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Every $v\in V$ is a finitely supported sequence, so for a finite set of indices $S$,Thus $E$ is a spanning set. If $F$ contained $v\notin E$, the displayed expression would give a nontrivial relationamong the distinct vectors of the finite subset $\{v\}\cup \{e^{(i)}:i\in S\}$ of $F$: the coefficient of $v$ is $1$. This contradicts the linear independence of $F$. It also covers $v=0$, whose singleton is not linearly independent. Hence $F=E$. In particular, $E$ is both a basis and a maximal linearly independent subset of $V$.

Write $\mathbf{u}=\mathbf{b}-\mathbf{a}$. Since the triangle is nondegenerate, $\mathbf{u}\ne 0$. Every point on the line $AB$ has a unique position vector $\mathbf{p}=\mathbf{a}+t\mathbf{u}$. The condition that $CP$ is orthogonal to $AB$ is the dot product equationIts coefficient of $t$ is $-\Vert \mathbf{u}{\Vert }^2\ne 0$, so it has exactly one solution. The foot of a perpendicular is thereforeThis is the orthogonal projection of $C$ onto the line $AB$, expressed using the Euclidean norm.

On the surface of revolution, a normal vector is proportional to $-H^{\prime }(r){\mathbf{e}}_r+{\mathbf{e}}_z$. Thus the normal force and gravity have no ${\mathbf{e}}_{\theta }$ component. The ${\mathbf{e}}_{\theta }$ component of Newton's second law, using the given linear drag force, isMultiplication by $r$ therefore givesThe printed target equation omits the factor $1/m$. The matching original PDF, page 5, gives both ${\mathbf{F}}_D=-\alpha \dot{\mathbf{r}}$ and $\dot{h}=-\alpha h$, so this is a source inconsistency. If its $\alpha$ in the target equation is intended to mean the drag coefficient per unit mass, put $\lambda =\alpha /m$ and write $\dot{h}=-\lambda h$. Equivalently, if the intended force was ${\mathbf{F}}_D=-m\alpha \dot{\mathbf{r}}$, the requested convention givesThese formulas determine $\dot{\theta }$ once $r(t)$ is known; they do not assume a constant radius.

Differentiate the Airy ordinary differential equation $y^{\prime \prime }=sy$ to obtain $y^{(3)}=y+s{y}^{\prime }$. Combining part (a)'s derivatives yieldsTo make this vanish for an arbitrary Airy profile, the time powers must agree: $a-1=4a$, giving $a=-1/3$. Their common coefficient must then vanish, giving $-1/3+b^{-1}=0$, hence $b=3$. ThereforeFor $t>0$ the two terms cancel exactly:This is an Airy similarity solution of the linear dispersive equation. The Airy equation here is the linear partial differential equation $u_t+u_{xxx}=0$, the reduced equation specified in the question; it is distinct from the profile's Airy ordinary differential equation. The construction works for every linear combination of its two independent profiles, including zero. No condition at $t=0$ is supplied, and this singular time is excluded. With real cube roots the final formula can also be used separately on $t<0$.

If $Pdx+Qdy=dF$ with continuous second partial derivatives of $F$, then $P=F_x$, $Q=F_y$. The symmetry of second derivatives requiresThis is a necessary condition for an exact differential. With continuously differentiable coefficients it is also locally sufficient, and is globally sufficient on a simply connected open domain; a domain with holes can require additional global information.

Seek a separated nonzero solution $u=X(x)Y(y)$. Division by $XY$ on intervals where the factors are nonzero gives $X^{\prime \prime }/X+Y^{\prime \prime }/Y=-25$. Write $X^{\prime \prime }+\lambda X=0$ and $Y^{\prime \prime }+(25-\lambda )Y=0$, with both factors zero at their endpoints $0,\pi$.

For $X$, a nonpositive $\lambda$ gives only the zero solution under the two endpoint conditions: linear functions vanish identically when $\lambda =0$, and the hyperbolic solutions do likewise for $\lambda <0$. For $\lambda >0$, the endpoints force $X=A\sin (mx)$ with positive integer $m$ and $\lambda =m^2$. Applying the same argument to $Y$ gives $Y=B\sin (ny)$ and $25-m^2=n^2$, with positive integer $n$.

The only positive integer pairs satisfying $m^2+n^2=25$ are $(3,4)$ and $(4,3)$. By superposition,To establish that this is the most general sufficiently regular solution, expand $u$ in the complete product Fourier sine series on the square. Substitution, or integration by parts against each sine mode, gives $(25-m^2-n^2)c_{mn}=0$ for every coefficient. Hence only the two displayed modes remain; all boundary terms vanish because of the given zero values. This is the eigenvalue-$25$ case of the Dirichlet eigensystem of a rectangle. The zero solution is included when $A=B=0$.

Use $\sin x=x-x^3/6+O(x^5)$ inside the exponential series:The two cubic contributions cancel, leavingThus the coefficient of $x^3$ is zero, even though the function itself is not an even function.

Factor out $9$ and use the binomial series with exponent $-1/2$:Thus the expansion through cubic order isThe coefficients of $x$ and $x^3$ vanish because the function is an even function. The underlying binomial expansion converges for $|x|<3$.

Differentiate the Fourier series term by term and collect cosine and sine terms. Substitution into the equation givesFor a smooth periodic solution, differentiation of the series is justified. Alternatively, multiply the differential equation by each trigonometric mode and integrate by parts; periodicity cancels the boundary terms and gives the same coefficient constraints without first differentiating an infinite sum.

Fourier orthogonality isolates a coefficient. Multiplication by $\sin (mx)$ and integration over $[0,2\pi ]$ annihilates the constant, all cosine terms and all sine terms except the one with $n=m$, whose squared integral is $\pi$. Thus the corresponding coefficients on the two sides must agree for each positive integer $m$.

Multiplication by $\cos (mx)$ works in the same way, while integration against the constant mode extracts $a_0$. This is the uniqueness of the Fourier coefficients; it does not follow from comparing a few point values of the series.

The constant coefficient is $a_0=0$. For every $n\ge 1$ the two mode equations arewhere ${\delta }_{n1}$ is the Kronecker delta. The first equation gives $a_n=-\sqrt{3}n{b}_n$, so the second becomes $(1+3n^2)b_n=4{\delta }_{n1}$. HenceNo other modes survive. This is the single-frequency case of the Fourier solution of a first-order equation with periodic forcing.

If the form is $dF$ for a scalar function with continuous second partial derivatives, then $p=F_x$ and $q=F_y$. Equality of the mixed partial derivatives therefore requiresThis is necessary for an exact differential. For continuously differentiable coefficients on a simply connected open domain it is also sufficient; on a domain with holes, the equality alone need not give a single-valued global potential.

Let $w=x+iy$. Then $y-ix=-iw$, and the complex trigonometric identity $\sin (-iw)=-i\sinh w$ givesThus the form is exact, with complex-valued scalar potential $F(x,y)=\cosh (x+iy)$. Differentiating that potential with respect to the two real variables checks both coefficients individually.

For the printed coefficients, $p=2\sin x\cosh y$ and $q={\sin }^{2}x\sinh y$. Their cross derivatives areThey are not equal as functions; their difference is $2\sin x(1-\cos x)\sinh y$. Hence the form is not exact. Agreement on certain lines or isolated points is insufficient, because exactness requires the derivative identity throughout an open domain.

Use completing the square after taking out the leading coefficient:Thus $a=3$, $b=-2$ and $c=-1$. Since a real square is nonnegative, the global minimum is $-1$, attained at $x=2$.

Differentiate the first ordinary differential equation and substitute the second:The original first equation also gives $bv=u^{\prime }-au$, soTherefore the requested coefficients areThey are the trace and determinant of the coefficient matrix. This derivation never divides by $b$ and is valid even if $b=0$. In that degenerate case the scalar second-order equation is a necessary consequence, while the original constraint $u^{\prime }=au$ must also be retained when reconstructing the coupled system.

The vector triple product identity isConsequently a choice of the requested coefficients isThese are the standard identity's coefficients, even if the particular three vectors are linearly dependent and a representation need not be unique.

The associated homogeneous linear differential equation has characteristic roots $2\pm i$, so its complementary function is $e^{2x}(A\cos x+B\sin x)$. For a particular integral, the method of undetermined coefficients suggests $y_p=a{x}^2+bx+c$. Substitution and comparison of coefficients givesso $y_p=25x^2+40x+22$. The general solution is thereforeThe first boundary condition gives $A=-21$. At $x=\pi /2$, the polynomial part already equals the assigned boundary value and the cosine vanishes, so $e^{\pi }B=0$. ThusThis is the unique solution satisfying the two boundary conditions; here the phrase “particular solution” means that selected member of the full family, rather than only the polynomial particular integral.

Expanding the quadratic form givesMatching coefficients of $xy$ and $y^2$ yields

Adjoin $\dot{\mu }=0$ and put $w=z+\mu$. The extended centre manifold is a graph $w=H(x,y,\mu )$ tangent to $w=0$ at the origin. The equations becomeHere $ϵ>0$ is fixed. The centre-manifold invariance equation, obtained by the chain rule, isWrite $H=H_2+O({\rho }^3)$, where $\rho =|x|+|y|+|\mu |$ and $H_2$ is a homogeneous quadratic polynomial. To quadratic order the centre-manifold invariance equation isThe terms $\mu x,\mu y,{\mu }^2$ have zero coefficients: their equations are homogeneous, and the operator $1+y{\partial }_x-x{\partial }_y$ is invertible on these quadratic polynomials. For $H_2=a{x}^2+bxy+c{y}^2$, comparison of coefficients givesHence $a=3/5$, $b=-2/5$ and $c=2/5$, givingSubstitution into the $y$ equation givesThe constant in the big O notation may depend on fixed $ϵ$. Thus the equations printed in part (b) are the retained terms of the local Taylor series, rather than an exact identity for the full reduced vector field.

For the Hopf bifurcation classification in part (a), set $\mu =0$ and use $x=r\cos \theta$, $y=-r\sin \theta$. The radial derivative is $\dot{r}=(x\dot{x}+y\dot{y})/r$. Its cubic angular mean isHere $⟨\cdot ⟩$ denotes the mean over one revolution. Removing the oscillatory cubic terms by a near-identity transformation gives this same cubic radial coefficient in the Hopf normal form. The linear radial coefficient is $5ϵ\mu /2$.