Use the chain rule with $x=r\cos \theta$, $y=r\sin \theta$. Then $U_r=u_x\cos \theta +u_y\sin \theta$ and $V_{\theta }=-r{v}_x\sin \theta +r{v}_y\cos \theta$. Substituting the Cartesian Cauchy-Riemann equations gives $V_{\theta }=r{U}_r$. Similarly $V_r=v_x\cos \theta +v_y\sin \theta =-U_{\theta }/r$. Thus, for $r>0$, the polar equations areThe restriction $r>0$ is needed because the polar chart degenerates at the origin.

The far radial derivative selects the growing frequency-one term $Ur\cos \theta$. The matching decaying harmonic is $B\cos \theta /r$. The no-penetration condition gives ${\partial }_r\varphi =(U-B/r^2)\cos \theta =0$ at $r=a$, so $B=U{a}^2$. Thus the potential flow around a circular cylinder isThe remaining single-valued exterior harmonics are excluded by the zero radial derivative at the cylinder and the prescribed derivative at infinity, apart from the arbitrary additive constant. In particular the logarithmic radial term would have a nonzero radial derivative on the cylinder. A circulation term proportional to $\theta$ is excluded by single-valuedness of the potential.

For a separated harmonic function $R(r)\Theta (\theta )$ in polar coordinates, single-valuedness forces integer angular frequencies. The radial factors for frequency $m\ge 1$ are $r^m,r^{-m}$, while frequency zero has a constant and $\ln r$. Finiteness at the origin excludes $r^{-m}$ and $\ln r$.

The source's angular factor is $\cos \theta$ in every term, as printed, rather than $\cos (n\theta )$. The power-law angular mode for the planar Poisson equation givesHence, assuming the series permits termwise differentiation, the general finite, single-valued solution isThe arbitrary harmonic series must itself converge to a finite function on the prescribed disk. If ${\alpha }_0\ne 0$, the source has a direction-dependent limit at the origin; the displayed potential still extends finitely there, but the equation at that point is understood weakly rather than as a smooth classical source.

For the special source, a particular potential is $r^3\cos \theta /8$. Its boundary cosine term is canceled by $-a^{2}r\cos \theta /8$, and the boundary sine term is supplied by $(r/a{)}^2\sin 2\theta$. The resulting Dirichlet boundary conditions has solutionIt is finite at zero, has the required boundary values, and its Laplacian is $r\cos \theta$.

Differentiate the integral formula using the Leibniz integral rule. Its two moving-endpoint terms cancel, leavingEach term before subtraction lies between zero and $m/(3x)$. ThereforeThe bounds are sharp at any fixed positive $x$: put $f=m$ below that point and zero above it for the upper bound, or reverse those choices for the lower bound. If continuity of the forcing is required, smooth approximations approach the same bounds.

The Green's function is nonpositive, so $f\ge 0$ gives $y\le 0$. If also $f\le m$, thenThe integral solution therefore givesBoth bounds are attained, by the constant forcings $f=m$ and $f=0$, respectively. In particular, boundedness of the solution does not require it to approach zero.

Use the drift-removing substitution for a Dirichlet Sturm-Liouville problem, $z(x)=e^{-2x}u(x)$. It gives $u^{\prime \prime }+\lambda u=0$ with $u(0)=u(1)=0$. Positive $\lambda =k^2$ requires $\sin k=0$; zero or negative $\lambda$ yields only the trivial solution under both endpoint conditions. HenceMultiplication of the original equation by $e^{4x}$ gives $-(e^{4x}{z}^{\prime }{)}^{\prime }-4e^{4x}z=\lambda e^{4x}z$. Thus the appropriate weighted inner product has weight $e^{4x}$, andMultiplying each eigenfunction by $\sqrt{2}$ gives weighted norm one.

Divide the equation by $\sqrt{1-x^2}$ to obtain its Sturm-Liouville form:Thus $p=\sqrt{1-x^2}$, $q=0$ and $w=(1-x^2{)}^{-1/2}$. For $L=-w^{-1}(pd/dx{)}^{\prime }$, the Lagrange identity for a Sturm-Liouville operator gives the endpoint form $[p({\bar{f}}^{\prime }g-\bar{f}{g}^{\prime }){]}_{-1}^1$. A natural self-adjoint choice isUnder $x=\cos \theta$, these become the ordinary zero-derivative conditions at $\theta =0,\pi$ for $-d^2/d{\theta }^2$. They therefore define the realization whose Neumann eigenfunctions of the Chebyshev differential operator are $T_n(x)=\cos (n\arccos x)$, with $n=0,1,\dots$.

Boundedness alone does not choose these endpoint conditions: $\sin (n\arccos x)$ is also bounded but generally has a nonzero endpoint value of $p{y}^{\prime }$. Requiring bounded $y^{\prime }$ is a sufficient stronger regularity condition for the displayed zero-flux limits. Other self-adjoint endpoint choices are possible; the boundary form and the operator domain must specify which is intended.

A regular singular point is a nonordinary point $x_0$ for which $(x-x_0)p(x)$ and $(x-x_0{)}^{2}q(x)$ extend analytically across $x_0$. These pole restrictions permit the Frobenius method.

An ordinary point of the normalized second-order equation is a point where both coefficient functions $p(x)$ and $q(x)$ are analytic in a neighborhood. Ordinary continuity is enough for a local existence theorem, but analyticity is the condition used for a convergent local power series solution.

Let $A{v}_1={\lambda }_1{v}_1$ and $A{v}_2={\lambda }_2{v}_2$, with both vectors nonzero and ${\lambda }_1\ne {\lambda }_2$. If $c_1{v}_1+c_2{v}_2=0$, apply $A-{\lambda }_{2}I$ to get $c_1({\lambda }_1-{\lambda }_2)v_1=0$. Hence $c_1=0$, and then $c_2=0$.

For the general independence of eigenvectors for distinct eigenvalues, use induction. In a relation ${\sum }_{j=1}^n{c}_j{v}_j=0$, applying $A-{\lambda }_{n}I$ gives ${\sum }_{j=1}^{n-1}{c}_j({\lambda }_j-{\lambda }_n)v_j=0$. Independence of the first $n-1$ vectors makes each coefficient vanish. Distinctness of the eigenvalues then gives $c_1=\cdots =c_{n-1}=0$, and the original relation gives $c_n=0$. No Hermitian assumption is required.

The convolution is $(u\ast v)(x)={\int }_{-\infty }^{\infty }u(t)v(x-t)dt$. In this transform convention, the convolution theorem says $\widehat{u\ast v}=\hat{u}\hat{v}$, and Parseval's identity saysTo match the integral's scale exactly, set $u=f/2$ and $h(x)=\frac{1}{2}{1}_{[-1,1]}(x)=g(2\pi x)/2$. Then $\hat{u}=1/(1+k^2)$ and $\hat{h}=\sin k/k$, using the value one at zero. Their convolution is the box smoothing of a two-sided exponential with both scales equal to one:Its transform is precisely $\sin k/[k(1+k^2)]$. Since $w$ is even, Parseval's identity reduces the requested integral toUse ${\int }_0^1\cosh xdx=\sinh 1$, ${\int }_0^1{\cosh }^{2}xdx=1/2+\sinh 2/4$ and ${\int }_1^{\infty }{e}^{-2x}dx=e^{-2}/2$. Simplifying givesThe apparent singularity at $k=0$ is removable, and the integrand decays as $O(|k{|}^{-6})$, so the integral converges.

For the two-sided exponential, split the integral at zero:For the interval indicator, the Fourier transform of an interval indicator givesAt $k=0$, the continuous limiting value is $4\pi$, the length of the interval. Changing the values at its two endpoints has no effect on the transform.

For the proposed vector, the first horizontal component of $Kq$ is $-k(\alpha +\beta )/2$. Since the corresponding component of $q$ is zero, a normal mode requires $\beta =-\alpha$. The vertical components then give $Kq=2kq$. With an arbitrary nonzero overall amplitude, choose $\alpha =1$ to obtainThe analogous pattern invariant under reflection across the vertical axis isDirect multiplication confirms $K{q}_x=2k{q}_x$. The two patterns give independent eigenvectors with the same eigenvalue, as expected from the square's rotation symmetry.

In a breathing mode, each displacement is proportional to its equilibrium position. A representative vector isMultiplying by the stiffness matrix gives $K{q}_b=4k{q}_b$. ThusPositive and negative amplitudes are uniform expansion and contraction about the center.

The first two vectors displace every particle identically in the horizontal or vertical direction. Every relative displacement is zero, so every spring has zero extension to first order. The third vector is $(y_i,-x_i)$ evaluated at the equilibrium points, which is an infinitesimal clockwise rotation. For any edge the relative displacement is perpendicular to that edge, again giving zero extension.

These are the rigid-motion zero modes of a spring framework: two translations and one infinitesimal rotation. Each has $Kq=0$, so $\boxed{\omega =0}$. In the linearized dynamics its amplitude has the form $A+Bt$, expressing free rigid motion rather than a restoring oscillation. Exact finite rigid rotations also preserve every spring length; the mode vector is their first-order tangent.

The Cartesian tensor law is $T_{ik}^{\prime }=Q_{ia}{Q}_{kb}{T}_{ab}$. In every complete tensor contraction, summation over a transformed index contracts a pair of $Q$ factors to a Kronecker delta. In detail,Thus all three are invariant under rotations. They are $\mathrm{tr}T$, $\mathrm{tr}(T^2)$ and $\mathrm{tr}(T^3)$. For a real symmetric tensor, orthogonal diagonalization givesExpanding ${\prod }_i(\lambda -{\lambda }_i)$ and using Newton's identities to recover the characteristic polynomial from traces of powers yieldsComparison identifies $\boxed{c_2=-1,c_1=\frac{1}{2},c_0=-\frac{1}{6}}$.

Let $Av=\alpha v$. The commutation relation gives $A(Bv)=B(Av)=\alpha Bv$, so $B$ preserves the eigenspace of $A$ with eigenvalue $\alpha$. Because it is a nondegenerate eigenvalue, the eigenspace is one-dimensional, and $Bv=\beta v$ for some real $\beta$. This includes $Bv=0$. Thus the eigenvectors of $A$ form a common eigenbasis, andWriting the corresponding unit-modulus eigenvalue of $U$ as $\zeta =e^{i\theta }$ givesThe pairing refers to the same common eigenvector; independent reorderings of the two real spectra would lose this information. The simultaneous diagonalization can also be constructed without requiring simple eigenvalues by diagonalizing $B$ within each eigenspace of $A$.

Since $A$ and $B$ are real symmetric matrices, $U^{†}=A-iB$. Expanding the unitarity condition givesEquality of the entrywise real and imaginary parts yieldsThese are the identities for the real and imaginary parts of a symmetric unitary matrix.