Example 3: Let us consider the simplest self-adjoint differential operator with q = 0:

Example 3D (Dirichlet boundary conditions): Let us consider the one-dimensional Schrödinger equation

If λ =0, the general solution of the differential equation \( y'' =0 \) is a linear function:

For positive λ = ω², we get two linearly independent solutions sin (ωx) and cos (ωx), so their linear combination gives us the general solution.

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

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

Using R, we find the Rayleigh quotient for this trial function:

However, if we take as a trial function u = sin(πx/2), the Rayleigh quotient for this function becomes

Example 3N (Neumann boundary conditions): We consider a similar Sturm--Liouville problem (SL for short), but with the second kind boundary conditions, also known as Neumann boundary conditions:

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:

We take another trial function:

Example 3T (third kind boundary conditions): Consider the Sturm--Liouville problem

From this figure, we see that the Sturm--Liouville problem (3T.1) 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).

When H is negative and its absolute value exceeds 1, the SL-problem (3T.1) has no negative eigenvalues.

Assuming λ = 0, we find the general solution to be a linear function y = cx, with some constant c. From the boundary condition at x = 1, we get

Assuming that λ = ω² is positive, the general solution of the given differential equation becomes

The transcendent equation (3T.3) could be solved only by a numerical solver using, for instance, Newton's method. Nevertheless, if we let \( \omega = \sqrt{\lambda} , \) then we see from their graphs that there exists an infinite number of roots of \( \tan \omega = -H\,\omega . \)

Since tangent function has vertical asymptotes at \( \omega = \frac{\pi}{2} + n\pi , \quad n=0,1,2,\ldots ; \) the roots ω_n of the equation \( \tan \omega = -H\,\omega \) approach

Now we estimate the principal eigenvalue with the Rayleigh quotient. For illustration of its application, we consider two cases when H = 1 and H = 2.

When H = 1, the principal eigenvalue is λ₁ ≈ 2.02876² = 4.11586, which we find with the aid of R:

Upon setting H = 1, we choose a trial function that satisfies the boundary boundary conditions and calculate the corresponding Rayleigh quotient

The square norm ∥u∥² of the trial function is

If we choose another trial function, and repeat calculations, it yields

The Rayleigh quotient for first eigenfunction \( \displaystyle y(x) = \sin \left( 2.0287\,x \right) \) is the principal eigenvalue:

When H = 2, the transcendent equation (3T.2) has infinite many roots. The smallest positive one, known as the principal eigenvalue can be determined only numerically:

Therefore, we know that the principal eigenvalue is λ₁ ≈ 1.8366² = 3.37309 (according to R). Its value can be estimated with the Rayleigh quotient. Hence, we choose a trial function u(x) = −x (2x − 10/3) that satisfy the gven boundary conditions

Then we can apply the Rayleigh quotient to this function and obtain

Example 3A (third kind boundary conditions): We consider another Sturm--Liouville problem

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:

		Upon choosing positive constant h and H to be 2 and 3, respectively, we plot functions in Eq.(3A.4): Plot[{-5*x/(1 + 6*x*x), Tanh[x]}, {x, 0, 10}, PlotStyle -> Thick]
Graphs of tangent hyperbolic and the linear function for H = −0.7.		Mathematica code

As it is seem 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

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:

		Upon choosing positive constant h and H to be 2 and 3, respectively, we plot functions in Eq.(3A.4): Plot[-{5*x/(1 - 6*x*x), Tan[x]}, {x, 0, 13}, PlotStyle -> Thick]
Graphs of tangent and the rational function for h = 2, H = 3.		Mathematica code

Example 4: We reconsider differential operators in the first two examples. So we start with the constant coefficient differential operator

If λ = 25/16, then the characteristic equation has a double root r₁ = r₂ = −5/2. Hence, the general solution becomes

If λ > 25/16, then

Then we find a few roots of Eq.(3.3). sol = FindRoot[x - 5*Tan[x]/2 == 0, {x, 4}] r1 = x /. sol 4.1726} sol = FindRoot[x - 5*Tan[x]/2 == 0, {x, 7}] r2 = x /. sol div> 7.53357} sol = FindRoot[x - 5*Tan[x]/2 == 0, {x, 10}] r3 = x /. sol 10.7674 sol = FindRoot[x - 5*Tan[x]/2 == 0, {x, 14}] r4 = x /. sol 13.96

Example 4D: We consider the Sturm--Liouville problem

If λ = −μ² < 0, then quadratic equation (4D.2) has two distinct real roots, and the general solution of Eq.(4D.1) can be written as

		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. FindRoot[-m Cosh[m] + 14 Sinh[m] - 3 m^2 Sinh[m] == 0, {m, 2}] {m -> 1.9942} Plot[-m Cosh[m] + 14 Sinh[m] - 3 m^2 Sinh[m], {m, 0, 2.5}, PlotStyle -> Thick]
Graphs of determinant (4D.3).		Mathematica code

Therefore, the given Sturm--Liouvlle problem has one negative eigenvalue

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

		Upon plotting two functions, r/(14 + 3r²) and tan(r), we see that they meet at discrete number of points. Hence, λ > 0 provides an eigenvalue. Plot[{x/(14 + 3 x^2), Tan[x]}, {x, 0, 13}, PlotStyle -> Thick]
Graphs of cotangent and the linear function for −ω.		Mathematica code

Example 4C: We consider the Sturm--Liouville problem

= 1 dictates

\[ y (1) = e^2 \left[ c_1 \cosh \omega + c_2 \sinh\omega \right] = 0. \]

Therefore, ω has to satisfy the transcendent equation

\[ -\omega\,\cosh\omega + 2\,\sinh\omega = 0 \qquad\Longrightarrow \qquad \tanh \omega = \frac{\omega}{2} , \]

which has no real root. Hence, we conclude that λ < −9 cannot provide a nontrivial solution.

If λ = −9, then the differential equation has the general solution (with two arbitrary constants c₁ and c₂)

\[ y = e^{2x} \left( c_1 + c_2 x \right) . \]

This function cannot satisfy two boundary conditions in Eq.(4C.1); therefore, we dismiss this case for providing an eigenvalue.

Now let λ > −9. Then the differential equation \( \displaystyle y'' - 4\,y' + 13\,y + \lambda\, y = 0 \) has the general solution:

\[ y = e^{2x} \left[ c_1 \cos (\omega\,x) + c_2 \sin (\omega\,x) \right] , \qquad \omega = \sqrt{9+\lambda} . \tag{4C.4} \]

Substituting this function (4C.4) into boundary condition, we obtain

\[ \omega\,\cos\omega = 2\,\sin\omega \qquad\Longrightarrow \qquad \omega = 2\,\tan\omega . \tag{4C.5} \]

		Upon plotting two functions, x and 2tan(x), we see that they have infinite many discrete roots. Hence λ > −9 provides an eigenvalue. Plot[{x, 2*Tan[x]}, {x, 0, 13}, PlotStyle -> Thick]
Graphs of tangent and the linear function for positive ω.		Mathematica code

Let ω_n denote the n-th root of transcendent equation (4C.5), then eigenvalues and eigenfunctions are expressed through these roots:

\[ \lambda_n = \omega_n^2 -9 , \qquad \phi_n (x) = e^{2x} \left[ 2\,\sin (\omega_n x) - \omega_n \cos (\omega_n x) \right] , \qquad n=1,2,3,\ldots . \tag{4C.6} \]

The corresponding principal eigenvalue is 9.27376 as Mathematica confirms:

sol = FindRoot[2*Tan[x] == x, {x, 4}];
omega1 = x /. sol;
lambda1 = omega1^2 - 9

9.27376

We choose a function that satisfies the mixed boundary conditions and calculate the corresponding Rayleigh quotient

\[ R[u] = \frac{\langle u, L\,u \rangle}{\langle u, u \rangle} = 1 \qquad \mbox{for}\quad u(x) = x^2 \left( x-1 \right) , \]

with

\[ \langle u, L\,u \rangle = \int_0^1 x^2 (x-1) \left( 2 - 14 x + 25 x^2 - 13 x^3 \right) {\text d}x = \frac{1}{105} \qquad \langle u, u \rangle = \int_0^1 x^4 (x-1)^2 {\text d}x = \frac{1}{105} . \]

u[x_] = x* x*(x - 1); Integrate[-u[x]*(D[u[x], x, x] - 4*D[u[x], x] + 13*u[x]), {x, 0, 1}]

1/105

Integrate[x^4*(x - 1)^2, {x, 0, 1}]

1/105

 

Integrating factor: μ(x)

Multiplying the given differential equation (4C.1) by the integrating factor, we reduce it to the following form:

\[ \frac{\text d}{{\text d}x} \left( e^{-4x} \frac{{\text d}y}{{\text d}x} \right) + 13\, e^{-4x} y + \lambda\,e^{-4x} y = 0 . \qquad y' (0) = 0, \quad y(1) = 0. \tag{4C.7} \]

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:

\[ \langle u, L\,u \rangle = \int_0^1 e^{-4x} x^2 (x-1) \left( 2 - 14 x + 25 x^2 - 13 x^3 \right) {\text d}x = \frac{3417 e^{-4} - 61}{1024} \qquad \langle u, u \rangle = \int_0^1 x^4 (x-1)^2 e^{-4x} {\text d}x = \frac{9-437 e^{-4}}{1024} , \]

and

\[ \langle u, u \rangle = \int_0^1 x^4 (x-1)^2 e^{-4x} {\text d}x = \frac{9- 437\,e^{-4}}{1024} \approx 0.000972721. \]

u[x_] = x* x*(x - 1); Integrate[-Exp[-4*x]*u[x]*(D[u[x], x, x] - 4*D[u[x], x] + 13*u[x]), {x, 0, 1}]

(-61 + 3417/E^4)/1024

Integrate[Exp[-4*x]*x^4*(x - 1)^2, {x, 0, 1}]

(9 - 437/E^4)/1024

The corresponding Rayleigh quotient becomes

\[ R[u] = \frac{\langle u, L\,u \rangle}{\langle u, u \rangle} = 1.5908 \qquad \mbox{for}\quad u(x) = x^2 \left( x-1 \right) , \]

((-12585 + E^8)/(131072 E^8))/((9 - 437/E^4)/1024)

(-12585 + E^8)/(128 (9 - 437/E^4) E^8)

N[%]

-0.0252697

 

Transfering to self-adjoint normal form:

WE make substitution:

\[ y = e^{2x} u \qquad \Longrightarrow \qquad y' = e^{2x} \left( 2\,u + u' \right) , \quad y'' = e^{2x} \left( 4\,u + 4\,u' + u'' \right) . \]

This yields the Sturm--Liouville problem for function u(x):

\[ u'' + 9\, u + e^{-2x} \lambda \, u = 0, \qquad 2\,u(0) + u' (0) =0, \quad u(1) = 1. \tag{4D.7} \]

We choose a trial function u(x) = (1 −x)² \) that satisfies the boundary conditions in Eq.(4D.7). Then

\begin{align*} F[u] &= \int_0^1 \left( u' \right)^2 {\text d} x -9 \int_0^1 \left( u \right)^2 {\text d}x = - \frac{7}{15} \approx -0.466667, \\ \| u \|^2_w &= \int_0^1 u^2 (x)\, e^{-2x} {\text d} x \approx 0.148499. \end{align*}

u[x_] = 1 - 2*x + x^2 ;
Integrate[(D[u[x], x])^2 , {x, 0, 1}] - 9* Integrate[(u[x])^2 , {x, 0, 1}]

-(7/15)

N[%]

-0.466667

Integrate[(u[x])^2 * Exp[-2*x], {x, 0, 1}]

1/4 - 3/(4 E^2)

N[%]

0.148499

So the Rayleigh quotient gives an approximation to the lowest eigenvalue as

\[ \lambda_0 \approx -3.14256 . \]

   ■

Example 5: We reconsider the Euler differential operator from Example 2; now we analyze the corresponding Sturm--Liouville problem:

syms r a
eqn = r*(r-1) + 3*r + a == 0
S = solve(eqn, r)

S =
 
 - (1 - a)^(1/2) - 1
   (1 - a)^(1/2) - 1

If λ < 4, then the indicial equation (5.2) has two real roots, so the general solution of Eq.(5.1) is

If λ = 4, The indicial equation becomes \( \displaystyle r(r+2) + 1 = (r +1)^2 = 0, \) and the general solution of ODE from Eq.(5.1) becomes

Let us consider the case when λ > 4. The indicial equation has two complex roots:

		Upon plotting the function in the treanscendent equation (5.3), we see that it has infinite many discrete roots. Click to see matlab code syms x eqn = tan(x*log(2)) == x; fplot(lhs(eqn), 'MarkerFaceColor', '#E19E27','DisplayName','tan(x) = y','LineWidth',2) hold on fplot(rhs(eqn), 'MarkerFaceColor', '#6D8DBD','DisplayName','x = y','LineWidth',2) hold off xlim([0,4*pi]) ylim([-20,20]) legend Copy The least positive root of Eq.(5.3) is about 1.34259 . Click to see matlab code syms x eqn = tan(x*log(2)) == x; V = vpasolve(eqn,x,[.1,1.5]) Copy V = 1.3425911645108164475265208619691 FindRoot[m*Cos[m*Log[2]] == Sin[m*Log[2]], { m, 6}]
Plot of the transcendent function from Eq.(5.3).		matlab code

syms r a
eqn2 = r*(r-1) -15/4*r + a == 0
S = solve(eqn2, r)

S =
 
19/8 - (361/16 - 4*a)^(1/2)/2
(361/16 - 4*a)^(1/2)/2 + 19/8

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

Example 6: Let us consider a Sturm--Liouville problem that cannot be solved analytically, but semi-analytically:

syms y(x) a
ode = diff(y,x,2) + (a + x)*y(x) == 0;
ySol(x) = dsolve(ode)

ySol(x) =
 
C1*airy(0, - a - x) + C2*airy(2, - a - x)

Hence, the general solution is a linear combination of Airy functions:

		Upon plotting the function in the treanscendent equation (6.3), we see that it has infinite many discrete roots. Click to see matlab code syms y(x) x y(x) = -1*airy(1,-1*(1+x))*airy(2,-1*(3+x))-airy(3,-1*(1+x))*airy(0,-1*(3+x)); fplot(y,[0,3], 'Linewidth', 2) title('Graph of the Wronskian') Copy
Plot of the transcendent function from Eq.(6.3).		Mathematica code

Example 7: Let us consider a familiar equation

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

Example 8: Let us consider two differential equations and corresponding initil conditions

Example 9: Now we consider another Sturm--Liouville problem with anti-periodic boundary conditions

1. Guenther, R.B. and Lee, J.W., Sturm-Liouville Problems Theory and Numerical Implementation, 2018, CRC Press.
2. Li H. and Torney D., A complete system of orthogonal step functions, Proceedings of The American mathematical Society, Vol. 132, No 12, pp. 3491--3504.
3. Lützen, J., Sturm and Liouville's Work on Ordinary Linear Differential Equations. The Emergence of Sturm-Liouville Theory, online.
4. Simmons, G.F., Differential Equations with Applications and Historical Notes, Third edition, CRC Press, Boca Raton, London, New York.
5. Swanson, C.A., Comparison and oscillation theory of linear differential equations, Academic Press, New York and London, 1968.
6. Titchmarsh, E.C., Eigenfunction expansions associated with second-order differential equations I, Clarendon Press, Oxford, 1962.
7. Whittaker, E.T. and Watson, G.N., Modern analysis, Cambridge University Press, 1950
8. Zettl, A., Computing continuous spectrum, in Trends and Developments in Ordinary Differential Equations, 393–406, Y. Alavi and P. Hsieh editors, World Scientific, 1994.
9. 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.

• https://www.youtube.com/watch?v=_F0ck1JncLE
• https://www.youtube.com/watch?v=Oz_cnaz9zSo

1. Convert the differential equation \( y'' + x\,y' + 3\, y =0 \) to a self-adjoint form.

   Answer: \( \mu = e^{x^2 /2} . \)

2. Show that the substitution \( y = x^{-1/2} z \) reduces Bessel's equation of order ν
   \[ x^2 y'' + x\,y' + \left( x^2 - \nu^2 \right) y = 0 \]
   to the form
   \[ z'' + \left( x^2 + \frac{1}{4} - \nu^2 \right) x^{-2} z = 0. \]
3. Reduce the following differential equations to self-adjoint form:
   1. \( \displaystyle \left( 1 - x^2 \right) y'' - x\,y' + \lambda\, y = 0 ; \)     (Chebyshev equation);
   2. \( \displaystyle \left( 1 - x^2 \right) y'' - 2x\,y' + \lambda\, y = 0 ; \)     (Legendre equation);
   3. \( \displaystyle x^2 y'' + x\,y' + \lambda\, y = 0 ; \)     (Euler equation);
   4. \( \displaystyle x^2 y'' + y' \,\tan x = 0 ; \)
4. For each of the following differential equations, y = x³ is one solution; find a second, linearly independent solution by quadratures.
   1. \( \displaystyle x^2 y'' - 4x\,y' + 6\, y = 0 ; \)
   2. \( \displaystyle x\, y'' - \left( x-2 \right) y' - 3\, y = 0 . \)
7. Consider the Sturm--Liouville problem
   \[ y'' + \lambda\,y = 0, \qquad y(0) - h\,y' (0) = 0, \qquad y' (1) = 0, \]
   where h is a given parameter.
   1. Show that for all real values of parameter h there is an infinite sequence of positive eigenvalues
   2. 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.
   3. Show that λ = 0 is not an eigenvalue for any h.
   4. Show that the given SL problem has negative eigenvalues iff h < 0.
Answer:
1. 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
   \[ \mu h\,\sinh (\mu ) + \cosh (\mu ) = 0 \qquad \longrightarrow \qquad - \mu h = \coth (\mu ) = \frac{\cosh \mu}{\sinh \mu} \]
   Upon plotting these two functions, we see that they do not intersect; therefore, λ < 0 is not an eigenvalue.
   Plot[{-0.5*m, Coth[m]}, {m, -3, 3}, PlotStyle -> Thick]

   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

   \[ y = c_1 \cos (\mu x) + c_2 \sin (\mu x) . \]
   The boundary conditions yield
   \begin{align*} c_1 - \mu c_2 h &= 0 \qquad \longrightarrow \qquad c_1 = \mu c_2 h, \\ -c_1 \sin \mu + c_2 \cos \mu &= 0 \qquad \longrightarrow \qquad \mu h\, \sin \mu = \cos \mu . \end{align*}
   This leads to the transcendent equation
   \[ \mu h = \cot \mu = \frac{\cos \mu}{\sin \mu} . \]
   Upon plotting two functions, we see that that there are infinite many eigenvalues.
   Plot[{1.5*m, Cot[m]}, {m, 0, 20}, PlotStyle -> Thick]
   with \( \mu_n \approx n\pi \) for large n.
2. When h grows, the straight line hμ has larger slope that crosses cotangent curve near the origin.

   When h approaches zero , the straight line hμ becomes closer to the abscissa that crosses the cotangent curve at μ = π/2.

3. It was shown previously that λ = 0 is not an eigenvalue for any h. Show that λ = 0 is an eigenvalue only when h = 1.
4. From the graph
   Plot[{-1.5*m, Cot[m]}, {m, 0, 20}, PlotStyle -> Thick]
   it is obvious that the straight line with negative slope crosses the cotangent line in the fourth quadrant.
12. Consider the eigenvalue problem for transverse vibration of an uniform elastic bar:
    \[ y^{(4)} + \lambda\,y = 0, \qquad y(0) - h\,y' (0) = 0, \qquad 0 < x < \ell , \]
    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"
    \begin{align*} & y = y' = 0 \qquad \mbox{clamped end}, \\ & y = y'' = 0, \qquad \mbox{simply supported or hinged end}, \\ & y'' = y''' = 0, \qquad \mbox{free end}. \end{align*}
    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.
    1. y(0) = y'(0) = 0,      y(ℓ) = y''(ℓ) = 0;
    2. y(0) = y''(0) = 0,      y(ℓ) = y'(ℓ) = 0;
    3. y(0) = y'(0) = 0,      y''(ℓ) = y'''(ℓ) = 0      (cantilevered bar).