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Introduction to Linear Algebra with Mathematica
Sturm--Liouville theory is actually a generalization for the infinite dimensional case of famous eigenvalue/eigenvector problems for finite square matrices that we discussed in Part I of this tutorial. Although a Sturm--Liouville problem can be formulated in operator form as L[ y ] = λy similar to the matrix eigenvalue problem Ax = λx, where the operator L is an unbounded differential operator and y is a smooth function.
The corresponding theory that is known as Sturm--Liouville theory originated in two articles by the French mathematician of German descent Jacques Charles François Sturm (1803--1855).
Sturm, J.C.F.,
"Mémoire sur les Équations différentielles linéaires du second ordre",
Journal de Mathématiques Pures et Appliquées, 1836, Vol. 1,
pp. 106-186. [sept. 28, 1833]
Sturm, , J.C.F.,
"Mémoire sur une classe des d'Équations à différences partielles".
Journal de Mathématiques Pures et Appliquées, 1836, Vol. 1,
pp. 373--444.
Following the same 1836 year, Sturm together with his friend, the
French
mathematician Joseph
Liouville (1809--1882), published very influential articles that provided crucial groundwork for the theory.
Sturm & Liouville,
"Démonstration d'un théorème de M. Cauchy relatif aux racines imaginaires des équations".
Journal de Mathématiques Pures et Appliquées, 1836, Vol. 1,
pp. 278--289.
Sturm & Liouville,
"Note sur un théorème de M. Cauchy relatif aux racines des équations simultanées", Comptes rendus de l'Académie des Sciences (English: Proceedings of the Academy of Sciences), 1837, Vol. 4, pp. 720--739.
Sturm & Liouville,
"Extrait d'un Mémoire sur le développement des fonctions en séries dont les différents termes sont assujettis à satisfaire à une même équation différentielle lindaire, contenant un paramètre variable",
Journal de Mathématiques Pures et Appliquées, 1837, Vol 2,
pp. 220--233.
Comptes rendus de l'Académie des Sciences ((English: Proceedings of the Academy of sciences), 1837, Vol. 4, pp. 675--677.
A classical Sturm–Liouville theory, named after Jacques Charles
François Sturm (1803--1855) and Joseph Liouville (1809--1882),
involves analysis of eigenvalues and eigenfunctions for a second order
linear differential operator (we use letter L to emphasize that this is a linear operator)
where I is the identity operator, and p(x) and q(x) are given continuous functions on some interval [𝑎, b]. Since p(x) follows the derivative operator, it should be a differentiable function, which we abbreviate as p(x) ∈ C¹[𝑎, b]. The linear self-adjoint operator \eqref{EqSturm.3} is referred to as the Sturm--Liouville operator.
Correspondingly, the Sturm--Liouville
theory is about a real second-order homogeneous linear self-adjoint differential equation with a parameter λ of the form
\begin{equation} \label{EqSturm.1}
\frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}y}{{\text d}x}
\right] - q(x)\, y + \lambda \,w (x)\,y(x) =0 , \qquad a < x < b ,
\end{equation}
subject to the homogeneous boundary conditions of the third kind
An operator L acting in a Hilbert space with inner product ⟨· , ·⟩ is called self-adjoint iff
\[
\langle L\,u , v \rangle = \langle u , L\,v \rangle
\]
for every pair of vectors u, v from the domain of operator
L.
Note: If boundary conditions \eqref{EqSturm.2} are not homogeneous, Sturm--Liouville theory is not applicable because the set of functions that satisfy these conditions is not a vector space. These two-point boundary value problems \eqref{EqSturm.1}, \eqref{EqSturm.2} occur in many mathematical applications.
Here y = y(x) is a nontrivial (meaning not identically zero) function of the free variable x∈[𝑎, b]. A positive function w(x) is called the "weight" or "density" function---it is not a part of the differential operator \eqref{EqSturm.3}. On the other hand,
functions p(x), p'(x), and q(x) are specified at the outset. In the simplest of cases, all coefficients are continuous on the finite closed interval [𝑎, b], and p(x) has continuous derivative. In this simplest case, a nontrivial function y(x) is called a solution of Eq.\eqref{EqSturm.1} if it is twice continuously differentiable on (𝑎, b) and satisfies the given equation at every point in (𝑎, b).
Lemma 1:
Any second order differential operator \( a(x)\,\texttt{D}^2 + b(x)\,\texttt{D} \) can be transferred to a self-adjoint form upon multiplying it by an integrating factor:
Let us consider an arbitrary second order differential equation
\[
M\left[ x, \texttt{D} \right] u = a(x)\,u'' + b(x)\, u' + c(x)\, u = 0 ,
\]
with some given continuous functions 𝑎(x), b(x), and c(x). We denote by μ(x) an integrating factor that reduces the differential operator M into exact form:
\[
\mu (x)\,M\left[ x, \texttt{D} \right] y = p(x)\, y'' + p' y' + c(x)\, y .
\]
Hence, we get two equations
\[
\mu (x)\,a (c) = p(x), \qquad \mu (x)\,b(x) = p' (x) = \mu' a + \mu \,a' .
\]
This yields the differential equation (which is separable) for the integrating factor:
\[
\frac{{\text d}\mu}{{\text d}x} = \mu \left[ b - a' \right] /a \qquad \Longrightarrow \qquad \mu (x) = \exp \left\{ \int \frac{b - a'}{a} \,{\text d} x \right\} .
\]
Lemma 2:
Any second order differential equation \( a(x)\,y'' +
b(x)\,y' + c(x)\, y =0 \) can be transferred to a normal form upon suitable substitution:
\[
u'' + q(x)\,u =0 .
\]
We make the Bernoulli substitution \( y = u(x)\,v(x) . \) Then its derivatives become
Remark:
Transformation of arbitrary differential operator of second order into an operator in self-adjoint form \( L\left[ x, \texttt{D} \right] = q(x)\,\texttt{I} - \texttt{D}\,p(x)\,\texttt{D} \) by means of an integrating factor is not unique---there are known many other options. It is a part of general approach of factorization of differential operator, discussed in tutorial I.
Since L ≠ L*, the given operator (1.1) is not self-adjoint.
We convert operator (1.1) into self-adjoint form by applying an integrating factor.
With 𝑎(x) = 1 and b(x) = 5, we can identify the integrating factor using Lemma 1.
Upon multiplication by the integrating factor \( \mu =
e^{5x}, \) we reduce the given differential operator to a self-adjoint form:
we can rewrite this Sturm--Liouville problem in operator form:
\begin{equation} \label{EqSturm.5}
L\left[ x, \texttt{D} \right] y = w\,\lambda y , \qquad B_a y = 0, \quad B_{b} y = 0.
\end{equation}
This boundary value problem has an obvious solution---the identically zero function. Since we are not after such a trivial solution, we need something more. The Sturm--Liouville problem (S-L, for short) consists of two parts: the first part is about finding values of parameter λ for which the problem has a nontrivial solution (not identically zero); such values are called eigenvalues. The second part includes determination of nontrivial solutions that are called eigenfunctions. Note that a Sturm--Liouville problem may have other constraints, not necessarily formulated above.
A Sturm--Liouville problem consists of a differential equation with a parameter (usually denoted by λ) subject to some additional conditions. These additional conditions could be homogeneous boundary conditions, periodic conditions or of some other type. The Sturm--Liouville problem involves finding nontrivial (not identically zero) solutions, called eigenfunctions and corresponding values of parameter λ, called eigenvalues. The set of all eigenvalues is called spectrum of the Sturm--Liouville problem. The pairof eigenvalue λ and corresponding eigenfunction u>, (λ, u), is called an eigenpair for the Sturm--Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2}.
The theory of Sturm--Liouville problems is well developed for second order differential equations. The most complete part of it is devoted to self-adjoint differential operators \eqref{EqSturm.3}, which is our objective. However, many results are known for not self-adjoint problems, problems with parameter λ in boundary conditions, and equations of order higher than two. It should be noted that spectrum for forth order (and higher) differential equations may have very complicated.
There are two kinds of Sturm--Liouville problems for second order differential operators. One is called classical or regular when p(x) > 0 and 1/p(x) > 0 for all points from the closed interval x∈[𝑎, b]. These assumptions are necessary to render the theory as simple as possible while retaining considerable generality. It turns out that these conditions are valid in many problems that we will consider shortly. If p(𝑎) = 0, or p(b) = 0, or p(𝑎) = 0 = p(b), the Sturm--Liouville problem is said to be singular. If conditions on coefficients p(x), q(x), and ρ(x) are held to make the Sturm--Liouville problem regular, but the interval is unbounded, then such problem is also referred to as singular.
Theorem 1:
If p(x) > 0, q(x) > 0, and \( \left. p\,u\,u' \right\vert_{x=a}^{x=b} \leqslant 0, \)
then classical Sturm--Liouville operator \eqref{EqSturm.3} is positive meaning that all its eigenvalues are positive. If q(x) ≥ 0, then its spectrum is nonnegative.
Let u be an eigenfunction corresponding to an eigenvalue λ for a classical Sturm--Liouville operator \eqref{EqSturm.3}. This means that u belongs to the domain of the operator \( L\left[ x, \texttt{D} \right] = q(x) \texttt{I} - \texttt{D}\,p(x)\,\texttt{D} , \) so u(x) is twice continuously differentiable and satisfies the homogeneous boundary conditions \eqref{EqSturm.2}. Then
\begin{align*}
\langle u, L\,u \rangle &= \langle u, \lambda\,u \rangle = \lambda\,\langle u, u \rangle = \lambda \,\| u \|^2
\\
&= \int_a^b \left( q\,u^2 - u\,\texttt{D}\,p(x)\,\texttt{D} \,u \right) {\text d}x .
\end{align*}
Jacques Charles François Sturm (1803--1855) was a French mathematician of Switzerland descent. Charles spent his adult life in Paris. His primary interests were fluid mechanics and differential equations. Sturm along with the Swiss engineer Daniel Colladon was the first to accurately determine the speed of sound in water. In mathematics, he won the coveted Grand prix des Scienes Mathematiques for his work in differential equations. Sturm held the chair of mechanics at the Sorbonne and was elected a member of the French Academy of Sciences. The asteroid 31043 Sturm is named for him. Sturm's name is one of the 72 names engraved at the Eiffel Tower.
Joseph Liouville was a French mathematician known for his work in analysis, differential geometry, and number theory and for his discovery of transcendental numbers---i.e., numbers that are not the roots of algebraic equations having rational coefficients. He was also influential as a journal editor and teacher. Joseph founded the Journal de Mathématiques Pures et Appliquées which retains its high reputation up to today.
In physics and other applications, many problems arise in the form of boundary
value problems involving second order ordinary differential equations, written
in self-adjoint form:
where \( \texttt{D} = {\text d}/{\text d}x \) is
the derivative operator and \( \texttt{I} = \texttt{D}^0 \) is the identity operator. Sturm was the first who generalized the well known
matrix problem for eigenvalues/eigenvectors to the linear differential
operators by considering the eigenvalue problem \eqref{EqSturm.5}
for functions y(x) subject to some boundary conditions. Now this problem is referred to as the Sturm--Liouville problem. Here p(x), q(x), and ρ(x) > 0 are specified continuous functions at the outset, which is usually some interval (finite or not) of real axis ℝ.
Example 3:
Let us consider the simplest self-adjoint differential operator with q = 0:
subject to the first kind boundary conditions, also known as Dirichlet boundary conditions:
\[
\psi (0) = 0 \qquad \psi (\ell ) = 0 .
\]
When a function is specified at the boundary, then the corresponding boundary conditions are named after G. Dirichlet.
We rewrite the corresponding Sturm--Liouville problem (SL problem for short) in the standard form by setting λ = 2m E/ ℏ:
Solutions of the differential equation \( y'' + \lambda \,y =0 \) depend on the sign of parameter λ.
When λ = −μ² is negative, the general solution is a linear combination of two exponential terms
Let us pretend that we don’t know the eigenpairs. Upon writing the equation for a particular eigenvalue λ_{n} with corresponding eigenfunction ϕ_{n}, we get
The goal is to estimate the eigenvalue λn without actually solving the differential equation, analytically or numerically. Let’s suppose that we wish to estimate the lowest eigenvalue λ₁, known as the principal eigenvalue. We’ll consider
the special case ℓ = 2. he exact answer is
\( \lambda_1 = \left( \frac{\pi}{\ell} \right)^2 . \) When ℓ = 2, it is
N[(Pi/2)^2]
2.4674
If we take \( u(x) = x(x-2) \) to satisfy the Dirichlet boundary conditions, then its Rayleigh quotient becomes
\[
R[u] = \frac{\langle u, L\,u \rangle}{\langle u, u \rangle} = \frac{F[u]}{\| u \|^2} = 2.5 .
\]
\[
\langle u, u \rangle = \| u \|^2 = \int_0^2 x^2 (x-2)^2 {\text d}x = \frac{16}{15} .
\]
2* Integrate[(2*x - 2)^2, {x, 0, 2}]
8/3
Integrate[x*x*(x - 2)^2, {x, 0, 2}]
16/15
8/3/(16/15)
5/2
So we see that 2.5 is larger than the principal eigenvalue, λ₁ ≈ 2.4674;.
Since we are trying to estimate the lowest eigenvalue,
we’ll use functions that have no zeroes in (0,1) – the idea is to use a function that approximates the
eigenfunction ϕ₁(x). We’ll try several functions, and denote each one as u_{t}(x) for “trial function”, or simply u(x). Another trial function can be chosen as
\[
u(x) = \begin{cases}
x, & \ \mbox{ for }\quad 0 < x < 1,
\\
2-x , & \ \mbox{ for }\quad 1 < x < 2.
\end{cases}
\]
Note that the derivative of this trial function is ±1 for x ≠ 1. As such, the numerator of the Rayleigh quotient becomes
\[
\int_0^2 \left( u' \right)^2 {\text d} x = \int_0^2 {\text d} x = 2.
\]
The denominator is
\[
\int_0^2 \left( u \right)^2 {\text d} x = \int_0^1 x^2 {\text d} x + \int_1^2 \left( 2 -x \right)^2 {\text d} x = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} .
\]
Therefore, the Rayleigh quotient associated with this trial function,
has value 3, which lies above the true value λ₁ ≈ 2.4674;.
However, if we take as a trial function u = sin(πx/2), the Rayleigh quotient for this function becomes
which has one node at x = 1. Hence, we expect that this trial function will peovide a good approximation for ϕ₂(x) = sin(πx), the eigenfunction corresponding to the second eigenvalue λ₂ = π² ≈ 9.8696.
Using Mathematica, we find the Rayleigh quotient for this trial function:
As usual, prime stands for the derivative (in Lagranhe notation), y' = dy/dx.
It could be shown similarly to the previous example that negative λ are not possible. So we need to consider only two cases: λ = 0 and λ = ω² > 0. The former gives us an eigenvalue λ = 0 to which corresponds a constant as an eigenfunction. To the latter corresponds the general solution
to the lowest eigenvalue λ_{0} = 0. However, any constant trial function will give the true value of λ_{0} = 0.
■
Example 3M:
There are two known Sturm--Liouville problems with mixed boundary conditions when on one end we have the Dirichlet condition while on the other end we have the Neumann condition. So we start with one of them:
It is clear from our previous discussion that eigenvalues must be positive in the given Sturm--Liouville problem.
The general solution (λ > 0) of the equation \( y'' + \lambda \,y = 0 \) subject to the Dirichlet boundary condition at the left end x = 0 is
When a product is zero, then at least one multiple must be zero. Obviously, c cannot be zero, otherwise we will have a trivial solution. Since λ > 0 according to our assumption, we get the condition
Therefore, the Rayleigh quotient gives an upper estimate of 3.5 to the true value of the principal eigenvalue λ = π²/16 ≈ 0.61685. So we take another trial function
\[
u[x] = \begin{cases}
x , & \quad\mbox{for} \quad 0 < x < 1,
\\
1, & \quad\mbox{for} \quad 1 < x < 2
\end{cases}
\]
Graphs of tangent function and the linear function.
Mathematica code
From this figure, we see that the Sturm--Liouville problem has no solutions for λ < 0 when H is positive. However, when H is negative and less than 1, there exists one negative eigenvalue (see figure below).
We plot two functions, 0.7 μ and tangent hyperbolic:
Let ω_{n} (n = 1, 2, 3, …) be the sequence of positive roots of the transcendent equation (3T.3). Then the eigenvalues and corresponding eigenfunctions are
The transcendent equation (3T.3) could be solved only by a numerical solver using, for instance, Newton's method. Nevertheless, if we let \( t = \sqrt{\lambda} , \) then we see from their graphs that there exists an infinite number of roots of \( \tan t = -t . \)
Since tangent function has vertical asymptotes at \( t = \frac{\pi}{2} + n\pi , \quad n=0,1,2,\ldots ; \) the roots t_{n} of the equation \( \tan t = -t \) approach
Note that this asymptotic formula is valid only for positive H. For negative H, the SL problem has a sequence of positive eigenvalues, but with different asymptotic behavior.
The principal eigenvalue is λ_{1} ≈ 2.02876, which we find with the aid of Mathematica:
FindRoot[Tan[k] == -k, {k, 1.6}]
{k -> 2.02876}
Upon setting H = 1,
we choose a trial function that satisfies the boundary boundary conditions and calculate the corresponding Rayleigh quotient
\[
R[u] = \frac{\langle u, L\,u \rangle}{\langle u, u \rangle} = \frac{35}{12} \approx 2.91667 \qquad \mbox{for}\quad u(x) = x \left( 2x-3 \right) ,
\]
with
\[
F[u] = \langle u, L\,u \rangle = \int_0^1 \left( 4x-3\right)^2 {\text d}x = \frac{7}{3} \qquad \langle u, u \rangle = \int_0^1 x^2 (2x-3)^2 {\text d}x = \frac{4}{5} .
\]
where h and H are some constants. In applications, these constant are usually positive numbers.
As usual, we start analyzing this SL problem with the case when λ = −μ² is negative. The general solution of the differential equation
\( y'' - \mu^2 y =0 \) is a linear combination of hyperbolic functions:
Graphs of tangent hyperbolic and the linear function for H = −0.7.
Mathematica code
As it is seen from the figure, the determinant is not zero for any positive values of parameters h and H. Therefore, the eigenvalue of the given problem cannot be negative. Of course, if at leas on of the parameters h or H is negative, it is possible that one eigenvalue is negative.
Let us consider the case when λ = 0. Substituting the general solution y = c₁ + c₂x into the boundary conditions, we obtain
\[
c_1 - h c_2 =0, \qquad c_1 + c_2 + H c_2 = 0 \qquad \Longrightarrow \qquad c_2 \left( 1 + h + H \right) = 0.
\]
Therefore, we conclude that λ = 0 is not an eigenvalue for positive h and H.
Finally, we consider the case of positive eigenvalues: λ = ω² > 0. The general solution of the corresponding differential equation
\( y'' + \omega^2 y =0 \) is a linear combination of trigonometric functions:
Graphs of tangent and the rational function for
h = 2, H = 3.
Mathematica code
The principal eigenvalue is λ_{1} ≈ 3.19291, which we find with the aid of Mathematica:
FindRoot[Tan[k] == k/(1 + 6*k^2), {k, 3}]
{k -> 3.19291}
■
Theorem 2:
Let u and v be linearly independent solutions of regular Sturm--Liouville problem \eqref{EqSturm.5} for the same value of λ. Then λ is an eigenvalue of the Sturm-Liouville problem \eqref{EqSturm.5} if and only if
Let w be a linear combination of these two functions:
\[
w = c_1 u(x) + c_2 v(x) .
\]
Function w(x) is a solution of the differential equation
\( L \left[ x, \texttt{D} \right] w = \lambda\,w(x) \) for Sturm--Liouville operator L of \eqref{EqSturm.3}. This function satisfies the homogeneous boundary condition if and only if
Here we use the more compact prime notation (due to Lagrange) for derivatives.
To solve this two-point boundary value problem (4.2), we first find the general solution to the differential equation \( y'' + 5\,y' + 4\,y + \lambda\,y =0 . \) The characteristic equation (4.2) is
Since the determinant of this system is \( r_2 e^{r_2} - r_1 e^{r_1} \ne 0 , \) the system has only the trivial solution. Therefore λ isn’t an eigenvalue of Eq.(4.2)
If λ = 25/16, then the characteristic equation has a double root
r₁ = r₂ = −5/2. Hence, the general solution becomes
\[
y = \left( c_1 + c_2 x \right) e^{-5 x/2} .
\tag{4.4}
\]
The boundary condition at x = 0 dictates that c₁ = 0. The boundary condition at x = 1 specifies c₂ = 0. Therefore, λ = 25/16 = 1.5625 is not an eigenvalue.
To estimate the princioal eigenvalue, we calculate integrals for the trial
function \( \displaystyle u(x) = x \left( \cosh (3x/2) + \sinh (3x/2) \right): \)
\begin{align*}
F[u] &= \langle u, L[u] \rangle = \frac{9}{4} \int_0^1 u(x)^2 {\text d} x + \int_0^1 \left( u' \right)^2 {\text d} x \approx 36.4902 ,
\\
\| u \|^2_w &= \int_0^1 u^2 (x) \, e^{5x/2}\,{\text d}x \approx 31.2409 .
\end{align*}
Upon plotting the determinant, we see that it has one real root μ = 1.9942.
Therefore, this Sturm--Liouville problem has one negative eigenvalue λ = −μ² = −3.97683.
The case λ = 0 cannot provide an eigenvalue because function
\( \displaystyle y = e^{2x} \left( 1 - 2x \right) \) cannot satisfy the boundary condition at x = 1.
Now we consider the case when λ = ω² > 0. The general solution of differential equation \( \displaystyle y'' - 4\,y' + 4\,y + \omega^2 y = 0 \) is
Substituting \( \displaystyle y = e^{2x} \left[ 2\,\sin (\omega x) - \omega\,\cos (\omega x) \right] \) into the boundary condition at x = 1, we obtain
Therefore, we see that the Rayleigh--Ritz formula does not work in this example because the operator L is not self-adjoint. So we transfer the differential equation (4D.1) into normal form by substitution:
This is classical Sturm--Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2},
with \( p(x) = e^{-4x} \) \( q(x) = -13\,e^{-4x} \) and weight function w(x) = p(x) > 0.
The Sturm--Liouville problem (4C.7) has exactly the same solution as was obtained previously because multiplication by a function μ(x) ≠ 0 does not change its solution. However, calculation of Rayleigh quotient will be different because the inner product should incorporate the weight function:
where L is the self-adjoint differential operator \eqref{EqSturm.3}.
The Lagrange identity is established by integration by parts and is left to the reader.
Let \( \displaystyle L\left[ x, \texttt{D} \right] = q(x)\,\texttt{I} - \texttt{D}\,p(x)\,\texttt{D} \) be the Sturm--Liouville differential operator from \eqref{EqSturm.3}. The expression
\begin{equation} \label{EqSturm.8}
R[u] = \frac{\langle u, L\,u \rangle}{\langle u , u \rangle_w} = \dfrac{p(x) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right)_{x=a}^{x=b} + \int_a^b \left[ p \left( u' \right)^2 + q \left( u \right)^2 \right] {\text d}x}{\int_a^b \left( u \right)^2 w(x)\,{\text d}x}
\end{equation}
is called the Rayleigh quotient. If u is a solution of the regular Sturm--Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2}. then the Rayleigh quotient is positive when p(x) > 0, q(x) ≥ 0.
The Rayleigh quotient is named in honour of John William Strutt, Lord Rayleigh (1842-1919), who made a
great number of contributions to the study of sound and wave phenomena in general. He was awarded
the 1904 Nobel Prize in Physics for the discovery of argon. He also developed a number of important approximation methods that have become fundamental tools in applied mathematics and physics. Presented below the Rayleigh--Ritz formula is one of them.
The principal eigenvalue, also known as the ground state energy, of
a Sturm-Liouville problem is the minimal eigenvalue λ_{0}. The principal eigenfunction
is the eigenfunction corresponding to the principal eigenvalue.
The following functional is associated with
the Sturm-Liouville operator \eqref{EqSturm.3}
whose corresponding Euler--Lagrange equation is given by the Sturm-Liouville
equation. A minimizer will yield the equation \( \displaystyle F'[u] = \lambda\,u , \) assuming that p(x) ≥ α > 0, q(x) ≥ 0. The functionalF[u] appears in the numerator of the Rayleigh quotient \eqref{EqSturm.8} and it is related to the eigenvalues of classical Sturm--Liouville problem. Indeed, consider equation \eqref{EqSturm.1}, written for any twice differentiable function u(x):
\[
\frac{\text d}{{\text d}x} \left[ p(x)\,\frac{{\text d}u}{{\text d}x}
\right] - q(x)\, u + \lambda \,w (x)\,u(x) =0 , \qquad a < x < b .
\]
Multiplying this equation by u(x) and integrating, we obtain
Integrating by parts in the first term, we get (assuming that u(x) in addition satisfies the boundary conditions \eqref{EqSturm.2})
\[
-\int_a^b \left[ p \left( u' \right)^2 + q\left( u \right)^2 \right] {\text d}x
+ \lambda \,\int_a^b w (x)\,u^2 (x)\,{\text d}x =0 .
\]
Integrating by parts in the first term, we get (assuming that u(x) in addition satisfies the boundary conditions \eqref{EqSturm.2})
\[
-\int_a^b \left[ p \left( u' \right)^2 + q\left( u \right)^2 \right] {\text d}x
+ \lambda \,\int_a^b w (x)\,u^2 (x)\,{\text d}x =0 .
\]
Any function having square integrable derivative (du/dx ∈ 𝔏²) and satisfying the boundary conditions \eqref{EqSturm.2} is called a trial function for the given Sturm--Liouville problem \eqref{EqSturm.5}. Actually, all eigenvalues are extrema of the Rayleigh quotient. So higher
eigenvalues can be evaluated by a process called deflation algorithm (sequential elimination of first determined eigenfunctions from the domain).
Theorem 3:
The principal eigenvalue λ_{0} of a regular Sturm-Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2}
satisfies the variational principle, known as the Rayleigh-Ritz formula:
\[
\lambda_0 = \inf_{u\in D, \ u\ne 0} R[u] ,
\]
where D = D(L) is the domain of the Sturm-Liouville operator \eqref{EqSturm.3} in the Hilbert space 𝔏² (this means that u is twice continuously differentiable functions on
[𝑎, b] that satisfy the corresponding boundary conditions).
Note that Theorem 3 is valid for trial functions that have only first integrable derivative---having second derivative is not needed for formula \eqref{EqSturm.8}.
Proof of this statement is a subject of the “Calculus of Variations,” and it relies on some facts that were established by mathematicians:
The domain of the Sturm--Liouville operator is dense in the Hilbert space 𝔏².
The set of the eigenfunctions possess
orthonormality property and is complete in the domain of the SL operator.
The eigenvalues of a SL problem are bounded from below.
The eigenfunction expansion of a
twice continuously differentiable function converges uniformly on [𝑎, b].
because every function u from the domain of the SL operator can be expanded into a uniformly convergent series with respect to the set of eigenfunctions. Since each function ϕ_{n} is an eigenfunction, we have
\[
\| u \|^2 = \langle u, u \rangle = \sum_{n\ge 0} c_n^2 ,
\]
which Parseval's identity (it is equivalent to completeness of the set of eigenfunctions).
Theorem 4: Properties of the regular Sturm--Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2}.
Suppose that functions p(x), p'(x), q(x), ρ(x) are continuous on [𝑎, b], and also p(x) and ρ(x) are positive. Then the Sturm–Liouville problem \eqref{EqSturm.5} has the following properties.
There exists an infinite number of real eigenvalues that can be arranged in increasing order λ_{1} < λ_{2} < λ_{3} < … λ_{n} < … such that λ_{n} →∞ as n→∞.
For each eigenvalue there is only one linearly independent eigenfunction (up until nonzero multiple). In other words, all eigenvalues are simple.
Eigenfunctions corresponding to different eigenvalues are linearly
independent.
The set of eigenfunctions { ϕ_{n} } corresponding to the set of eigenvalues is
orthogonal with respect to the weight function w(x) on the interval x ∈ [𝑎, b]:
Note:
In Sturm--Liouville problems that do not involve self-adjoint operators, the eigenvalues and eigenfunctions may have similar properties to those outlined in Theorem 4. However, the theory of such operators is more complicated, so we restrict our attention to Sturm--Liouville operators.
Since the proof of this part requires a solid knowledge of many topics, we just outline the idea to convince the reader that the statement is true.
It is known that the domain D = D(L) of the Sturm-Liouville operator \eqref{EqSturm.3} is dense in the Hilbert space 𝔏². Then the principal eigenvalue λ_{0} can be determined by the Rayleigh-Ritz formula:
\[
\lambda_0 = \inf_{u\in D, \ u\ne 0} R[u] .
\]
We denote by u_{0} an eigenfunction corresponding to the principal eigenvalue.
Then the next eigenvalue λ₁ can be found from the formula
This eigenvalue λ₁ is larger than λ₀ because the infinimum is taken over a smaller space from which one dimensional space spanned on the principal eigenfunction is removed.
We denote by u₁ an eigenfunction corresponding to the eigenvalue
λ₁ and remove from the domain D of the Sturm--Liouville operator two dimensional subspace spanned on two eigenfunctions, u₀ and u₁. This allows us to evaluate the next eigenvalue:
\[
V_n = W_n^{\perp} \cap D \subset V_{n-1} \subset V_{n-2} \subset \cdots \subset V_1 \subset V_0 = D .
\]
Because of this nesting, every next eigenvalue λ_{n} is larger than all previously obtained eigenvalues because it is determined by the Rayleigh-Ritz formula:
Since the domain D of the Sturm-Liouville operator \eqref{EqSturm.3} is dense in the infinite dimensional Hilbert space 𝔏², the number of eigenvalues is infinitely numerable.
Sturm did not prove by himself the existence of eigenvalues for regular Sturm--Liouville problems---it was done more than 70 years later, at the beginning of twenteen century. Usually proof of part (a) is accomplished by transfering the boundary value problem for unbounded Sturm--Liouville operator into equivalent integral equation that is actually the inverse operator for the Sturm--Liouville operator. The corresponding integral equation defines a bounded compact operator, for which Hilbert-Schmidt theorem is applied.
Let α and β be arbitrary numbers from the closed interval [𝑎, b] where Sturm--Liouville problem \eqref{EqSturm.1}, \eqref{EqSturm.2} is given. Let u(x) and v(x) be two eigenfunctions corresponding to an eigenvalue λ.
Integrating the Lagrange identity over interval {α, β], we obtain
\begin{align*}
0 &= \left\langle L\left[ x, \texttt{D} \right] u , v \right\rangle -
\left\langle u , L\left[ x, \texttt{D} \right] v \right\rangle = \int_{\alpha}^{\beta} \frac{\text d}{{\text d}x} \left[ p(x) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right) \right] {\text d} x
\\
&= p(\beta ) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right)_{x=\beta} - p(\alpha ) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right)_{x=\\alpha} .
\end{align*}
Observing that the expression in parenthesis is just the Wronskian f two functions,
Since λ ≠ μ, we obtain orthogonality of eigenfunctions ψ and φ.
When zero is an eigenvalue, we usually start labeling the eigenvalues at 0 rather than at 1 for convenience. That is we label the eigenvalues 0 = λ_{0} < λ_{1} < λ_{2} < ···. It is convenient arrangjng eigenvalues in an increasing order because it enables us to arrange the corresponding egenfunctions
Here j denotes the imaginary unit, so j² = −1.
In this case, the general solution of the differential equation
\( \displaystyle x^2 y'' + 3x\,y' - 3\,y + \lambda\,y = 0 \) is
\[
y = x^{-1} \left[ c_1 \cos \left( \omega\, \ln x \right) + c_2 \sin \left( \omega\, \ln x \right) \right] , \qquad x \ge 0.
\]
Upon substituting the general solution into the given boundary conditions, we find c₁ = 0 and ω should be a colution of the transcendent equation
It is not hard to see that the case when 4 ≥ λ does provide an eigenvalue. So we assume that λ > 4 and we set λ = 4 + ω². Then the general solution of the Euler equation
Therefore, ω must be a root of the transcendent equation (5.3) that was considered previously. This leads to exactly the same eigenvalues and eigenfunctions (5.4).
End of Example 5
■
Example 6:
Let us consider a Sturm--Liouville problem that cannot be solved analytically, but semi-analytically:
\[
y'' + \left( 1+ x \right) y + \lambda\,y = 0, \qquad y'(0) = 0, \quad
y (2) = 0.
\tag{6.1}
\]
To find the general solution, we ask Mathematica for help:
Now we use Rayleigh quotirent to estimate this principal eigenvalue from above. First, we start with a trial function u(x) = x²(x - 2). Calculations show
So the numerical value of the Rayleigh quotient corresponding to this function is 1.25, which is a good approximation for λ₂ ≈ 1.60799, but not for the lowerst eigenvalue λ₁ ≈ 0.705506. Let us try another trial function:
which closer to the second eigenvalue λ₂ ≈ 1.60799, but not to λ₁ ≈ 0.705506. Now we finish our estimations with another trial function
u(x) = x⋅sin(πx/2). Then
Sturm’s Separation Theorem:
If u_{1}(x), u_{2}(x) are two linearly independent solutions of the differential equation \( \displaystyle \left( p(x)\, u' \right)' + q\, u_i = 0 \) and α, β are two consecutive zeros of u_{1}(x), then u_{2}(x) has a zero on the open interval (α, β).
The main idea of this proof is based on the property of the Wronskian of two linearly independent solutions not to vanish. So if u₁(x) and u₂(x) are two linearly independent solutions of the second order differential equation, then their Wronskian
has a constant sign in the interval (α, ^beta;). Two linearly independent solutions of the second order linear differential equation cannot have a common zero; for
if they do, then the Wronskian will vanish at that point, which is impossible.
We now assume that x₁ and x₂ are successive zeros of
y₂ and show that y₁ vanishes between these points.
The Wronskian clearly reduces to
y₁ dy₂/dx at x₁ and x₂,
so both factors y₁ and dy₂/dx are ≠ 0 at
each of
these points. Furthermore, dy₂/dx(x₁) and
dy₂/dx(x₂) must have opposite signs, because if
y₂ is increasing at x₁, it must be
decreasing at x₂, and vice versa. Since the
Wronskian has constant sign, y₁(x₁) and
y₁(x₂) must also have opposite signs,
and therefore, by continuity, y₁(x) must vanish at some
point between x₁ and x₂. Note that y₁
cannot vanish more than once between x₁ and x₂; for
if it does, then the same argument shows that y₂ must vanish
between these zeroes of y₁, which contradicts the original
assumption that x₁ and x₂ are successive zeroes of
y₂.
It is known that differential equation (7.1) has two linearly independent solutions y₁(x) = cos(x) and y₂(x) = sin(x) through which the general solution is expressed. However, we are going to show how their properties can be squeezed out of differential equation (7.1) and corresponding initial conditions.
Accordingly, let y = c(x) be the solution of Eq.(7.1) subject to the initial conditions y(0) = 1, y'(0) = 0, and
y = s(x) be defined as the solution of Eq.(7.1) determined by the
initial conditions s(0) = 0 and s′(0) = 1. If we try to sketch the graphs of these functions, we know from the initial conditions that s(x) starts from the origin, but c(x) starts from point (0,1). We know that s(x) is increasing because its derivative at the origin is 1 (positive). From the equation itself we have s″(x) = −s(x), so when the curve is above the x-axis, s″(x) is
a negative number that increases in magnitude as the curve rises. Since s″(x)
is the rate of change of the slope s′(x), this slope decreases at an increasing
rate as the curve lifts, and it must reach 0 at some point x = m.
As x continues
to increase, the curve falls toward the x-axis, s′(x) decreases at a decreasing
rate, and the curve crosses the x-axis at a point we can define to be π. Since
s″(x) depends only on s(x), we see that the graph between x = 0 and x = π is
symmetric about the line x = m, so m = π/2 and s′(π) = −1. A similar argument
shows that the next portion of the curve is an inverted replica of the first
arch, and so on indefinitely.
On the other hand, slope of c(x) beginning at 0. since by equation (7.1) we know that c″(x) = –c(x), the same reasoning as before shows that
the curve bends down and crosses the x-axis. It is natural to conjecture that
the height of the first arch of s(x) is 1, that the first zero of c(x) is π/2, etc.; but
to establish these guesses as facts, we begin by showing that
\[
s' (x) = c(x) \qquad\mbox{and} \qquad c' (x) = -s(x) .
\tag{7.2}
\]
To prove the first statement, we start by observing that (7.1) yields y‴+ y′ = 0
or (y′)″ + y′ = 0, so the derivative of any solution of (7.1) is again a solution. Thus, s′(x) and c(x) are both solutions of (7.1), and it suffices to show that they have the same values and the same derivatives
at x = 0. This follows at once from s′(0) = 1, c(0) = 1 and s″(0) = –s(0) = 0, c′(0) = 0.
The second formula in (7.2) is an immediate consequence of the first, for
c′(x) = s″(x) = –s(x). We now use (7.2) to prove
\[
c(x)^2 + s(x)^2 = 1.
\tag{7.3}
\]
Since the derivative of the left side of (7.3) is
\[
2s(x)\,c(x) - 2c(x) \,s(x) ,
\]
which is 0, we see that s(x)² + c(x)² equals a constant, and this constant must be
1 because s(0)² + c(0)² = 1. It follows at once from (7.3) that the height of the first
arch of s(x) is 1 and that the first zero of c(x) is π/2. This result also enables
us to show that s(x) and c(x) are linearly independent, for their Wronskian is
Among other things, it is easy to see from the above results that the positive zeros of s(x) and c(x) are, respectively, π, 2π, 3π, . . . and π/2, π/2 + π,
π/2 + 2π, . . . .
<[>
There are two main points to be made about the above discussion.
First, we
have extracted almost every significant property of the functions sin x and
cos x from equation (7.1) by the methods of differential equations alone, without
using any prior knowledge of trigonometry. Second, the tools we did use
consisted chiefly of convexity arguments (involving the sign and magnitude
of the second derivative) and the basic properties of linear equations set forth.
It goes without saying that most of the above properties of sin x and cos x
are peculiar to these functions alone. Nevertheless, the central feature of
their behavior—the fact that they oscillate in such a manner that their zeroes
are distinct and occur alternately—can be generalized far beyond these particular functions.
End of Example 7
■
Sturm’s Comparison Theorem:
For i = 1,2, let u_{i}(x) be a nontrivial solution of the differential equation \( \displaystyle \left( p_i(x)\, u'_i \right)' + q_i u_i = 0 \) on α ≤ x ≤ β. Suppose further that the coefficients are continuous and for x ∈ [α, β]
Then if α, β are two consecutive zeros of u_{1}(x), the open interval (α, β) will contain at least one zero of u_{2}(x).
It was Mauro Picone who in 1909 disposed of the case p₁ ≠ p₂.
For simplicity, we denote by u and v eigenfunctions corresponding i₁ and i₂,
respectively.
Suppose to the contrary that v does not vanish in (&alha;, β). It may be supposed without loss of generality that v(x) > 0 and also u(x) > 0 in (&alha;, β).
Multiplication of the equation
Since the integrand on the left side is the derivative of
\( p(x)\, W[u,v) = p \left( u' v - v' u \right) , \)
and q₁ > q₂ by hypothesis, it follows that
However, u(α) = u(β) = 0 by hypothesis, and since u(x) > 0 in (α, β), u'(α) > 0
and u'(β) < O. Thus the left member of Eq.(P.1) is negative, which is a contradiction.
Sturm’s proofs of course do not meet the standards of modern rigor. They meet
the standards of his time, and are in fact correct in method and can without too
much trouble be made rigorous. The first efforts to do this are due to Maxime Bôcher
in a series of papers in the Bulletin of the AMS and are also contained in
his book. Bôcher remarks that “the work of Sturm may, however, be
made perfectly rigorous without serious trouble and with no real modification of
method”. The conditions placed on the coefficients were to make them piecewise
continuous. Bôcher used Riccati equation techniques in some of his proofs; it should be noted
that Sturm mentions the Riccati equation, but does not employ it in his proofs.
Riccati equation techniques in variational theory go back at least to Legendre who
in 1786 gave a flawed proof of his necessary condition for a minimizer of an integral
functional. A correct proof of Legendre’s condition using Riccati equations can be
found in Bolza’s 1904 lecture notes. Bolza attributes this proof to Weierstrass.
Bôcher. M., The theorems of oscillation of Sturm and Klein, Bull. Amer. Math. Soc.
4 (1897–1898), 295–313, 365–376.
Bôcher, M., Leçons sur les méethodes de Sturm dans la théorie des équations différentielles linéaires, et leurs déeveloppements modernes, Gauthier-Villars, Paris, 1917.
Bolza, O., Lectures on the Calculus of Variations, Dover, New York, 1961.
Example 8:
Let us consider two differential equations and corresponding initil conditions
Since these initial value problems are impossible to solve analytically, we apply Mathematica and its dedicated command NDSolve for numerical approximation.
As it is seen from the graph, between any two zeroes of u(x), there is always a zero of v(x), but not vice versa.
End of Example 8
■
Sturm--Liouville problem with periodic conditions
Another important class of Sturm--Liouville problems provide second order differential equations with periodic boundary conditions. As an example, we consider the eigenvalue problem
We are going to show that eigenvalues of this Sturm--Liouville problem with
periodic conditions possess almost all properties as regular Sturm--Liouville
problems, except that they are not simple having multiplicity two.
You will observe the difference in behavior of the eigenvalues between the regular Sturm--Liouville problem and periodic problems is due to
the fact that the eigenvalues of a regular problem are simple, whereas for the periodic case they
can have multiplicity two.
Although Sturm--Liouville problems with periodic conditions provide different kind of conditions compared to problems with homogeneous boundary conditions, it can be analyzed in a similar way as we did for regular eigenvalue/eigenfunction problems. First observation that you need to make is to prove that the operator generated by periodic conditions \eqref{EqSturm.7} is self-adjoint. Thus, we have to demonstrate the validity of the following identity:
\[
\left\langle L\left[ \texttt{D} \right] u , v \right\rangle = \left\langle u , L\left[ \texttt{D} \right] v \right\rangle \qquad \mbox{or} \qquad \left\langle L\left[ \texttt{D} \right] u , v \right\rangle - \left\langle u , L\left[ \texttt{D} \right] v \right\rangle = 0
\]
for any two functions u and v from the domain of the
corresponding operator. This means that these functions have two first continuous derivatives and satisfy the periodic conditions.
Using integration by parts and Lagrange's identity, we get
\[
\left\langle L\left[ \texttt{D} \right] u , v \right\rangle - \left\langle u , L\left[ \texttt{D} \right] v \right\rangle = \int_0^T \frac{\text d}{{\text d}x} \left[ p(x) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right) \right] {\text d}x = \left[ p(x) \left( v\, \frac{{\text d}u}{{\text d}x} - u \,\frac{{\text d}v}{{\text d}x}\right) \right]_{x=0}^{x=T} .
\]
and the operator becomes self-adjoint. Therefore, we expect that the
Sturm--Liouville problem with periodic conditions has only real eigenvalues and
its corresponding eigenfunctions are orthogonal.
Solutions of the differential equation \( y'' + \lambda \,y =0 \) depend on the sign of parameter λ. If λ = −μ^{2} is negative, the equation has two exponential linearly independent solutions \( y_1 = e^{\mu x} \quad\mbox{and}\quad y_2 = e^{-\mu x} \) that also sometimes can be written through hyperbolic sine and cosine functions. These functions cannot be periodic, so negative λ is not an eigenvalue.
If λ = 0, the general solution of the differential equation \( y'' =0 \) is a linear function \( y = a + b\,x , \) which could be periodic only when b = 0. So zero is an eigenvalue corresponding to an eigenfunction which is a constant function in this case.
When λ = ω^{2} is positive, then the general solution becomes
where 𝑎_{n}, b_{n} are some real constants. These arbitrary constants indicate that the eigenfunction is two-dimensional. Of course, you can organize it in one-dimensional array upon introducing positive and negative indices:
Observe that the differential operator \( -\texttt{D}^2 \) in Sturm--Liouville problem \eqref{EqSturm.7} is a square of another differential operator of order one: \( \quad -\texttt{D}^2 = \left( -{\bf j}\texttt{D} \right)^2 . \) Correspondingly, we consider a Sturm--Liouville problem for first order differential operator:
Following the previous example, it is clear that the given anti-periodic problem has a nontrivial solution only when &lambda is positive; so we let λ = μ². Then the general solution of the differential equation (A1.1) is
\[
y(x) = a\,\cos \mu x + b\,\sin \mu x ,
\]
for some constants 𝑎 and b. To satisfy the anti-periodic conditions, we should have
This eigenfunction relects that the eigenvalue has multiplicity two.
■
Buckling of an Elastic Column
We consider a buckling problem as an example of boundary value problems that lead to the fourth order differential equations. Therefore, we consider an elastic column of length ℓ, one end ot it is clumped, but another one is simply supported. Let y(x) be the deflection of the column at point x from its equilibrium position.
The general solution of the forth order differential equation
\( y^{(iv)} + k^2 y'' =0 \) is
\[
y (x) = a + bx + c\,\cos kx + d\, \sin kx,
\]
with some constants 𝑎, b, c, and d. The boundary conditions dictate that 𝑎 = c = 0, and the eigenvalues k_{n} (n = 1, 2, 3, …) are roots of the transcendent equation
\[
\sin (k\ell ) = k\ell\,\cos (k\ell ).
\]
The eigenfunctions corresponding to the eigenvalues are
When k = 0, the general solution becomes
\( y = a + bx + c x^2 + d x^3 . \) The boundary conditions are satisfied only when 𝑎 = b = c = d = 0. Hence λ = 0 is not an eigenvalue.
Suppose that a straight elastic bar (beam, rod, or column) of length ℓ is positioned horizontally and is
anchored at its base. Experimentally one can take a thick metal wire. A small compressive force
of magnitude P acts horizontally on pin ended column. downward on the free end of the bar as in Figure 1.1.
The equation governing the shape of the bar is
Euler's critical load is the compressive load at which a slender column will suddenly bend or buckle. It is given by the formula tat was derived in 1757 by Leonhard Euler:
\[
P_{cr} = \frac{\pi^2 E\.I}{(K\ell )^2} ,
\]
where
P_{cr} is Euler's critical load (longitudinal compression load on column),
E is Young's modulus of the column material (usually expressed in gigapascals),
I is minimum area moment of inertia of the cross section of the column (second moment of area),
If buckling occurs, it must be possible to ﬁnd a solution (or solutions)
to the governing equations diﬀerent from the obvious solution w(x) = 0 for 0 ≤ x ≤ ℓ, the
so-called trivial solution. Other solutions, if any exist, are called nontrivial. They are
with some nonzero constant A. The smallest
eigenvalue determines the minimum compressive force needed to buckle a beam of given ﬂexural rigidity EI.
The Euler model predicts that buckling can occur and does occur only at the
eigenvalues λ_{n} and that the corresponding buckled equilibrium
states are multiples of sin(nπx/&ell:). Actually, once the bar
has buckled, a new model is needed because the physical situation has become
nonlinear. Nevertheless, even in the nonlinear regime the linear problem
above, which is the linearization of an appropriate nonlinear model, still
predicts the values of P at which buckling can occur. Problems of this
sort are called bifurcation (branching) problems because nonlinear
states branch from a stable linear state at certain critical values, the eigenvalues
of the linearized problem. The eigenfunction w_{n} corresponding to the branch point determined
by the eigenvalue λ_{n} approximates the shape of the nonlinear buckled responses of small amplitude that occur near the branch point.
Simmons, G.F., Differential Equations with Applications and Historical Notes, Third edition, CRC Press, Boca Raton, London, New York.
Swanson, C.A., Comparison and oscillation theory of linear differential equations, Academic Press, New York and London, 1968.
Titchmarsh, E.C., Eigenfunction expansions associated with second-order differential equations I, Clarendon Press, Oxford, 1962.
Whittaker, E.T. and Watson, G.N., Modern analysis, Cambridge University Press, 1950
Zettl, A., Computing continuous spectrum, in Trends and Developments in Ordinary Differential Equations, 393–406, Y. Alavi and P. Hsieh editors, World Scientific, 1994.
Zettl, A., Sturm-Liouville problems, in Spectral Theory and
Computational Methods of Sturm-Liouville problems, 1–104, Lecture
Notes in Pure and Applied Mathematics 191, Marcel Dekker, Inc., New
York, 1997.
Show that for all real values of parameter h there is an infinite sequence of positive eigenvalues
Show that the principal eigenvalue (a smallest one) approaches zero as h approaches infinity. Show that the principal eigenvalue approaches π/2 as h approaches zero from above.
Show that λ = 0 is not an eigenvalue for any h.
Show that the given SL problem has negative eigenvalues iff h < 0.
Answer:
Suppose that λ < 0, then denoting −λ = μ², we get the general solution of \( y'' - \mu^2 y = 0 \) is
\[
y = c_1 \cosh (\mu x) + c_2 \sinh (\mu x) .
\]
From the boundary condition at x = 0, we get c₁ = μc₂h. Using this value, we obtain from another boundary condition at x = 1 that
If λ = 0, the general solution of \( y'' = 0 \) is q linear function c₁ + c₂x. From the boundary condition at x = 1, it follows that c₂ = 0. Another boundary condition at the origin leads to c₁ = 0. Therefore, λ = 0 is not an eigenvalue.
Assuming λ = &mu² > 0, we have the general solution
where y is the transverse displacement and λ = mω²/(EI); m is the mass per unit length of the rod, E is Young's modulus, I is the moment of inertia of the cross section about an axis through the centroid perpendicular to the plane of vibration, and ω is the frequency of vibration. Hence, for a bar whose material and geometrical properties are given, the eigenvalues determine the natural frequencies of vibration. Boundary conditions at each end are usually one of the following types"
For each of the following three cases, find the form of the eigenfunctions and the equation satisfied by the eigenvalues of this fourth order boundary value problem. Determine λ₁ and λ₂, the two eigenvalues of smallest magnitude. Assume that the eigenvalues are real and positive.