-difference matrix is skew-symmetric, so its contribution has zero real part even for complex grid values. The homogeneous Dirichlet boundary conditions remove the endpoint terms.

Therefore the semidiscretization is stable for every real $\alpha$ and every positive $h$, with the mesh-independent bound $\Vert u(t){\Vert }_h\le \Vert u(0){\Vert }_h$. No restriction on $|\alpha |h$ is needed for this energy estimate.

Consider a bounded Lipschitz domain $\Omega$ and homogeneous Dirichlet boundary conditions forAssume $A$ is a bounded real symmetric matrix field with ${\xi }^{T}A(x)\xi \ge a_0|\xi {|}^2$ uniformly, $c\ge 0$ is bounded, and $f$ defines a bounded linear functional on $V=H_0^1(\Omega )$, the zero-boundary Sobolev space. The weak formulation isThe Cauchy-Schwarz inequality gives continuity $|a(u,v)|\le M\Vert u{\Vert }_V\Vert v{\Vert }_V$. Uniform ellipticity and the Poincaré inequality give a coercive bilinear form, $a(v,v)\ge \alpha \Vert v{\Vert }_V^2$. The Lax-Milgram theorem therefore supplies a unique weak solution, with $\Vert u{\Vert }_V\le \Vert \ell {\Vert }_{V^{\prime }}/\alpha$. Its proof represents $a$ by a bounded linear operator $T$ using the Riesz representation theorem; coercivity gives an injective operator with closed image, and a zero orthogonal complement makes that image all of $V$. This yields existence, uniqueness and the displayed estimate.

For a conforming finite element space $V_h\subseteq V$, the Galerkin method finds $u_h\in V_h$ such that $a(u_h,v_h)=\ell (v_h)$ for every $v_h\in V_h$. With a basis ${\varphi }_1,\dots ,{\varphi }_N$, this is the linear system $KU=F$, where the stiffness matrix is $K_{ij}=a({\varphi }_j,{\varphi }_i)$ and $F_i=\ell ({\varphi }_i)$. Its positive-definite matrix property proves unique discrete solvability. Local support gives a sparse matrix through elementwise assembly.

For the symmetric problem, the Ritz method minimizes $J(v)=a(v,v)/2-\ell (v)$ over $V_h$. Differentiating in every direction $v_h$ gives precisely the Galerkin method equations. Conversely, if $u_h$ solves them, $J(u_h+w_h)-J(u_h)=a(w_h,w_h)/2$, proving it is the unique minimum. The same argument over $V$ gives $J(v)-J(u)=\Vert v-u{\Vert }_a^2/2$, where $\Vert v{\Vert }_a=\sqrt{a(v,v)}$ is the energy norm. Thus the Ritz method chooses the best trial function in that norm.

Subtracting the continuous and discrete equations gives Galerkin orthogonality, $a(u-u_h,v_h)=0$. For arbitrary $v_h\in V_h$ and $e=u-u_h$, it follows that $a(e,e)=a(e,u-v_h)$. Coercivity and continuity prove the Céa lemma:In the symmetric energy norm, the Cauchy-Schwarz inequality improves the constant to one. Equivalently, Galerkin orthogonality gives the Pythagorean identity $\Vert u-v_h{\Vert }_a^2=\Vert u-u_h{\Vert }_a^2+\Vert u_h-v_h{\Vert }_a^2$. These estimates establish stability and show that approximation properties of the trial space determine convergence.

For continuous piecewise-linear elements on a shape-regular mesh in dimensions at most three, a finite element interpolation estimate gives $\Vert u-I_{h}u{\Vert }_{H^1}\le Ch\Vert u{\Vert }_{H^2}$ when $u\in H^2(\Omega )$. Combining it with the Céa lemma yields $O(h)$ energy norm error. More generally, degree-$p$ elements give $O(h^p)$ error in $H^1$ under the corresponding regularity and approximation assumptions. Mere mesh refinement cannot supply an order whose required solution regularity is absent.

An additional order in the L2 norm follows under a dual elliptic regularity assumption. Solve $a(v,z)=(e,v{)}_{L^2}$ and assume $\Vert z{\Vert }_{H^2}\le C\Vert e{\Vert }_{L^2}$. The Aubin–Nitsche duality argument uses Galerkin orthogonality to obtainTherefore $\Vert e{\Vert }_{L^2}\le Ch\Vert e{\Vert }_{H^1}$, giving $O(h^2)$ for the piecewise-linear case with $u\in H^2$. This improvement explicitly requires regularity of the dual problem.

As a concrete example, solve $-u^{\prime \prime }=1$ on $(0,1)$ with zero endpoint values. On a uniform mesh of spacing $h$, the piecewise-linear hat functions give $K_{jj}=2/h$, $K_{j,j\pm 1}=-1/h$ and $F_j=h$. The discrete equations are $2U_j-U_{j-1}-U_{j+1}=h^2$, whose solution is $U_j=x_j(1-x_j)/2$. The exact solution is $u(x)=x(1-x)/2$, so $u_h$ is its piecewise-linear nodal interpolant. On each interval $[x_j,x_{j+1}]$,Integrating the squared error and squared derivative error gives the explicit rateswhere the $H_0^1$ norm here is $\Vert v^{\prime }{\Vert }_{L^2}$.

For nonsymmetric coercive problems, the Galerkin method, Galerkin orthogonality and the Céa lemma still apply, but the symmetric quadratic minimization interpretation of the Ritz method generally does not. Nonconforming trial spaces, inexact integration and noncoercive equations need additional arguments beyond the conforming coercive theory proved here.

Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous Robin boundary conditionFor $0<\lambda <1$, Green's first identity givesThe left side is nonnegative, so both sides vanish and $w=0$. For $\lambda =0$, the Dirichlet boundary condition $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\le \lambda <1$.

For a regular spherically symmetric solution, the Laplacian equation isRegularity at the origin gives $u^{\prime }=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yieldsAt $\lambda =1$ the Neumann boundary condition would require $u^{\prime }(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require ${\int }_{V}6dV=0$.

For a normalized wavefunction obeying the infinite-wall Dirichlet boundary condition, integration by parts gives the energy expectationThus every energy eigenvalue is nonnegative.

For $0<E<U_0$, the time-independent Schrödinger equation and the wall conditions givewhere $k=\sqrt{2mE}/\hslash$ and $l=\sqrt{2m(U_0-E)}/\hslash$. Continuity of $\psi$ and ${\psi }^{\prime }$ at the finite potential step givesDividing and rearranging yields the bound-state quantization condition

Varying the gauge-fixed Polyakov action gives the bulk wave equation and the boundary termIt vanishes in each target direction either through the Neumann boundary conditionor through the Dirichlet boundary conditionwhich fixes the endpoint positions.

Let $L_h=D_{xx}+\alpha D_x$, where the centered second-difference matrix $D_{xx}$ is real symmetric negative definite under homogeneous Dirichlet boundary conditions, and the centered first-difference matrix $D_x$ is real skew-symmetric. HenceThe semidiscrete energy method gives $\Vert \mathbf{u}(t){\Vert }_2\le \Vert \mathbf{u}(0){\Vert }_2$ for every real $\alpha$, so the scheme is stable.

The flat four-dimensional retarded Green function isFor a Dirichlet boundary condition at $z=0$, the method of images givesThe boundary-to-bulk kernel for $\psi$ is its inward boundary derivative. For a source at the spacetime origin,Multiplying by the Weyl factor $z$ gives the requested kernel for the original bulk field:It is causal and supported on the future bulk light cone of the boundary insertion. Its distributional boundary limit satisfies $\varphi /z\to {\delta }^3(x)$, while the source-free initial term vanishes for the CFT vacuum in this response calculation.

Put $\eta =\psi -\varphi$. The common Dirichlet boundary condition gives $\eta =0$ on $S$. Expanding the energy functional,Integration by parts and the field equation make the cross term zero:Consequently

For $m>0$, equality holds exactly when $\eta =0$. For $m=0$, equality holds exactly when $\nabla \eta =0$, so $\eta$ is constant on the connected region; its zero boundary value then again forces $\eta =0$. The minimizing property therefore proves uniqueness for the Dirichlet problem in both cases:

Inside the infinite square well, the time-independent Schrödinger equation has Dirichlet boundary conditions at $0$ and $a$. Its normalized energy eigenstates and energy eigenvalues are

Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

Let $L_h$ be the spatial matrix, so the forward Euler method isThe conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha +\beta +\gamma +\delta <4a_{\max }$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous Dirichlet data remove some unknown neighbors.

The Gershgorin circle theorem and symmetry therefore place every eigenvalue of $L_h$ inIf $\mu =k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ wheneverorSince the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the Gershgorin stability bound for a variable-coefficient diffusion stencil.

The zero-boundary Sobolev space isOn it defineIf this quadratic form vanishes, then $\nabla u=0$. The Poincaré inequality givesso $u=0$. It is therefore an inner product, and its norm is equivalent to the usual $H^1$ norm.

A function $u\in H_0^1(\Omega )$ is a weak solution whenThis follows from integration by parts and incorporates the homogeneous Dirichlet boundary condition through membership in $H_0^1$.

Extend a grid vector $v$ by zero to the Dirichlet boundary. Pairing contributions along undirected nearest-neighbour edges gives the discrete energy identitywhere $E$ includes edges from an interior node to a boundary node. This is the three-dimensional version of summation by parts for the seven-point Dirichlet Laplacian.

The right side is nonpositive. If it vanishes, every pair of neighbouring values agrees; connectivity of the grid and the zero boundary values then imply $v=0$. Thus $v^{T}Av<0$ for every nonzero $v$, so $A$ is negative definite. In particular, zero is not an eigenvalue, and therefore

Take the real $L^2$ inner product of the equation with $u$. The homogeneous Dirichlet boundary conditions and integration by parts giveThus $\Vert u(t){\Vert }_{L^2}\le \Vert u_0{\Vert }_{L^2}$, and applying the same estimate to the difference of two solutions gives continuous dependence on the initial data. The constant-advection term is skew-symmetric under these boundary conditions and contributes no energy. Hence