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Let $h_1,\dots ,h_n$ be a basis of homogeneous solutions of a linear ordinary differential operator $L$ of order $n$. Let $B_1,\dots ,B_n$ be independent homogeneous linear boundary conditions, which may be coupled. Suppose a fundamental solution $G_0(x,\xi )$ satisfies $L_x{G}_0=\delta (x-\xi )$. Define the boundary matrix $M_{ij}=B_i{h}_j$ and $b_i(\xi )=B_i{G}_0(\cdot ,\xi )$. If $M$ is invertible, thenThe added terms are homogeneous solutions, so the Dirac delta source is preserved. Applying the boundary operators gives $b-M{M}^{-1}b=0$. Any two such Green functions differ, for fixed source position, by a homogeneous solution with coefficient vector in the null space of $M$; invertibility therefore proves uniqueness. A singular boundary matrix signals a nontrivial homogeneous solution satisfying the endpoint conditions, where a compatibility condition and a nonunique inverse may arise. For $L=D^2$, one convenient fundamental solution is $(x-\xi )H(x-\xi )$.

For a harmonic function between $\tau =a$ and $\tau =b$, $0<a<b$, in bipolar coordinates, suppose $\Phi (\sigma ,a)=0$ and $\Phi (\sigma ,b)=F(\sigma )$. Write the periodic boundary values as a Fourier seriesSeparation of variables givesEach radial factor solves $M^{\prime \prime }-n^{2}M=0$ and has endpoint values zero and one; the mean mode solves $M^{\prime \prime }=0$. Sufficiently regular boundary data give the required Fourier series solution. Uniqueness follows from the maximum principle for a harmonic function with zero Dirichlet boundary conditions on both circles. The formula follows the coordinate values; $\tau =b$ is the geometrically smaller circle.

A solution exists when at least one object satisfies all the stated conditions. In first-order logic, this is expressed by existential quantification. Uniqueness additionally requires that any two solutions are equal.

One integral form of the Gronwall inequality states that a continuous nonnegative function $u$ satisfying $u(t)\le A+{\int }_0^{t}b(s)u(s)ds$, with $A\ge 0$ and continuous $b\ge 0$, obeysSet $v(t)=A+{\int }_0^{t}b(s)u(s)ds$. Since $u\le v$ and $v^{\prime }\le bv$, an integrating factor gives the result; if $A=0$, replace it by $\epsilon >0$ and let $\epsilon$ decrease to zero. The zero case proves uniqueness when the difference between two solutions satisfies such an estimate.

For $Ly=y^{\prime \prime }+y^{\prime }/x-y/x^2$ on $x>0$, impose $y(0)=0$ by a continuous limit and $y(x)\to 0$ at infinity. The adapted homogeneous solutions are $x$ at zero and $1/x$ at infinity, with Wronskian $-2/x$. The displayed Green function is continuous and its first derivative has jump one at $x=s$, so $L_{x}G(x,s)=\delta (x-s)$ with ordinary source measure $ds$. For continuous $f$ with ${\int }_0^{\infty }(1+s^2)|f(s)|ds<\infty$,This solves $Ly=f$ with the stated endpoint limits. No nonzero combination of $x,1/x$ meets both homogeneous endpoint conditions, proving uniqueness. Multiplying $L$ by $x$ gives the self-adjoint expression $(x{y}^{\prime }{)}^{\prime }-y/x$, explaining the weighted reciprocity $xG(x,s)=sG(s,x)$.

On the unit square, these data determine only the zero solution of the heat equation $u_y=u_{xx}$. The squared $L^2$ norm decreases because its derivative is $-2\int u_x^{2}dx$, and its initial value is zero. For Laplace's equation, $\sin (\pi x/2)\sinh (\pi y/2)$ is a nonzero solution with the same three data; adding zero Dirichlet data on the top edge restores uniqueness by Green's first identity. For the wave equation $u_{yy}=u_{xx}$, $\sin (\pi x/2)\sin (\pi y/2)$ is a counterexample; the missing datum is initial velocity. Adding $u_y(x,0)=0$ restores uniqueness by conservation of $\frac{1}{2}\int (u_x^2+u_y^2)dx$. Thus the type of equation determines what information is needed on an unprescribed edge.

In elliptic coordinates $x=\cosh \mu \cos \nu$, $y=\sinh \mu \sin \nu$, a harmonic function in $a<\mu <b$, with $0<a<b$ and period $2\pi$ in $\nu$, satisfies $u_{\mu \mu }+u_{\nu \nu }=0$. Prescribe $u(b,\nu )=f(\nu )$ and $u_{\mu }(a,\nu )=g(\nu )$. Expand both boundary functions in Fourier series, writing their constant terms as $f_0,g_0$. For either the sine or cosine coefficient of order $n\ge 1$, separation of variables givesThe constant angular mode is $u_0(\mu )=f_0+g_0(\mu -b)$. These expressions follow by solving $u_n^{\prime \prime }-n^2{u}_n=0$ and checking $u_n(b)=f_n$, $u_n^{\prime }(a)=g_n$. Their Fourier series gives the solution for sufficiently regular boundary data.

The phrase normal derivative needs care: the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$, so the outward unit normal derivative on the inner ellipse is ${\partial }_{n}u=-h^{-1}{u}_{\mu }$. Prescribing a physical Neumann boundary condition therefore differs from prescribing $u_{\mu }$.

For uniqueness, subtract two solutions to obtain $v$. Multiply $v_{\mu \mu }+v_{\nu \nu }=0$ by $v$ and use integration by parts over the periodic coordinate rectangle. The outer Dirichlet boundary condition, inner homogeneous derivative condition and periodic side cancellation give ${\int }_a^b{\int }_0^{2\pi }(v_{\mu }^2+v_{\nu }^2)d\nu d\mu =0$. Thus $v$ is constant, and the outer condition makes it zero.

Multiplying the equation for $v=y^{\prime }$ by the integrating factor $e^x$ gives $v(x)=-e^{-x}{\int }_x^1{e}^{t}f(t)dt$. Integrating once more and reversing the order of integration gives the Green's functionIt is continuous across $x=t$, its first derivative has jump $+1$, and it satisfies the mixed Dirichlet boundary condition and Neumann boundary condition. Both homogeneous solutions, $1$ and $e^{-x}$, are eliminated by these two conditions, giving uniqueness. For $f=e^{-x}$ the solution is $-x{e}^{-x}$.

Suppose $-u^{\prime \prime }+u+q(u)=f$ and $-v^{\prime \prime }+v+q(v)=g$ on a bounded interval, the solutions have the same Dirichlet boundary conditions, and $q$ is a nondecreasing function. Multiplying the difference equation by $w=u-v$ and using integration by parts givesThe nonlinear integral is nonnegative. Young's inequality for products gives the displayed energy estimate. It proves uniqueness and continuous dependence on forcing. The same argument holds for nodal values when integration by parts is replaced by discrete summation by parts.

The first iterate isOn the prescribed box, $F(x,y)=1+x^2+y^2$ is continuous, $|F|\le 102$, and $|F(x,y)-F(x,v)|\le 20|y-v|$. The initial value is distance nine from the nearer horizontal edge. Thus on $|x|\le h=9/102=3/34$, the Picard map keeps functions within the box: its values differ from one by at most $102h=9$.

Successive differences satisfy, by induction,The Weierstrass M-test proves uniform convergence on this local interval. The uniform limit is continuous, and the Lipschitz estimate allows passage to the limit inside the integral, giving the displayed integral equation. Differentiation gives $y^{\prime }=1+x^2+y^2$, $y(0)=1$. The same local Lipschitz bound proves local uniqueness.

Direct differentiation gives $u^{\prime }={\sec }^2(x+\pi /4)=1+u^2$, $u(0)=1$, and $u\to +\infty$ as $x\uparrow \pi /4$. On the common positive existence interval put $w=y-u$. It satisfies $w^{\prime }=(y+u)w+x^2$, $w(0)=0$, so the integrating-factor formula givesThus $y\ge u$. Let $x_{\ast }>0$ be the maximal forward existence endpoint. The solution is strictly increasing because $y^{\prime }\ge 1$. If $x_{\ast }$ were finite and $y$ remained bounded, its finite limit and the smooth right side would allow local continuation, contradicting maximality. Comparison with $u$ prevents existence through $\pi /4$. ConsequentlyThere is no contradiction with Picard's theorem, which gives local existence and uniqueness. Its box estimate gives only a short guaranteed interval, and the Lipschitz constant for $y^2$ is not globally bounded in $y$.

First solve for $v=y^{\prime }$. Multiplication by $e^x$ and use of $v(1)=0$ giveIntegrating from zero to $x$, and using $y(0)=0$, yields $y(x)=-{\int }_0^x{e}^{-s}{\int }_s^1{e}^{t}f(t)dtds$. Reversing the order of integration givesThus $G(x,t)$ equals $G_1$ for $t<x$ and $G_2$ for $x<t$. It is continuous at $x=t$, vanishes at $x=0$, and has $G_x(1,t)=0$. Its derivative jumps from $-1$ to zero across $x=t$, so the jump is $+1$, matching the coefficient of $y^{\prime \prime }$. Both pieces satisfy $G_{xx}+G_x=0$ away from the jump. These facts also verify the Green representation directly. The homogeneous solutions are $A+B{e}^{-x}$, and the two boundary conditions force $A=B=0$, proving uniqueness.

For $f(t)=e^{-t}$, the formula givesHence $y(x)=-x{e}^{-x}$. Indeed $y^{\prime }=(x-1)e^{-x}$ and $y^{\prime \prime }=(2-x)e^{-x}$, which verify the equation and both boundary conditions.

Assume the vector field and multiplier are continuously differentiable, as required for Green's theorem and uniqueness of trajectories. A nonconstant periodic trajectory of a planar autonomous system is a simple closed orbit: uniqueness prevents crossings, and one traversal can be taken over its least period. Let $\Omega$ be its bounded interior. Along the orbit $dx=fdt$, $dy=gdt$, soGreen's theorem identifies this with ${\iint }_{\Omega }[{\partial }_x(\varphi f)+{\partial }_y(\varphi g)]dxdy$, which is strictly positive by hypothesis, a contradiction. Thus there is no nonconstant periodic orbit. This standard periodic-orbit criterion does not by itself exclude stationary solutions.

For the first concrete system choose $\varphi =-1$. ThenThere are also no equilibria: $f=0$ requires $y=1$, but then $g=-e^{-x}-x^2<0$. Consequently it has no periodic solution, including constant ones.

For the second system the origin is outside the domain of the printed vector field. For $r>0$,Thus for an admissible initial point with $\rho =\sqrt{a^2+b^2}>0$,where $a=\rho \cos \alpha$, $b=\rho \sin \alpha$. These formulas are valid while $r>0$, in particular for every $t\ge 0$. Trajectories rotate counterclockwise, spiral outward when $0<r<1$, and inward when $r>1$, approaching the unit circle. On $r=1$ they give $x(t)=\cos (\alpha +t)$, $y(t)=\sin (\alpha +t)$, a periodic orbit with least period $2\pi$. It is an attracting limit cycle. There is no contradiction with the preceding criterion: this vector field is singular at the origin and has the exhibited closed orbit.

The geometry of a strictly convex normed space proves uniqueness whenever a best approximation exists. If distinct $p_1,p_2$ both have best error $E>0$, then their two residuals have norm $E$. The defining midpoint inequality givescontradicting best approximation because the midpoint lies in $A$. If $E=0$, both approximants equal $u$.

The printed existence assertion is false for an arbitrary vector subspace. For a concrete counterexample take $B={\ell }^2$ with its strictly convex Hilbert norm, let $A=c_{00}$ be its finite-support sequences, and take $u=(2^{-j}{)}_{j\ge 1}$. Truncations in $A$ tend to $u$, so the best-error infimum is zero, but $u\notin A$ and that infimum is not attained. If the finite-dimensional assumption of part (b) is retained, it supplies existence and the argument above supplies uniqueness.

For the estimate, suppose $Y$ is a projection operator onto $A$, as needed for a finite operator norm. Since $Yp=p$,The triangle inequality and operator norm giveThe estimate actually holds without attainment: apply it with any $q\in A$ instead of $p$ and take the infimum. Its proof applies in every normed space. This is strict convexity separates uniqueness from existence in best approximation.

The polynomial $T_6$ isSubtract its monic version from $x^6$:This has degree four and hence belongs to the usual space $P_5$. The error is $T_6/32$, with norm $1/32$ and alternating extrema at the seven points $\cos (j\pi /6)$, $j=0,\dots ,6$. Part (a), followed by uniqueness, provesand that this is the unique best approximation of degree at most five.

If “degree five” is taken literally to exclude degree four, no best approximation exists in that nonclosed class: $p+ϵx^5$ approaches the error $1/32$ as $ϵ\to 0$, but uniqueness in $P_5$ excludes attainment by any degree-five polynomial. This is why the standard at-most-degree convention matters here.

The endpoint chord is $l(x)=1-x$. Since the square-root function is concave, the gap $g(x)=\sqrt{1-x}-(1-x)$ is nonnegative. Put $t=\sqrt{1-x}$ to obtain $g=t-t^2$, whose maximum is $1/4$ at $t=1/2$, or $x=3/4$.

Shift the chord upward by half that maximum. ThenThe errors at $x=0,3/4,1$ are respectively $-1/8,+1/8,-1/8$. These are three alternating extrema, so equioscillation proves optimality and uniqueness for linear polynomials. This is minimax linear approximation of a square-root endpoint profile.

Let $V_h$ be the continuous piecewise linear space including both endpoint hat functions. Robin conditions are natural conditions and do not restrict the endpoint values of this space. The finite element problem is $u_h\in V_h$, $a(u_h,v_h)=\ell (v_h)$ for all $v_h\in V_h$. coercivity restricted to $V_h$ gives existence and uniqueness.

Expanding $u_h={\sum }_{j=0}^N{U}_j{\varphi }_j$ gives $\mathsf{A}U=F$, whereThe stiffness contribution has diagonal $2/h$ at interior nodes, $1/h$ at endpoints, and adjacent entries $-1/h$. For completeness, the weighted mass contribution has interior diagonal $2h{x}_i^2/3+h^3/15$, left endpoint diagonal $h{a}^2/3+a{h}^2/6+h^3/30$, right endpoint diagonal $h{b}^2/3-b{h}^2/6+h^3/30$, and adjacent entry $h{x}_i^2/6+h^2{x}_i/6+h^3/20$. All more distant entries vanish. The Robin contribution adds one to each endpoint diagonal.

Symmetry follows from symmetry of the bilinear form. If $V\ne 0$, nodal independence makes $v_h={\sum }_i{V}_i{\varphi }_i$ nonzero, and $V^T\mathsf{A}V=a(v_h,v_h)\ge \gamma \Vert v_h{\Vert }_{H^1}^2>0$. Thus the matrix is symmetric positive definite.

The Sobolev space and norm areIn one dimension these functions have continuous representatives and well-defined endpoint traces. integration by parts gives the weak problem: find $u\in H^1(a,b)$ such that $a(u,v)=\ell (v)$ for all $v\in H^1(a,b)$, whereIndeed the integration boundary term is $-u^{\prime }(b)v(b)+u^{\prime }(a)v(a)$; substituting $u^{\prime }(a)=u(a)-A$ and $u^{\prime }(b)=B-u(b)$ gives the two positive boundary terms on the left.

If $0\notin [a,b]$, put $m={\min }_{[a,b]}{x}^2>0$. Then $a(v,v)\ge \Vert v^{\prime }{\Vert }_2^2+m\Vert v{\Vert }_2^2\ge \min (1,m)\Vert v{\Vert }_{H^1}^2$. If $0\in [a,b]$, the supplied trace inequality givesThus the form is coercive in both cases. The trace theorem and boundedness of $x^2$ give continuity of $a$ and $\ell$, and the Lax-Milgram theorem gives existence and uniqueness.

A scalar Tonelli existence theorem states: if $f(x,y,p)$ is continuous, convex in $p$, and has a uniform superlinear lower bound $f(x,y,p)\ge \Theta (|p|)-C$ with $\Theta (r)/r\to \infty$, and at least one fixed-endpoint admissible absolutely continuous function has finite energy, then the energy attains its infimum in the fixed-endpoint $W^{1,1}$ class. One may take $\Theta$ nonnegative and convex. The lower bound makes derivatives uniformly integrable; endpoint control then gives uniform compactness of a minimizing sequence, and convexity supplies the required lower semicontinuity.

For the specified functional, the integrand is continuous, nonnegative, and convex in the derivative, withThe affine function $u_0(x)=2x-1$ has the required endpoints and finite energy. Thus a global minimizer exists. In fact any finite-energy minimizer belongs to $W^{1,52}$, since its derivative's 52nd power is integrable.

It is unique. The admissible class is convex, and the integrand is strictly convex in $(y,p)$: $|p{|}^{52}$ and $|y{|}^{2011}$ are strictly convex, with the latter multiplied by $e^x>0$. If two distinct admissible finite-energy functions minimized the energy, their midpoint would have strictly smaller energy. Distinct continuous representatives differ on a set of positive measure, so the strict integral inequality really applies. This contradiction proves uniqueness.

Start with a fundamental solution of the second derivative operator:Here $H$ is the Heaviside step function and the last equality holds in the sense of distributions. The factor $x-\xi$ removes a jump in the function itself, while its first derivative has a unit jump. Adding any homogeneous solution preserves the Dirac delta source.

Choose $y_1(x)=x-2$ and $y_2(x)=x-3$. They are independent homogeneous solutions, with $B_1{y}_1=0$, $B_2{y}_2=0$, $B_1{y}_2=-1$, and $B_2{y}_1=1$. Since $1<\xi <2$, the fundamental solution has $B_1\tilde{g}=1$ and $B_2\tilde{g}=2-\xi$. Imposing the two boundary conditions on the proposed combination therefore gives $-A_2+1=0$ and $A_1+2-\xi =0$. ConsequentlyEquivalently,At $x=\xi$ both expressions equal ${\xi }^2-3\xi +1$ and the right slope minus the left slope is $\xi -(\xi -1)=1$. At the endpoints, $g(1,\xi )=-\xi$ and $g_x(2,\xi )=\xi$, while $g_x(1,\xi )=\xi -1$ and $g(2,\xi )=1-\xi$. These explicitly verify both coupled boundary conditions. The homogeneous equation has only affine solutions; its two boundary conditions give $2a+b=3a+b=0$, hence $a=b=0$, proving uniqueness of the Green function.

The reciprocity defect is $g(x,\xi )-g(\xi ,x)=2(x-\xi )$. It records how the coupled boundary conditions affect self-adjointness, even though the differential expression $D^2$ is formally self-adjoint. The construction is an instance of the boundary-matrix construction of an ordinary differential Green function.

Apply energy uniqueness for a linear elastic traction problem. Subtract two solutions with the same loading, writing their displacement difference as $\mathbf{v}$. Linearity makes its body force and boundary traction zero. The preceding energy identity therefore yields ${\int }_{B}2W(e(\mathbf{v}))dV=0$. Positive definiteness and continuity imply $e(\mathbf{v})=0$ everywhere. By the permitted zero-strain characterization,Thus the solution is unique modulo infinitesimal rigid translations and rotations, on each connected body. Conversely adding such a displacement changes neither strain nor stress, so these are exactly the unconstrained modes. This is conditional uniqueness; existence of a pure-traction solution additionally requires balanced resultant force and torque.