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Take $z$ upward along the straight rod. The part above height $z$ has weight $\lambda g(L-z)$, so, with tensile force positive, the internal tension is compressive:For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies areThe Euler-Lagrange equation for this functional iswhich is exactly

Clamping at the bottom fixes displacement and slope:At the free upper end, bending moment and transverse force vanish:Since $\sigma (L)=0$, the four boundary conditions are

Set $u=X^{\prime }$. Integrating the field equation once and using the free-end shear condition givesor, with $s=L-z$,Introduce the dimensionless similarity coordinateand write $u={\eta }^{1/3}F(\eta )$. Direct substitution reduces the equation toThis is the Bessel differential equation of order $1/3$, so

The free-moment condition is $u^{\prime }(L)=0$. As $\eta \to 0$,The second term has nonzero limiting $z$ derivative, so the free-end condition forces $b=0$. The free-shear condition $u^{\prime \prime }(L)=0$ then follows from the differential equation. At the clamp, $u(0)=0$, givingLet $j_{-1/3,1}$ be the smallest positive zero of this Bessel function. The first self-buckling threshold isor equivalently

For $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, a singular point $x_0$ is a regular singular point if $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. The Bessel differential equation has $xP=1$ and $x^{2}Q=x^2-{\nu }^2$, so zero is regular singular.

Use the Frobenius method with $y=x^s{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$. Matching coefficients gives $(s^2-{\nu }^2)a_0=0$ and $[(s+n{)}^2-{\nu }^2]a_n+a_{n-2}=0$, with missing negative-index coefficients zero. The regular root is $s=\nu$. Then $a_1=0$, all odd coefficients vanish, and the Frobenius series for Bessel functions has recurrenceFor noninteger $\nu$, the root $s=-\nu$ supplies an independent even-power series $y_2=x^{-\nu }{\sum }_{m\ge 0}{b}_{2m}{x}^{2m}$ with $b_{2m}=-b_{2m-2}/[4m(m-\nu )]$ and $b_0\ne 0$. None of these denominators vanishes. With standard normalization this is $J_{-\nu }(x)$, independent of the Bessel function of the first kind $J_{\nu }(x)$. The real powers here are understood for $x>0$.

For integer $\nu \ge 0$ and the stated normalization $a_0=1/(2^{\nu }\nu !)$, iteration givesThe ratio of consecutive term magnitudes is $|x{|}^2/[4(m+1)(m+\nu +1)]\to 0$. Thus the series converges absolutely for every finite complex $x$, defining an entire function for integer $\nu$. At integer order, $J_{-\nu }=(-1{)}^{\nu }{J}_{\nu }$; an independent solution is instead the Bessel function of the second kind.

Termwise differentiation of the convergent Frobenius series and collection of powers in the Bessel differential equation gives the lowest coefficient $({\sigma }^2-{\nu }^2)a_0$, the next coefficient $[(\sigma +1{)}^2-{\nu }^2]a_1$, and, for $n\ge 2$, the coefficient $[(\sigma +n{)}^2-{\nu }^2]a_n+a_{n-2}$. Since $a_0\ne 0$, the stated choices $\sigma =\pm \nu$, $a_1=0$ and the recurrence make all coefficients vanish.

For $\nu =1/2$, the two resulting even-coefficient series areandThese are linearly independent on $x>0$ and give the Bessel functions of order one half, up to normalization. The specified condition $a_1=0$ is sufficient; in the negative-exponent branch at this particular order the next coefficient equation alone would allow $a_1$ to be free, adding a multiple of the positive-exponent solution.

For the Legendre differential equation, substitution gives the full conditionsThus $\sigma =0$ permits arbitrary $a_1$, while $\sigma =1$ requires $a_1=0$. This Frobenius recurrence for the Legendre equation separates the even and odd series. The nearest singular points of the differential equation are $\pm 1$, so the series about zero has radius at least one, and the generic nonterminating solutions have radius exactly one. Particular cancellations can give a larger radius.

Termination occurs when the degree reaches a nonnegative integer $N$ with $\lambda (\lambda +1)=N(N+1)$. For $\lambda \ge 0$ this givesChoose the even branch for even $N$ or the odd branch for odd $N$ to obtain a Legendre polynomial. The recurrence then terminates and gives a solution for all real $x$, with infinite radius of convergence. A nonzero entire solution at other $\lambda \ge 0$ is impossible: each nonterminating parity has coefficient ratio tending to one in the variable $x^2$, and so radius one.

Let $Y=\hat{y}$. Transform the three terms of the order-zero Bessel differential equation using part (a):Adding and dividing by $i$ givesOn $-1<\omega <1$, separation of variables gives $Y^{\prime }/Y=\omega /(1-{\omega }^2)$, hence $Y=C/\sqrt{1-{\omega }^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extensionThis is a genuine distributional weak solution: the differential equation can be written $[(1-{\omega }^2)Y{]}^{\prime }+\omega Y=0$, and $(1-{\omega }^2)Y=C\sqrt{1-{\omega }^2}{1}_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint Dirac delta terms.

Fourier inversion now givesThis integral is smooth at the origin and solves the original equation. Its value there is $C/2$. Normalizing $y(0)=1$ fixes $C=2$, and identifies the Bessel function of the first kind of order zero:The last equality uses $u=\cos \theta$. Without normalization, the regular solution is an arbitrary constant times this integral; this is the Fourier transform of the Bessel function J0 pair.

Regularity qualification. The source specifies the differential equation but no condition at its singular point $t=0$. The transformed equation separately has solutions $C_{\pm }/\sqrt{{\omega }^2-1}$ on the two exterior intervals. Thus it does not, by itself, determine the support or a single global multiplicative constant. The compactly supported choice above constructs the regular Bessel branch. The other independent branch, the Bessel function of the second kind, is singular at the origin. Also $J_0$ is not absolutely integrable on the real line, so its transform is properly taken as a tempered distribution, rather than relying on the stronger classical integrability assumptions of part (a).

Put $G(p)=\mathcal{L}[J_0](p)$, initially for real $p>0$. Besides the Laplace transform of a derivative, use multiplication by the original variable in the formTransforming the Bessel differential equation, with its given initial conditions, yieldsThe constant terms cancel, leaving $(1+p^2)G^{\prime }+pG=0$. This first-order ordinary differential equation integrates toThe initial value theorem for Laplace transforms gives ${\lim }_{p\to \infty }pG(p)=J_0(0)=1$, so $C=1$. Thus the Laplace transform of the Bessel function J0 isFor complex $p$ with $\mathrm{Re}p>0$, choose the analytic square-root branch agreeing with the positive square root on the positive real axis.

For the forcing $k=J_0$, part (b) now gives $F(p)=(1+p^2{)}^{-3/2}$. On the other hand, the same multiplication rule givesTaking an integral from zero divides the Laplace transform by $p$. By uniqueness of the Laplace transform,A direct check also explains the cancellation. This expression has $f^{\prime }=x{J}_0$ and $f^{\prime \prime }=J_0+x{J}_0^{\prime }$. The Bessel differential equation says $(x{J}_0^{\prime }{)}^{\prime }=-x{J}_0$, so integration from zero gives $x{J}_0^{\prime }=-{\int }_0^{x}t{J}_0(t)dt=-f$. Hence $f^{\prime \prime }+f=J_0$, and both initial conditions hold.

Let $y$ be a smooth constrained minimizer. For any smooth variation $h$ vanishing at the endpoints, use the normalized admissible curve $y_{\epsilon }=(y+\epsilon h)/\sqrt{J[y+\epsilon h]}$. Since $J[y]=1$, differentiating at zero givesIn particular this holds for every compactly supported test function $h$ in $(0,1)$. Integration by parts and the fundamental lemma of the calculus of variations givewhich is the required Bessel differential equation with $\boxed{\lambda =I[y]}$. The lemma used here says that a continuous function whose integral against every compactly supported smooth test function is zero must itself vanish. No multiplier-existence theorem is needed: the explicit normalization produces the variation directly.

The endpoint $0$ causes no divergent energy for these variations: smoothness and $y(0)=h(0)=0$ imply $y,h=O(x)$. Also $J[v]>0$ for every nonzero admissible $v$, and $v/\sqrt{J[v]}$ has unit $J$ and energy $I[v]/J[v]$. Thus minimizing $I$ with $J=1$ is equivalent to minimizing the scale-invariant Rayleigh quotient $I/J$. The zero function is excluded whenever this quotient is written.

Set $t=\sqrt{\lambda }x$ and $w(t)=y(t/\sqrt{\lambda })$. The chain rule turns the equation into the parameter-one Bessel differential equation. By the regular-at-zero solution property supplied in the paper,For a nonzero solution, $c\ne 0$, so $y(1)=0$ forces $J_k(\sqrt{\lambda })=0$. Conversely, every positive zero $z$ supplies the admissible eigenfunction $J_k(zx)$ with eigenvalue $z^2$.

Here is a direct justification of the infimum characterization that does not assume a minimizer exists. Let $u(x)=J_k(z_{k,1}x)$ and choose its sign so that $u>0$ on $(0,1)$. The first positive zero exists by the oscillatory large-argument behavior of the Bessel function of the first kind. At zero, its Frobenius series for Bessel functions gives $u(x)=C{x}^k(1+O(x^2))$. At one, its zero is simple by uniqueness for a regular second-order ordinary differential equation. Expanding $y=u(y/u)$ and using the equation for $u$ gives the Ground-state representation of a Sturm-Liouville energy:For completeness, apply the identity first on $[\epsilon ,1-\epsilon ]$. The discarded boundary term is $[x{u}^{\prime }{y}^2/u]$. It tends to zero at both ends: $x{u}^{\prime }/u\to k$ and $y=O(x)$ at zero, while $u^{\prime }/u=O((1-x{)}^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, soThe infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.

Put $x=r\cos \theta$. The function $e^{ikx}$ solves the two-dimensional Helmholtz equation, $({\nabla }^2+k^2)e^{ikx}=0$. Expand its angular dependence as a Fourier series,Substitution into the polar-coordinate Laplacian,givesWith $\rho =kr$, this is the Bessel differential equation. Regularity of the original plane wave at the origin excludes the singular second-kind solution. Thus $F_m(r)$ is a constant multiple of the regular first-kind function, andFor integer negative orders, $J_{-m}=(-1{)}^m{J}_m$, so this indexing still supplies the regular radial solution. The radial equation fixes the form of each term; the small-radius plane-wave expansion fixes its coefficient.

The incident plane wave is periodic in $\theta$, so expand it as a Fourier series ${\sum }_m{R}_m(r)e^{im\theta }$. Substitution into the Helmholtz equation, with $\rho =kr$, gives the radial Bessel differential equationRegularity at the origin selects the Bessel function of the first kind, rather than the singular second solution. ThusThe coefficients $a_m$ are independent of $r$; their values follow from matching near the origin in the next part. The completed identity is the Jacobi-Anger expansion.