Original profile residuals, coordinates and network error bounds

A numerical profile gives predicted fields and their equation errors. To use those errors, we need exact maps back to the original temperature, pressure, velocity, domain and derivatives. This lesson constructs those maps and then proves continuous error bounds for a specified finite network.

Part 1 retains all six original residuals, both curl signs, both integration constants and the exceptional exponent. Part 2 carries the entire system through the sinh coordinates, keeping its metric, Jacobian, gradient factors and sampling measures. Part 3 proves what finite observations can and cannot determine, then calculates explicit derivatives of every network layer and uses the actual grid to bound the errors between samples.

Five solved exercises test the maps on complete fields. They include the original physical pressure and time, two incompatible predictions, exact measure comparisons, and an explicitly specified six-layer network. The last exercise derives every pure second derivative of all six residuals and proves a stronger grid estimate with both coordinate contributions.

The human source is Wang, Lai, Gómez-Serrano and Buckmaster; the original author-source files and passage locators are identified in each part. These calculations are tools for the remaining profile and stability arguments. They do not certify missing trained parameters or a source blowup theorem.

Part 1. The original six profile residuals and complete temperature and pressure reconstruction

This editorial derivation compares the original author sources of Yongji Wang, Ching-Yao Lai, Javier Gómez-Serrano and Tristan Buckmaster, arXiv:2201.06780v3: BoussinesqPRL_v5.tex, lines 106–139 and 169–203, and BoussinesqSupp_v5.tex, lines 180–259. The intact source archive and file hashes are recorded in the private source-use ledger. The main text uses positive oriented curl; the included supplement writes negative curl. Both print the same numerical origin condition. The exact change of sign below changes that condition too. No convention used by unavailable trained weights is inferred here.

The original physical/profile dictionary, including the dynamic clock, pressure and both original cylindrical domains, is proved in lesson 46, CP1–CP20 and DR1–DR42. That link is in the existing private course repository. This note proves the full stationary residual map on the original upper half-plane. It does not construct or certify the numerical profile or its nonlinear stability.

The diagram shows the actual horizontal and vertical paths of RQ4 and RQ12, with the illustrative rectangle \(A=2\), \(B_0=1.5\) and point \(y=(1.2,1.0)\). The triangle is the exact integration region \(0\le t\le s\le y_2\), whose area gives the retained factor \(y_2^2/2\) in RQ18–RQ19. It depicts the proved maps, not a computed profile. Its reproducible Python and SVG sources accompany it.

RQ1. Both original signs and all six errors

Let \(H=\{(y_1,y_2):y_2\geq0\}\), and fix the original exponent \(\lambda>-1\). Write \(\partial_i=\partial_{y_i}\). Take smooth real functions \(U_1,U_2,\Omega,\Phi,\Psi\) on \(H\), smooth to its boundary. Let \(\varepsilon\in\{1,-1\}\) indicate the original sign: \(+1\) for the main text and \(-1\) for the included supplement. Put

and retain the six residuals

No residual in this definition is assumed zero. The supplement's six equations are exactly the case \(\varepsilon=-1\), before changing spatial coordinates. The fields \(\Phi,\Psi\) are initially independent data; their being a temperature gradient is a conclusion only when its precise compatibility condition holds.

The corresponding complete stationary physical-profile equations, including the original force sign and pressure, have errors

Thus the main source force is \((0,\Theta)\); the negative-curl presentation with the unchanged right side \(f_1\) requires force \((0,-\Theta)\). This follows from the curl calculation below, not from a change in the name of the temperature.

RQ2. Reconstruct the scalar and retain its compatibility error

For a real constant \(C\), define the actual finite-path integrals

The horizontal integral is oriented when \(y_1<0\). All these integrals are on finite compact segments, so no condition at infinity is needed. Differentiation and the fundamental theorem of calculus give

Indeed \(\partial_1\Psi=\partial_2\Phi-f_6\); its integral from zero to \(y_2\) is \(\Phi(y_1,y_2)-\Phi(y_1,0)-I\), giving the second identity. Therefore \(f_6=0\) gives the prescribed gradient everywhere. Conversely any smooth scalar with gradient \((\Phi,\Psi)\) has \(f_6=0\) by equality of mixed derivatives. Any two such scalars differ by a constant: integrate their zero gradient on the same two segments. Thus (RQ4) proves both existence and the complete constant freedom for the compatible data, without an unproved global integration step.

Compute \(T\) in (RQ3) using \(\Theta_C\). Differentiating a scalar along \(V\) adds its full derivative \(\partial_iV_j=(1+\lambda)\delta_{ij}+\partial_iU_j\). The two terms \(1-\lambda\) and \(1+\lambda\) sum to two. Using (RQ5) in each component gives the exact identities

For example, the first differentiated equation is

which is exactly the first line. The second is

as in the second line. These calculations retain the effect of incompatible gradient data.

Integration of the actual derivatives in (RQ6) gives a particularly direct scalar error formula:

There is no choice of contour hidden in this expression: it is the stated horizontal segment followed by the stated vertical segment. In particular, when the wall value \(U_2(y_1,0)=0\) holds, both \(I\) and \(\partial_1I\) vanish on that wall, while \(V_2(y_1,0)=0\). The first integrand in (RQ7) then reduces exactly to \(f_2(s,0)\), even if \(f_6\ne0\).

RQ3. Reconstruct the full pressure and its vector error

Set

We now prove, with no residual set to zero,

Expand the curl of the advective term. Its terms containing second derivatives of \(U\) are \(V\cdot\nabla\omega\). The remaining first-derivative products are

Subtracting the curl of \(\lambda U\) leaves \(\omega+V\cdot\nabla\omega+f_4\omega\). By (RQ2), \(\varepsilon\omega=\Omega-f_5\); by (RQ5), \(\partial_1\Theta_C=\Phi-I\). Consequently

which is (RQ9). Both the divergence error and the derivative of the curl-reconstruction error are indispensable in this exact formula.

For any constant \(P_0\), reconstruct

Then \(P_2=K_2\), and

The first equality follows by adding and subtracting the vertical integral of \(\partial_2K_1\) and using the fundamental theorem of calculus. This proves the pressure formula, not merely a necessary curl condition. It also specifies the complete momentum error for arbitrary initial data. Any other pressure solving \(\nabla P=K\) differs by a constant, by the same finite-segment argument used for the temperature.

RQ4. Exact solution correspondence and the exceptional exponent

Now take exactly compatible stationary data, meaning that all six functions in (RQ2) vanish on \(H\). This is a statement about the correspondence of two equation systems, not an assumption that a computed profile solves either of them. Equations (RQ4)–(RQ13) construct the full fields and prove

For \(\lambda\ne1\), the unique constant giving \(T=0\) is

For \(\lambda=1\), that value of \(T\) is independent of \(C\). It vanishes if and only if the displayed scalar product is zero; when it does vanish, every \(C\) is allowed. This includes the exceptional exponent without dividing by zero or dropping a scalar equation.

In the source class, \(U_1,\Omega,\Phi\) are odd in \(y_1\), \(U_2,\Psi\) are even, and \(U_2(y_1,0)=0\). Thus \(U(0,0)=0\) and \(\Phi(0,0)=0\). The scalar product is zero. The source reconstruction uses \(C=0\) if \(\lambda\ne1\), and allows all constants if \(\lambda=1\). The constructed \(\Theta_C\) is even in \(y_1\): the integral of the odd \(\Phi(s,0)\) from zero is even, and the vertical integral of the even \(\Psi\) is even. The vector \(K\) has odd first component and even second component, so (RQ12) gives an even pressure.

When \(C\) is increased by \(c\), the full change of the fields and equations is

The pressure increment follows from the actual vertical term in (RQ12). At \(\lambda=1\) it gives the complete additional solution freedom. It does not change \(\Phi\) or \(\Psi\), so those variables alone cannot encode it.

Conversely a smooth full solution of (RQ3) with \(M=T=D=0\) gives six zero residuals on setting \(\Omega=\varepsilon\operatorname{curl}U\), \(\Phi=\Theta_1\), \(\Psi=\Theta_2\). The fourth, fifth and sixth vanish from their definitions. Formula (RQ11) with \(I=0\) gives \(f_1=0\). Differentiating \(T=0\) by (RQ6) gives \(f_2=f_3=0\). The reconstruction returns the original \(\Theta\) by choosing \(C=\Theta(0,0)\), and the original \(P\) by choosing \(P_0=P(0,0)\). We have proved the exact correspondence in both directions, including both constant freedoms and the exceptional exponent.

RQ5. Finite-domain error bounds with every contributing residual

For \(A,B_0>0\), let \(Q_{A,B_0}=[-A,A]\times[0,B_0]\), and take suprema on this closed rectangle. Let \(\|\cdot\|_\infty\) mean the scalar absolute supremum, or the Euclidean vector supremum as appropriate. From the actual integrals and \(\partial_2I=f_6\),

The complete pressure error obeys

This is the triangle inequality applied to (RQ8) and (RQ13), retaining \(\int_0^{y_2}\int_0^s dt\,ds=y_2^2/2\) for the contribution of \(I\). It includes the nonlinear product and both curl-error derivatives. The rectangle is convex along every integration segment used above. Under the exact source wall condition, (RQ7) gives the sharper scalar bound

The last term again retains the nested integral \(y_2^2/2\). Each factor is the length of its actual segment, not a rescaled unit domain. These are estimates in terms of continuous suprema of the stated fields. They are not supplied by a finite collection of sampled residual values. The next numerical comparison must control those quantities on the original domain and retain the coordinate multipliers used to do so.

RQ6. Coupled sign and full spatial scale maps

Keep \(U,P,\lambda,y\) fixed and change

Then \(M,D\) are unchanged, \(T\) changes sign, \(f_1,f_2,f_3,f_5,f_6\) change sign and \(f_4\) is unchanged. Every assertion follows by substitution into (RQ2)–(RQ3), including the product \(\varepsilon\Theta\). Also \(I,B,C\) change sign, while \(K\) and the reconstructed \(P\) remain the same. The origin derivative transforms as

Thus the printed value \(-1\) in one sign convention becomes \(+1\) in the other under this exact same-velocity correspondence. Keeping \(-1\) in both conventions selects different constrained data; a sign change of the curl variable alone is not the full map.

Finally let \(\kappa\ne0\), put \(H_\kappa=\kappa H\), and define every field on its actual image domain by

Here \(V^\kappa(y)=\kappa V(y/\kappa)\). Chain rule, retaining the factor \(1/\kappa\) in each derivative, gives

For example, each term of \(f_2\) contains either one derivative of \(U^\kappa\), whose factors cancel, or \(V^\kappa\) times one derivative of \(\Phi^\kappa\), whose factors also cancel. The compatibility error has no velocity factor and retains \(\kappa^{-1}\). The original pressure gradient has exactly the factor \(\kappa\) required in the full momentum equation. The inverse is the same map with \(1/\kappa\). If \(\kappa>0\), the image is the original upper half-plane. If \(\kappa<0\), it is the lower half-plane, with its boundary still mapped to the same line. No negative scalar scale is a self-map of the original upper half-plane. The supplement's scalar-variable scaling therefore has the full meaning (RQ22)–(RQ23), including velocity, temperature, pressure, anchor and actual image domain.

These proofs establish exact maps and error bounds for the original profile equations. They do not supply a profile, an error bound for given trained weights, a tail estimate on an infinite domain, or the remaining source stability argument. The next calculation is the complete hyperbolic-sine coordinate and metric map for all six residuals and the continuous quantities in (RQ17)–(RQ19).

Part 2. The complete hyperbolic-sine profile map and its original loss measures

This editorial calculation uses the original author files of Yongji Wang, Ching-Yao Lai, Javier Gómez-Serrano and Tristan Buckmaster, arXiv:2201.06780v3, BoussinesqPRL_v5.tex lines 106–139 and 169–203 and BoussinesqSupp_v5.tex lines 180–259. It continues the complete six-residual reconstruction, RQ1–RQ23. The original author archive is retained intact. The comparison below is an exact change of coordinates and measures; it is not a certificate for the computed profile or its nonlinear stability.

The two grids show the same points before and after \(S\), using \(L_1=L_2=3\) solely for this illustration. The curve is the exact factor \(\operatorname{sech}^2 z_i\). SG19–SG23 prove the gradient and area relations shown; no computed profile is plotted. Reproducible Python and SVG sources accompany the figure.

SG1. Original domain, inverse, metric and all scalar derivatives

The original domain is \(H_y=\{y_2\geq0\}\). Define

Since \(c_i>0\), each coordinate is strictly increasing with range the whole real line, and the displayed inverse is smooth. In particular, \(S\) is a smooth bijection of these closed half-planes, including their boundary. Its full differential, metric and oriented area are

The metric follows by expanding \(|DS\,\xi|^2\) for an arbitrary coordinate tangent vector \(\xi\); the Jacobian is its positive determinant. Every original scalar derivative is represented by

Here and below the derivative of a residual means the derivative of the fully evaluated residual, including every dependence of its predicted fields and their derivatives on the input. It is not a partial derivative holding the network output fixed.

A second differentiation retains the derivative of \(a_i\):

Indeed \(\partial_{z_i}a_i=-a_i\tanh z_i\), and \(Ja_i^2=c_j/c_i\) for \(j\ne i\); differentiating the latter gives the same first-order term. The off-diagonal derivatives commute because \(a_1\) and \(a_2\) depend on different coordinates.

For positive lengths \(L_1,L_2\), the exact finite domain map is

The original upper wall maps to \(z_2=0\). The source's reported full numerical square instead maps \([-L_1,L_1]\times[-L_2,L_2]\) to \([-\sinh L_1,\sinh L_1]\times[-\sinh L_2,\sinh L_2]\). Its two areas are \(4L_1L_2\) and \(4\sinh L_1\sinh L_2\). Its negative second-coordinate part is not part of the original upper half-plane; a field defined there is additional data or an explicitly chosen extension. The stated parity in the first coordinate alone supplies no such extension in the second.

SG2. Cartesian components, coordinate components and full momentum

Keep the original exponent \(\lambda>-1\) and the sign \(\varepsilon\in\{1,-1\}\) of RQ1–RQ3. Let \(V=(1+\lambda)y+U\). The composed Cartesian components are \(\widehat U_i=U_i\circ S\). The components with respect to the coordinate tangent vectors \(\partial_{z_i}S=c_i e_i\) are different, with the exact invertible map

Thus \(|U|^2\circ S=\sum_i c_i^2(u^i)^2\), and for every scalar \(q\),

This proves why the Cartesian components in the author's equations may be composed without being called coordinate velocity components.

For the full stationary momentum error \(M\), temperature error \(T\) and divergence error \(D\) defined in RQ3, the exact formulas in composed Cartesian components are

For the last equality \(Ju^i=c_j\widehat U_i\), where \(j\ne i\); its prefactor has zero \(z_i\)-derivative. If instead momentum is written in the coordinate basis, every original component is divided by \(c_i\), and differentiation of \(\widehat U_i=c_i u^i\) gives

The term \(\tanh z_i\,v^iu^i\) comes exactly from \(a_i v^i(\partial_{z_i}c_i)u^i\); it cannot be removed in that basis. Likewise the pressure coefficient is \(a_i^2\), and the force coefficient is \(a_i\). These bases have the exact residual-norm relation \(|\widehat M|^2=\sum_i c_i^2(m^i)^2\).

The original oriented curl also has both equivalent full expressions:

The second follows because \(c_2\) is constant in the first derivative and \(c_1\) in the second. It retains the original orientation; the sign \(\varepsilon\) enters the definition of \(\Omega\), not this oriented curl.

SG3. All six equation residuals, compatibility and boundary data

Let \(F_j=f_j\circ S\), where all six original \(f_j\) are defined in RQ2. Every term is

Substitution of SG3 and SG7 in each of the six source expressions proves these identities, term by term, without setting any residual to zero. In particular the original gradient one-form and its compatibility error become

Indeed the coefficient of its exterior derivative is \(c_2\partial_{z_1}\widehat\Psi -c_1\partial_{z_2}\widehat\Phi=-JF_6\). Thus testing \(\partial_{z_2}\widehat\Phi =\partial_{z_1}\widehat\Psi\) would test a different one-form.

The wall and origin data map exactly to

The origin equality uses \(c_1(0)=1\). At a general point the factor \(a_1\) remains. The source side \(y_1=\pm d\) maps to \(z_1=\pm\operatorname{arsinh}d\); the upper side \(y_2=d\) maps to \(z_2=\operatorname{arsinh}d\). Derivative conditions on these sides keep all the factors in SG13. Values of \(\Phi\) and \(\Psi\) are simply composed, not multiplied by \(c_i\).

SG4. The complete reconstructed fields and error integrals in these coordinates

The finite-path maps RQ4 and RQ12 retain their full integration factors:

Here \(K=\lambda U-V\cdot\nabla U+\varepsilon\Theta_C e_2\) is the original Cartesian vector of RQ8. The changes of variables \(s=\sinh t\) in each finite original integral give SG14, including the orientation for negative first endpoints. For example, \(D_1\widehat\Theta_C=\widehat\Phi-\widehat I\) and \(D_2\widehat\Theta_C=\widehat\Psi\); no gradient component has been discarded.

The full curl-error integrand and momentum error are

The scalar error, under the exact source wall condition, is

This is RQ7 with the wall cancellation already proved there, followed by the same change of variables. The exact nested integral now has value

Consequently the original rectangle bounds RQ18–RQ19 have \(A=\sinh L_1\), \(B_0=\sinh L_2\), not \(L_1,L_2\). Every derivative \(\partial_{y_i}f_5\) in them is exactly \(a_i\partial_{z_i}F_5\). Suprema of a composed scalar have the same values on the two corresponding compact domains because \(S\) is bijective. This proves the complete use of the earlier residual bounds in the source's computational coordinates.

SG5. Sampled gradient losses and their actual comparison constants

Take corresponding sample points \(y^{(m)}=S(z^{(m)})\), with the same number \(N\) and the same weights \(1/N\). For a fixed original residual \(f_k\), its value loss is unchanged:

Its original-coordinate gradient loss, however, is

Thus the main text's gradient in its \(y\)-formula and a Euclidean \(z\)-gradient in the supplement are related by these exact factors. Writing the sample location as \(z\) does not by itself specify which gradient a program uses. SG19 states both meanings explicitly.

If every sample satisfies \(|z_i^{(m)}|\le L_i\), then

This follows term by term from \(\operatorname{sech}^2 L_i\le\operatorname{sech}^2 z_i\le1\) and the nonnegative squared derivatives. In particular, on the source neighborhood \(|z|<1\),

These positive finite comparison constants do not make the two losses equal. They prove exactly what the bounded near-origin region permits. There is no positive uniform lower constant on the whole coordinate half-plane: at \(z=(M,0)\) take the scalar \(F(z)=z_1\). Then \(|\nabla_zF|^2=1\) and its original-coordinate gradient square is \(\operatorname{sech}^2 M\), which tends to zero as \(M\to\infty\). This is a counterexample for comparison of general scalar derivative norms, not a claimed computed equation residual. All these comparisons require matched points and weights; different sample sets or different counts cannot silently be substituted.

SG6. Full area, boundary and average measures

For every nonnegative measurable function, or every integrable function, the original change of variables gives

The square roots are exactly \(\cosh(\operatorname{arsinh}y_i)\), positive on the whole original domain. In particular uniform area in \(z\) corresponds to the explicit nonuniform density in the third line, and uniform original area corresponds to the fourth line. These formulas keep both original area denominators.

On the horizontal and vertical sides respectively, the exact boundary line measures are

Here \(b\) is the fixed original height of the chosen horizontal side and \(a\) the fixed original first coordinate of the vertical side. The use of these letters in this display does not replace the coordinate factors \(a_i\). The formulas follow from the one-dimensional derivatives of \(\sinh\) and include the wall case \(b=0\).

SG1–SG23 give the complete coordinate, component, residual and measure maps needed to interpret the source losses. Nothing in a small finite sample average supplies the continuous suprema used in RQ18–RQ19. The next calculation constructs the exact fibers of that finite observation map, then derives explicit derivative bounds for finite hyperbolic-tangent networks to turn sample information into continuous finite-domain bounds when their actual parameters are supplied.

Part 3. Finite profile observations and complete network derivative bounds

The original profile equations and all six residuals are defined and reconstructed in RQ1–RQ23. Their complete computational-coordinate map is proved in SG1–SG23. The human source is Yongji Wang, Ching-Yao Lai, Javier Gómez-Serrano and Tristan Buckmaster, arXiv:2201.06780v3, original BoussinesqPRL_v5.tex lines 169–203 and included BoussinesqSupp_v5.tex lines 206–259. The source describes five separate networks with six hidden layers of thirty units and hyperbolic-tangent activation. It also describes parity projection and finite value and residual-gradient losses. This note retains those distinctions.

We first construct the exact family of smooth fields invisible to any given finite collection of sampled derivatives. That family is not asserted to belong to a fixed finite network architecture. We then prove explicit bounds for the stated type of finite architecture, including all layers and all original coordinate factors. No actual trained parameter file is supplied or certified by these formulas.

The left panel gives explicit disjoint supports at \(p=(0.6,0.85)\) and \(\rho p=(-0.6,0.85)\), radius \(0.12\), inside \([-1.5,1.5]\times[0,1.6]\). The sample coordinates are \(\{-1.2,0,1.2\}\times\{0,0.4,1.5\}\). The right panel evaluates the exact derivative formulas NB17–NB18 for the fully specified network NB16. These illustrate the two respective function classes; neither is identified with the source-trained profile. Reproducible Python and SVG sources accompany the figure.

NB1. The smooth finite-observation fiber

Fix a finite set \(\Sigma\subset H_y\), containing every location used by all value, derivative, boundary and origin samples under consideration. If different residuals use different sets, take their finite union. Fix any smooth original quintuple \((U_1,U_2,\Omega,\Phi,\Psi)\) and its exponent \(\lambda>-1\). For the parity conclusion, take the source class: \(U_1,\Omega,\Phi\) odd in \(y_1\), \(U_2,\Psi\) even, with \(U_2(y_1,0)=0\).

Choose \(p=(p_1,p_2)\) with \(p_1,p_2>0\), outside \(\Sigma\) and its reflection in the first coordinate. Such a point exists since an open rectangle in this quadrant contains infinitely many points and the excluded set is finite. If a finite source rectangle \([-A,A]\times[0,B]\), with \(A,B>0\), is specified, choose \(0<p_1<A\), \(0<p_2<B\). Choose \(\delta>0\) smaller than one quarter of the positive minimum of \(p_1,p_2,A-p_1,B-p_2\) and the distances from \(p\) to every point in \(\Sigma\cup\rho\Sigma\), omitting the last minimum if that set is empty. Here \(\rho(y_1,y_2)=(-y_1,y_2)\). If no rectangle was initially specified, choose one containing \(p\) in its interior first. The two closed balls of radius \(\delta\) at \(p\) and \(\rho p\) are disjoint, strictly above the wall, inside this rectangle, and disjoint from \(\Sigma\).

Define the complete cutoff

The denominator is positive for every real \(s\): simultaneous zeros would require \(s\ge1\) and \(s\le1/4\). The function \(\eta\) is smooth at zero because each of its derivatives for \(t>0\) is \(e^{-1/t}\) times a polynomial in \(1/t\), and all such products tend to zero there. To see the last assertion without a growth assumption, put \(x=1/t\); for an integer \(m\ge0\), \(e^x\ge x^{m+1}/(m+1)!\) implies \(x^me^{-x}\le(m+1)!/x\to0\). Repeated differentiation and the zero one-sided derivatives therefore prove the smooth extension. Consequently \(\chi\) is smooth, equals one for \(s\le1/4\) and zero for \(s\ge1\).

Put

These are smooth and supported in the two specified balls. The function \(h\) is odd in \(y_1\), so the first component of \(W_h\) and \(Z_h\) are odd, and the second component of \(W_h\) is even. Direct mixed differentiation gives

For any real amplitude \(m\), define the actual perturbed fields

They preserve the original parity, the wall condition, every boundary value outside the two balls and the original far-field behavior. They also preserve every derivative of every field at every point in \(\Sigma\): each perturbation vanishes on an open neighborhood of that point. As a result every sampled derivative of every local residual is exactly unchanged, not merely small. In particular all the source finite value and residual-gradient losses are unchanged. Equations (NB3) show the stronger global facts

Near \(p\), the first cutoff is one and the reflected term is zero, so \(h(y)=|y-p|^2/4\) on an open neighborhood. Since the dimension is two, at the exact center

Therefore the full first residual, including its nonlinear advection, obeys the exact formula

Indeed its velocity perturbation and its vorticity-gradient perturbation both vanish at \(p\), whereas the undifferentiated vorticity changes by \(\varepsilon m\); the right-side field \(\Phi\) is unchanged. Thus the continuous supremum of \(|f_1^{[m]}|\) on the rectangle is unbounded as \(|m|\to\infty\), while all those finite observations agree exactly. This proves the limitation in the full smooth source parity class without claiming a singular solution or a network counterexample.

The observation fiber contains an infinite-dimensional affine space of such fields. In a small open subrectangle disjoint from the finite set and its reflection, choose countably many pairwise disjoint closed balls with positive radii tending to zero, also disjoint from their reflected balls. For example, inside a ball of radius \(r\) at a point \(q\) in that subrectangle, centers \(q+(r2^{-n},0)\), \(n\ge2\), and radii \(r2^{-n-3}\) have the required pairwise separation. Choose \(r\) small enough that the containing ball and its reflection remain disjoint and interior. Repeat NB1–NB2 there. Any finite linear combination remains smooth and compactly supported away from the samples. The resulting pairs \((W_h,Z_h)\) are linearly independent: evaluation of \(Z_h\) at each associated center isolates its coefficient by NB6. Only finite linear combinations are used; no unproved smoothness of an infinite sum is needed.

NB2. Why the fixed finite network is a different, explicitly related class

Every finite real network built from affine maps and \(\tanh\) is real analytic on the real input plane. The denominator \(\cosh x\) never vanishes there, so \(\tanh x=\sinh x/\cosh x\) is real analytic on a neighborhood of each real point; finite composition and addition preserve that property. Its odd or even first-coordinate projection is real analytic too.

A nonzero compactly supported smooth function such as NB2 cannot equal such an analytic output on the connected open half-plane. If equality held, the analytic function would vanish on a nonempty open subset outside its support. Along any straight segment from that subset to an interior point, its restriction is a real analytic function of one real variable on a neighborhood of the segment. An analytic function vanishing on an interval vanishes on the entire connected interval: its zero neighborhood extends past any interior endpoint by the convergent Taylor series there, all of whose derivatives are zero by continuity from the zero side. Compactness of the segment then continues this identity to its other endpoint. The half-plane is convex, so every interior point is reached in this way. This contradicts NB6.

Thus NB1–NB7 describe an exact obstruction and its smooth observation fiber, but do not by themselves settle injectivity or error control in a specific finite architecture. We now derive the actual finite-network derivative bounds needed to compare that architecture with the same continuous residuals.

NB3. Every layer, bias and derivative of a finite tanh network

Let the network input be \(z=(z_1,z_2)\). For a network with \(L\) hidden layers, define every layer by

All weights and biases are arbitrary finite real numbers; no value is assigned to a source-trained parameter. The source architecture is the special case \(L=6\), \(d_0=2\), \(d_1=\cdots=d_6=30\). If the actual input is \(y\) instead, NB8 and the derivative proof apply in that variable first, followed by SG1–SG4. This does not infer an input convention from a symbol in a loss display.

NB8 explicitly uses an affine output. If an implementation puts a final \(\tanh\) after that affine map, its derivatives are obtained by one additional application of NB12–NB14 with its actual output weights and bias; the proof below covers that composition as well.

Write \(t=\tanh x\). Direct differentiation gives

Since \(-1<t<1\), their absolute bounds are respectively

For the second, maximize \(2r(1-r^2)\) on \(0\le r\le1\); its derivative is \(2-6r^2\), and its value at \(r=1/\sqrt3\) is the displayed constant. For the third put \(s=t^2\in[0,1]\). The polynomial \(-2+8s-6s^2\) has endpoints \(-2,0\) and its only interior critical point at \(s=2/3\), with value \(2/3\). These observations prove all three absolute bounds, including their extrema, without a numerical maximization.

For each hidden component and coordinate indices \(k,m,n\in\{1,2\}\), define nonnegative derivative bounds recursively. The input values are

Given the previous layer, first retain its complete weighted sums

The letters in NB12 are derivative bounds; the actual affine bias in NB8 is the separately indexed \(\beta_i^{(\ell)}\). Set

These bound the absolute first, second and third mixed partial derivatives of \(a_i^{(\ell)}\) on the entire input plane. Here is the induction. Differentiation of the affine expression NB8 gives the sums of previous derivatives, bounded by NB12. For a scalar \(b(z)\), the complete composition derivatives are

Each follows by the product and chain rules, also when indices repeat. Apply NB10 and the triangle inequality to get exactly NB13. NB11 starts the induction. All biases in NB8 remain in the network; their derivatives vanish and the uniform bounds NB10 are valid at their shifted arguments.

For the output, the complete bounds are

The first uses \(|\tanh|\le1\); the others follow by differentiating the affine output and using the proved layer bounds. For the exact odd or even projection \(q_\pm(z)=[q(z)\pm q(\rho z)]/2\), the same bounds hold. Every reflected derivative has the appropriate factor \(-1\) for each first-coordinate derivative, of absolute value one, and the factor one half cancels the sum of the two equal bounds. Thus the source parity construction is fully included.

NB4. An exact six-layer example of the bounds

Take the stated six-layer, thirty-unit shape. In every hidden layer keep only the first unit active, put its incoming first-channel weight equal to one, and all other weights and all biases equal to zero. The first layer's first channel is the first input coordinate. The output reads that active unit with weight one. The resulting exact function is

It is independent of \(z_2\) and already odd in \(z_1\). For derivatives in the first coordinate, the recurrence gives

The first two follow immediately from NB13. For the third, its increment is \(3C_2B_{\ell-1}+2=3C_2^2(\ell-1)+2\); summing all increments from one to \(\ell\) gives NB17. In particular

At the origin, direct induction in NB14 gives \(q(0)=0\), \(q_1(0)=1\), \(q_{11}(0)=0\), and \(q_{111}(0)=-12\): each of the six odd compositions adds \(-2\) to the third derivative there. All derivatives involving the second coordinate vanish. This is a fully specified example in the actual layer shape, not the source-trained profile or a solution claim.

NB5. Retain the original derivative factors in all residual bounds

For each predicted Cartesian scalar \(q\in\{\widehat U_1,\widehat U_2,\widehat\Omega, \widehat\Phi,\widehat\Psi\}\), use NB15 to set \(Q_1(q)=\max_k Q_{1,k}\), \(Q_2(q)=\max_{k,m}Q_{2,km}\) and \(Q_3(q)=\max_{k,m,n}Q_{3,kmn}\), retaining the individual arrays when a sharper value is needed. These are explicit bounds, not replacement definitions of the original fields. For the original derivative \(D_iq=a_i\partial_iq\) of SG3, differentiation gives

The full factors are displayed before taking their bounds. They satisfy \(|a_i|\le1\), \(|a_i'|\le1/2\), \(|a_i''|\le1\). For the middle bound, its square is \(t^2(1-t^2)\le1/4\); the last follows from \(|2t^2-1|\le1\) and \(\operatorname{sech}\le1\). Consequently

Work on the actual coordinate rectangle \(Z_L\) of SG5. Put \(p=1+\lambda\) and retain the Cartesian drift components \(A_i=p\sinh z_i+\widehat U_i\). Their full bounds are

We use \(A_iD_iq\), exactly equal to \(v^i\partial_iq\), in the following calculation. The original drift and the original derivative factor thus remain identifiable rather than being replaced by a different transport field.

The full derivative of the first residual is

It has the explicit bound

For the next two residuals, set \(G_1=\widehat\Phi\), \(G_2=\widehat\Psi\), take \(i\in\{1,2\}\) and \(j=3-i\). Their complete original expression and derivative are

The resulting explicit bounds are

Finally differentiating the last three expressions of SG11 gives

For every \(r=1,\ldots,6\), these formulas prove \(|\partial_kF_r|\le H_{rk}\) on the complete original computational rectangle. The sign \(\varepsilon\) in \(F_5\) has absolute value one; no sign choice is imposed on the fields themselves. Every quantity on the right is determined by the specified weights, biases, exponent and actual domain lengths, via the proved finite sums.

NB6. From actual point data to continuous finite-domain error bounds

Let sample points \(z^{(m)}\in Z_L\) cover the rectangle in the following precise sense: for each \(z\in Z_L\) there is one sample with \(|z_k-z_k^{(m)}|\le h_k\) for \(k=1,2\). For a tensor grid containing both endpoints in each coordinate, this holds with half of its largest mesh gap in each coordinate: choose the nearest coordinate node in each interval and use their tensor product. No covering property is inferred merely from a sample count.

The straight segment between \(z\) and that sample stays in the rectangle. The fundamental theorem of calculus along the segment gives

This proves a continuous bound for any particular network whose parameters and sample values are supplied exactly or bounded by verified intervals. All terms are explicit. If only the exact average loss \(\ell_r=N_r^{-1}\sum_{m=1}^{N_r}|F_r(z^{(m)})|^2\) is available, nonnegativity of every summand gives the full conversion

The factor \(\sqrt{N_r}\) cannot be dropped. If the full source cost is

and all its terms are nonnegative, then \(\ell_r\le6J_{\rm cost}/\gamma\) and NB28 becomes \(\max_m|F_r(z^{(m)})|\le\sqrt{6N_rJ_{\rm cost}/\gamma}\). For the stated \(\gamma=1/10\), this is \(\sqrt{60N_rJ_{\rm cost}}\), before adding the two coverage terms in NB27. It is a bound on the actual value samples entering that cost, not samples belonging to a different loss.

To use RQ18–RQ19, transfer these continuous bounds by SG14–SG17. The original lengths remain \(\sinh L_1\), \(\sinh L_2\), and the original derivative in the pressure error is \(a_i\partial_iF_5\). Its absolute bound is supplied by NB26 with the full factor of SG3 retained before taking a supremum. Wall and anchor errors require their actual values too: RQ6 supplies the scalar value at the origin, and the sharper scalar estimate applies only to a wall condition that is exact. If the wall is merely sampled, use the full RQ7 horizontal integrand or bound its wall error; do not silently replace sampled impermeability by exact impermeability.

NB27 is the concrete next operation after the smooth observation obstruction: finite-network parameters give proved derivative bounds, and an actual covering radius turns finite observations into continuous finite-domain errors. It does not establish the missing source-trained parameter bounds, infinite-domain tails, an exact nearby profile or its nonlinear stability. Those are distinct subsequent calculations with their original hypotheses and equations retained.

Five solved exercises

These exercises use the exact fields and signs of RQ1–RQ23, the coordinate map SG1–SG23 and the network bounds NB1–NB29. Write \(H=\{(y_1,y_2):y_2\ge0\}\), \(\lambda>-1\) and \(\varepsilon\in\{-1,1\}\). The original velocity is Cartesian. The residuals, scalar and pressure are the original objects of those parts; no coordinate factor or additive constant is absorbed into them.

Exercise 1. An entire affine family, including pressure and physical time

Take real constants \(\alpha,K,C,P_0\) and the fields

Find the pressure, all six profile errors, the physical fields and every restriction needed for these fields to solve the original equations. Check the source wall, parity, origin and far-field conditions.

Solution. The full drift is

Thus the first component of \(-\lambda U+V\cdot\nabla U\) is \(\alpha(1+\alpha)y_1\), and the second is \((\alpha^2-\alpha)y_2\). Subtracting \(\varepsilon\Theta e_2\) and integrating the two required pressure derivatives gives

Direct differentiation now gives the original momentum residual \(M=0\) and divergence \(D=0\). The scalar residual and all six predicted-field residuals are

Indeed, the curl of \(U\) is zero, its divergence is \(\alpha-\alpha=0\), both predicted scalar gradients are constant, and \(2+\partial_2U_2=2-\alpha\). The scalar equation follows by adding \((1-\lambda)(C+Ky_2)\) to \((1+\lambda-\alpha)Ky_2\). Consequently the complete equations hold exactly when

Both conditions are necessary because a constant and the function \(y_2\) are linearly independent on \(H\); they are sufficient by EX2–EX3. In particular the first condition is not detected by the six derivative residuals alone, and \(C\) is unrestricted when \(\lambda=1\).

For physical time \(t<1\), retain \(\tau=1-t>0\) and \(y=x/\tau^{1+\lambda}\). The original similarity map gives

To check momentum directly, \(\partial_tu_1=\alpha x_1/\tau^2\) and \(u\cdot\nabla u_1=\alpha^2x_1/\tau^2\). Their sum cancels \(\partial_1p\). Similarly the second acceleration is \((\alpha^2-\alpha)x_2/\tau^2\); adding \(\partial_2p\) leaves exactly \(\varepsilon\theta\). The time-dependent pressure constant is retained and has zero spatial gradient.

The wall condition \(U_2(y_1,0)=0\) and the original horizontal parities hold: \(U_1,\Omega,\Phi\) are odd in \(y_1\), and \(U_2,\Psi,\Theta,P\) are even. But the required source anchor \(\partial_1\Omega(0)=-1\) fails because it is zero here. The far-field condition \(\nabla U\to0\) forces \(\alpha=0\), and \(\Psi\to0\) forces \(K=0\). Neither restores the anchor. Moreover, if \(\alpha\ne0\), the original kinetic energy is infinite: on each \([-R,R]\times[0,R]\),

This family solves exactly the equations under EX4. It does not supply the source's anchored decaying profile or a finite-energy blowup example.

Exercise 2. Two incompatible predictions and their exact reconstructed fields

Calculate every error and the RQ finite-path reconstruction for each of the following quintuples. What does a successful reconstructed momentum equation say about the separately predicted fields?

Solution. For A, direct use of the six original definitions gives

The two advection contributions to \(f_2\) are each \((1+\lambda)y_1y_2\), in addition to \(2y_1y_2\). The path correction is \(I=y_1y_2\). Since \(\Phi(y_1,0)=0\) and \(\Psi=0\), the reconstructed scalar is \(\Theta=0\). The entire pressure curl integrand is \(B=f_1+I=0\), so \(P=P_0\) and \(M=T=D=0\). Nevertheless \(\Phi\ne\partial_1\Theta\) away from its zero set. RQ's exact identity is \(\partial_1\Theta=\Phi-I\), which records precisely this mismatch. For example, \((2+\partial_1U_1)I+V\cdot\nabla I=(4+2\lambda)y_1y_2\), so the full formula for \(\partial_1T\) cancels EX8 exactly.

For B, divergence is \(y_2-y_2=0\), whereas the actual curl is \(-y_1\). Thus

Retaining both contributions to each component of \(V\cdot\nabla U\) gives the pressure target and its curl integrand

The horizontal path contribution to pressure is zero. Its vertical contribution and the resulting residual are exactly

Both equal direct differentiation; also \(M_1=-\varepsilon\int_0^{y_2}B(y_1,s)\,ds\), retaining the sign twice. Since \(\lambda>-1\), the factor multiplying \(|y_1|\) is nonnegative and increasing in \(y_2\ge0\). On the original rectangle \([-A,A]\times[0,B_0]\), with \(A,B_0>0\), its exact supremum is

The generic RQ18 bound instead yields \(A((\lambda+2)B_0+B_0^2)\): it first bounds the full drift by its rectangle supremum. Integrating the actual drift as in EX11 retains the smaller second term. Both quintuples obey the wall and horizontal parities, but neither obeys the origin vorticity anchor. Neither example changes a claim about a profile satisfying all six equations and all the original source conditions.

Exercise 3. Two different averages of an actual residual

Let \(U=0\), \(\Omega=y_1/(\lambda+2)\), and \(\Phi=\Psi=0\). Calculate the original and coordinate averages of the first residual and its gradient on the corresponding rectangles.

Solution. The denominator is nonzero because \(\lambda>-1\). The complete residual list is

For \(y_i=\sinh z_i\), set

Write \(d_i=\sinh L_i\). The original area is \(2d_1d_2\). Integrating the full scalar numerator, or using the full Jacobian, gives

The unweighted coordinate numerators and area \(2L_1L_2\) instead give

The original scalar mean is strictly larger than the coordinate scalar mean for every \(L_1>0\). To prove this without discarding a factor, put \(f(s)=\sinh^2s\), \(g(s)=\cosh s\) on \([0,L_1]\). Their exact difference is

Expansion of the double integral gives the equality, because \(\int_0^{L_1}g=\sinh L_1\). Strict increase of both functions makes the integrand positive off the diagonal, except on sets of area zero. These are averages of the same realized residual, under the two specified measures. The original-coordinate gradient under uniform coordinate sampling still has mean one:

Thus the derivative factor and the sampling measure have separate, explicit roles. No finite sampled average is equated to one of these integrals without a further quadrature calculation.

Exercise 4. A specified network that satisfies the anchor

Use the six-hidden-layer, thirty-unit architecture of NB8. In one channel take all six successive weights equal to one, its first input equal to \(z_1\), and all other weights and biases zero. Take the final output weight to be one. This gives

Use four all-zero networks for \(\widehat U_1,\widehat U_2, \widehat\Phi,\widehat\Psi\), and negate the output weight of EX19 to obtain \(\widehat\Omega=-q\). Compute the anchor, all residuals and explicit continuous error bounds on \(Z_L\).

Solution. Each layer is odd and strictly increasing. Each finite preactivation has positive \(\tanh'\), so their composition has \(q'>0\), \(q(0)=0\), \(q'(0)=1\). All prescribed horizontal parities hold, and the velocity wall condition is exact. Since \(\partial_{y_1}=\operatorname{sech}z_1\partial_{z_1}\), the origin anchor is exactly \(\partial_{y_1}\Omega(0)=-1\). With \(p=1+\lambda>0\), the entire residual list is

In particular \(F_1(0)=0\), but \(\partial_1F_1(0)=-(\lambda+2)\). At every positive \(z_1\), both nonzero terms of \(F_1\) are negative. Thus the network does not solve the profile equations despite its exact anchor. Monotonicity and oddness give \(\|F_5\|_\infty=q(L_1)>0\).

Differentiate before combining the coordinate factors:

The second line is the exact identity of the full first line; no original derivative factor has been discarded. NB18 gives \(|q'|\le1\) and \(|q''|\le B_6=8/\sqrt3\). Write \(t_L=\tanh L_1\). Since \(1-r^2+B_6r\) has derivative \(B_6-2r>0\) for \(0\le r\le t_L<1\), we obtain

For a tensor grid containing both rectangle endpoints, let \(\Delta_1\) be its largest first-coordinate gap. The full NB27 bound becomes

The exact value \(q(L_1)\) is sharper for \(F_5\) and is already attained at endpoint nodes. These bounds use actual parameters; no trained source weights have been substituted. RQ reconstructs \(\Theta=0\), \(P=P_0\) and \(U=0\) from this example with \(C=0\), because \(B=F_1-F_5-p\tanh z_1\partial_1F_5=0\). That exact reconstructed solution still has curl zero, different from the network's predicted \(-q\). The nonzero \(F_5\) records it.

Exercise 5. A sharper bound between grid nodes

Use the network derivative arrays of NB15 and the actual coordinate rectangle \(Z_L\). Derive explicit bounds for both pure second derivatives of every original residual. Prove a cellwise estimate with the complete quadratic mesh terms and apply it to Exercise 4.

Solution. Retain the definitions \(D_i=a_i\partial_i\), \(a_i=\operatorname{sech}z_i\), \(A_i=(1+\lambda)\sinh z_i+\widehat U_i\), and every constant \(Q_0,Q_1,Q_2,Q_3,E_{ki},M_i,N_{ik}\) from NB15–NB21. For any one of the five predicted scalar components \(q\), the complete second derivative is

Here \(|a_i'|\le1/2\) cancels the displayed factor two only after the full derivative is shown. Both Kronecker symbols are retained. These are bounds on the original \(D_iq\), not on a replacement derivative. Every product is differentiated by

For the first residual this gives the exact formula

Its continuous bound is

For \(G_1=\widehat\Phi\), \(G_2=\widehat\Psi\), \(i\in\{1,2\}\) and \(j=3-i\), the complete formula is

Applying EX24 and NB20–NB21 to each displayed factor proves

Finally the last three equations are linear in the indicated fields and original first derivatives. Their exact second derivatives are

Since \(|\varepsilon|=1\), their bounds are

We have proved \(|\partial_k^2F_r|\le K_{rk}\) on the entire rectangle, for all six original residuals and both coordinate directions. Mixed derivatives were not set to zero; the next argument explains why the two pure derivatives suffice for this particular bound.

For a twice continuously differentiable scalar \(f\) on \([a,b]\), with \(a<b\), its linear interpolant is

Taylor's formula with an integral remainder around \(x\) gives

For completeness, the first remainder follows by integrating \(f'(s)-f'(x)=-\int_s^xf''(t)\,dt\) from \(a\) to \(x\) and reversing the order over \(a\le s\le t\le x\); the second follows over \(x\le t\le s\le b\). Both kernels in the last line are nonnegative. Their full integral is

It follows that \(|I_{ab}f-f|\le(b-a)^2\|f''\|_\infty/8\).

On a grid cell \([a_1,b_1]\times[a_2,b_2]\), write \(\Delta_i=b_i-a_i>0\) and let \(I_i\) interpolate in coordinate \(i\) alone. The exact decomposition is

The weights of \(I_1\) are nonnegative and sum to one, so it does not increase a supremum bound. Applying EX30 in each coordinate therefore proves

The last step uses the four nonnegative product weights of the bilinear interpolant, whose sum is one. Applying this argument to every cell gives the same bound with the largest actual coordinate gaps and the maximum over all grid nodes. The constants can instead be bounded per cell for a tighter estimate. No derivative in a mixed direction is needed in this interpolation identity.

The coefficient \(1/8\) and the sum of its two contributions are sharp for the stated scalar \(C^2\) class. Given \(k_1,k_2\ge0\), take the full polynomial

All four corner values vanish, \(\partial_i^2F=-k_i\), and its cell-center value is exactly \((k_1\Delta_1^2+k_2\Delta_2^2)/8\). This proves scalar sharpness, without claiming that this polynomial is a residual of the fixed network architecture.

For Exercise 4, put \(t=\tanh z_1\). Differentiating the complete identity EX21, or differentiating its first line twice through the original factors, gives

Here the three product contributions use, respectively, \(|t''|\le4/(3\sqrt3)\), \(|t'|\le1\), \(|t|\le t_L\), and the full NB18 bounds \(1,8/\sqrt3,116/3\). The first bound on \(t''\) follows by maximizing \(2r(1-r^2)\) on \([0,1]\): its derivative vanishes at \(r=1/\sqrt3\), giving the displayed value; both endpoint values are zero. Thus no extra network derivative has been assumed. Combining the two independently proved estimates,

For \(0<\Delta_1<4H/K_*\), the quadratic contribution in the first line is strictly smaller. Both constants are positive by \(p>0\). The second line remains improvable to the exact endpoint value from Exercise 4.

For a supplied network, EX24–EX31 give a proved finite-domain continuous error estimate using its parameters and grid values. SG and RQ carry this estimate to the original scalar and pressure errors, retaining \(\sinh L_i\), the original derivative factors, and the wall and origin terms. The estimate does not supply missing trained parameters, infinite-domain tails, a nearby exact profile or nonlinear stability. Those require their own complete calculations.

The left panel shows the complete scalar polynomial EX32 on \([-2,1]\times[0,2]\), with \(k_1=3\), \(k_2=2\): its four corner values are zero and its center value is exactly \(35/8\). This is the scalar sharpness example, not a fixed-network residual. The right panel plots the two proved error contributions in EX34 for the actual six-layer network EX19, with \(\lambda=1/2\) and \(L_1=1\). Their crossing is exactly \(4H/K_*\), about \(0.433\); the curves are bounds, not observed interpolation errors. Reproducible figure source and vector rendering. The human profile and architecture source is Wang, Lai, Gómez-Serrano and Buckmaster, arXiv:2201.06780v3, with the exact original author-source locators recorded in RQ, SG and NB. The exercises and interpolation derivations here are supplementary calculations; no novelty claim is made.

Continue with the profile and stability arguments

The reconstructed fields and every continuous error contribution are now explicit. The grid estimates apply to the actual finite parameters and domain supplied to them. Infinite-domain control and stability still need their own estimates and computational inputs.

The next comparisons retain the exact real Burgers branch, the CCF anchor and the full approximate-profile defect in its mode identities. The series then continues through the original analytic and computational stability arguments, Alpöge–Buckmaster, OpenAI and retained workbench constructions.