Exercises — Bessel's equation and Bessel functions (intro, physical relevance)

WHAT we compare: the standard form has the constant ${\nu }^2$ subtracted from $x^2$. Here that constant is $25$. WHY: in $(x^2-{\nu }^2)$ the number is always ${\nu }^2$, never $\nu$ — it is order-squared. So ${\nu }^2=25\Rightarrow \nu =5$ (we take $\nu \ge 0$ by convention). General solution: two independent solutions are needed for a 2nd-order linear ODE, so $y(x)=c_1{J}_5(x)+c_2{Y}_5(x).$

$J_3(0)=0$ (finite). $Y_3$ diverges. For integer order $n\ge 1$ the second kind behaves like $Y_n(x)\sim -\frac{(n-1)!}{\pi }{\left(\frac{2}{x}\right)}^n$ near $0$, i.e. like $x^{-3}$ here — it shoots to $-\infty$. Why it matters: on any region that contains the centre $r=0$ we must set the coefficient of $Y_3$ to zero.

WHAT we do: multiply every term by $x^2$ to clear denominators. WHY: the standard form starts with $x^2{y}^{\prime \prime }$; the given equation is that form divided by $x^2$. $x^2{y}^{\prime \prime }+x{y}^{\prime }+(x^2-4)y=0\Rightarrow {\nu }^2=4\Rightarrow \nu =2.$ Solution: $y=c_1{J}_2(x)+c_2{Y}_2(x)$.

With $\nu =1$: $J_1(x)={\sum }_{m\ge 0}\frac{(-1{)}^m}{m!\Gamma (m+2)}{\left(\frac{x}{2}\right)}^{2m+1}$. Here $\Gamma (m+2)=(m+1)!$ (the Gamma function on integers is a factorial).

• $m=0:\frac{1}{0!1!}{\left(\frac{x}{2}\right)}^1=\frac{x}{2}$.
• $m=1:\frac{-1}{1!2!}{\left(\frac{x}{2}\right)}^3=-\frac{x^3}{16}$. So $J_1(x)\approx \frac{x}{2}-\frac{x^3}{16}$. At $x=0.5$: $0.25-\frac{0.125}{16}=0.25-0.0078125=0.2421875$. (True value $J_1(0.5)=0.24227$ — our two terms already agree to 4 decimals.)

WHY this equation is $\nu =\frac{1}{2}$: the constant is $\frac{1}{4}=(\frac{1}{2}{)}^2$. $J_{1/2}(x)=\sqrt{\frac{2}{\pi x}}\sin x$ is proportional to $\frac{\sin x}{\sqrt{x}}$, so it must satisfy the ODE. Direct check: let $y=x^{-1/2}\sin x$. $y^{\prime }=-\frac{1}{2}{x}^{-3/2}\sin x+x^{-1/2}\cos x.$ $y^{\prime \prime }=\frac{3}{4}{x}^{-5/2}\sin x-x^{-3/2}\cos x-x^{-1/2}\sin x.$ Substitute: $x^2{y}^{\prime \prime }=\frac{3}{4}{x}^{-1/2}\sin x-x^{1/2}\cos x-x^{3/2}\sin x,$ $x{y}^{\prime }=-\frac{1}{2}{x}^{-1/2}\sin x+x^{1/2}\cos x,$ $(x^2-\frac{1}{4})y=x^{3/2}\sin x-\frac{1}{4}{x}^{-1/2}\sin x.$ Sum of $x^{1/2}\cos x$ terms: $-1+1=0$. Sum of $x^{3/2}\sin x$: $-1+1=0$. Sum of $x^{-1/2}\sin x$: $\frac{3}{4}-\frac{1}{2}-\frac{1}{4}=0$. All cancel ⇒ it is a solution. ✓

General radial solution (written back in the physical variable $r$, since $x=kr$): $R(r)=c_1{J}_0(kr)+c_2{Y}_0(kr)$. Finiteness at $r=0$: $Y_0\to -\infty$ there (see the definition callout above), so $c_2=0$ — this is the Sturm-Liouville theory boundedness condition at the singular endpoint. Edge condition $R(a)=R(1)=0$: $J_0(k\cdot 1)=0\Rightarrow k={\alpha }_{0,j}$, the zeros of $J_0$. Lowest two: $k_1=2.405,k_2=\mathrm{5.520.}$

$\frac{f_2}{f_1}=\frac{{\alpha }_{0,2}}{{\alpha }_{0,1}}=\frac{5.520}{2.405}=2.2952\dots$ A string's overtones are $2,3,4,\dots$ (integers), giving harmony. Here the ratio $2.295$ is not an integer, so the overtones clash — the sound is inharmonic, a metallic "thud" rather than a clear pitch.

(a) First a needed fact — WHY $J_0^{\prime }(x)=-J_1(x)$. Differentiate the first-kind series term by term. The general identity $\frac{d}{dx}(x^{-\nu }{J}_{\nu }(x))=-x^{-\nu }{J}_{\nu +1}(x)$, taken at $\nu =0$, gives directly $J_0^{\prime }(x)=-J_1(x)$ (this is proven from the series in the parent note's Frobenius section). Now the reading: because $J_0^{\prime }=-J_1$, the slope of $J_0$ is zero exactly where $J_1$ crosses zero, so $J_0$'s turning points line up with $J_1$'s zeros. Between $x=0$ and its first zero at $2.405$, $J_0$ falls from $1$ (no turning point yet), while $J_1$ rises from $0$, reaches its first maximum near $x\approx 1.84$, then heads back down. On the figure the amber dot marks that $J_1$ peak, and it sits to the left of the dotted cyan line at the first zero of $J_0$ ($x=2.405$). (b) For large $x$, $J_{\nu }(x)\approx \sqrt{\frac{2}{\pi x}}\cos (\cdots )$. The envelope is $\sqrt{2/(\pi x)}$ (the amber dashed curves), which decays like $1/\sqrt{x}$. Physically the wave's energy spreads over an ever-larger ring (circumference $\propto x$), so amplitude${}^2\propto 1/x$ ⇒ amplitude $\propto 1/\sqrt{x}$. That is the shrinking you see.

(i) Separate time and space. Write $u(r,\theta ,t)=v(r,\theta )T(t)$. Substituting into $u_{tt}=c^2{\nabla }^{2}u$ and dividing by $c^{2}vT$ (this is Separation of variables): $\frac{T^{\prime \prime }}{c^{2}T}=\frac{{\nabla }^{2}v}{v}=\text{constant}.$ Why the constant must be negative (call it $-k^2$): the left side depends only on $t$, the right only on space, so both equal a single constant. If that constant were $+k^2>0$, then $T^{\prime \prime }=c^2{k}^{2}T$ gives growing/decaying exponentials $e^{\pm ckt}$ — a drum that runs away or dies, not one that rings. A vibrating membrane must oscillate in time, which needs $T^{\prime \prime }=-c^2{k}^{2}T\Rightarrow T=\cos (ckt),\sin (ckt)$. That physical demand forces the sign to be $-k^2$: ${\nabla }^{2}v+k^{2}v=0,T^{\prime \prime }+c^2{k}^{2}T=0.$ (ii) Radial modes. For radially-symmetric modes ($\nu =0$) the Laplacian in polar and cylindrical coordinates gives $v_{rr}+\frac{1}{r}{v}_r+k^{2}v=0$; with $x=kr$ this is Bessel order $0$: $v(r)=c_1{J}_0(kr)+c_2{Y}_0(kr).$ Boundedness at $r=0$ kills $Y_0$ ($c_2=0$); fixed rim $v(a)=0$ gives $J_0(ka)=0\Rightarrow k_j={\alpha }_{0,j}/a$. Mode shape: $v_j(r)=J_0({\alpha }_{0,j}r/a)$. (iii) Fundamental frequency. Angular frequency ${\omega }_1=c{k}_1=c{\alpha }_{0,1}/a$; frequency $f_1={\omega }_1/(2\pi )$: $f_1=\frac{c{\alpha }_{0,1}}{2\pi a}=\frac{100\times 2.405}{2\pi \times 0.15}=\frac{240.5}{0.94248}=255.2\text{Hz}.$

The centre $r=0$ is no longer in the region, so $Y_0$ need not be dropped — both survive: $R(r)=c_1{J}_0(kr)+c_2{Y}_0(kr).$ Two boundary conditions $R(b)=0$ and $R(a)=0$: $c_1{J}_0(kb)+c_2{Y}_0(kb)=0,c_1{J}_0(ka)+c_2{Y}_0(ka)=0.$ A nonzero $(c_1,c_2)$ exists only if the determinant vanishes: $J_0(ka)Y_0(kb)-J_0(kb)Y_0(ka)=0.$ The roots $k$ of this transcendental equation give the annular drum's frequencies. Why the contrast with L4.1: the number of surviving constants is dictated by whether the singular point $r=0$ is inside the domain.

WHAT we choose and WHY: the coefficient $e^{2x}$ grows exponentially; substituting $t=e^x$ turns exponential growth in $x$ into polynomial behaviour in $t$, which is what Bessel's equation (a polynomial-coefficient ODE) needs. With $t=e^x$: $\frac{dt}{dx}=e^x=t$, so $\frac{dy}{dx}=t\frac{dy}{dt}$. $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(t\frac{dy}{dt}\right)=t\frac{d}{dt}\left(t\frac{dy}{dt}\right)=t\left(\frac{dy}{dt}+t\frac{d^{2}y}{d{t}^2}\right)=t^2{y}_{tt}+t{y}_t.$ Also $e^{2x}=t^2$. The ODE $y^{\prime \prime }+e^{2x}y=0$ becomes $t^2{y}_{tt}+t{y}_t+t^{2}y=0,$ which is exactly Bessel's equation of order $\nu =0$ in the variable $t$. Hence $y=c_1{J}_0(t)+c_2{Y}_0(t)=c_1{J}_0(e^x)+c_2{Y}_0(e^x).$

WHY the recurrence: it lets us climb from known half-integer orders to the next without touching the series. Set $\nu =\frac{1}{2}$: $J_{-1/2}+J_{3/2}=\frac{2(\frac{1}{2})}{x}{J}_{1/2}=\frac{1}{x}{J}_{1/2}$. So $J_{3/2}=\frac{1}{x}{J}_{1/2}-J_{-1/2}=\sqrt{\frac{2}{\pi x}}\left(\frac{\sin x}{x}-\cos x\right).$ Evaluate at $x=\pi /2$: $\sin (\pi /2)=1$, $\cos (\pi /2)=0$, so the bracket is $\frac{1}{\pi /2}-0=\frac{2}{\pi }$. Prefactor $\sqrt{\frac{2}{\pi \cdot \pi /2}}=\sqrt{\frac{4}{{\pi }^2}}=\frac{2}{\pi }$. $J_{3/2}(\pi /2)=\frac{2}{\pi }\cdot \frac{2}{\pi }=\frac{4}{{\pi }^2}=0.40528\dots$

WHY orthogonality matters: it is what lets you expand an arbitrary initial drum shape as ${\sum }_j{c}_j{J}_0({\alpha }_{0,j}r)$ — the Bessel analogue of Fourier series, guaranteed by the self-adjoint Sturm–Liouville form of Bessel's equation. Diagonal case $i=j=1$: $N_1=\frac{1}{2}{J}_1({\alpha }_{0,1}{)}^2=\frac{1}{2}(0.5191{)}^2=\frac{1}{2}(0.26946)=\mathrm{0.13473.}$

Multiply to standard form before reading $\nu$ ::: yes — leading term must be $x^2{y}^{\prime \prime }$ The slot in $(x^2-\square )$ holds ::: ${\nu }^2$, not $\nu$ Domain contains $r=0$? drop ::: $Y_{\nu }$ (it diverges there) Annulus (hole at centre)? keep ::: both $J_{\nu }$ and $Y_{\nu }$ Orthogonality weight for $J_{\nu }$ on a disc ::: $w(r)=r$ Drum fundamental frequency ::: $f_1=c{\alpha }_{0,1}/(2\pi a)$ Separation constant sign for a vibrating drum ::: $-k^2$ (so time part oscillates, not grows)