G. Operators and Differential Equations

Consider the differential equation $$\[ y''' +p\mathopen{}\left( t\right)\mathclose{}y'' +q\mathopen{}\left( t\right)\mathclose{}y' +r\mathopen{}\left( t\right)\mathclose{}y= γ\mathopen{}\left( t\right)\mathclose{} \]$$ with $\( y= y\mathopen{}\left( t\right)\mathclose{} \)$. What is $\( y \)$ given that $\( p \)$, $\( q \)$, $\( r \)$, and $\( γ \)$ belong to $\( \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{} \)$? You can make the system first-order by passing to a linear system: $$\[ \begin{bmatrix}y \\ y' \\ y'' \end{bmatrix} ' = \begin{bmatrix}y' \\ y'' \\ y''' \end{bmatrix}= \begin{bmatrix}0 \\ 0 \\ γ\mathopen{}\left( t\right)\mathclose{}\end{bmatrix} ' + \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ {-}r\mathopen{}\left( t\right)\mathclose{} & {-}p\mathopen{}\left( t\right)\mathclose{} & {-}q\mathopen{}\left( t\right)\mathclose{}\end{bmatrix} ' \begin{bmatrix}y \\ y' \\ y'' \end{bmatrix} ' \]$$ Or, more compactly, $\( \mathbf{f}= \begin{bmatrix}y \\ y' \\ y'' \end{bmatrix} \)$ satisfies $\( \mathbf{f}' = \mathbf{g}+A\mathopen{}\left( \mathbf{f}\right)\mathclose{} \)$. Then $$\[ \int _{a}^{t}{} \mathbf{f}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x= \mathbf{f}\mathopen{}\left( t\right)\mathclose{}-\mathbf{f}\mathopen{}\left( a\right)\mathclose{}= \int _{a}^{t}{}g\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x+\int _{a}^{t}{} A\mathopen{}\left( x\right)\mathclose{}\mathbf{f}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \text{;} \]$$ that is, $\( \mathbf{f}-T\mathopen{}\left( \mathbf{f}\right)\mathclose{}= \mathbf{f}\mathopen{}\left( a\right)\mathclose{}+\mathbf{G} \)$ where $\( \mathbf{G}\mathopen{}\left( t\right)\mathclose{}= \int _{a}^{t}{}g\mathopen{}\left( x\right)\mathclose{}\,\mathrm{d}x \)$ and $\( T\mathopen{}\left( \mathbf{f}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{a}^{t}{} A\mathopen{}\left( x\right)\mathclose{}\mathbf{f}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \)$. So we ought to have $\( \mathbf{f}= {\mathopen{}\left(1-T\right)\mathclose{}}^{-1}\mathopen{}\left( \mathbf{f}\mathopen{}\left( a\right)\mathclose{}+G\right)\mathclose{} \)$.

General setup: Fix $\( A\in \mathrm{M}_n\mathopen{}\left( \mathrm{C}\mathopen{}\left( \mathopen{}\left[a, b\right]\mathclose{}\right)\mathclose{}\right)\mathclose{} \)$ (i.e., $\( A : \mathopen{}\left[a, b\right]\mathclose{} \to \mathcal{L}\mathopen{}\left( {\mathbb{C}}^{n}\right)\mathclose{} \)$ is continuous). Let $\( M= \max_{ x\in \mathopen{}\left[a, b\right]\mathclose{} }{}\mathopen{}\left\lVert{}A\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} \)$. Define $$\[ H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathopen{}\left[a, b\right]\mathclose{}}^{2}\right)\mathclose{}= \mathopen{}\left\{\, \mathbf{f}= \begin{bmatrix}{f}_{1} \\ {f}_{2} \\ \vdots \\ {f}_{n}\end{bmatrix}\,\middle\vert\, , {f}_{i}\in \mathrm{L}^{\mathrm{2}}\mathopen{}\left( {\mathopen{}\left[a, b\right]\mathclose{}}^{2}\right)\mathclose{}, , i\in \mathopen{}\left\{\, 1, 2, \dotsb, n\,\right\}\mathclose{}, \,\right\}\mathclose{} \text{.} \]$$ This is a Hilbert space with inner product $\( \mathopen{}\left\langle{}\mathbf{f}, \mathbf{g}\right\rangle\mathclose{}= \sum_{j}{} \mathopen{}\left\langle{}{f}_{j}, {g}_{j}\right\rangle\mathclose{} = \sum_{j}{} \int _{a}^{b}{} {f}_{j}\overline{{g}_{j}} \)$; or, think $$\[ H= \mathopen{}\left\{\, \mathbf{f} : \mathopen{}\left[a, b\right]\mathclose{} \to {\mathbb{C}}^{n}\,\middle\vert\, , \int _{a}^{b}{} {\mathopen{}\left\lVert{}\mathbf{f}\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{}}^{2} \,\mathrm{d}x\lt \infty, \,\right\}\mathclose{} \]$$ with $\( \mathopen{}\left\langle{}\mathbf{f}, \mathbf{g}\right\rangle\mathclose{}= \int _{a}^{b}{} \mathbf{f}\mathopen{}\left( x\right)\mathclose{}\cdot \mathbf{g}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \)$ as inner product where $\( \mathbf{f}\mathopen{}\left( x\right)\mathclose{}\cdot \mathbf{g}\mathopen{}\left( x\right)\mathclose{} \)$ is the pointwise dot product.

Define $\( T : H \to T\mathopen{}\left( H\right)\mathclose{} \)$ by $\( \mathopen{}\left(T\mathopen{}\left( \mathbf{f}\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( t\right)\mathclose{}= \int _{a}^{t}{} A\mathopen{}\left( x\right)\mathclose{}\mathbf{f}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \)$. For $\( {t}_{1} \)$ and $\( {t}_{2} \)$ in $\( \mathopen{}\left[a, b\right]\mathclose{} \)$ (say $\( {t}_{1}\lt {t}_{2} \)$), $$\[ \mathopen{}\left\lVert{}T\mathopen{}\left( \mathbf{f}\mathopen{}\left( {t}_{2}\right)\mathclose{}\right)\mathclose{}-T\mathopen{}\left( \mathbf{f}\mathopen{}\left( {t}_{1}\right)\mathclose{}\right)\mathclose{}\right\rVert\mathclose{}= \mathopen{}\left\lVert{}\int _{{t}_{1}}^{{t}_{2}}{} A\mathopen{}\left( x\right)\mathclose{}\mathbf{f}\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x\right\rVert\mathclose{}\leq M\int _{{t}_{1}}^{{t}_{2}}{} \mathopen{}\left\lVert{}f\mathopen{}\left( x\right)\mathclose{}\right\rVert\mathclose{} \,\mathrm{d}x\leq M\mathopen{}\left\lVert{}\mathbf{f}\right\rVert\mathclose{}\sqrt{ \mathopen{}\left\lvert{}{t}_{2}-{t}_{1}\right\rvert\mathclose{} } \text{.} \]$$ So, $\( T\in \mathcal{L}\mathopen{}\left( H\right)\mathclose{} \)$ and, furthermore, if $\( \mathopen{}\left({\mathbf{f}}_{k}\right)\mathclose{} \)$ is a bounded sequence in $\( H \)$, then each of the entries of the vector sequence $\( \mathopen{}\left( T\mathopen{}\left( {\mathbf{f}}_{k}\right)\mathclose{} \right)\mathclose{} \)$ is a bounded equicontinuous sequence of $\( {\mathbb{C}}^{n} \)$-valued functions. So, repeated application of Arzela's Theorem shows that $\( \mathopen{}\left( T\mathopen{}\left( {\mathbf{f}}_{k}\right)\mathclose{} \right)\mathclose{} \)$ has a uniformly (hence $\( \mathrm{L}^{\mathrm{2}} \)$) convergent subsequence.

We turn now to an extended example illustrating another way that compact operators can arise from ODE problems, in this case boundary value problems rather than initial value problems. The solution operator here will turn out to be a trace-class operator whose trace we can easily calculate. Fix a positive continuous function $\( r= r\mathopen{}\left( x\right)\mathclose{} \)$ on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$. Let $\( {D}_{0}= \mathopen{}\left\{\, f\in \mathrm{C}^{2}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{}\,\middle\vert\, , f\mathopen{}\left( 0\right)\mathclose{}= 0= f\mathopen{}\left( 1\right)\mathclose{}, \,\right\}\mathclose{} \)$. Define $\( L : {D}_{0} \to \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ by $\( L\mathopen{}\left( f\right)\mathclose{}= {r}^{-1}\mathopen{}\left( f'' \right)\mathclose{} \)$. The operator $\( L \)$ is injective because of the boundary conditions. Define $\( K : \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \to {D}_{0} \)$ by $$\[ \mathopen{}\left(K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-} \int _{0}^{x}{} \int _{0}^{s}{} g\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t \,\mathrm{d}s +\mathopen{}\left(\int _{0}^{1}{} g\mathopen{}\left( t\right)\mathclose{}\mathopen{}\left(1-t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\right)\mathclose{}x \text{.} \]$$ The first integral is $\( {-} \int _{0}^{x}{} \int _{t}^{x}{} g\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}s \,\mathrm{d}t \)$ by Fubini, giving $$\[ \mathopen{}\left(K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= {-} \int _{0}^{x}{} \mathopen{}\left(x-t\right)\mathclose{}g\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t +\mathopen{}\left(\int _{0}^{1}{} g\mathopen{}\left( t\right)\mathclose{}\mathopen{}\left(1-t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t\right)\mathclose{}x \text{.} \]$$ From the first equation above, we see that $\( \mathopen{}\left(K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}'' = {-}rg \)$, and the second equation gives $\( \mathopen{}\left(K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 0\right)\mathclose{}= 0= \mathopen{}\left(K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( 1\right)\mathclose{} \)$, so $\( K \)$ does indeed map to $\( {D}_{0} \)$. Further, $\( L\mathopen{}\left( K\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}= g \)$ for all $\( g\in \mathrm{C}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$, and, because $\( L \)$ is injective, $\( K\mathopen{}\left( L\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}= f \)$ for all $\( f\in {D}_{0} \)$.

Consider $\( H= \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}, r \, \mathrm{d}x\right)\mathclose{} \)$, using Lebesgue measure weighted by $\( r \)$. The inner product is given by $\( \mathopen{}\left\langle{}f, g\right\rangle\mathclose{}= \int _{0}^{1}{} f\mathopen{}\left( x\right)\mathclose{}\overline{g\mathopen{}\left( x\right)\mathclose{}}r\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \)$. Notice that the map $\( U : H \to \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ defined by $\( U\mathopen{}\left( f\right)\mathclose{}= {r}^{\frac{1}{2}}\mathopen{}\left( f\right)\mathclose{} \)$ is a unitary map (that is, an isomorphism of Hilbert spaces) with $\( U^{*}\mathopen{}\left( h\right)\mathclose{}= {U}^{-1}\mathopen{}\left( h\right)\mathclose{}= {r}^{{-}\frac{1}{2}}\mathopen{}\left( h\right)\mathclose{} \)$.

In this inner product we have $\( \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{}= \mathopen{}\left\langle{}f, L\mathopen{}\left( g\right)\mathclose{}\right\rangle\mathclose{} \)$ and $\( \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, f\right\rangle\mathclose{}= 0 \)$ for $\( f \)$ and $\( g \)$ in $\( {D}_{0} \)$. Indeed the boundary conditions give $$\[ \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, g\right\rangle\mathclose{}-\mathopen{}\left\langle{}f, L\mathopen{}\left( g\right)\mathclose{}\right\rangle\mathclose{}= {-} \int _{0}^{1}{} f'' \overline{g} +\int _{0}^{1}{} f\overline{g}'' = \int _{0}^{1}{}\mathopen{}\left(f\overline{g}'' -f'' \overline{g}\right)\mathclose{}= \int _{0}^{1}{} \mathopen{}\left(f\overline{g}' -f' \overline{g}\right)\mathclose{} ' = \mathopen{}\left( \mathopen{}\left(f\overline{g}' -f' \overline{g}\right)\mathclose{} \right|\mathclose{}_{0}^{1}= 0 \]$$ and $$\[ \mathopen{}\left\langle{}L\mathopen{}\left( f\right)\mathclose{}, f\right\rangle\mathclose{}= {-} \int {} f'' \overline{f} = {-} \int {} \mathopen{}\left( \mathopen{}\left(f' \overline{f}\right)\mathclose{} ' -f' \overline{f}' \right)\mathclose{} = \mathopen{}\left( {-}f' \overline{f} \right|\mathclose{}_{0}^{1}+\int _{0}^{1}{} {\mathopen{}\left\lvert{}f' \right\rvert\mathclose{}}^{2} = \int _{0}^{1}{} {\mathopen{}\left\lvert{}f' \right\rvert\mathclose{}}^{2} \text{.} \]$$ The operator $\( K \)$ extends to $\( H \)$. Namely the extended $\( K \)$ is $\( {-}T+{ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{} \)$, where $\( {ξ}_{j}\mathopen{}\left( x\right)\mathclose{}= {x}^{j} \)$ and $\( T \)$ is the integral operator on $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}, r \, \mathrm{d}x\right)\mathclose{} \)$ with kernel $\( k\mathopen{}\left( x, t\right)\mathclose{}= {χ}_{\mathopen{}\left[0, x\right]\mathclose{}}\mathopen{}\left( x-t\right)\mathclose{} \)$. Thus $\( K\in \mathcal{HS}\mathopen{}\left( H\right)\mathclose{} \)$. Notice that $\( T \)$ maps $\( H \)$ to $\( \mathrm{C}^{1}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ because $\( \mathopen{}\left(T\mathopen{}\left( g\right)\mathclose{}\right)\mathclose{}\mathopen{}\left( x\right)\mathclose{}= \int _{0}^{x}{} \int _{0}^{s}{} g\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( t\right)\mathclose{} \,\mathrm{d}t \,\mathrm{d}s \)$ by Fubini, and the inside integral is a continuous function of $\( s \)$.

We remark that $\( \operatorname{Ker}\mathopen{}\left( K\right)\mathclose{}= \mathopen{}\left\{\, 0\,\right\}\mathclose{} \)$. (Indeed, if $\( f\in H \)$ and $\( K\mathopen{}\left( f\right)\mathclose{}= 0 \)$, then $\( \mathopen{}\left(K\mathopen{}\left( f\right)\mathclose{}\right)\mathclose{}'= 0 \)$, so $\( \int _{0}^{x}{}f\mathopen{}\left( r\right)\mathclose{}= \mathopen{}\left\langle{}f, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{} \)$ for all $\( x\in \mathopen{}\left[0, 1\right]\mathclose{} \)$, which makes $\( f= 0 \)$ a.e.) Thus, the eigenvalues of the positive compact operator $\( K \)$ are precisely the reciprocals of the eigenvalues of the differential operator $\( L \)$.

Thanks to the providential Corollary III.43, the calculation of $\( \operatorname{Tr}\mathopen{}\left( K\right)\mathclose{} \)$ as an integral involving $\( r \)$ is straightforward. One checks easily that $\( UT U^{*}= \mathrm{M}_{\sqrt{r}}{S}^{2}\mathrm{M}_{\sqrt{r}} \)$, where $\( S \)$ is the Volterra operator on $\( \mathrm{L}^{\mathrm{2}}\mathopen{}\left( \mathopen{}\left[0, 1\right]\mathclose{}\right)\mathclose{} \)$ and $\( \mathrm{M}_{\sqrt{r}} \)$ is multiplication by $\( \sqrt{r} \)$. We then have $$\[ \operatorname{Tr}\mathopen{}\left( T\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( \mathrm{M}_{\sqrt{r}}{S}^{2}\mathrm{M}_{\sqrt{r}}\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( {S}^{2}\mathrm{M}_{r}\right)\mathclose{}= 0 \]$$ by Proposition III.29. Since $\( K= {-}T+{ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{} \)$, it follows that $$\[ \operatorname{Tr}\mathopen{}\left( K\right)\mathclose{}= \operatorname{Tr}\mathopen{}\left( {ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\right)\mathclose{}= \mathopen{}\left\langle{}{ξ}_{1}, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= \int _{0}^{1}{} x\mathopen{}\left(1-x\right)\mathclose{}r\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x \text{.} \]$$

We're on a roll—let's find the analogous formula for $\( \operatorname{Tr}\mathopen{}\left( {K}^{2}\right)\mathclose{} \)$.

Now $$\[ {K}^{2}= {T}^{2}-T\mathopen{}\left({ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\right)\mathclose{}-\mathopen{}\left({ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\right)\mathclose{}T+{\mathopen{}\left({ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\right)\mathclose{}}^{2} \text{.} \]$$ Since $\( \operatorname{Tr}\mathopen{}\left( {\mathopen{}\left(ξ\otimes ν\right)\mathclose{}}^{2}\right)\mathclose{}= \mathopen{}\left\langle{}ξ, ν\right\rangle\mathclose{}\operatorname{Tr}\mathopen{}\left( ξ\otimes ν\right)\mathclose{}= {\mathopen{}\left\langle{}ξ, ν\right\rangle\mathclose{}}^{2} \)$, we get $$\[ \operatorname{Tr}\mathopen{}\left( {K}^{2}\right)\mathclose{}= 0-2\operatorname{Tr}\mathopen{}\left( T{ξ}_{1}\otimes \mathopen{}\left({ξ}_{0}-{ξ}_{1}\right)\mathclose{}\right)\mathclose{}+{\mathopen{}\left\langle{}{ξ}_{1}, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}}^{2}= {\mathopen{}\left\langle{}{ξ}_{1}, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}}^{2}-2\mathopen{}\left\langle{}T{ξ}_{1}, {ξ}_{0}-{ξ}_{1}\right\rangle\mathclose{}= {\mathopen{}\left(\int _{0}^{1}{} x\mathopen{}\left(1-x\right)\mathclose{}r\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}x\right)\mathclose{}}^{2}-2\int _{0}^{1}{} \int _{0}^{x}{} \mathopen{}\left(1-x\right)\mathclose{}\mathopen{}\left(x-t\right)\mathclose{}tr\mathopen{}\left( t\right)\mathclose{}r\mathopen{}\left( x\right)\mathclose{} \,\mathrm{d}t \,\mathrm{d}x \text{.} \]$$

The final result can be stated entirely in ODE terms.

In the case $\( r\equiv 1 \)$, the (normalized) eigenfunctions form the sine basis, and $\( {λ}_{n}= {n}^{2}{π}^{2} \)$. The formulas specialize to $$\[ \sum_{n}{}\frac{1}{{n}^{2}}= \frac{{π}^{2}}{\mathrm{6}} \]$$ and $$\[ \sum_{n}{}\frac{1}{{n}^{4}}= {π}^{4}\mathopen{}\left(\frac{1}{\mathrm{36}}-\frac{1}{\mathrm{60}}\right)\mathclose{}= \frac{{π}^{4}}{\mathrm{90}} \text{,} \]$$ as one would hope.

It is highly unlikely that the eigenvalues can be found exactly for general $\( r \)$. They can, however, be approximated for a given $\( r \)$ using a numerical ODE solver as follows. For $\( λ\gt 0 \)$, denote by $\( {f}_{λ} \)$ the solution of the initial value problem $\( f'' +λrf= 0 \)$, $\( f\mathopen{}\left( 0\right)\mathclose{}= 0 \)$, $\( f' \mathopen{}\left( 0\right)\mathclose{}= 1 \)$ on $\( \mathopen{}\left[0, 1\right]\mathclose{} \)$. Let $\( Φ\mathopen{}\left( λ\right)\mathclose{}= {f}_{λ}\mathopen{}\left( 1\right)\mathclose{} \)$. Use the ODE solver to calculate $\( Φ\mathopen{}\left( λ\right)\mathclose{} \)$ for a great many $\( λ \)$, or for not so many $\( λ \)$ and then calculate an interpolating function. Anyway, you can draw the graph of $\( Φ \)$ fairly well. As intuition would predict, it oscillates back and forth across $\( 0 \)$ with diminishing amplitude and increasing distance between the successive zeros. These zeros are of course the eigenvalues of $\( L \)$, which can be searched for and zoomed in on.

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