Put $v(r,t)=ru(r,t)$. Multiplication of the radial wave equation by $r$ gives $v_{tt}=c^2{v}_{rr}$ for $r>0$. In the characteristic coordinates $\xi =r+ct$, $\eta =r-ct$, this becomes $v_{\xi \eta }=0$. Integrating first in $\xi$ and then in $\eta$ gives $v=f(\xi )+g(\eta )$. ConsequentlyThe two terms are incoming and outgoing radial disturbances. If the field is finite at the origin, their numerators must cancel there; arbitrary independent choices generally produce a $1/r$ singularity.

Let $\alpha =\pi (m+1)/n$, which lies strictly between zero and $\pi$. On the lower ray $z=x$, the integral is $I_R={\int }_0^R{x}^m/(1+x^n)dx$. On the upper ray $z=x{e}^{2\pi i/n}$, the positive boundary orientation runs from $R$ down to zero. Since $z^n=x^n$, that contribution is $-e^{2i\alpha }{I}_R$. The circular contribution vanishes by part (b); henceUse $1-e^{2i\alpha }=-2i{e}^{i\alpha }\sin \alpha$ to obtainThe result is positive as expected. Convergence near zero follows from $m\ge 0$, so both endpoint limits used in the contour argument are justified.

On the circular arc, $|z|=R$, the reverse triangle inequality gives $|1+z^n|\ge R^n-1$. The arc length is $2\pi R/n$, so the ML inequality givesThe strict assumption $n>m+1$ is exactly what makes the exponent negative; it also ensures convergence of the positive-real-axis integral at infinity.

The sector contains exactly one simple pole, at $z_0=e^{i\pi /n}$. All roots of $1+z^n$ are simple, and its residue issince $z_0^n=-1$. For the positively oriented boundary, the residue theorem givesThe vertex at zero is harmless: $f$ is analytic there for nonnegative integer $m$, or a small indent may be removed because its integral is $O({\rho }^{m+1})$.

Let $m=\dim U$, $n=\dim V$, $r=\mathrm{rank}\alpha$ and $r_{\ast }=\mathrm{rank}{\alpha }^{\ast }$. The isomorphism of part (c) preserves dimensions, so the formulas of part (a), applied also to ${\alpha }^{\ast }:V^{\ast }\to U^{\ast }$, giveIf $m>0$, cancellation proves $\mathrm{rank}\alpha =\mathrm{rank}{\alpha }^{\ast }$. If $m=0$, both maps are zero and the equality remains true. This argument derives dual-rank equality from the requested subspace correspondence without assuming it beforehand.

Here $\Phi$ acts on $\beta :V\to U$, so it is the same dualization construction with the source and target interchanged: ${\Phi }_{V,U}:L(V,U)\to L(U^{\ast },V^{\ast })$. This is the natural family interpretation of the symbol used in part (b).

If $\beta \alpha =0$, composition reversal gives ${\alpha }^{\ast }{\beta }^{\ast }=0$, so ${\beta }^{\ast }\in M^r({\alpha }^{\ast })$. Conversely, if $\delta \in M^r({\alpha }^{\ast })$, the isomorphism in part (b) writes $\delta ={\beta }^{\ast }$ uniquely. Then ${\alpha }^{\ast }{\beta }^{\ast }=0$ implies $(\beta \alpha {)}^{\ast }=0$; injectivity of dualization gives $\beta \alpha =0$. ThusSimilarly, $\alpha \beta =0$ is equivalent to ${\beta }^{\ast }{\alpha }^{\ast }=0$, givingEquivalently, the reverse implications follow by taking double duals and using the canonical double-dual identification. The equalities are of subspaces under an isomorphism, not only inclusions.

The dual map is defined by ${\alpha }^{\ast }(\lambda )=\lambda \circ \alpha$. Suppose ${\alpha }^{\ast }=0$. Then $\lambda (\alpha u)=0$ for every $u\in U$ and every $\lambda \in V^{\ast }$. The dual space separates vectors: a nonzero vector can be included in a basis and a coordinate functional takes value one on it. Thus $\alpha u=0$ for every $u$, and $\alpha =0$. This proves injectivity of $\Phi$.

Its source and target have the same finite dimension:An injective linear map between equal finite dimensions is surjective by the rank-nullity theorem. Therefore $\Phi$ is an isomorphism. The inverse is also described by taking the dual again and identifying each space canonically with its double dual.

The zero-composition subspaces of a linear map are subspaces because the defining equations are homogeneous linear equations in $\beta$: they hold for the zero map and are preserved by addition and scalar multiplication. A map with $\beta \alpha =0$ vanishes on $\mathrm{im}\alpha$, so it factors uniquely through the quotient vector space $V/\mathrm{im}\alpha$. Conversely every map from that quotient to $U$ gives such a $\beta$. ThereforeThe condition $\alpha \beta =0$ says exactly that the image of $\beta$ lies in $\ker \alpha$. ConsequentlyBoth dimension formulas use the fact that the space of maps between spaces of dimensions $p,q$ has dimension $pq$: matrix units in any chosen bases supply a basis.

Let $N(t)$ count roots in the open unit disk. A boundary crossing can occur only at the real values $0,\pm 1,\pm 2$ found above, so the count is constant between them. Also $N(-t)=N(t)$, because the polynomial is odd. At $t=0$, $p(z)=z(z^4+1)$ has one root inside and four on the circle, giving $N(0)=1$.

For small positive $t$, the root near zero stays inside. At each boundary root ${\zeta }^4=-1$, $p^{\prime }(\zeta )=-4$, so implicit differentiation gives $z(t)=\zeta -t/4+O(t^2)$. The radial crossing test for polynomial root counts isThe two roots with positive real part enter and the other two leave. Hence $N(t)=3$ for $0<t<1$.

At $t=1$, the boundary roots are $\zeta =e^{\pm i\pi /3}$. Since $\zeta p^{\prime }(\zeta )=5t-4\zeta$, their radial derivative is $\mathrm{Re}(1/(5-4\zeta ))=1/7>0$. Both leave as $t$ increases, so the count becomes one; at $t=1$ itself they are excluded, also leaving one interior root. At $t=2$, the remaining boundary root $z=1$ has radial derivative $1/6>0$ and leaves. For $t\ge 2$ no interior root is possible, since $|p(z)|\le |z{|}^5+|z|<2$ in the open disk. Combining these facts and odd symmetry givesThis also gives the corresponding winding numbers of the image contour away from its crossings. Boundary roots are excluded at the crossing values; using a nearby winding number at $t=0$ would incorrectly count two extra roots.

The image contour has polar representation $2\cos 2\theta e^{3i\theta }$. It is invariant under complex conjugation and under reflection through the origin. As $\theta$ runs from zero to $2\pi$, the successive real-axis visits areat angles $0,\pi /4,\pi /3,2\pi /3,3\pi /4,\pi ,5\pi /4,4\pi /3,5\pi /3,7\pi /4,2\pi$. To specify the crossing direction, differentiate:Its imaginary component is $6$ at the point $2$, $-6$ at $-2$, $3$ at either visit to $1$, and $-3$ at either visit to $-1$. At a visit to zero it is $-4\cos \theta$, nonzero for each of the four angles. Therefore all these visits are transverse crossings of the real axis, even though different arcs share the same image point.

Write $z=e^{i\theta }$. The trigonometric sum formulas giveThus the imaginary part is zero either when $\theta =k\pi /3$, $k=0,\dots ,5$, or when $\theta =\pi /4+k\pi /2$, $k=0,\dots ,3$. These two lists are disjoint. At the latter four points $p=0$. At the former six points the real values, in increasing angular order, areEquivalently, $z=1$ maps to $2$, $z=-1$ to $-2$, $z=e^{\pm i\pi /3}$ to $1$, $z=e^{\pm 2i\pi /3}$ to $-1$, and the four roots of $z^4=-1$ map to zero. These are all the unit-circle points with real image.

For a closed piecewise smooth contour $\Gamma$ in a domain, null-homologous there and avoiding the poles of a meromorphic function $F$, the residue theorem iswhere $n(\Gamma ,a)=(2\pi i{)}^{-1}{\int }_{\Gamma }dz/(z-a)$ is the winding number and the sum runs over enclosed poles with their indices. Suppose now $f$ is meromorphic and has neither zeros nor poles on $\Gamma$. At a zero or pole $a$, write $f(z)=(z-a{)}^{m}h(z)$ with $h$ holomorphic and nonzero, where $m>0$ for a zero and $m<0$ for a pole. Thenso its residue is $m$. Substituting into the residue theorem and changing variables along the image contour gives the argument principle:For a positively oriented simple boundary, this is the number of zeros minus poles inside, counted with multiplicity. The winding-number form also applies to self-intersecting image paths.

The far-field derivative fixes the growing $n=1$ cosine coefficient to $U$ and removes all other growing modes. The possible logarithmic term has derivative $D_0/a$ at the wall, so the zero wall derivative removes it. For each decaying Fourier mode, the wall condition relates its coefficient to its growing partner: the $n=1$ cosine partner has coefficient $U{a}^2$, and every other partner vanishes. Thus the potential flow around a circular cylinder hasIndeed, ${\varphi }_r=U(1-a^2/r^2)\cos \theta$, which is zero at $r=a$ and tends to $U\cos \theta$ at infinity. The additive constant does not change the velocity. The equipotential curves satisfy $(r+a^2/r)\cos \theta =\text{constant}$. They approach vertical straight lines far away and meet the cylinder radially wherever the tangential velocity is nonzero. The two stagnation points are at $\theta =0,\pi$.

Use separation of variables, $\varphi =R(r)\Theta (\theta )$. After multiplying the Laplace equation by $r^2/(R\Theta )$, the two separated equations areSingle-valuedness requires $2\pi$-periodicity. Thus the nonconstant angular modes have $\lambda =n^2$, $n=1,2,\dots$, and angular factors $\cos n\theta ,\sin n\theta$. Their radial factors are $r^n,r^{-n}$. The zero mode has constant angular factor and radial solution $A+B\log r$; a term proportional to $\theta$ would be multivalued.

Finiteness at the origin removes the logarithm and negative powers. In the disk the general regular harmonic expansion isFor the exterior, regularity at every finite radius allowsCoefficients must give convergent harmonic expansions on the domain and the intended boundary regularity at $r=a$. If “finite” is intended to mean bounded as $r\to \infty$, then $D_0=C_n=E_n=0$ and only the constant and decaying modes remain. Infinity is not included in the printed interval; moreover part (b)'s uniform background flow necessarily has an unbounded potential. Stating both interpretations avoids discarding that background mode.

At a simple root of a polynomial $x_i$ of $\omega$, differentiating the product gives ${\omega }^{\prime }(x_i)={\prod }_{k\ne i}(x_i-x_k)\ne 0$. Consequently the Lagrange basis polynomials can be writtenDivide the polynomial interpolation identity from part (a) by $\omega (x)$, at points outside the nodes. This gives the partial fraction decompositionEvery denominator of $\omega$ is simple, and the rational function is proper because $\mathrm{deg}p<\mathrm{deg}\omega$, so no polynomial part is missing.

The Lagrange basis polynomials satisfy ${\ell }_i(x_j)={\delta }_{ij}$: for $j\ne i$, the product has a zero factor, whereas for $j=i$ every factor is one. The polynomialhas degree of a polynomial at most $n$ and has $n+1$ distinct roots of a polynomial. The root bound for a polynomial implies $q=0$. Hence the interpolation identity holds for every polynomial of degree at most $n$.

The composition $g(z)=e^{-f(z)}$ is an entire function, andBy Liouville's theorem, $g$ is constant. Since the exponential function never vanishes,Thus $f$ is constant. Notice that constancy of an exponential alone need not be inverted by choosing a global logarithm; differentiating supplies the conclusion directly.

Fix any $z_0\in \mathbb{C}$. The Cauchy estimate on the circle $|z-z_0|=R$ applies for every $R>0$ because $f$ is an entire function. It givesTherefore $f^{\prime }(z_0)=0$. Since $z_0$ was arbitrary and $\mathbb{C}$ is connected, $f$ is constant. This is the sublinear case of the polynomial growth theorem for entire functions.

Take any positive definite inner product $⟨,{⟩}_0$ on $U$ and define the inner product adapted to an idempotent linear mapIt is symmetric and bilinear. If its quadratic value at $u$ is zero, positive definiteness of the original inner product forces both $\varphi u=0$ and $(I-\varphi )u=0$, hence $u=0$. Thus it is positive definite.

Using ${\varphi }^2=\varphi$ and $(I-\varphi )\varphi =0$ givesThe map is therefore self-adjoint for this new inner product and remains idempotent, so it is an orthogonal projection for the explicitly constructed inner product. Geometrically the construction makes its complementary kernel and image orthogonal.

The finite-dimensional inner product space has the orthogonal decomposition $U=W\oplus W^{\perp }$. Define $\tau (w+z)=w$ for $w\in W$, $z\in W^{\perp }$. This is a linear map with image $W$, satisfies ${\tau }^2=\tau$, and is self-adjoint by taking inner products between decomposed vectors. Hence it is an orthogonal projection.

Any other orthogonal projection with image $W$ has kernel $W^{\perp }$ by part (a); it is identity on $W$ by idempotence and zero on its kernel. Its values are therefore forced to agree with $\tau$, proving uniqueness.

If ${\tau }_2{\tau }_1=0$, every $w_1\in W_1$ has $w_1={\tau }_1{w}_1$, so ${\tau }_2{w}_1=0$ and $w_1\in W_2^{\perp }$. Conversely, if $W_1\subset W_2^{\perp }$, every image ${\tau }_{1}u$ is killed by ${\tau }_2$. Therefore