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For the real coefficients of the stated self-adjoint Sturm-Liouville problem, define the weighted inner productSince $w>0$, a nonzero eigenfunction has positive squared norm. Let $Ly=-(p{y}^{\prime }{)}^{\prime }+qy$. Integration by parts and the Dirichlet boundary conditions giveTaking $u=v$ to be an eigenfunction shows that its eigenvalue is real, since $\lambda \int w|u{|}^2=\int (p|u^{\prime }{|}^2+q|u{|}^2)$ is real. For two eigenfunctions with eigenvalues $\lambda ,\mu$, the same identity givesHence distinct eigenvalues imply orthogonality in the weighted inner product. This is the orthogonality of Sturm-Liouville eigenfunctions. The calculation assumes the regularity needed for the stated self-adjoint boundary problem, so all boundary terms and integrals are defined.

Set $s=\log x$ and $Y(s)=y(e^s)$, so $0\le s\le 1$. By the chain rule, the equation becomes $Y^{\prime \prime }+2Y^{\prime }+(1+\lambda )Y=0$. The substitution $Y=e^{-s}v$ reduces it further toMultiplication by $\overline{v}$ and integration by parts gives $\lambda {\int }_0^1|v{|}^2={\int }_0^1|v^{\prime }{|}^2>0$ for a nonzero solution, ruling out zero and negative eigenvalues. For $\lambda =k^2>0$, the first boundary condition gives $v=A\sin (ks)$, and the second gives $k=n\pi$, $n=1,2,\dots$. Thus all eigenpairs arewith arbitrary nonzero scalar multiples of each eigenfunction. These are the complete list since all solutions of the constant-coefficient differential equation have been considered; the transformed sine system is also complete in the corresponding weighted Hilbert space.

Use the Fourier transform $\tilde{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$. For a rapidly decaying smooth function, integration by parts gives $\widetilde{f^{\prime }}=ik\tilde{f}$, and differentiation under the integral gives $\widetilde{xf}=id\tilde{f}/dk$. Both are justified here by Gaussian decay. Since $f^{\prime }=-2a^{2}xf$,The Gaussian integral supplied in the hint gives $\tilde{f}(0)=\sqrt{\pi }/a$. Solving this first-order equation yieldsThe calculation states the transform normalization and the two differentiation properties explicitly.

Write $F(p)=\mathcal{L}[f](p)$. Assume $f$ and $f^{\prime }$ have right limits at zero and sufficient regularity and exponential-order bounds for the following integration by parts; take $\mathrm{Re}p$ large enough to kill the boundary terms at infinity. First,Apply the same calculation to $f^{\prime }$:This is the Laplace transform of a derivative formula with both initial-data terms retained.

Take $T>0$ and a positive prescribed distance $L$; if $L=0$, the zero-speed journey has zero cost. Minimize $E[V]={\int }_0^T(a{V}^2+b{\dot{V}}^2)dt$ subject to $V(0)=0$ and the isoperimetric constraint ${\int }_0^{T}Vdt=L$. Introduce a constant Lagrange multiplier $\lambda$ and the integrand $a{V}^2+b{\dot{V}}^2-\lambda V$. For a variation $\eta$ with $\eta (0)=0$, integration by parts givesThe interior Euler-Lagrange equation isWhen terminal speed is free, the same calculation also imposes the natural boundary conditions for a free endpoint:One cannot drop this boundary term when interpreting an extremal as a genuine minimum. If $b=0<a$, the interior condition makes $V$ constant; with a continuous zero initial speed and $L>0$ it cannot be attained. In that case Cauchy-Schwarz inequality gives the infimum $a{L}^2/T$, approached by speeds nearly constant except in a short initial layer. For $b>0$, the derivative cost controls such layers and the natural terminal condition matters in parts (b) and (c).

Here $b>0$ because $a=0$ and the two cost coefficients are not both zero. The interior Euler-Lagrange equation makes $V$ quadratic. Imposing each displayed pair of endpoint data and the distance constraint gives the following formal profiles and their costs:Thus choice (3) is the best strategy, and its true minimum cost isThe qualification from part (b) applies to row (1): its formal quadratic is not a minimizer with free terminal speed, since $\dot{V}(T)\ne 0$. Row (2) is a genuine minimizer for its fixed zero terminal speed, and row (3) is the genuine free-terminal-speed minimizer, satisfying its natural condition $\dot{V}(T)=0$.

A direct lower bound makes the comparison rigorous. For every admissible speed with $V(0)=0$,by integration by parts. The Cauchy-Schwarz inequality gives $L^2\le (T^3/3){\int }_0^T{\dot{V}}^{2}dt$. Equality requires $\dot{V}$ to be proportional to $T-t$, giving exactly row (3). Row (1)'s zero initial acceleration excludes equality; however short initial layers, as in part (b), approach that same infimum. Hence choice (1) has infimum $3b{L}^2/T^3$ but does not attain it, while choice (3) does. This distinction resolves the boundary-condition issue in the printed minimization request.

The function is even, so its Fourier series has $b_n=0$. In the convention $f\sim a_0/2+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)$,Twice applying integration by parts gives, for $n\ge 1$,Therefore the periodic Fourier representation iswith periodic continuation. The periodic extension is continuous, including the identified endpoints, and piecewise smooth, so the Fourier series converges to it. The coefficients also give absolute convergence and uniform convergence by comparison with $\sum n^{-2}$.

At $x=\pi$, the cosine cancels the alternating coefficient sign; at $x=0$, it is identically one. These evaluations yieldThe first identity solves the Basel problem.

Write $B=-A=-{\partial }_x^2+1+x^2$. Smooth compactly supported functions lie in $D(A)$ and are dense in L2 space, so $A$ is densely defined.

For $\lambda >0$, use the formon the one-dimensional harmonic oscillator form domain $X$. It is bounded and coercive in the $X$ norm. The Lax-Milgram theorem gives a unique $u\in X$ satisfying $a_{\lambda }(u,v)=⟨g,v⟩$ for every $v\in X$. Thus $-u^{\prime \prime }+(1+\lambda +x^2)u=g$ in distributions.

We must verify the full specified operator domain, not just the form domain. The equation first gives $u\in H_{\mathrm{loc}}^2$. For compactly supported $H^2$ functions, integration by parts gives the graph estimate for the harmonic oscillator, obtained fromApply this to $v={\chi }_{R}u$, with ${\chi }_R=1$ on $[-R,R]$, supported in $[-2R,2R]$, and derivatives bounded by $C/R$ and $C/R^2$. Its right-hand source is ${\chi }_{R}g-2{\chi }_R^{\prime }{u}^{\prime }-{\chi }_R^{\prime \prime }u$, uniformly bounded in L2 space. The estimate and the Fatou lemma give $x^{2}u\in L^2$; the original equation then gives $u^{\prime \prime }\in L^2$, and hence $u\in H^2$. Thus $u\in D(A)$, proving that $\lambda -A$ is onto. The energy identity proves injectivity and the resolvent of the shifted harmonic oscillator estimateTo prove closedness, suppose $u_n\to u$ and $A{u}_n\to v$ in L2 space. For one fixed $\lambda >0$,Therefore $u$ belongs to the range of the resolvent, namely $D(A)$, and $Au=v$.

All hypotheses of the Hille-Yosida theorem now hold; the resolvent bound is stronger than $1/\lambda$. Let $S(t)$ be its contraction semigroup. The exponentially shifted semigrouphas generator $A+I={\partial }_x^2-x^2$. It is continuous in L2 space, takes the initial value ${\psi }_0$, and satisfies $\boxed{\Vert \psi (t){\Vert }_2\le e^t\Vert {\psi }_0{\Vert }_2}$. For ${\psi }_0\in D(A)$, its time derivative exists in L2 space and equals $(A+I)\psi (t)$, so the equation holds as a strong L2 space solution. For general initial data it is the semigroup solution, without an unproved pointwise differentiability claim.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

Choose a smooth cutoff function $\eta$ supported in $B_2$, equal to one on $B_1$, with $0\le \eta \le 1$ and $|D\eta |\le C$. For $i<n$, put $v_i=D_{i}u$. The zero Dirichlet boundary condition implies that the tangential boundary derivative $v_i$ vanishes on the flat face. Differentiating the Poisson equation gives $\Delta v_i=D_{i}f$. Test this equation with ${\eta }^2{v}_i$. The boundary terms vanish because $v_i=0$ on the flat face and $\eta =0$ near the curved face. Integration by parts in the tangential direction moves $D_i$ off $f$, yieldingThis is the essential Caccioppoli inequality step: it involves $f$, rather than $Df$. Expanding $D_i({\eta }^2{v}_i)$ and using the Cauchy-Schwarz inequality and the Young inequality givesThe term $f\eta (D_i\eta )v_i$ is bounded by $C{f}^2+C|D\eta {|}^2{v}_i^2$. Absorbing the first term and summing over $i<n$ proves the hinted estimate:All mixed second partial derivatives are now controlled, since $D_{ni}u=D_{in}u$. The remaining normal derivative is recovered from the equation:The Cauchy-Schwarz inequality bounds its squared L2 norm by $n$ times the sum of the squared L2 norms on the right. Including the already controlled zeroth and first partial derivatives, and taking square roots, gives the second-derivative estimate at a flat Dirichlet boundary:The full second-order Sobolev norm is controlled without any norm of $Df$.

Use a signed power test for a Laplacian eigenfunction. For $\gamma \ge 2$, test the Dirichlet Laplacian eigenfunction equation with $|u{|}^{\gamma -2}u$. Its derivative is $(\gamma -1)|u{|}^{\gamma -2}Du$; at $\gamma =2$ the test function is just $u$. Integration by parts gives the exact energy estimateSet $v=|u{|}^{\gamma /2}$. It belongs to the zero-boundary Sobolev space and obeys the Sobolev chain rule:For $\gamma >2$, the power map is $C^1$ with bounded derivative on the bounded range of $u$. For $\gamma =2$, the absolute value map is Lipschitz, and $D|u|=\mathrm{sgn}(u)Du$ almost everywhere; on the zero set, use the fact that a gradient of a Sobolev function vanishes on a level set. These facts justify the identity even when $u$ changes sign.

Apply the given Sobolev inequality with the Sobolev conjugate exponent $2\kappa =2n/(n-2)$:Testing with a signed power raises the integrability exponent from $\gamma$ to $\gamma \kappa$. Absorbing the factor $1/4$ into $C(n)$ gives precisely the requested inequality.

Multiply the Dirichlet Laplacian eigenfunction equation by $u$ and use integration by parts. The zero Dirichlet boundary condition removes the boundary term and givesThe denominator is positive because the eigenfunction is nonzero. The numerator is also positive: if it were zero, the given Sobolev inequality for $u\in W_0^{1,2}(\Omega )$ would force $u=0$. Therefore the Dirichlet Laplacian eigenvalue has the positive Rayleigh quotientEvery nonzero Dirichlet eigenfunction in this problem has a positive eigenvalue.

Choose a constant positive Hermitian form on $V$. Its underlying flat Kähler metric is translation-invariant and descends through the Euclidean lattice to the compact complex torus $X$. The Dolbeault Hodge decomposition gives a unique Dolbeault Laplacian-harmonic representative of every Dolbeault cohomology class. By the Kähler Laplacian identity, these are also Hodge Laplacian-harmonic.

Choose a constant coframe on $V$. A lifted harmonic form is a sum of constant wedge basis forms with smooth periodic coefficients. For the flat metric, the Hodge Laplacian acts coefficientwise by the scalar constant-coefficient Laplacian. Each coefficient is a harmonic function on a compact flat real torus, hence is constant: integration by parts gives $\int |Da{|}^2=0$. Conversely every constant-coefficient form is harmonic. Thus the harmonic representatives are exactly the translation-invariant differential forms of type (p, q).

The constant $(1,0)$ covectors form $V^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(V,\mathbb{C})$; the constant $(0,1)$ covectors form the space of antilinear maps $\overline{V^{\ast }}={\mathrm{Hom}}_{\overline{\mathbb{C}}}(V,\mathbb{C})$. Sending a constant wedge form to its cohomology class givesThe map is intrinsic and does not depend on the auxiliary flat metric: that metric proves every class has exactly one constant representative, while the inclusion of constant forms is defined directly by the quotient. The Dolbeault cohomology of a complex torus consists of constant forms, and in particularFor indices outside $0,\dots ,g$, both the exterior-power expression and the cohomology group vanish.

For a continuous vector semimartingale $X$ and $f\in C^{1,2}$, the Itô formula isThe Itô product rule, or stochastic integration by parts, isThe additional second-order term is the quadratic covariation.

For general semimartingales with càdlàg paths, use left limits in the integrands. The Itô formula for semimartingales with jumps, for time-independent $f\in C^2$, isHere $X^c$ denotes the continuous local martingale part; the continuous bracket is that of these parts. For time-dependent $f\in C^{1,2}$ add $\int {\partial }_{s}f(s,X_{s-})ds$ and insert $s$ in the other evaluations of $f$. The integration by parts formula becomesFor continuous paths the jump sum vanishes, recovering the first two formulas.

Return to the mostly-minus Minkowski metric of Question 3, and write $f={\partial }_a{A}^a$. The field $\xi (x)$ is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation $A_a\mapsto A_a+{\partial }_a\lambda$, with $\xi$ inert, the electromagnetic field tensor is invariant but $f\mapsto f+\square \lambda$. The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying $\square \lambda =0$, with the same boundary conditions imposed before and after the transformation. The action itself is undefined at $\xi =0$; that value can only be considered as a limiting gauge.

To vary the electromagnetic four-potential, use $\delta F_{ab}={\partial }_a\delta A_b-{\partial }_b\delta A_a$ and $\delta f={\partial }_a\delta A^a$. The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable $\xi$, the derivative must act on $1/\xi$ as well as on $f$:Since $\xi$ enters algebraically, its Euler-Lagrange equation isFor a real field this gives $f=0$. Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, ${\partial }_a{F}^{ab}=0$, ${\partial }_a{A}^a=0$, and therefore $\square A^b=0$. There is no independent kinetic equation that determines $\xi$.

For the momentum-space equation at a prescribed constant $\xi \ne 0$, use $A_a(x)=\int d^{4}p{e}^{-ip\cdot x}{A}_a(p)/(2\pi {)}^4$. Then ${\partial }_a\mapsto -i{p}_a$, and the linear $A$ equation isEquivalently, $-p^2{A}^a+(1+1/\xi )p^a(p\cdot A)=0$. If the separately varied $\xi$ equation is also imposed on a real classical solution, then $p\cdot A=0$ and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding $\xi$ fixed and integrating over $A$ is a Gaussian functional integral; integrating over the original variable $\xi$ as well is a different constrained problem.

For $p^2\ne 0$, the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy $P_T^2=P_T$, $P_L^2=P_L$, $P_T{P}_L=0$ and $P_T+P_L=I$. The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define $G_{ab}$ by $K^{ac}{G}_{cb}={\delta }_b^a$. The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor $i{G}_{ab}$ and a Feynman i-epsilon prescription for its simple and double poles. If the name $G$ is used for the vacuum two-point function, include that factor $i$ consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter $\alpha =-\xi$; in particular, $\xi =-1$ gives Feynman gauge and $G_{ab}=-{\eta }_{ab}/p^2$.

Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any $\xi \ne 0$, and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when $p_b=0$. At $p^2=0$, the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting Landau gauge, $\xi \to 0$, the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.

Use Lie algebra generators with $[T_b,T_c]=f^a{}_{bc}{T}_a$, and define the adjoint gauge covariant derivative by $(D_{\mu }X{)}^a={\partial }_{\mu }{X}^a+f^a{}_{bc}{A}_{\mu }^b{X}^c$. Varying the gauge field strength gives $\delta F_{\mu \nu }=D_{\mu }\delta A_{\nu }-D_{\nu }\delta A_{\mu }$. Antisymmetry followed by integration by parts therefore givesFor variations vanishing at the boundary, the Yang-Mills equations areThe Jacobi identity for the commutators of gauge covariant derivatives, using $[D_{\mu },D_{\nu }]X=[F_{\mu \nu },X]$, gives the gauge-theory Bianchi identity:In the second form of the gauge-theory Bianchi identity ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{\epsilon }^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. The identity is a geometric consequence of the definition of curvature, not a second dynamical field equation.

For the usual non-Abelian Yang-Mills theory, the quantum running coupling becomes strong at low energies. Dimensional transmutation generates a scale absent from the classically scale-invariant equations. The expected confining dynamics and massive colour-singlet spectrum involve confinement and a mass gap, rather than freely propagating weakly coupled coloured waves. These nonperturbative effects are not captured by simply solving the classical equations; this is not a claim of a mathematical proof of the Yang-Mills mass gap.

Write the gauge condition as ${\chi }^a[A]=f^a[A]$ to distinguish it from the structure constants. Under an infinitesimal gauge transformation, ${\delta }_{\omega }{A}_{\mu }=D_{\mu }\omega$. Define the Faddeev-Popov operator by ${\mathcal{M}}^{ab}=\delta {\chi }^a[A+D\omega ]/\delta {\omega }^b{|}_{\omega =0}$. In a Euclidean convention, the gauge-fixing and ghost additions can be chosen asA nonlocal kernel would require a double integral; a local functional $\chi$ instead makes $\mathcal{M}$ a local differential operator. Choosing a local gauge functional thus retains a local gauge-fixed action and the ordinary local interaction structure of perturbative quantum field theory.

The fields $c^a$ and ${\overline{c}}^a$ are independent Grassmann-valued fields in the Adjoint representation, Lorentz scalars with ghost numbers $+1$ and $-1$. They are the Faddeev-Popov ghost fields: their functional integral produces $\det \mathcal{M}$, the Faddeev-Popov determinant correcting the volume element along a gauge orbit. They are internal fields, not physical asymptotic particles. Their statistics supply a minus sign for every closed ghost loop. No auxiliary field is needed in the displayed gauge-fixing representation.

In axial gauge, ${\chi }^a=n^{\mu }{A}_{\mu }^a$, soWith the Fourier transform convention $e^{ik\cdot x}$, the ghost Feynman rules in the original normalization of $A$ areMomentum conservation accompanies the vertex; each ghost loop has the extra minus sign. With a canonically normalized field $A=g{A}_{\mathrm{can}}$, the vertex is instead $-g{f}^a{}_{bc}{n}_{\mu }$. The gauge fixing term is quadratic and introduces no further interaction vertex.

The Euclidean quadratic kernel of $A$ is $g^{-2}[k^2{\delta }_{\mu \nu }-k_{\mu }{k}_{\nu }+{\xi }^{-1}{n}_{\mu }{n}_{\nu }]$. For $n\cdot k\ne 0$, its inverse isThis follows by multiplication with the kernel: the $k^2\delta -kk$ piece gives $I-n{k}^T/(n\cdot k)$, and the ${\xi }^{-1}nn$ piece supplies the missing $n{k}^T/(n\cdot k)$. The strict axial-gauge propagator is thereforeIt obeys $n^{\mu }{D}_{\mu \nu }=0$. Every ghost attachment to an internal gauge propagator therefore vanishes in the strict limit; equivalently, $\mathcal{M}=n\cdot \partial$ on the gauge slice. Thus the interacting Faddeev-Popov ghosts decouple. Residual gauge transformations and the poles at $n\cdot k=0$ require compatible boundary conditions and an axial-gauge pole prescription.

After Wick rotation, the corresponding Minkowski propagator has ${\delta }_{\mu \nu }$ replaced by ${\eta }_{\mu \nu }$ and the usual overall factor $-i$, with a compatible causal prescription. The fixed $n$ makes the gauge-fixed propagator nonmanifestly Lorentz covariant. Nevertheless, gauge invariance makes physical observables independent of this gauge-choice vector, so their perturbative predictions are Lorentz invariant. Already at tree level, contraction with conserved external currents removes all terms containing $k_{\mu }$ or $k_{\nu }$, leaving $g^{2}j\cdot j^{\prime }/k^2$. The quantum Ward identities give the analogous cancellation in complete physical amplitudes; arbitrary gauge-dependent Green functions need not be independent of $n$.

The odd symmetries of a zero-dimensional polynomial model preserve both the action and the flat integration measure: their coefficients have zero superdivergence. The confining polynomial weight removes boundary terms at infinity. Thus the integral of an odd-symmetry derivative is zero, a supersymmetric Ward identity for this finite-dimensional integral in this zero-dimensional supersymmetric field theory.

Since $g(z)$ is a holomorphic function, the barred symmetry obeys ${\overline{V}}_1({\overline{\theta }}_{2}g(z))=-P(z)g(z)$. Applying the supersymmetric Ward identity gives the requested unnormalized insertion:One can also verify the result after Berezin integration, without invoking the symmetry terminology:The final integration by parts uses ${\partial }_{\bar{z}}(g{P}^{\prime })=0$ and rapid decay. The derivative on $W^{\prime }$ in the insertion is essential.