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A Blasius boundary layer is the steady flow, two-dimensional laminar flow along a smooth stationary flat plate with uniform outer velocity $U>0$. Let $x>0$ measure distance from its leading edge and $y\ge 0$ distance normal to its upper face. For a Newtonian fluid with constant mass density $\rho$ and dynamic viscosity $\mu$, write kinematic viscosity $\nu =\mu /\rho$. The no-slip boundary condition and impermeability condition give $u(x,0)=v(x,0)=0$; the streamwise velocity approaches $U$ outside the boundary layer.

For example, choose $U=10\mathrm{m}{\mathrm{s}}^{-1}$ and $\nu =1.5\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. At $x=0.30\mathrm{m}$, the local Reynolds number is ${\mathrm{Re}}_x=Ux/\nu =2.0\times 10^5$ and the boundary layer thickness is about $3.29\mathrm{mm}$. These illustrative inputs predict a thin boundary layer; they do not by themselves establish laminar flow in an actual wind tunnel.

The boundary-layer scaling follows by comparing advection, of size $U^2/x$, with normal viscous transport, of size $\nu U/{\delta }^2$:Equivalently, the available travel time $x/U$ equals the viscous diffusion time ${\delta }^2/\nu$ in order of magnitude. Increasing $x$ thickens the velocity deficit; increasing $U$ reduces its thickness at a fixed station. This boundary-layer scaling requires ${\mathrm{Re}}_x\gg 1$. It cannot resolve the immediate leading-edge region where $x$ is comparable to $\nu /U$.^[ref]

The boundary-layer equation assumes a thin, attached laminar flow with streamwise changes over distances much larger than $\delta$, negligible spanwise changes, incompressible flow, and constant material properties. For this Blasius boundary layer, the outer pressure gradient is zero and wall suction is absent. Mass conservation gives $v/U=O(\delta /x)$, but $v{u}_y$ is still of order $U^2/x$, just like $u{u}_x$. Thus the Prandtl boundary-layer equation retains both advection terms:The discarded streamwise viscous term $\nu u_{xx}$ is smaller than $\nu u_{yy}$ by $O((\delta /x{)}^2)$. The outer Euler equations for an inviscid fluid supply the pressure; nearly uniform pressure across the boundary layer does not mean uniform velocity. Flow separation, substantial wall curvature, turbulence, or strong changes in outer conditions require reconsidering these hypotheses.

Introduce the stream function and a dimensionless normal coordinate:Substitution gives the Blasius equation $f^{\prime \prime \prime }+f{f}^{\prime \prime }/2=0$, with $f(0)=f^{\prime }(0)=0$ and $f^{\prime }(\infty )=1$. Its numerical boundary value problem gives $f^{\prime \prime }(0)\simeq 0.332057$. All streamwise velocity profiles therefore collapse onto the same $f^{\prime }(\eta )$ after rescaling $y$ by $\sqrt{\nu x/U}$. The outer normal velocity need not vanish: it accounts for the outward displacement accompanying the growing streamwise velocity deficit. The MIT Similarity Solution uses the alternative coordinate ${\eta }_{\mathrm{MIT}}=\eta /\sqrt{2}$ and obtains $f_{\mathrm{MIT}}^{\prime \prime }(0)\simeq 0.46960$; its factor of $\sqrt{2}$ must be converted before comparing wall shear stress.

The historical Motion of fluids with very little viscosity by L. Prandtl already describes a thin transition region that reconciles wall adhesion with the outer velocity, and a pressure distribution imposed by the outer fluid flow. Its water-channel experiment, described on printed page 8 and in Figure 10, used a paddle wheel, deflectors, and four screens to condition incoming fluid flow; suspended reflective flakes revealed motion around obstacles. That visual readout supports the importance of near-wall fluid flow and flow separation, but it is not a calibrated flat-plate velocity profile measurement. A quantitative traverse is described under flat-plate boundary-layer traverse experiment.

In an elastic boundary value problem, a dead load is a prescribed force independent of the incremental displacement. A dead reference traction has $\delta t^0=0$, even when the current face changes orientation. This differs from a follower load, whose direction follows the moving body, and from the structural-engineering use of the same phrase for permanent self-weight. Under pure dead tractions, a constant incremental translation remains a zero-frequency mode unless a displacement condition fixes it.

For real $\beta$, the displayed boundary value problem with $v(\xi )\to 0$ as $\xi \to \infty$ has a solution exactly when $\beta >0$. Put $a={\beta }^{1/3}>0$. Its unique real solution isIndeed the Euler differential operator has characteristic roots $0,a,-a/2\pm i\sqrt{3}a/2$. Decay retains only the conjugate pair, and the two endpoint conditions determine their sine combination. For negative $\beta$ only one root has negative real part, so the two boundary conditions are incompatible. At zero every solution is a cubic polynomial in $\log \xi$, excluding nonzero decaying solutions. Complex exponents thus produce an oscillatory but decaying power-law envelope.

A degenerate ordinary differential equation has a highest-derivative coefficient that vanishes somewhere in its domain. Its solutions can have lower regularity at that point than in the interior. A boundary value problem normally imposes endpoint values by continuous extension while requiring the differential equation in the interior. Requiring it pointwise at a degenerate endpoint with all derivatives finite can impose an extra condition and destroy existence. In a square-root-degenerate endpoint layer, the product of a vanishing square-root coefficient with a diverging second derivative has a finite nonzero interior limit.

The boundary value problem $y^{\prime \prime }+\alpha y(c){\delta }_c=f$ on $(0,\ell )$ with Dirichlet boundary conditions has a unique solution for every continuous forcing and prescribed endpoint values exactly when $\alpha \ne {\alpha }_{\ast }:=\ell /(c(\ell -c))$. The left-normalized homogeneous solution iswhere $(t{)}_{+}=\max (t,0)$. Continuity at $c$ and the transmission conditions for a delta coefficient give this formula. The final boundary condition fixes the initial slope unless the displayed endpoint value is zero. This is a one-dimensional kernel obstruction, analogous to the Fredholm alternative.

At resonance the normalized null function isFor $y(0)=a$ and $y(\ell )=b$, the compatibility condition isIndeed, integrate $\varphi L_{\alpha }y-y{L}_{\alpha }\varphi$ on both sides of $c$. The internal terms cancel by the matching derivative jumps, leaving $[\varphi y^{\prime }-{\varphi }^{\prime }y{]}_0^{\ell }$. This equals the right-hand side. The initial-slope construction also proves sufficiency: when compatible there is a one-parameter family $y_0+t\varphi$; otherwise no solution exists.

Electrostatics studies time-independent electric charge distributions and their electric fields. In vacuum, its local equations in SI units areHere $\rho$ is charge density, ${ϵ}_0$ is vacuum permittivity, and $\Phi$ is electric potential. Gauss's law supplies the divergence equation; the static form of Faraday's law of induction supplies the zero curl. The Poisson equation becomes Laplace's equation where there is no electric charge. Prescribed electric charges give electric fields through Coulomb's law and the superposition principle; mobile electric charges on an electrical conductor instead require solving for their distribution together with the electric potential. ^[ref].

The useful distinction is between prescribed sources and prescribed boundary conditions. On a grounded electrical conductor, an external connection fixes its electric potential relative to an electrical reference potential and allows electric charge to enter or leave. On an isolated electrical conductor, total electric charge is fixed and its constant electric potential is an unknown. A neutral isolated electrical conductor can therefore have a nonzero local surface charge density through electrostatic induction. Equal electric potentials also do not imply equal electric charges: charge sharing between distant connected spheres gives a simple geometric example. ^[ref].

Electrostatics is useful because a spatial boundary value problem replaces the need to follow every moving charge carrier during charge relaxation. It describes the settled state of a capacitor and explains electrostatic shielding of sensitive equipment. Its assumptions must match the apparatus: stationary boundaries and sources, negligible time variation of the magnetic field, and enough time for charge relaxation. A metal carrying a maintained electric current through finite electrical resistance generally needs a nonzero internal electric field by Ohm's law; it is not a conductor in electrostatic equilibrium. Static macroscopic electric charge also does not mean that every electron is motionless. ^[ref].

The antisymmetric four-node stencil at offsets $-3h/2,-h/2,h/2,3h/2$ about a midpoint approximates the third derivative. Taylor expansion gives leading error $h^2{u}^{(5)}/8$. For a third-order boundary value problem on nodes $0,\dots ,n+1$, use rows $j=1,\dots ,n-1$, prescribed endpoint values and one consistent one-sided derivative condition. This avoids a ghost node while providing the correct number of equations. For instance a known right-endpoint first derivative is approximated by $(3U_{n+1}-4U_n+U_{n-1})/(2h)$.

An initial value problem combines an ordinary differential equation with an initial condition, for example $y^{\prime }=f(t,y)$ and $y(t_0)=y_0$. The task is to find a solution of a differential equation on an interval containing $t_0$ which has that starting value. For a continuous linear differential equation $u^{\prime }=A(t)u$, every prescribed finite vector $u_0$ determines a unique solution on the coefficient interval. A boundary value problem can instead prescribe values at distinct endpoints.

For the cooling initial value problem $y^{\prime }=-k(y-T_a)$, $y(t_0)=y_0$, where $k>0$ and $T_a$ are constant, the unique solution isTake $T_a=20$ degrees Celsius, $y_0=80$ degrees Celsius, $t_0=0$, and $k=0.1$ per minute. Then $y(10)=20+60e^{-1}\approx 42.07$ degrees Celsius. Reaching $30$ degrees Celsius requires $60e^{-0.1t}=10$, so $t=10\log 6\approx 17.92$ minutes. The time constant $1/k=10$ minutes reduces the excess temperature by a factor $e^{-1}$; it does not reduce the absolute temperature by that factor.

Uniqueness needs assumptions. If $f$ is a continuous function on an open neighborhood of $(t_0,y_0)$ and satisfies a local Lipschitz condition in $y$, uniformly in $t$, the Picard-Lindelöf theorem gives a unique local solution. Concretely, a finite $L$ bounds $|f(t,u)-f(t,v)|\le L|u-v|$ in a sufficiently small rectangle. Cooling satisfies this with $L=k$. Local existence does not promise existence for all future time: $y^{\prime }=y^2$, $y(0)=1$, has $y(t)=1/(1-t)$ on $(-\infty ,1)$ and finite-time blowup at $t=1$.

Continuity alone is insufficient for uniqueness. Consider the dimensionless initial value problem involving a square root and an absolute value:Besides $y(t)\equiv 0$, each delay $a\ge 0$ gives a solution for $t\ge 0$:At $t=4$, delays $a=0,1,2$ give $y=4,9/4,1$, despite identical starting data. These are valid solutions of a differential equation: for $a>0$, both one-sided derivatives at the join are zero; for $a=0$, the right derivative at the starting time is zero. After the join, $y_a^{\prime }=(t-a)/2=\sqrt{|y_a|}$ because $t>a$. Squaring alone would also suggest a parabola before the join, but its negative derivative there would violate the equation. Since $y^{\prime }\ge 0$, a forward solution starting from zero stays nonnegative.

The right-hand side is a continuous function but fails the Lipschitz condition at zero: for $h>0$, $|\sqrt{|h|}-0|/|h|=1/\sqrt{h}$ is unbounded. Thus the Peano existence theorem still supplies local existence, while the Picard-Lindelöf theorem does not supply uniqueness. Related waiting-time examples and a sufficient uniqueness condition appear in MIT 18.034, Lecture 2: Fundamental Principles.

A brief proof outline explains the mechanism: rewrite the initial value problem as the integral equation $y(t)=y_0+{\int }_{t_0}^{t}f(s,y(s))ds$. On a short enough interval, the Lipschitz condition makes the right-hand side a contraction mapping on a closed set of continuous functions near $y_0$. The contraction mapping theorem gives a unique fixed point; the fundamental theorem of calculus turns it back into a solution.

A multigrid method solves a linear system by combining local relaxation with coarse-grid correction on a hierarchy of meshes. The key is to divide the work by scale: a suitable relaxation removes rapidly oscillating approximation error, while coarse-grid correction addresses slowly varying approximation error that relaxation barely changes. A broad error hump occupies many fine-grid points but fewer coarse-grid points; it becomes easier to resolve by relaxation after enough coarsening. This complementary action is why multigrid methods are useful for large finite difference method and finite element method systems.

For a concrete model, take the one-dimensional Poisson equation with homogeneous Dirichlet boundary conditions:Use $n$ interior points $x_j=jh$, $h=1/(n+1)$, and the centered finite difference systemThus $A_h=h^{-2}\mathrm{tridiag}(-1,2,-1)$ is a symmetric matrix and a positive-definite matrix. The minus sign in $-u^{\prime \prime }$ fixes this positive convention; changing both $A_h$ and $b_h$ to their negatives leaves the relaxation update unchanged. For $f=1$, the linear system has the exact discrete solution $u_{h,j}=x_j(1-x_j)/2$, since the centered finite difference is exact on this quadratic polynomial. In general, the exact discrete solution need not equal the sampled continuous solution: solving the linear system removes algebraic approximation error, not the global discretization error.

Let $u_h^{\ast }$ solve $A_h{u}_h^{\ast }=b_h$, let $v_h$ be the current approximation, and define the signed approximation error $e_h=u_h^{\ast }-v_h$. The linear-system residual is $r_h=b_h-A_h{v}_h$, so $A_h{e}_h=r_h$. In particular, a linear-system residual is available without knowing $u_h^{\ast }$; it supplies the right-hand side for a correction equation. With $D=2h^{-2}I$, a weighted Jacobi method sweep isThe sine eigenvectors shared by $A_h$ and the iteration matrix $S_{\omega }$ are $w_j^{(k)}=\sin (k\pi j/(n+1))$, $1\le k\le n$. Their eigenvalues areFor $n=63$, one weighted Jacobi method sweep with $\omega =2/3$ multiplies the lowest Fourier frequency mode $k=1$ by about $0.999197$, but the highest Fourier frequency mode $k=63$ by about $-0.332530$. Four sweeps retain about $99.68\%$ of the first amplitude and $1.22\%$ of the second. The sign reversal of the second amplitude is harmless when its magnitude shrinks. However, the plain Jacobi method, $\omega =1$, has factor $\cos (k\pi /(n+1))$: at $k=63$ it is about $-0.998795$. A nearly alternating mode near the grid's Nyquist limit therefore persists almost as strongly as the smoothest mode. Plain Jacobi method convergence on a fixed Dirichlet boundary condition grid does not imply effective smoothing.^[ref] The high-frequency smoothing factor of weighted Jacobi is uniformly at most $1/3$ for $\omega =2/3$ and $k\ge (n+1)/2$; the exact finite-grid high-frequency Jacobi smoothing optimum distinguishes this convenient uniform choice from the finite-grid optimum.^[ref]

One two-grid cycle applies a few weighted Jacobi method sweeps, computes the linear-system residual, performs coarse-grid correction, and applies a few more sweeps. The first relaxation makes the remaining approximation error easier to represent on a coarse mesh; the last relaxation damps oscillatory components left after the transferred correction. In a geometric multigrid V-cycle, replace the expensive exact coarse solve by one recursive cycle on the coarse correction equation, starting its correction from zero. Repeat until a tiny coarsest linear system can be solved directly, then return through the levels by adding each interpolated correction and post-smoothing. The descent and return give the cycle its V shape.^[ref]

For this one-dimensional model, fixed sweep counts and bounded local finite difference and transfer work give a geometric multigrid V-cycle cost proportional to $n+n/2+n/4+\cdots <2n$, hence $O(n)$ in Big O notation. A fixed error reduction also needs a contraction factor bounded below one independently of $h$; per-cycle work alone does not establish that. Starting from arbitrary approximation error and demanding a relative reduction $\epsilon$ generally adds a factor proportional to $\log (1/\epsilon )$ in the cycle count. Full multigrid uses coarse solutions as initial guesses on successively finer meshes to target the global discretization error efficiently. The two-grid Poisson factor with weighted Jacobi illustrates favorable convergence for this model. Other partial differential equations, strong anisotropy, or poorly chosen coarse spaces and relaxation can defeat these choices; a linear-work guarantee for arbitrary linear systems does not follow.

A historical example of the multilevel viewpoint is Achi Brandt's April 1977 paper, which analyzes adaptive meshes and reports numerical experiments on linear and nonlinear boundary value problems. This is a documented development of the approach, not a claim that the paper was its first invention.^[ref] The figure above is generated from the displayed eigenvalues, with lines joining discrete nodes; it contains no experimental measurements.

Neumann boundary data are prescribed values of the outward normal derivative of a function on the boundary of a boundary value problem. They enter a Neumann boundary condition as ${\partial }_{n}u=g_N$. Directional derivatives defined using a fixed coordinate direction must be converted to the outward-normal convention: for example, on the bottom of an upper domain, ${\partial }_{n}u=-u_y$.

A boundary value problem with prescribed outward normal derivative, or the associated weighted flux, on the boundary. Its boundary data are Neumann boundary data. For an elliptic divergence-form operator without a zeroth-order term, integration over the domain imposes a compatibility condition between total source and boundary flux. A connected domain generally leaves a constant undetermined, fixed by a mean or reference value.

An oblique derivative boundary condition prescribes a linear combination of the outward normal derivative and the tangential derivative on a boundary. Its strictly oblique case has a nonzero normal coefficient; a zero normal coefficient is a limiting tangential condition. Both trace directions must be tracked when constructing a global relation. On unbounded domains, regularity at corners and a growth condition at infinity are separate parts of an admissible boundary value problem.

For indentation $F=b{x}^{1/3}$, the common layer displacement is $H=b{x}^{1/3}/2$. The boundary layer stream function $\psi =x^{2/3}G(z{x}^{-1/3})$ gives velocity $U=x^{1/3}{G}^{\prime }$ and pressure $P=\bar{p}{x}^{2/3}$. Substitution into the Prandtl boundary-layer equation yields the displayed equation, with $G(0)=G^{\prime }(0)=0$ and $G^{\prime }\sim 6\eta +3b$. This is a conditional similarity construction: the constant $\bar{p}$ must permit a solution of the boundary value problem.

For $(L_0+ϵL_1)u=f(ϵ)$ and $Bu=g(ϵ)$, expand all three quantities in powers of $ϵ$. The order-$n$ problem has $L_0{u}_n=f_n-L_1{u}_{n-1}$ and $B{u}_n=g_n$, with $u_{-1}=0$. In particular, Taylor series boundary coefficients are generally nonzero even when the differential operator depends linearly on the parameter. If the homogeneous realization of $L_0$ is invertible and boundary data admit smooth lifts, the same inverse solves each successive inhomogeneous boundary value problem. Induction establishes unique coefficients at every order. This formal conclusion does not alone assert convergence of the full power series; convergence needs additional bounds on the inverse and the perturbation.

Use the original scanned boundary condition. The scan fixes the normal-field amplitude at the corner $x=z=0$. The supplied TeX drops the condition $x=0$ and would instead prescribe a constant normal field on the entire bottom. Those are different boundary-value problems; the solution below uses the legible original scan.

Choose a positive integer $n$, put $k=n\pi /a$, and seek a separated decaying mode. A suitable flux function isThis is a decaying linear force-free magnetic arcade. The first and third curl components then satisfy the force-free equation automatically; its second component requires $-({\partial }_x^2+{\partial }_z^2)A={\lambda }^{2}A$. ThusDirect differentiation gives $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=\lambda \mathbf{B}$. Both horizontal components vanish at $x=0,a$, and $B_z(0,0)=B_0$. The magnetic field on the bottom is $B_0\cos kx$, not a uniform $B_0$.

A strictly decaying real normal mode requires $q>0$, soFor the fundamental, single-arcade choice $n=1$, this becomes $|\lambda |<\pi /a$. Equality gives no vertical exponential decay, while a larger magnitude gives an oscillatory rather than decaying vertical dependence for that mode. The data do not explicitly fix the horizontal mode number. If arbitrarily fine higher-mode arcades are permitted, every finite real $\lambda$ admits some integer $n$ with $n\pi /a>|\lambda |$. The fundamental restriction must therefore not be advertised as a necessary bound for every possible field satisfying only the stated pointwise and wall data.

For a fundamental field line, $A$ is constant along its projection in the $(x,z)$ plane. If $0<s<1$ denotes its normalized flux-function value, its arch isThe maximum height is $q^{-1}\log (1/s)$ at $x=a/2$. In the horizontal projection,Thus the vertical projections are arches and the horizontal projections are straight sheared segments joining their paired footpoints. Positive $\lambda$ gives positive horizontal slope, negative $\lambda$ reverses the shear, and $\lambda =0$ gives an unsheared potential arcade. Higher $n$ repeats such cells across the width.

The transcription's alternative uniform-bottom condition would be inconsistent with exponential decay and zero side flux for $B_0\ne 0$: solenoidality gives $d{\int }_0^a{B}_{z}dx/dz=-B_x(a,z)+B_x(0,z)=0$, whereas a uniform nonzero bottom would give a conserved flux $a{B}_0$ that cannot decay to zero. This integral check is an independent reason not to silently solve that mistranscribed problem.

If $u_1,u_2$ satisfy the same boundary value problem, their difference $w=u_1-u_2$ satisfies $Lw=0$ in $D$ and $w=0$ on the boundary, by linearity. Part (b)(ii) givesApply it again to $-w$, which also satisfies $L(-w)=0$ and zero Dirichlet boundary conditions. This gives $-w\le 0$. Hence $w=0$ throughout $\overline{D}$, provingThis proves uniqueness whenever a solution exists; existence is the separately supplied assumption. When the drift coefficients vanish, the same argument uses part (a).

For the linear differential operator $L=D^2+4D+5$, the characteristic equation is $m^2+4m+5=0$, with roots $-2\pm i$. Take the fundamental system of solutionsThe Wronskian is never zero, so variation of parameters is valid on the whole interval. Write a particular solution as $a(x)u(x)+b(x)v(x)$, and impose the auxiliary condition $a^{\prime }u+b^{\prime }v=0$. Substitution then gives $a^{\prime }{u}^{\prime }+b^{\prime }{v}^{\prime }=f$. These two equations giveChoosing both integrals to start at zero makes the resulting particular solution and its first derivative vanish there:A general solution is $Au+Cv+y_p$. The boundary condition at zero sets $A=0$. The condition at $\ell =\pi /2$ givesbecause $\sin \ell =1$ and $\sin (\ell -t)=\cos t$. Combining the two integrals yields the Green's function representationIndeed, for $t<x$ the kernel is $-e^{2(t-x)}\sin t\cos x$, since $\sin (x-t)-\sin x\cos t=-\cos x\sin t$; for $x<t$ it is $-e^{2(t-x)}\sin x\cos t$.

For fixed interior $t$, the Green's function vanishes at both endpoints, is continuous at $x=t$, solves the homogeneous equation away from $t$, and satisfiesThis derivative jump is the normalization required by the leading coefficient one in $L$. Finally, a homogeneous solution satisfying the first boundary condition is $C{e}^{-2x}\sin x$, and the second boundary condition at $\pi /2$ forces $C=0$. Thus the boundary value problem has exactly one solution for every continuous forcing, justifying the unique inverse represented by this kernel.

For this forcing the integral in part (a) simplifies toConsequently the unique boundary value problem solution isIt vanishes at zero and at $\pi /2$. To check the differential equation directly, set $y=e^{-2x}w$. Then $Ly=e^{-2x}(w^{\prime \prime }+w)$. For $w=1-\sin x-\cos x$, one has $w^{\prime \prime }+w=1$, as required.

Let $L_{0}u=(x+1)u^{\prime \prime }+x{u}^{\prime }$. Matching powers in the regular perturbation expansion and in the Taylor series of the left endpoint value gives the complete recursive boundary value problemThe higher-order left endpoint values are not zero. The parameter-dependent boundary data in a regular perturbation hierarchy rule applies here.

First prove uniqueness for any one of these forced problems. The difference $v$ of two solutions obeys $L_{0}v=0$ and homogeneous Dirichlet boundary conditions. The positive integrating factor $r_0=e^x/(x+1{)}^2$ givesMultiplication by $\overline{v}$ and integration by parts yieldHence $v^{\prime }=0$, and the endpoint values force $v=0$. The kernel is trivial: zero is not an eigenvalue of this homogeneous Dirichlet boundary value problem.

Existence is also explicit. For arbitrary smooth forcing $g$ and endpoint values $u(0)=a$, $u(1)=b$, two integrations of $(p{u}^{\prime }{)}^{\prime }=r_{0}g$ givewhereSince $H(1)=2-3/e>0$, the unique choicealways meets the second endpoint value. Starting from the smooth $y_0$, induction with $g=-y_{n-1}$, $a=2/n!$, $b=0$ proves existence and uniqueness at every order.

Equivalently, the Fredholm alternative has no compatibility obstruction because the homogeneous kernel vanishes; the corresponding adjoint operator has no zero eigenvalue either. The direct construction above proves this without computing any particular $y_n$ beyond the leading term.

The homogeneous solution is $A\cos x+B\sin x$. The left boundary condition removes $A$, and the right boundary condition gives $B\sin l=0$. Thus the kernel of a linear map associated with this boundary value problem isThe Fredholm alternative now gives the complete classification:For necessity at $l=m\pi$, multiply by the homogeneous solution $\sin x$ and use the Lagrange identity for a Sturm-Liouville operator to obtain ${\int }_0^{l}f\sin xdx=0$. For sufficiency, the particular solutionsatisfies $z(0)=0$ and $z^{\prime \prime }+z=f$. At the other endpoint,which is zero under the compatibility condition. All solutions are then $z+B\sin x$, with $B$ arbitrary. The assumed Lipschitz continuity of $f$ in particular makes $f$ continuous, so the displayed integral defines a classical solution. The length $l=0$ is excluded: it would not define a nonempty interval for this boundary value problem.