Write $\rho$ for the mass density. A material volume contains the same fluid elements at all times, so conservation of mass requiresby the Reynolds transport theorem. Since the initial material volume can be chosen arbitrarily, continuity of the integrand forces it to vanish pointwise. Thus the continuity equation isFor positive mass density, the incompressibility condition $D\rho /Dt=0$ reduces this to $\nabla \cdot \mathbf{u}=0$. Mass conservation itself does not require incompressible flow.

Let $V(t)$ be a smoothly moving material volume, carried by a smooth velocity field $\mathbf{u}$, and let $f$ be continuously differentiable. The material-volume form of the Reynolds transport theorem statesEquivalently, the divergence theorem giveswhere $\mathbf{n}$ is the outward unit normal vector. The second integral measures the flux carried across the instantaneous boundary by its motion.

For the proof, use the particle map $\mathbf{x}=\mathbf{X}(\mathbf{a},t)$ from $V(0)$, with positive Jacobian determinant $J=\det (\partial \mathbf{X}/\partial \mathbf{a})$. Euler's volume identity isIt is the Jacobian evolution of a smooth flow: if $F=\partial \mathbf{X}/\partial \mathbf{a}$, differentiation gives $F_t=(\nabla \mathbf{u})F$, and differentiation of the determinant gives $J_t=J\mathrm{tr}(F^{-1}{F}_t)=J\nabla \cdot \mathbf{u}$. The change of variables formula now moves the integral to the fixed reference volume:Finally, $Df/Dt+f\nabla \cdot \mathbf{u}={\partial }_{t}f+\nabla \cdot (f\mathbf{u})$, giving the equivalent boundary form.

For a scalar fluid property $f(\mathbf{x},t)$, the Eulerian time derivative ${\partial }_{t}f$ measures change at a fixed Eulerian coordinate. The material derivative, also called the Lagrangian time derivative, measures change while following one fluid element. If its Lagrangian trajectory is $\mathbf{X}(\mathbf{a},t)$, where $\mathbf{a}$ is its fixed Lagrangian coordinate, then ${\partial }_t\mathbf{X}=\mathbf{u}(\mathbf{X},t)$.

The chain rule therefore givesThe last term accounts for entering a region where the property differs. For example, a spatially varying but time-independent velocity field has zero Eulerian time derivative, while a moving fluid element can have nonzero material acceleration. The same formula applies componentwise to a vector field.

Multiplication of the preceding displacement formula by $-ik$ givesFor the supplied kernels, $\widehat{f_{1,x}}=-ik{e}^{-qy}$ and ${\hat{f}}_2=i(k/q)e^{-qy}$. The convolution theorem therefore givesSince $f_{1,x}(s,y)=-2ys/[\pi (s^2+y^2{)}^2]$ and $f_2(s,y)=s/[\pi (s^2+y^2)]$, combine the two fractions and divide by two:The sign is tied to the positive forward Fourier phase used here. This is a displacement derivative, so an additive displacement reference drops out. At $y>0$ the kernel is regular near $s=0$; the usual decay/integrability of the applied load controls the tails. A boundary-limit singular integral, if wanted at $y=0$, would require its own principal-value interpretation.

The source uses the forward Fourier phase $e^{ikx}$, so $\widehat{{\partial }_{x}A}=-ik\hat{A}$. For a nonzero wavenumber put $q=|k|$. Biharmonicity becomes $({\partial }_y^2-q^2{)}^2\hat{A}=0$. The decaying solutions are $(C+Dy)e^{-qy}$. The normal-stress condition gives $-k^2\hat{A}(k,0)=-\hat{P}(k)$, hence $C=\hat{P}/k^2$. The shear condition gives ${\hat{A}}_y(k,0)=0$, so $D=qC$. ConsequentlyIts second derivatives are $\widehat{A_{xx}}=-\hat{P}(1+qy)e^{-qy}$ and $\widehat{A_{yy}}=\hat{P}(qy-1)e^{-qy}$. The given strain/stress inversion therefore yieldsIntegrating in $y$ with the stated decaying-mode condition givesThis is the Fourier solution of a loaded elastic half-space.

The zero wavenumber needs separate treatment: these displayed divisions by $k^2$ and $|k|$ are not formulas at $k=0$. In particular a nonzero resultant $\hat{P}(0)=\int Pdx$ is incompatible with an ordinary decaying zero-mode displacement of the stipulated kind; two-dimensional net loading produces a logarithmic displacement reference issue. Balanced loads or an appropriate distributional/reference-displacement formulation are needed for a global transform solution. The nonzero modes and the displacement-gradient kernel below remain the correct formal calculation; the singular zero mode must not be silently assigned a finite value.

The Airy stress function determines the in-plane stress through its second derivatives. If two functions give the same stress, their difference $B$ has $B_{xx}=B_{xy}=B_{yy}=0$. On a connected domain this implies that its gradient is constant. Thereforeis exactly the local affine gauge freedom. It also preserves biharmonicity. An imposed decay condition for $A$ can fix this freedom; adding an arbitrary biharmonic function would usually change the physical stress and is not a gauge transformation.

Differentiating the displacement givesThe isotropic stress is diagonal, withThe two transverse traction-free conditions set the latter entries to zero. Hence $\lambda =2\nu (\lambda +\mu )$ andSubstitution shows $E=2\mu (1+\nu )$; solving for the two Lamé parameters givesTo determine the allowed interval without assuming signs not given, put $K=\lambda +2\mu /3>0$. Then $\nu =(3K-2\mu )/[2(3K+\mu )]$, soTherefore $\boxed{-1<\nu <1/2}$. The endpoints would require a degenerate elastic modulus, so they are excluded by the strict material inequalities.

For a surface with unit normal $\mathbf{n}$, traction is $t_i={\tau }_{ij}{n}_j$. Thus ${\tau }_{ij}$ is the $i$th force-per-area component on a plane whose positive normal is ${\mathbf{e}}_j$. Diagonal entries are normal stresses and off-diagonal entries are shear stresses, with tensile normal stress positive in this convention.

The symmetric strain has six independent components and so does the symmetric stress. A general linear map between these two six-dimensional spaces therefore hasEquivalently choose the minor symmetries $C_{ijkl}=C_{jikl}=C_{ijlk}$. Components antisymmetric in the strain indices have no observable effect and are omitted from this representative.

A further distinction is important: if a strain energy density exists, $C_{ijkl}={\partial }^{2}W/(\partial e_{ij}\partial e_{kl})$ has the major symmetry $C_{ijkl}=C_{klij}$. Then the six-by-six map is symmetric and has $6\cdot 7/2=21$ independent coefficients. Thus 36 follows from the two stated tensor symmetries alone; 21 additionally uses the energy/reciprocity assumption, as in the variational setting later in the paper. These are the minor and major symmetries of linear elasticity.

Take $F(p,q)=p^2+pq-1$. Parametrize the initial line by $x_0=s>0$, $y_0=0$, $u_0=2s^2$. Initial compatibility gives $p=4s$, and $F=0$ then gives $q=1/(4s)-4s$. The characteristic formula in part (a) yieldsSince $y>0$, take $\tau >0$. To select the branch connected smoothly to the initial data, stop each characteristic before its coordinate map folds. Its Jacobian determinant isso this branch has $0<\tau <2s^2$. Eliminating $\tau =y/(4s)$ givesFor fixed $y>0$, the right side defining $x-y$ is strictly increasing on that selected branch. At its lower endpoint it equals $3y^{1/3}/4$, and it tends to infinity as $s\to \infty$. Thus it gives exactly one $s$ for every point inSubstituting that unique $s$ into the formula for $u$ gives the required classical solution. As $y\downarrow 0$ at fixed $x>0$, $s\to x$, and $u\to 2x^2$ as required.

Without the pre-fold branch restriction, the algebraic equation also has a smaller positive root. That root tends to zero as $y\downarrow 0$, and its $u$ tends to zero rather than $2x^2$, so it does not solve the prescribed initial-boundary problem. The fold $J=0$ is the characteristic caustic; uniqueness here is for the classical branch continued from the given initial data before that fold.

Write $F_p,F_q$ evaluated at the fixed pair $(p(s),q(s))$. Differentiating in $\tau$ gives $x_{\tau }=F_p$, $y_{\tau }=F_q$, and $u_{\tau }=p{F}_p+q{F}_q$, proving $u_{\tau }=p{x}_{\tau }+q{y}_{\tau }$.

Differentiate the other parameter. The difference of the required sides isThe first bracket is zero by the prescribed initial compatibility condition, and the second is zero by differentiating $F(p(s),q(s))=0$. Thus $u_s=p{x}_s+q{y}_s$ as well. These calculations use the differentiability of the supplied parametric functions.

If the coordinate Jacobian determinant $J=x_s{y}_{\tau }-y_s{x}_{\tau }$ is nonzero, the inverse function theorem gives local coordinates $(x,y)$. The chain rule then makes $(u_x-p,u_y-q)$ solve a homogeneous two-by-two linear system with determinant $J$. ThereforeThis is the method of characteristics for a first-order equation depending only on the gradient.

For a fixed observation point $(\xi ,\tau )$, use the backward adjoint Green function with $L^{\ast }=-{\partial }_t-{\partial }_x^2+1$. It satisfies $L^{\ast }G=0$ for $t<\tau$, decays with its spatial derivative as $|x|\to \infty$, and has terminal distribution $G(x,{\tau }^{-};\xi ,\tau )=\delta (x-\xi )$. Extending it by zero for $t>\tau$ gives $L^{\ast }G=\delta (x-\xi )\delta (t-\tau )$.

For $0<t<\tau$, integration by parts provesEvaluating this constant pairing at the initial and terminal times yieldsBy translating the heat kernel with constant absorption and reversing the time difference,The terminal singularity is understood as a distribution; the representation follows for smooth decaying data and extends to standard integrable data by the heat-kernel approximation argument.

Use the Fourier transform convention $\hat{u}(k,t)={\int }_{\mathbb{R}}{e}^{-ikx}u(x,t)dx$. Then the equation and Dirac delta initial condition becomeso $\hat{u}=e^{-(k^2+1)t}$. The inverse Fourier transform givesComplete the square as $-t{k}^2+ikx=-t(k-ix/(2t){)}^2-x^2/(4t)$ and use the permitted complex-shift Gaussian integral. Thus the heat kernel with constant absorption isIts mass is $e^{-t}$, consistent with constant absorption, and it tends to the Dirac delta as a distribution as $t\downarrow 0$.

Assume a classical solution continuous up to the relevant boundary. At an interior positive maximum, $u_t=0$ and $u_{xx}\le 0$, hence $Lu=u_t-u_{xx}+u>0$, a contradiction. At a positive maximum on the top edge $t=\tau$, away from the spatial endpoints, the left time derivative satisfies $u_t\ge 0$, while again $u_{xx}\le 0$; the same contradiction applies. Thus a positive maximum can occur only on the initial or lateral parabolic boundary.

For two solutions with the same data, their difference $w$ satisfies $Lw=0$ and vanishes on that parabolic boundary. The weak parabolic maximum principle prohibits a positive maximum of $w$ and also of $-w$. Therefore $w=0$, proving at most one solution. The positive reaction term is what gives the immediate strict contradiction at a positive maximum.

In the moving stretched coordinates, ${\partial }_t={\partial }_{\tau }-s^{\prime }(\tau ){\epsilon }^{-1}{\partial }_{\xi }$ and ${\partial }_x={\epsilon }^{-1}{\partial }_{\xi }$. Multiplying the damped Burgers equation by $\epsilon$ givesDiscarding the order-$\epsilon$ terms yields the leading Viscous Burgers travelling shock equation $\boxed{(u-s^{\prime })u_{\xi }=u_{\xi \xi }}$. Integrate across a nonconstant transition whose derivative vanishes at either end:Multiplication by $u$ and a further integration by parts giveThe first identity is the Rankine-Hugoniot condition, giving $s^{\prime }=(u_{-}+u_{+})/2$. Substitution into the second makes its left side $(u_{+}-u_{-}{)}^3/12$, so the viscosity-selected entropy shock has $u_{-}>u_{+}$.

After the first characteristic collapse, the shock is fed by the left constant initial state and the right zero state. Their damped values are $u_{-}=2e^{-\tau }$ and $u_{+}=0$. Therefore $s^{\prime }=e^{-\tau }$. Matching to $s(\log 2)=0$ yieldsThese matching states are consistent with the characteristic feet: the left foot is $s-2(1-e^{-\tau })=-3/2+e^{-\tau }<-1$, while the right foot is $s>0$. The formula describes the leading thin-layer position in the small-viscosity limit.

For zero viscosity, the method of characteristics gives $u=u_0(a)e^{-t}$ and $x=a+u_0(a)(1-e^{-t})$. On the sloping part of the initial profile, $u_0(a)=-2a$, so $x=a(2e^{-t}-1)$. While this map is increasing, inversion givesThe sloping interval shrinks from width one to zero while its height decreases from two to one. Its spatial derivative is $-2e^{-t}/(2e^{-t}-1)$ and diverges as $2e^{-t}-1\downarrow 0$. Thus the first shock formation by characteristic intersection occurs atFor $0\le t<t_{\ast }$ the profile consists of a left plateau, a straight descending ramp ending at zero, and a right zero plateau. At $t_{\ast }$ the ramp becomes a vertical jump. This is a damped Burgers equation; viscosity smooths that jump into a thin transition.

For $m=0$ the two normalizations agree, and the Laguerre polynomials are orthonormal with weight $e^{-x}$. Take real $-1<h<1$. Their expansion coefficients for the requested function areSubstitute Rodrigues' formula and integrate by parts $n$ times. Differentiating $e^{-xh/(1-h)}$ transfers a factor $[h/(1-h){]}^n$ after the two sign factors cancel. ThereforeThe integral is $n!(1-h{)}^{n+1}$ by the Gamma function. Boundary terms vanish throughout this range.

The resulting generating function for Laguerre polynomials isOne can also verify convergence and equality directly: Rodrigues' formula gives $L_n(x)={\sum }_{j=0}^n\left(\genfrac{}{}{0ex}{}{n}{j}\right)(-x{)}^j/j!$, and summing ${\sum }_{n\ge j}\left(\genfrac{}{}{0ex}{}{n}{j}\right)h^n=h^j/(1-h{)}^{j+1}$ reduces the double series to the exponential on the left. The absolute-value double series converges for fixed $x$ and $|h|<1$, justifying the interchange; the formula then also holds for complex $h$ in that disk.

There is a normalization conflict in the source: the printed Rodrigues' formula gives the standard Generalized Laguerre polynomial $P_n=L_n^{(m)}$ with $P_n(0)=\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$, which is generally not one. The norm requested here corresponds to that Rodrigues normalization. A solution normalized to one at zero is ${\hat{P}}_n=P_n/\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$.

Set $w=x^m{e}^{-x}$ and $p=x^{m+1}{e}^{-x}$. The equation becomes $(p{P}_n^{\prime }{)}^{\prime }+nw{P}_n=0$. Multiply the equations for $P_n,P_k$ by the opposite polynomial, subtract and integrate. The boundary term $p(P_n^{\prime }{P}_k-P_k^{\prime }{P}_n)$ vanishes at zero and infinity, soThis proves orthogonality for $k\ne n$ under either normalization.

For the squared norm, use the printed Rodrigues' formula and integrate by parts $n$ times:All boundary terms vanish: the required derivatives have a positive power of $x$ at zero and exponential decay at infinity. The leading coefficient is $(-1{)}^n/n!$, so $P_n^{(n)}=(-1{)}^n$. The Gamma function integral givesFor the source's alternative unit-at-zero normalization the correct answer instead isBoth cannot equal the printed norm when $m,n>0$. This distinction is the normalization of generalized Laguerre polynomials.

After division by $x$, the ordinary differential equation has coefficients $P(x)=(m+1)/x-1$ and $Q(x)=n/x$. At zero, $xP\to m+1$ and $x^{2}Q\to 0$, making zero a regular singular point. The indicial equation isso the indicial exponents are $\boxed{0,-m}$, coincident when $m=0$.

At infinity put $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation isThe coefficient of $Y^{\prime }$ already fails the regular-singularity test at $t=0$. Thus $\boxed{x=\infty \text{ is an irregular singular point}}$, including when $n=0$.