A locally integrable function has weak derivative in direction when is locally integrable andThe test functions have compact support inside the domain, so no boundary values enter this definition. Such is unique almost everywhere and is denoted .
For , the first-order Sobolev space consists of the Lp space classes whose first weak derivatives are also in that Lp space:This Sobolev norm is equivalent to ; either convention is acceptable. When , this is the Hilbert space .
The zero Dirichlet boundary condition must be part of the definition. Thus is a weak solution precisely when , equivalently its Sobolev trace vanishes on , andHere is the indicator function of the measurable set . One may initially require the identity only for test functions : both sides are continuous in the Sobolev norm, and the definition of the zero-trace Sobolev space then extends it to all . This formulation uses first weak derivatives and encodes without assuming classical second derivatives.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 10 2 a Solution Created 2026-10-03 Updated 2026-10-14
Write . Use a submultiplicative Sobolev algebra norm for which the Fourier projections are contractive, and its compatible matrix/column norms. The numeric hypothesis is understood in this algebra norm; the unscaled Fourier Sobolev norm need not have multiplication constant one.
Let and denote the strictly positive and strictly negative Fourier subspaces. The equation is a contraction on , so it has a unique solution. Put and . The equation says , and . Similarly the contraction gives with . No invertibility of these solutions has yet been assumed.
Now belongs to both Fourier subalgebras, hence is a constant matrix . Taking constant coefficients on either side shows . We claim is invertible. If , then the Hardy compression annihilates , because . But on the Hardy Sobolev space has inverse given by a Neumann series, since the perturbation has norm less than one. Thus , and its constant coefficient is , forcing . Finite-dimensionality proves the claim.
Consequently is a left inverse of , and is a left inverse of . Matrix entries are functions in a commutative algebra, so pointwise finite-dimensional inversion makes these left inverses two-sided inverses in their respective Fourier subalgebras. SetThen , both factors are invertible in the stated algebras, and . If two normalized factorizations exist, lies in and is constant; evaluating the analytic side at zero makes it . This proves uniqueness. The argument is near-identity Sobolev loop factorization and, crucially, proves invertibility rather than treating the contraction solution itself as automatically invertible.
Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 10 6 Solution Created 2026-10-03 Updated 2026-10-14
On the torus , choose normalized Haar measure, so that , , is an orthonormal basis of . A periodic Sobolev space is defined, for any real , byFor negative , the coefficients describe distributions rather than necessarily functions. For a nonnegative integer , the weight is comparable to , so this norm is equivalent to the sum of the squared norms of all weak derivatives through order . Fourier truncation proves that smooth trigonometric polynomials are dense. The pairing identifies the continuous dual of with , giving Sobolev duality.
Two Fourier estimates explain the usefulness of these spaces. If , the Cauchy-Schwarz inequality givesUniform convergence of the differentiated Fourier series proves the Sobolev embedding . If , a bounded subset of has uniformly small high-frequency tails in :Finite-dimensional Fourier truncation therefore proves the compact embedding , the periodic version of Rellich-Kondrachov compactness theorem. In addition, for multiplication is continuous , making it a Sobolev algebra. Indeed the weighted coefficient of a product is bounded by the sum of the convolutions of with and of with ; Cauchy-Schwarz puts unweighted coefficients in , and the case of Young's convolution inequality finishes the estimate.
For eigenfunction expansions, distinguish ellipticity from self-adjointness. An arbitrary non-self-adjoint elliptic operator need not have an orthonormal basis of eigenfunctions. A useful definite setting iswith smooth real periodic coefficients, , real , and uniform ellipticityTake the self-adjoint realization in with domain . Its construction and spectral properties can be seen from the associated form on ,For a sufficiently large constant , is a positive coercive form: , and . The Lax-Milgram theorem supplies the unique weak solution of for ; the form is Hermitian, which gives symmetry of its inverse on .
The regularity input is the periodic elliptic estimateHere is why an elliptic operator gains two derivatives. With coefficients frozen at a point, its principal symbol dominates , and the estimate follows by multiplying Fourier coefficients by that symbol. On a sufficiently small coordinate patch, the difference of the leading coefficients from these constants has arbitrarily small supremum norm. In the integer-order differentiated estimate, this absorbs the highest derivatives into the left side; derivatives falling on coefficients contribute only lower-order terms. A partition of unity reduces the variable-coefficient problem to such patches. Its commutators with have order one, controlled bywhich follows by splitting low and high Fourier modes. Absorbing these terms gives the displayed estimate. For a weak solution, the same localization with difference quotients first bounds its second weak derivatives for ; differentiating and repeating then gives for nonnegative integer . Interpolation and duality extend the scale to real orders. In particular is bounded.
Let . Compactness of makes a compact operator. Symmetry of the form makes self-adjoint, and positivity of makes positive and injective. Its range is dense because . The spectral theorem for compact self-adjoint operators consequently gives an orthonormal basis withElliptic regularity bootstraps every to a smooth function. Each eigenvalue has finite multiplicity. Thus a self-adjoint uniformly elliptic second-order operator on the torus has a complete smooth orthonormal eigenfunction expansion with discrete eigenvalues tending to infinity.
Writing , every has in , with . Iterating the elliptic estimate shows that the graph norm of is equivalent to the norm. Interpolation and duality give the more general spectral descriptionThe comparison constants depend on the operator and . In particular smooth functions have coefficients decaying faster than every power of , and their expansions converge in every Sobolev norm and hence in every norm.
This gives a practical way to solve elliptic equations. The equation is solvable if and only if whenever ; then one may take on the other eigenspaces and add an arbitrary vector in . The Sobolev description proves the two-derivative gain and the Fredholm alternative, rather than leaving them as formal series manipulations. Similarly the solution of the heat equation isFor every the exponential decay at high eigenvalues places this solution in every Sobolev space, hence it is smooth.
For the Laplacian the eigenfunctions are precisely with eigenvalues . The number with is the number of lattice points in a radius- ball, asymptotic to its volume by comparison with unit cubes. For the general self-adjoint operator above, bounds on its form above and below by positive multiples of plus constants, together with the Courant–Fischer min-max principle, imply . Thus the growth reflects dimension and the two-derivative order. Compactness of the torus is essential: on , translated bumps prevent the compact embedding, and the Laplacian instead has continuous spectrum described by the Fourier transform.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 13 2 a Solution Created 2026-10-03 Updated 2026-10-14
The space is the closure of in the Sobolev norm; membership imposes the homogeneous Dirichlet condition in the weak sense, without assuming that the boundary is smooth. A weak solution is an element of that space satisfyingRepeated indices are summed. The minus sign comes from integration by parts in the divergence equation. Boundedness of the coefficients, Hölder's inequality and density extend this identity to every . Symmetry permits interchanging the derivative indices in the left side.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 13 2 b Solution Created 2026-10-03 Updated 2026-10-14
Choose an upper ellipticity bound for the symmetric coefficient matrix, possible because its entries are bounded. ThenThe Poincaré inequality on a bounded open set for zero-boundary functions states . ConsequentlyThe symmetric positive bilinear form is therefore an inner product, and its square-root norm is equivalent to the usual Sobolev norm. Its triangle inequality follows from the Cauchy-Schwarz inequality for that inner product. completeness of in its original norm makes it a Hilbert space in the norm as well.
The linear functional is bounded, sinceThe Hilbert-space Riesz representation theorem states that every bounded linear functional is the inner product with a unique element. Applied in the inner product it gives a unique with for all , which is precisely the weak-solution identity. Alternatively uniqueness follows directly by testing the difference of two solutions with itself: . Thus .
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 13 2 c Solution Created 2026-10-03 Updated 2026-10-14
The required functional, with the sign appropriate to a positive divergence operator, isIt is differentiable on , with . A critical point consequently satisfies the weak equation. We now prove existence by the direct method in the calculus of variations, rather than invoke the representation theorem again.
The ellipticity and Poincare estimates implyThus the infimum is finite and any minimizing sequence is bounded in the Sobolev norm. A bounded sequence in a Hilbert space has a weakly convergent subsequence; write in . The equivalent norm has the same continuous linear functionals and hence the same weak convergence. A Hilbert norm is Weak lower semicontinuity, giving . The term converges to because it is a bounded linear functional. ThereforeThe limit attains the infimum. Differentiating at its minimum, for every , gives . This produces the desired weak solution. Strict positivity of also makes the functional strictly convex, consistent with uniqueness.
Past exam of the mathematics course of the University of Cambridge 2008 iii Paper 88 5 d Solution Created 2026-10-03 Updated 2026-10-14
The nullspace in the Fourier diffraction theorem already prevents uniqueness, and the rapid decay of evanescent high-frequency information prevents stable recovery. Both make this an ill-posed inverse problem.
Let be the linear Born approximation operators mapping a real contrast supported in a known region to the measured fields on the two observation planes. For data , a suitable Tikhonov reconstruction from two-plane scattering data minimizesA Sobolev norm penalty can instead express a smoothness preference. Reality, known support and can be imposed as constraints. For the unconstrained complex Hilbert-space problem with the displayed quadratic penalty, write and . Its Tikhonov normal equation isTikhonov regularization selects a stable preferred estimate; it does not establish that the original two-plane data uniquely determine the true contrast.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 12 6 iii Solution Created 2026-10-03 Updated 2026-10-14
The affine class is nonempty, and part (i) gives a finite comparison energy . Since , its energy infimum satisfies . Choose a minimizing sequence with . ThenThe gradient bounds also control the full Sobolev norm. Indeed , so the Poincaré inequality givesThus is bounded in the Hilbert space . Its weak compactness yields a weakly convergent subsequence . The closed linear subspace is weakly closed, hence and . By part (ii),The direct method in the calculus of variations therefore provesNeither a smooth boundary nor a classical trace theorem is needed: the boundary condition is encoded by the closed zero-boundary Sobolev space.
Past exam of the mathematics course of the University of Cambridge 2009 iii Paper 72 1 a Solution Created 2026-10-03 Updated 2026-10-14
Multiply by a test function vanishing at the endpoints and apply integration by parts. The positive operator is , so the forcing changes sign. The weak formulation of this forced Airy boundary-value problem is: find such thatHere is the zero-boundary Sobolev space and can be equipped with . The interval Poincaré inequality states , so this norm makes a Hilbert space and is equivalent to its usual Sobolev norm. The form is symmetric and satisfiesThe Lax-Milgram theorem says that a bounded coercive bilinear form on a Hilbert space and a bounded linear functional determine exactly one weak solution. All its hypotheses have just been verified. Moreover, gives on the interval; the right-hand side is then continuous, so this weak solution is the classical solution with the prescribed endpoint values.
Since the form is symmetric, the equivalent Ritz method minimizesIndeed, stationarity gives the weak equation andwith equality only at . This proves the minimization equivalence, including the sign of the linear term.
Restricting to a conforming finite element space gives a unique discrete solution and Galerkin orthogonality, for every . The Céa lemma states that if the continuity and coercivity constants are and , then . For this symmetric problem the exact best-approximation constant in the energy norm is one. The piecewise-linear finite element interpolation estimate on regular finite element meshes therefore gives convergence of a numerical method of the Ritz method, with first-order error in the energy norm.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 69 1 1 Solution Created 2026-10-03 Updated 2026-10-14
Let and use the clamped second-order Sobolev spaceHere weak derivatives up to order two are square integrable, with Sobolev norm . The boundary values are the continuous traces given by the Sobolev trace theorem, not arbitrary pointwise representatives. This space is the closure of , so it is a closed Hilbert space. The derivative boundary conditions are essential, and alone is insufficient.
Define the symmetric bilinear form and bounded linear functionalThen the variational problem isFor complex data replace by its sesquilinear version, take the real part of the forcing term and use absolute squares in ; the same argument applies.
Its first variation is . Thus stationary points satisfy for every , the weak formulation of a clamped fourth-order equation. For smooth coefficients and functions, twice integrating the highest term and once the middle term gives the differential equation, with all boundary terms zero because at both ends. With only the stated continuous coefficients, this manipulation is instead understood distributionally: compactly supported test functions give as a distribution. This avoids assuming nonexistent classical derivatives of or .
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 7 2 e Solution Created 2026-10-03 Updated 2026-10-06
Take the positive integer and put . From the preceding estimate and the absent zero Fourier mode, for ,Here is the Riemann zeta function, and its displayed series is finite because . One may replace the last derivative norm by the given full mixed Sobolev norm to obtain the requested constant depending only on and .
An absolutely summable sequence of Fourier coefficients gives a uniformly and absolutely convergent Fourier series, with supremum bounded by the sum of their absolute values. Its sum agrees almost everywhere with by uniqueness of Fourier coefficients for integrable periodic functions. HenceThis is the uniform phase-mixing bound from velocity derivatives: uniform convergence for the continuous representative of the density, with rate . No uniform decay of the full phase-space distribution is asserted. The free-transport phase mixing acts by shifting nonzero spatial modes to large velocity frequency. If a convention allows , that endpoint needs separate assumptions or an argument: the harmonic series in this proof would diverge.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 10 2 2 Solution Created 2026-10-03 Updated 2026-10-06
We establish a quantitative almost global existence for wave equations estimate. Take , fix an integer , and use the commutation vector fields for the wave equation from the preceding part. Put and define the commuted wave energyAll these L2 norms are finite on any smooth existence interval by finite propagation speed. At , the polynomial coefficients of the vector fields are bounded on the fixed compact support of the Cauchy data. Whenever a higher time derivative occurs, use and its differentiated versions to express it in terms of initial spatial derivatives. Every term contains at least one factor of ; consequentlyfor a constant depending only on finitely many derivatives and the support radius of .
The commutators are constant linear combinations of translations. Together with and the Leibniz rule, this shows that each commuted source is a finite linear combination of productsThis statement includes the extra copies of the original source produced by the scaling vector field. In each product put the factor with fewer commutations in the Lp norm and the other in the L2 norm. The lower order is at most . Applying the Klainerman-Sobolev inequality to costs at most two additional commutations; commuting those past introduces only lower-order translations. Since ,The inhomogeneous wave energy estimate now givesLet . Use a bootstrap argument with up to the smaller of and the maximal existence time. The energy estimate improves this toFor each fixed ,Choose so that for every . Then , a strict improvement. A continuity argument closes the bootstrap argument.
Finally, the translation terms in control ordinary spatial Sobolev norms of . The missing L2 norm of satisfiesThus the full local-existence Sobolev norms remain bounded on this finite interval. The smooth continuation criterion for semilinear wave equations extends the solution past any finite endpoint before . To see smooth persistence explicitly, the tame Sobolev product estimate gives . Ordinary differentiated wave energy estimates therefore bound each higher derivative energy by its initial value times . This is finite on the interval already controlled by the base commuted wave energy; no separate is needed for each derivative order. ThereforeFor the zero solution is global. The same energy estimate in fact permits an exponential lower bound for the lifespan, which is stronger than any fixed inverse power.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 10 4 6 Solution 2026-10-06
Small data global regularity for wave maps holds in four spatial dimensions. Smallness is measured in sufficiently high weighted Sobolev norms relative to a constant map, with localized compatible Cauchy data. The vector field method for wave equations gives derivative decay . This is time-integrable, so commuted wave energy estimates close a small-data bootstrap argument for the derivative-quadratic semilinear wave equation. Higher regularity persists. No smallness of energy alone is asserted.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 10 4 7 Solution 2026-10-06
Small data global regularity for wave maps also holds in three spatial dimensions. For localized compatible data small in high weighted Sobolev norms, the key is the classical null condition for wave equations. Each derivative contraction is a null form for wave equations, vanishing for parallel null derivatives. The vector field method for wave equations exploits derivatives tangent to the light cone and weighted energy estimates to close the bootstrap argument; ordinary decay alone is insufficient.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 5 3 1 a Solution Created 2026-10-03 Updated 2026-10-06
For , the first-order Sobolev space isThe weak derivative is unique as an equivalence class modulo equality almost everywhere. We use the standard Sobolev norm . Pointwise values of an arbitrary representative have no intrinsic meaning; the one-dimensional Sobolev representative supplies the continuous representative used later.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 5 3 1 b Solution Created 2026-10-03 Updated 2026-10-06
Reflect across the two endpoints of into a slightly larger interval, giving an extension, and multiply by a smooth cutoff function equal to one near . Convolve the resulting function on with a mollifier. The convolutions are smooth; differentiation commutes with convolution of the weak derivative. Continuity of translations in proves both and in , hence convergence in the Sobolev norm.
To obtain convergence almost everywhere, select a sequence with . Then , so the summands tend to zero almost everywhere. Thus a smooth approximating sequence converges both in and almost everywhere. This is density of smooth functions in a Sobolev space; smooth functions need not vanish at the endpoints.
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 4 29E a Solution Created 2026-09-24 Updated 2026-10-06
For smooth periodic initial data, Fourier expansion reduces the equation to . Therefore define the Fourier multiplier solution operators byTheir Sobolev norm satisfies , so they extend to contractive linear operators on . Multiplication of the Fourier factors gives and . Finallyby dominated convergence theorem, dominated by four times the summable norm terms. This proves a strongly continuous contraction semigroup for every requested ; the semigroup property extends continuity from zero to other times. For smooth data, termwise derivatives converge and recover the original Cauchy problem for a partial differential equation uniquely.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 105 2 a Solution Created 2026-10-03 Updated 2026-10-06
For an integer , the Sobolev space iswhere is a weak derivative. For one may use the Sobolev norm ; for use the maximum of the finitely many essential-supremum norms.
For , the Sobolev inequality is , with Sobolev conjugate exponent . For , Morrey's inequality supplies a continuous representative satisfyingAt the representative is Lipschitz continuous. At the critical exponent , first-order Sobolev regularity gives every finite embedding for , with an inhomogeneous norm, but generally no embedding. The one-dimensional endpoint is an exception.
For the proof of Morrey's inequality, start with a smooth and write for its average on a ball. Averaging the fundamental theorem of calculus along a line segment and changing radial variables givesThe last step is the Holder inequality; integrability of the kernel to power is exactly . For , translate the averaging ball along the segment from to . The fundamental theorem of calculus along a line segment and the Holder inequality giveCombining the two point-to-average bounds and this average-to-average bound proves the required Hölder estimate. The point-to-average bound with , together with , gives the supremum estimate. Density of smooth functions in a Sobolev space then gives a uniformly convergent sequence of smooth representatives, preserving both bounds. For , mollification gives the Lipschitz version.
For the decay conclusion assume . The representative is uniformly continuous. If along points escaping to infinity, the Hölder bound gives a radius , independent of , on which . A subsequence has disjoint radius- balls, each contributing at least to , a contradiction. This is uniformly continuous integrable functions vanish at infinity.
The finite- restriction is necessary. If the printed range includes , its decay assertion is false: belongs to but does not tend to zero.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 203 2 c Solution Created 2026-10-03 Updated 2026-10-05
For real smooth functions of compact support in a planar domain , use the Dirichlet inner productChanging the positive normalization factor does not change orthogonality. For complex functions, insert complex conjugation in the second factor to obtain the corresponding Hermitian form.
Let be a conformal bijection, and let be smooth functions of compact support on . Its real Jacobian matrix is , where is a rotation. By the chain rule,The change of variables formula has Jacobian determinant , so this factor cancels:This proves conformal invariance of the planar Dirichlet inner product. By completion it is also an isometry between the corresponding Dirichlet energy spaces. It asserts invariance of the energy form, not of the inhomogeneous Sobolev norm, whose term has a different transformation rule.