Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 78 (Waves in Fluids)

Let $\chi =H(F)$ use the Heaviside step function, and ${\delta }_s=|\nabla F|\delta (F)$, so that ${\mathbf{n}}_s=\nabla F/|\nabla F|$ points from the body into the fluid. The kinematic boundary condition gives $F_t+\mathbf{u}\cdot \nabla F=0$ on the surface. In the exterior, mass conservation and the Euler equations for an inviscid fluid areProducts with $\chi$ are understood distributionally, using ${\partial }_t\chi =F_t\delta (F)$ and ${\partial }_i\chi =F_i\delta (F)$. With ${\rho }^{\prime }=\rho -{\rho }_0$, the extended continuity equation isThe corresponding extended momentum equation isbecause its advective surface term is $\rho u_i(F_t+\mathbf{u}\cdot \nabla F)\delta (F)=0$. Differentiate continuity in time and substitute the divergence of momentum. Subtracting $c_0^2{\nabla }^2({\rho }^{\prime }\chi )$ gives the Ffowcs Williams-Hawkings equationwhere the inviscid Lighthill stress tensor isIf $p$ denotes total pressure, this tensor includes the constant ambient pressure $p_0$. Its contribution $p_0{\nabla }^2\chi$ cancels the corresponding surface term exactly. For radiation integrals it is convenient to subtract $p_0$ from both terms, making $T_{ij}=\rho u_i{u}_j+(p^{\prime }-c_0^2{\rho }^{\prime }){\delta }_{ij}$ with $p^{\prime }=p-p_0$. The displayed source identity is unchanged by this consistent pressure-gauge subtraction.

Assume no incoming sound and no free initial acoustic field. Put $R_y=|\mathbf{x}-\mathbf{y}|$, ${\tau }_y=t-R_y/c_0$, $Q={\rho }_0{u}_n{\delta }_s$, and $P_i=p^{\prime }{n}_{s,i}{\delta }_s$. Convolution with the retarded acoustic Green function gives the three source integrals, for an observer outside the body:These are the surface thickness, surface loading and volume quadrupole contributions. The retarded surface Dirac delta distributions must remain inside the integrals: integrating them over a moving retarded surface produces the moving-surface retarded Jacobian, and is not generally the same as simply substituting one retarded time into an instantaneous surface integral.

For an acoustic compact-source approximation, suppose the dominant sources have size $\ell$, source frequency scale $\omega$, and $\omega \ell /c_0\ll 1$. Also require $r=|\mathbf{x}|\gg \ell$ and the acoustic far field $\omega r/c_0\gg 1$. Then $R_y\simeq r-\mathbf{n}\cdot \mathbf{y}$, $1/R_y\simeq 1/r$, and the leading source integrals use the common time $\tau =t-r/c_0$. Slow source motion relative to $c_0$ is implicit in this leading retardation approximation. DefineThus $m$ is net displaced mass per unit time, $\mathbf{f}$ is the fluid pressure force on the body, and $S_{ij}$ is the integrated excess Lighthill stress tensor. The minus sign in $\mathbf{f}$ fixes the requested loading convention; the force on the fluid is opposite. On the radiating $1/r$ terms, ${\partial }_{x_i}\simeq -(n_i/c_0){\partial }_t$, givingSpatial derivatives of $1/r$ produce near-field terms of orders $r^{-2}$ and $r^{-3}$, omitted here. Neglecting source delays is a separate approximation from discarding those observer near-field terms. If a leading integrated source vanishes by symmetry, its first nonzero delay correction can be as important as another retained multipole.

For the translating sphere, write its center as $\mathbf{X}(t)=a_1\cos (\omega t)\mathbf{e}$, with $\mathbf{e}$ a unit vector, and its velocity as $\mathbf{V}=\dot{\mathbf{X}}$. Let $\mathcal{V}=4\pi a^3/3$, $\eta =a\omega /c_0\ll 1$ and $d=a_1/a\ll 1$. Impermeability gives $u_n=\mathbf{V}\cdot {\mathbf{n}}_s$. Since its volume is constant,in the leading compact formula. The individual local thickness sources have scale ${\rho }_0(a_1/r){\eta }^2$, but their net monopole cancels.

At first order in amplitude and leading order in $\eta$, the exterior incompressible velocity potential is $\varphi =-a^3\mathbf{V}\cdot {\mathbf{r}}_s/(2r_s^3)$, where ${\mathbf{r}}_s$ is measured from the mean center. It has ${\partial }_{r_s}\varphi =\mathbf{V}\cdot {\mathbf{n}}_s$ at $r_s=a$. Linear Bernoulli equation gives $p^{\prime }=-{\rho }_0{\varphi }_t=({\rho }_{0}a/2)\dot{\mathbf{V}}\cdot {\mathbf{n}}_s$ there. Since $\int n_{s,i}{n}_{s,j}dS=(4\pi a^2/3){\delta }_{ij}$,This is the added mass of a sphere. The explicit leading loading contribution isIt scales as ${\rho }_0(a_1/r){\eta }^3$. In homentropic small-amplitude flow the volume stress is quadratic to leading order, $T_{ij}=O({\rho }_0{V}^2)$ over volume $O(a^3)$, so the volume acoustic quadrupole has scaleIt is smaller than the linear dipole by $O(d\eta )=O(V/c_0)$ and can contain a second harmonic.

For completeness, the exact thickness-source integral has a nonzero first delay correction even though $m=0$. Its first spatial moment isExpand its integrand at ${\tau }_y=\tau +\mathbf{n}\cdot \mathbf{y}/c_0$. The leading nonzero translating-sphere acoustic thickness dipole is thereforeIt too scales as ${\rho }_0(a_1/r){\eta }^3$. Thus zero net-volume monopole does not mean zero thickness radiation. The full leading linear sound, summing thickness and loading, isThe leading compact equation gives the requested zero $\dot{m}$ and added-mass $\dot{f}$ terms; this extra delay calculation is necessary to obtain the complete linear field when its nominal monopole has cancelled.

Suppress the common time factor $e^{i\omega t}$ and invert the stated Fourier transform with $(2\pi {)}^{-1}\int \Phi (k,y)e^{-ikx}dk$. Use limiting absorption principle $k_0\mapsto k_0-i0$, and choose $\gamma =\sqrt{k^2-k_0^2}$ with positive real part for the evanescent spectrum. For $|k|<k_0$ its outgoing continuation is $\gamma =+i\sqrt{k_0^2-k^2}$. The pole $-k_0$ lies above the inversion contour and $+k_0$ below it. These prescriptions fix both radiation and pole signs.

The even interior scattered transform solves ${\Phi }_{yy}-{\gamma }^2\Phi =0$, so it is $A(k)\cosh (\gamma y)$. The exterior transform is $B(k)e^{-\gamma (|y|-b)}$. Let $D^{-}(k)$ be the transform of their common upward normal derivative on $y=b$. On $x>0$ the rigid plates set both derivatives to zero; on $x<0$ continuity requires them to agree. Hence this derivative is supported on the negative half-line and is analytic below the transform contour. It follows thatThe difference between exterior and interior scattered traces on $y=b$ is $-D^{-}/L$, where $L=\gamma \sinh (\gamma b)e^{-\gamma b}$. At the open negative half-line the total traces must match, so this difference equals the incident wave $e^{i{k}_{0}x}$. Its negative-half transform is $-i/(k+k_0)$. The positive-half trace difference has an unknown upper-half-plane transform $C^{+}$, giving the Wiener-Hopf equationFactor $L=L^{+}{L}^{-}$, divide by $L^{-}$, and subtract the forcing pole using $C_0=L^{+}(-k_0)$:The left side is analytic below and the right side above; the pole in the last quotient is removable. Analytic continuation makes their common value entire. The finite-energy edge condition and algebraic factor growth make this entire remainder vanish: the allowed normal-velocity edge singularity is at most inverse-square-root, so its half-line transform is $O(|k{|}^{-1/2})$, while $L^{\pm }=O(|k{|}^{1/2})$. The same edge condition excludes a growing polynomial remainder. This pole subtraction in a Wiener-Hopf equation yieldsThe corresponding exterior result for the Wiener-Hopf solution for an open parallel-plate waveguide isBoth forms are useful: the first makes the common normal derivative transparent, and the second removes the apparent factor singularity at a grazing saddle.

The final unheaded request has a sign error in its printed exponent. To see this from the derived solution rather than an outgoing-decay assertion alone, simplify the interior transform toNear $k=-k_0$, $C_0/L^{+}(k)=1+O(k+k_0)$ and $e^{-\gamma b}\cosh (\gamma y)=1-b\gamma +O({\gamma }^2)$ for fixed $|y|<b$. ThusFor $x<0$, inversion of the first term gives $-e^{i{k}_{0}x}$, cancelling the continued incident potential. The remaining branch-point term uses $\gamma \sim i\sqrt{2k_0}(k+k_0{)}^{1/2}$ on the specified contour, so it is $b\sqrt{2k_0}(k+k_0{)}^{-1/2}$. Its inverse transform at $x=-r$ is proportional to $e^{-i{k}_{0}r+i\pi /4}/\sqrt{\pi r}$. Therefore the upstream decay of an open parallel-plate waveguide isThe amplitude decays as $(-x{)}^{-1/2}$, not $(-x{)}^{1/2}$. The positive exponent is present in the original PDF, so it cannot be proved under the stated outgoing problem. The incident pole cancellation and the explicit branch expansion give the corrected conclusion.

Put $x=r\cos \theta$, $y=r\sin \theta$, and $s=|\sin \theta |$. From the exterior transform, the inverse integral has the amplitudemultiplying $e^{-ikr\cos \theta -\gamma rs}$. Apply the supplied method of steepest descent formula at $k_s=k_0\cos \theta$. On the outgoing sheet, $\gamma (k_s)=i{k}_{0}s$, so $-i\sinh (i{k}_{0}bs)=\sin (k_{0}bs)$. The far-field scattered potential isRestore $e^{i\omega t}$ for the physical time-harmonic potential. This is cylindrical outgoing spreading, with angular directivity determined by the kernel factor and aperture. The ordinary saddle formula applies for a fixed non-grazing direction, away from a coalescing pole or branch point. Its finite upstream angular limit is consistent with the separate branch-point calculation in the root solution; at grazing directions that separate calculation supplies the appropriate justification.

A plane wave returning down the guide has spatial factor $e^{-i{k}_{0}x}$ and arises from the pole at $k=k_0$, lying below the inversion contour. For $x>0$ closing below is clockwise, so the inverse transform is minus $i$ times the residue. Since $\gamma \sinh (\gamma b)\sim 2b{k}_0(k-k_0)$, the interior residue isHence the reflected plane-wave amplitude, relative to the incident amplitude one, isEquivalently, because $L^{+}$ carries the simple zero at $k_0$, $\mathcal{R}=L^{+}(-k_0)/(2k_0(L^{+}{)}^{\prime }(k_0))$. Other guided modes, if propagating at the chosen frequency, come from the additional kernel zeros; they do not alter which residue gives the plane-mode amplitude. Evanescent terms near the aperture are also separate. A multiplicative renormalization of the factors cancels in the product defining $\mathcal{R}$.

For zero viscosity the Inviscid Burgers equation is $q_X-q{q}_{\theta }=0$. Along a characteristic curve, $dq/dX=0$ and $d\theta /dX=-q$. Labelling a characteristic by its initial position $\eta$ givesWhere $f$ is differentiable, $q_{\theta }=f^{\prime }(\eta )/(1-X{f}^{\prime }(\eta ))$. The first characteristic crossing occurs at $X=1/\max f^{\prime }$ if the maximum positive slope exists. A multivalued continuation is not a physical wave: rewrite the equation as the conservation law $q_X+{\partial }_{\theta }(-q^2/2)=0$ and replace the overturning part by an entropy-admissible shock wave. The Rankine-Hugoniot conditions give its trajectoryThe states on either side are supplied by the characteristics meeting the shock. Conservation supplies the equal-area construction, and the entropy condition requires characteristics to enter the shock. Since the flux is concave, a jump with $q_L<q_R$ is compressive.

For the rising step, $q_L=0$ and $q_R=U>0$, the left characteristics have slope zero and the right ones slope $-U$. They overlap immediately. The Burgers Riemann problem with negative flux is solved byIt is a shock, not a rarefaction wave; the minus sign in the nonlinear flux is decisive.

For positive viscosity, substitute $q=2ϵ{\partial }_{\theta }\log \psi$ into the viscous Burgers equation. Writing $w=\log \psi$ givesAn $X$-dependent multiplicative factor in $\psi$ removes the possible integration function. The Cole-Hopf transformation therefore gives ${\psi }_X=ϵ{\psi }_{\theta \theta }$. Normalize its initial value continuously at zero:Split the heat kernel integral at zero. Completing the square in the positive-half integral yieldsWhen differentiating, the Gaussian terms from the two complementary error functions cancel, becauseThus the exact viscous Burgers step solution with negative flux isFor $X>0$ its denominator is positive, its tails tend to $0$ and $U$, and it approaches the initial step as $X\downarrow 0$. At $\theta =-UX/2$ it equals $U/2$. Its vanishing-viscosity limit gives the entropy shock above. At large $U\sqrt{X/ϵ}$ near the moving shock center it approaches the travelling profile $U/[1+e^{-U(\theta +UX/2)/(2ϵ)}]$, with thickness of order $ϵ/U$.

Factor the forward carrier as $E=e^{ikx}U$. Substitution into the scalar Helmholtz equation gives the exact envelope equationIf $|U_{xx}|\ll 2k|U_x|$, drop the second longitudinal derivative to obtain the parabolic wave equationTypical sufficient conditions are a transverse width $w$ with $kw\gg 1$, longitudinal envelope and material variation on scales much longer than $k^{-1}$, and weak refractive contrast so that factoring out the free-space carrier leaves a slowly varying envelope. A paraxial ordering has transverse propagation angles $O(\eta )$ and ${ϵ}_{\mathrm{rel}}-1=O({\eta }^2)$, with $U_x=O(k{\eta }^{2}U)$ and $U_{xx}$ higher order. A large refractive contrast may require a different, medium-adapted carrier even when the ray direction is nearly axial.

The paraxial approximation retains transverse diffraction and slowly accumulated refraction, but neglects backward propagation, strong reflection from rapid material changes, wide-angle propagation and, in this scalar model, polarization coupling. The supplied vector Helmholtz equation is itself a model assumption: in a source-free inhomogeneous dielectric, Maxwell equations additionally require $\nabla \cdot (ϵ\mathbf{E})=0$, and generally supply a longitudinal-gradient term when the vector wave equation is expanded. An independent scalar component is justified only with a compatible polarization and the relevant small-gradient approximation.

Assume the relative permittivity is real and the field decays sufficiently at transverse infinity, or impose boundary conditions with zero transverse flux. Then $L(x)$ is self-adjoint on its transverse domain by integration by parts. The scalar parabolic wave equation givesThis is norm conservation for the scalar parabolic wave equation. It remains true when $L$ depends explicitly on $x$; its instantaneous self-adjointness suffices. Equivalently, the local balance isSince $|E|=|U|$, the invariant can also be writtenFor a unit transverse polarization in a fixed reference medium with impedance $Z_0=\sqrt{{\mu }_0/{ϵ}_0}$, the leading forward paraxial Poynting vector has $S_x\simeq |E{|}^2/(2Z_0)$. Thus the scalar model conserves its leading reference-normalized power $P_0=\mathcal{N}/(2Z_0)$. This is the intended consequence of the Hermitian-operator hint. Complex permittivity would add absorption or gain and invalidate this norm invariant.

Two qualifications are necessary for the printed physical interpretation. First, $(1,0,1)$ is neither a unit vector nor transverse to nearly axial propagation: its longitudinal electric component is as large as its transverse one. In a homogeneous source-free medium, the example $\mathbf{E}=(1,0,1)e^{ikx}$ satisfies the componentwise Helmholtz equation but violates $\nabla \cdot \mathbf{E}=0$. It therefore is not the stated near-axial electromagnetic plane wave. The same scalar calculation formally conserves $\int |\mathbf{U}{|}^2=2\mathcal{N}$ for that fixed vector, but does not make it a physical Maxwell field. A unit transverse polarization, for example $(0,0,1)$, is the consistent paraxial choice, with only small longitudinal corrections.

Second, for varying local refractive index $n(x)=\sqrt{{ϵ}_{\mathrm{rel}}(x)}$, the quoted local plane-wave impedance relation gives $Z(x)=Z_0/n(x)$ and hence a flux proportional to $n(x)|E{|}^2$, not merely $|E{|}^2$. The scalar norm proof alone cannot make this exactly constant for arbitrary $n(x)$. For example a transversely uniform envelope obeying the printed parabolic equation hasso its intensity is constant while the locally inferred plane-wave flux changes with $n$. This supplies a direct obstruction to the literal exact-flux claim within the supplied scalar assumptions.

The consistent physical statement follows from Poynting theorem: for a lossless Maxwell solution and no lateral energy leakage,For a slowly varying transverse plane wave, the corresponding forward WKB field has amplitude proportional to $n^{-1/2}$, making $n|E{|}^2$ constant. In one dimension this is also encoded in the exact conserved flux $(2{\mu }_0\omega {)}^{-1}\mathrm{Im}(E^{\ast }{E}_x)$. Thus the requested physical conservation holds for the corrected lossless Maxwell description, while the given scalar equation proves the reference-normalized paraxial invariant above. The supplied fixed longitudinal polarization and arbitrary local-impedance inference cannot be used to claim more.

Fix time dependence $e^{-i\omega t}$ and use pressure amplitude for $\psi$. Write the incident acoustic plane wave as ${\psi }_i=A{e}^{i\alpha x-i\beta z}$, where $\alpha =k\cos \theta$, $\beta =k\sin \theta >0$. “Perfectly reflecting” specifies zero absorbed energy but does not uniquely select a boundary condition: a rigid wall has a Neumann boundary condition, whereas a pressure-release surface has a Dirichlet boundary condition. We first give the rigid-wall result, and then the alternative pressure-release result.

For the rigid surface the normal-velocity condition, without an immaterial normalizing factor, isExpand $\psi ={\psi }^{[0]}+{\psi }^{[1]}+{\psi }^{[2]}+\cdots$ in the height amplitude, scaling a fixed differentiable profile. The flat total field is ${\psi }^{[0]}=A{e}^{i\alpha x}(e^{-i\beta z}+e^{i\beta z})$. Taylor expansion givesLet $\hat{h}(q)=\int h(x)e^{-iqx}dx$ and $\beta (q)=\sqrt{k^2-q^2}$ on the outgoing sheet: it is positive real for propagating modes and positive imaginary for evanescent modes. Fourier transforming this boundary derivative gives $2A(k^2-\alpha q)\hat{h}(q-\alpha )$. The outgoing angular spectrum therefore yields the small-height Neumann scattering fieldFor an infinite stationary surface the Fourier quantities may be interpreted as random distributions or through a finite illumination window and its limiting mean. The correction is linear in $h$ and $h^{\prime }$, so zero mean height and stationarity giveOnly the rough correction has zero mean at first order; the flat reflected wave remains.

The first nontrivial coherent correction depends on correlation, not just the r.m.s. height. Define $C_h(s)=⟨h(x+s)h(x)⟩$, its power spectrum of surface height $S_h(q)=\int C_h(s)e^{-iqs}ds$, and ${\sigma }^2=(2\pi {)}^{-1}\int S_h(q)dq$. At second order the flat-field third normal derivative and mixed derivative vanish at zero, leavingFor each first-order outgoing Fourier component, its derivative data are $g_1(q)=2A(k^2-\alpha q)\hat{h}(q-\alpha )$, its second normal derivative is $i\beta (q)g_1(q)$, and its tangential derivative is $q{g}_1(q)/\beta (q)$. Averaging their products with $h,h^{\prime }$ gives the mean derivativeDividing by the specular outgoing derivative $i\beta$ therefore gives oblique coherent reflection from a stationary rigid rough surface:This second-order expression requires the relevant spectral moments to exist and the boundary expansion to remain controlled. Small $|kh|$ alone does not control arbitrarily large slopes or high-frequency roughness. The given stationarity and $\sigma$ determine the first-order mean above, but do not determine this spectrum-weighted correction numerically.

For a pressure-release surface, instead impose $\psi (x,h)=0$. The flat reflected coefficient is $-1$, and Taylor expansion gives ${\psi }^{[1]}(x,0)=2i\beta Ah{e}^{i\alpha x}$. Thus small-height Dirichlet scattering givesThe zero-mean first correction again vanishes on averaging. At second order ${\psi }^{[2]}(x,0)=-h{\partial }_z{\psi }^{[1]}(x,0)$, givingIf the roughness spectrum is concentrated near $q=\alpha$ after the incident tangential shift, the additional long-correlation approximation yields $r_N\simeq 1-2{\beta }^2{\sigma }^2$ and $r_D\simeq -1+2{\beta }^2{\sigma }^2$. A local Gaussian random-height phase model would then give $r_N=e^{-2{\beta }^2{\sigma }^2}$ and $r_D=-e^{-2{\beta }^2{\sigma }^2}$, but Gaussian statistics and that local approximation were not supplied. They must not be inferred from stationarity and r.m.s. height alone.

Translation invariance of the stationary surface statistics implies that a single incident tangential wavenumber retains that wavenumber in its coherent mean. Hence, away from the surface, writeThe unknown effective reflection coefficient $r_b$ is referenced to the mean plane $z=0$. It may be the rigid or pressure-release coefficient just derived, or a more general rough-boundary response. Lossless microscopic reflection does not require $|r_b|=1$: diffuse scattering removes power from the coherent specular field.

For the added planar upper interface, keep the same tangential wavenumber $\alpha$. Let ${\beta }_2=\sqrt{k_2^2-{\alpha }^2}$ on its outgoing sheet. The normal acoustic impedances in the layer and upper medium areContinuity of pressure and normal velocity gives the reflection coefficient for upward incidence from the layer and transmission coefficient for downward incidence from above:If a separate impedance sheet rather than an ordinary fluid interface is intended, its specified effective impedance replaces the upper termination; a sheet impedance has not otherwise been given. These formulas correspond to the stated two-fluid interface.

Let the upper incident pressure amplitude at $z=d$ be $I$. If $D$ and $U$ denote downward and upward layer amplitudes at that plane, their interface and lower-reflection relations areSolving, or summing the geometric series of repeated round trips, gives the coherent acoustic layer above a rough reflector:The expression describes the field at heights where the common homogeneous layer exists; perturbative mean fields near the moving rough boundary need their physical trace evaluated separately. For an upper incident wave normalized as $e^{i\alpha x-i{\beta }_{2}z}$, set $I=e^{-i{\beta }_{2}d}$. With a matched upper medium, $r_{12}=0$ and $t_{21}=1$, the formula reduces to the single lower-reflection field. Zeros of the denominator give ideal layer resonances, interpreted with outgoing continuation or small loss.

This is a coherent specular description: a single effective $r_b$ closes the downward/upward mean amplitudes. Diffuse paths reflected back from the upper interface can modify the mean response; they must either be neglected in this approximation or included in the effective lower reflection coefficient for the actual layered environment. Without a full roughness spectrum and boundary type, a unique numerical mean field cannot be specified beyond this general coefficient form.

Use the Euclidean norm and its induced operator norm. The spectral theorem for real symmetric matrices gives $\Vert A\Vert ={\lambda }_n$ and $\Vert A^{-1}\Vert =1/{\lambda }_1$. If $e=y_{\delta }-y$, then the direct inverse has error $A^{-1}e$, soFor nonzero $y$, the relation $\Vert y\Vert =\Vert Ax\Vert \le {\lambda }_n\Vert x\Vert$ also givesThis is the spectral condition number of a positive-definite matrix. The worst relative amplification is attained by exact data along an eigenvector for ${\lambda }_n$ and noise along one for ${\lambda }_1$. A large $\kappa$ therefore makes the reconstruction sensitive to relative data errors. Absolute sensitivity additionally depends on ${\lambda }_1$ itself: scaling every eigenvalue equally leaves $\kappa$ unchanged but changes $1/{\lambda }_1$. Since the matrix is finite dimensional and positive definite, the inverse exists and is continuous; the issue is ill-conditioning rather than mathematical nonexistence or discontinuity.

For $\alpha >0$, Tikhonov regularization minimizes the strictly convex quadraticDifferentiating in every direction gives the Tikhonov normal equation $(A^{T}A+\alpha I)v=A^T{y}_{\delta }$. Symmetry of $A$ therefore givesThe spectral filter suppresses unstable division by small eigenvalues. Put $e_i=(y_{\delta }-y,u_i)$ and $x_i=(x,u_i)$. The exact noise-bias decomposition for linear regularization isBy orthonormality and the triangle inequality,The scalar gain satisfies $\lambda /({\lambda }^2+\alpha )\le 1/(2\sqrt{\alpha })$, because $(\lambda -\sqrt{\alpha }{)}^2\ge 0$. Hence the Tikhonov stability bound givesThis explicitly separates data noise from regularization bias. For a known bound $X\ge \Vert x\Vert >0$, minimizing this last bound gives an a priori regularization parameter choiceOne data-based bound is $X=(\Vert y_{\delta }\Vert +\delta )/{\lambda }_1$. More generally, $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$ make the robust estimate converge to zero.