β1/α1 < 0 and β2/α2 > 0. These restrictions on the signs follow from
rather general physical arguments. Consider, for instance, the simple prob- lem of longitudinal oscillations of a spring of lengthl with one end rigidly fixed at x = 0 and the other end attached elastically to a support with an equilibrium location of x = l. A given point on the spring, which initially has an equilibrium positionx at time t = 0, will have the loca- tion ˜x = x + u(x, t) at subsequent times, where u(x, t) is the longitudinal displacement from equilibrium at location x and time t. The boundary condition atx = 0 for the displacement u(x, t) is
u(0, t) = u|x=0 = 0 (2.32)
and has the simple physical meaning that the end atx = 0 is not moving and the displacement is equal to zero for all timest.
Now let us discuss the boundary condition at the other of the spring endx = l, which is attached to a support in such a way that it moves due to both internal forces and an external elastic force of attachment. The internal elastic force in the spring, F (x, t), will obey Hooke’s law. At x = l, this force is equal to
F (l, t) = k∂u ∂x x=l = kux|x=l, (2.33)
wherek is Young’s modulus and k > 0. Here we denote the partial deriv- ative of the functionu(x, t) with respect to x as ux. When the right end is free there is no external force and the boundary condition atx = l has the simple form
ux|x=l= 0. (2.34)
In the case of elastic attachment, the right end experiences small displace- ments; however, we will assume these displacements are small enough that we may still formulate the boundary condition as a condition on the spring at the locationx = l. The external elastic force acting on this end is di- rected against the force of tension in the spring given in Equation (2.33), and again assuming a linear Hooke’s law force, will be proportional to the displacement,u(x, t). Thus, we have an external force, γu|x=l, where the coefficientγ may be referred to as the rigidity of attachment, and we have γ > 0. Therefore, the boundary condition at x = l is given by
kux|x=l= −γu|x=l (2.35)
or
kux+γu|x=l= 0. (2.36)
This boundary condition coincides with the second case of Equation (2.31). From Equation (2.36), we have γ/k > 0; otherwise, we would have a physical inconsistency in the formulation of the problem. This sign cor- responds to the positive sign of the ratio in the second case of Equation (2.31);β2/α2 > 0.
Suppose now we let the left end atx = 0 also be elastically attached. The elastic restoring force in the spring at this point isF (x, t), as given by
F |x=0= kux|x=0. (2.37)
The elastic attachment force acting on this end is directed against the force of tension in the spring, and its magnitude is γu|x=0. Thus, the boundary
92 2. Sturm-Liouville Theory
condition atx = 0 is
kux− γu|x=0= 0. (2.38)
This boundary condition coincides with the first case of Equation (2.31). Because −γk < 0, from the physical arguments given above, we have β1/α1< 0.
Often it is the case that physical problems have nonhomogeneous boundary conditions in which the right sides of Equation (2.31) are not zero. In such cases, they must be converted to homogeneous boundary conditions by introducing some auxiliary function—a task we will have to perform many times in the following chapters, since Sturm-Liouville problems are defined to have only homogeneous boundary conditions.
Let us give one more physical example that has boundary conditions similar to Equations (2.31) and (2.36). Consider the cooling of a uniform rod with insulated lateral surfaces where the ends exchange heat with the environment, which has a temperatureθ (t). Consider a segment of the rod (x1, x1+ ∆x) of length ∆x. During a unit of time this segment obtains an
amount of heat,cρS∆xut, wherec is specific heat capacity, ρ is the mass density,S is the cross sectional area of the rod, u(x, t) is the temperature of the rod at location x and time t, and the subscript t denotes a time derivative. This amount of heat is equal to the heat this segment obtains (or loses) during a unit of time through cross sections atx1 andx1+ ∆x,
given by
−κS ux|x=x1 + κSux|x=x1+∆x.
The coefficientκ is called the thermal conductivity. Thus, we have cρ∆xut =−κ ux|x=x1 +κ ux|x=x1+∆x. (2.39)
Let us discuss with more detail the signs in the right side of Equation (2.39). If, atx1 the gradient of temperature isux > 0, then in the region x1< x < x1+ ∆x, the temperature is higher than at x1and heat is leaving
the segment ∆x (where the heat flux is directed in the negative direction along thex-axis). In this case the first term on the right of Equation (2.39) should be taken with a minus sign. Ifux < 0, then the temperature for x < x1(to the left ofx1) is higher than in the regionx1 < x < x1+ ∆x,
and the heat flows into ∆x directed along the positive x-axis. In this case, the first term on the right of Equation (2.39) still should be taken with a minus sign becauseux < 0. Similar consideration can be carried out to justify the sign of the second term in the right side of Equation (2.39).
Using Equation (2.39), we can obtain the boundary conditions at the ends of the rod that are in contact with the environment having temperature θ (t). Considering the end at x = l first, we write the analogue of Equation (2.39) for a segment (l − ∆x, l). At the left end, x = l − ∆x and heat flux coincides with the first term in Equation (2.39): −κ ux|x=l−∆x. Heat passing per unit time into this segment through the right end atx = l is SH u|x=l− θ(t), where H > 0 is the heat exchange coefficient. This expression is positive if the temperature of the rod atx = l is higher than the temperature of the environment (i.e., the heat flux is directed along the positive x-axis and the segment ∆x loses heat), and negative in the opposite situation. Therefore, Equation (2.39) in the vicinity of the right end becomes
cρ∆xut=−κ ux|x=l−∆x− H u|x=l− θ(t) . (2.40) Taking the limit as ∆x → 0 in Equation (2.40) we obtain the boundary condition for the rod’s end atx = l:
κux+H [u − θ(t)]|x=l = 0. (2.41)
This condition looks like the second condition in Equation (2.31) and con- dition (2.36) if we set θ (t) = 0. Also we see that H/κ > 0. Similarly, considering a segment (0, ∆x) we obtain the boundary condition for the left end of the rod atx = 0. Heat passing per unit time into this segment through the right end atx = ∆x is equal to Sκ ux|x=∆x. The amount of heat obtained through the left end atx = 0 is −SH u|x=0− θ(t). This heat is positive if the rod temperature is less than the temperature of the environment (i.e., the rod is heating and heat flux is directed along the x-axis). Thus, the equation for the segment (0, ∆x) of the rod is
cρ∆xut =−H u|x=0− θ(t) + κ ux|x=∆x, (2.42)
or, taking the limit ∆x → 0,
κux− α [u − θ(t)]|x=0 = 0. (2.43) In the caseθ (t) = 0, the boundary condition in Equation (2.43) coincides with the first of Equations (2.31) and with condition (2.38). Also, we see that−H/κ < 0. When θ(t) 6= 0, Equations (2.41) and (2.43) are non- homogeneous boundary conditions, and to solve boundary value problems we have to reduce them to being homogeneous.
94 2. Sturm-Liouville Theory
From these two examples, it is clear that the signs of the ratios of the coefficients in the boundary conditions in Equation (2.31) are very impor- tant. Otherwise, we could face physically senseless situations such as heat flowing spontaneously from a cool body to a warm environment.
Next, we briefly discuss similar restrictions on the signs of the coef- ficients in the boundary conditions for two- and three-dimensional prob- lems. Suppose the domain for the rectangular coordinatex is the interval [a, b] and for coordinate y is the interval [c, d]. Many physical problems (e.g., rectangular oscillating membranes, heat conductivity, various elec- trostatics problems, etc.) reduce to separate Sturm-Liouville problems for the variablesx and y. In these cases, we may formulate boundary condi- tions for all four boundaries independently. For the variablex they are still given by Equation (2.31), and for the variabley they are in a similar form:
α3u′+β3u|y=c = 0,
α4u′+β4u|y=d = 0, (2.44)
with identical restrictions on the signs of the coefficients: β3/α3 < 0, β4/α4> 0.
For three-dimensional cases in Cartesian coordinates, where the vari- ablez varies in the intervale, f, we have two more boundary conditions, similar to Equations (2.31) and (2.44):
α5u′+β5u|z=e = 0, α6u′+β6u|z=f = 0,
(2.45) with the restrictionsβ5/α5< 0 and β6/α6> 0.
If we consider a circular domain with radius R, r ≤ R, it is more convenient to use polar coordinates (r, ϕ). The boundary condition at the boundaryr = R in general is
α u′+β u
r=R= 0, (2.46)
with the sign restrictions onα and β given by β /α > 0. If the domain is defined by the conditionr ≥ R, then obviously the sign of the ratio of the coefficients in Equation (2.46) should be the opposite, and we have β /α < 0. Similarly, for an annulus domain, a ≤ r ≤ b, the conditions are
α1u′+β1u|r=a = 0, α2u′+β2u|r=b = 0,
(2.47) with the signsβ1/α1< 0 and β2/α2> 0.
For a cylinder of radius R and height l, the homogeneous boundary conditions in cylindrical coordinates (r, ϕ, z) on the cylinder’s surface are
α1u′+β1u|r=R= 0, α2u′+β2u|z=0= 0, α3u′+β3u|z=l= 0.
(2.48)
Coefficient ratios in Equations (2.48) have sign restrictions similar to the situations discussed previously and are given byβ1/α1 > 0, β2/α2 < 0,
andβ3/α3> 0.
For a sphere of radius R, in spherical polar coordinates (r, ϕ, θ), the homogeneous boundary condition on the sphere’s surface is
α u′+β u
r=R= 0, (2.49)
where the restriction on the sign of the ratio of coefficients is given by β /α > 0.
2.3
Examples of Sturm-Liouville Problems
Example 2.1. Solve the equation
y′′(x) + λy(x) = 0 (2.50)
on the interval [0, 1] with boundary conditions
y(0) = 0 and y(1) = 0. (2.51)
Solution. First, comparing Equation (2.50) with Equations (2.5) and (2.6), it is clear that we have a Sturm-Liouville problem with linear operator L = −d2/dx2 and functionsq (x) = 0 and p (x) = r(x) = 1. (Note that L can be taken with either a negative or positive sign, as is clear from Equations (2.1), (2.5) and (2.6).)
Let us discuss the casesλ = 0, λ < 0, and λ > 0 separately. If λ = 0, then a general solution to Equation (2.50) is
y(x) = C1x + C2
and, from the boundary conditions of Equation (2.51), we haveC1=C2=
0 (i.e., there exists only the trivial solutiony(x) = 0). If λ < 0, then y(x) = C1e √ −λx+C 2e− √ −λx ,
96 2. Sturm-Liouville Theory
and the boundary conditions of Equation (2.51) again give C1 = C2 =
0, and therefore the trivial solution y(x) = 0. Thus, we have only the possibilityλ > 0, in which case we write λ = µ2withµ real, and we have a general solution of Equation (2.50) given by
y(x) = C1sinµx + C2cosµx.
The boundary conditiony(0) = 0 requires that C2 = 0, and the bound-
ary condition y(1) = 0 gives C1sinµ = 0. From this we must have
sinµ = 0 and µn = nπ since the choice C1 = 0 again gives the trivial
solution. Thus, the eigenvalues are
λn =µ2n =n2π2, n = 1, 2, . . . , (2.52)
and the eigenfunctions areyn(x) = Cnsinnπx, where for n = 0 we have the trivial solutiony0(x) = 0. It is obvious that we can restrict ourselves to
only positive values ofn since negative values do not give new solutions in the case that the constantsCn are arbitrary. These eigenfunctions are orthogonal over the interval [0, 1] since we can easily show that
1
Z
0
sinnπx · sin m πxdx = 0 for m 6= n. (2.53)
The orthogonality of eigenfunctions follows from the fact that the Sturm- Liouville operator,L, is Hermitian for the boundary conditions given in Equation (2.51).
Reading Exercise. Directly verify thatL is Hermitian.
Hint. Evaluate
1
Z
0
yn(x)Lym(x)dx
integrating by parts twice.
The eigenfunctions may be normalized by writing
Cn2 1 Z 0 sin2nπxdx = Cn2·1 2 = 1,
which results in the orthonormal eigenfunctions
yn(x) =√2 sinnπx, n = 0, 1, 2, . . . . (2.54) Thus, we have shown that the boundary value problem consisting of Equa- tions (2.50) and (2.51) has eigenfunctions that are sine functions. In other words, the expansion in eigenfunctions of the Sturm-Liouville problem for
solutions to Equations (2.50) and (2.51) is equivalent to the trigonometric Fourier sine series.
Reading Exercise. Suggest alternatives to boundary conditions (2.51) that will result in cosine functions as the eigenfunctions for equation (2.50).
Example 2.2. Determine the eigenvalues and corresponding eigenfunctions for the Sturm-Liouville problem
y′′(x) + λy(x) = 0, (2.55)
y′(0) = 0, y(1) = 0. (2.56)
Solution. As in Example 2.1, check as a reading exercise that the parameter λ must be positive in order to have nontrivial solutions. Thus, we may writeλ = µ2so that we have oscillating solutions given by
y(x) = C1sinµx + C2cosµx.
The boundary conditiony′(0) = 0 givesC1 = 0, and the boundary condi-
tion y(1) = 0 gives C2cosµ = 0. If C2 = 0, we have a trivial solution;
otherwise, we haveµn = (2n + 1)π/2, for n = 0, 1, 2, . . .. Therefore, the eigenvalues are
λn =µ2n = (2n + 1)
2π2
4 , n = 0, 1, 2, . . . , (2.57)
and the eigenfunctions are yn(x) = Cncos
(2n + 1)πx
2 , n = 0, 1, 2, . . . . (2.58)
We leave it to the reader to prove that the eigenfunctions in Equation (2.58) are orthogonal on the interval [0,1]. The reader may also normalize these eigenfunctions to find the normalization constantCn, which is equal to√2.
98 2. Sturm-Liouville Theory
Figure 2.1. The functions tanµn and−µn/5 plotted against µ. The eigenvalues
of the Sturm-Liouville problem in Example 2.3 are given by the intersections of these lines.
Example 2.3. Determine the eigenvalues and eigenfunctions for the Sturm- Liouville problem
y′′(x) + λy(x) = 0, (2.59)
y(0) = 0, y′(1) + 5y(1) = 0. (2.60)
Solution. As in the previous examples, nontrivial solutions exist only when λ > 0 (the reader should verify this as a reading exercise). Letting λ = µ2
we obtain a general solution as
y(x) = C1sinµx + C2cosµx.
From the boundary conditiony(0) = 0, we have C2 = 0. The other
boundary condition givesµ cos µ + 5 sin µ = 0. Thus, the eigenvalues are given by the equation
tanµn =− µn
5 . (2.61)
From Figure 2.1, we see directly that there are an infinite number of discrete eigenvalues. The eigenfunctions
yn(x) = Cnsinµnx, n = 0, 1, 2, . . . (2.62) are orthogonal so that
1
Z
0
sinµnx · sin µmxdx = 0 for m 6= n. (2.63)
The orthogonality condition shown in Equation (2.63) follows from the general theory as a direct consequence of the fact that the operator L = −d2/dx2is Hermitian for the boundary conditions (2.60).
The normalized eigenfunctions are yn(x) = p 2õn
2µn− sin 2µn
sinµnx. (2.64)
Example 2.4. Solve the Sturm-Liouville problem
y′′+λy = 0, 0< x < l, (2.65)
on the interval [0,l] with periodic boundary conditions
y(0) = y(l), y′(0) =y′(l). (2.66)
Solution. Again, verify as a reading exercise that nontrivial solutions exist only whenλ > 0 (for which we will have oscillating solutions, as before). Lettingλ = µ2, we can write a general solution in the form
y(x) = C1cosµx + C2sinµx.
The boundary conditions in Equations (2.66) give
C1(cosµl − 1) + C2sinµl = 0,
−C1sinµl + C2(cosµl − 1) = 0. (2.67)
This system of homogeneous algebraic equations has a nontrivial solution only when its determinant is equal to zero:
cosµl − 1 sin µl − sin µl cosµl − 1 = 0 (2.68) which yields cosµl = 1. (2.69)
The roots of Equation (2.69) are
λn = 2πn
l
2
, n = 0, 1, 2, . . . (2.70)
100 2. Sturm-Liouville Theory
With these values ofλn, Equations (2.67) forC1andC2have two linearly
independent nontrivial solutions given by
C1= C1(1) C2(1) ! =1 0 and C2= C1(2) C2(2) ! =0 1 . (2.71)
Substituting each set into the general solution, we obtain the eigenfunc- tions
yn(1)(x) = cospλnx and y(2)n (x) = sin
p
λnx. (2.72) Therefore, for the eigenvalueλ0 = 0, we have the eigenfunctiony0(x) = 1
and a trivial solutiony(x) ≡ 0. Each nonzero eigenvalue, λn, has two lin- early independent eigenfunctions so that for this example we have twofold degeneracy.
Collecting the above results, we have that this boundary value prob- lem with periodic boundary conditions has the following eigenvalues and eigenfunctions: λn = 2πn l 2 , n = 0, 1, 2, . . . (2.73) y0(x) ≡ 1, yn(x) = ( cos2πnl x, sin2πnl x, (2.74) (two expressions in Equation (2.74) reflect twofold degeneracy),
ky0k2 =l, kynk2= l
2 ( forn = 1, 2, . . .). In particular, whenl = 2π we have
λn =n2, y0(x) ≡ 1, yn(x) =
cosnx, sinnx.
From this result, we see that the boundary value problem consisting of Equations (2.65) and (2.66) results in eigenfunctions for this Sturm- Liouville problem which allows an expansion of the solution equivalent to
the complete trigonometric Fourier series expansion.
Example 2.5. In the following chapters of this book, we will encounter a number of two-dimensional Sturm-Liouville problems. Here we present a simple example for future reference.
Consider the equation ∂2u ∂x2 +
∂2u ∂y2 +k
2u = 0, (2.75)
where k is a real constant that determines the function u(x, y) with in- dependent variables in domains 0 ≤ x ≤ l, 0 ≤ y ≤ h. Define the two-dimensional Sturm-Liouville operator in a fashion similar to that for the one-dimensional case.
Solution. We have L = − d 2 dx2 − d2 dy2. (2.76)
Let the boundary conditions be Dirichlet type so that we have
u(0, y) = u(l, y) = u(x, 0) = u(x, h) = 0. (2.77)
Reading Exercise. By direct substitution into Equation (2.75) and using boundary conditions (2.77), check that this Sturm-Liouville problem has the eigenvalues k2nm = 1 π2 n2 l2 + m2 h2 (2.78) and the corresponding eigenfunctions
unm = sin nπx
l sin
m πy
h . (2.79)
In the case of a square domain where l = h, the eigenfunctions unm andum n have the same eigenvalues,knm = km n, which is a degeneracy
reflecting the symmetry of the problem with respect tox and y.
Example 2.6. Find the Green’s function of the operatorL given by Ly = −y′′− 2y′, y(0) = y(1) = 0
on the interval [0, 1].
102 2. Sturm-Liouville Theory
Solution. This Sturm-Liouville problem has the form
Ly = −y′′− 2y′=λy, (2.80)
y(0) = y(1) = 0, (2.81)
for which it is easy to check that a nontrivial solution exists only forλ > 1. In this case a general solution of Equation (2.80) is
y (x) = e−x h C1cos p λ − 1x +C2sinpλ − 1x i . (2.82)
Substituting Equation (2.82) into the boundary conditions of Equation (2.81) gives, for nontrivial solutions,
C1 = 0 and sin
p
λ − 1
= 0.
The second relation gives the eigenvalues,λn, of the Sturm-Liouville prob- lem as λn =π2n2+ 1, n = 1, 2, 3, . . . . (2.83) Whenn = 0, sin p λn− 1x
= 0 for any value ofx; thus, n = 0 corre- sponds to a trivial solution,y0(x) = 0. The eigenfunctions yn(x), corre-
sponding to the eigenvaluesλn in Equation (2.83), are yn(x) = e−xsin
p
λn− 1x
=e−xsinnπx. (2.84)
To find the Green’s functionG(x, x′), substitute Equations (2.83) and
(2.84) into Equation (2.29), which gives G x, x′ = ∞ X n=1 sin (πnx) sin (πnx′) π2n2+ 1 e −xe−x′ . (2.85) Example 2.7.
1. Find the Green’s function for the equation
y′′+y = f (x), 0≤ x ≤ 1, (2.86)
if the functiony(x) satisfies boundary conditions
y(0) = 0, y(1) = 0. (2.87)
2. Use this Green’s function to find the solution of the boundary prob- lemy′′+y = f (x) if f (x) = x.
Solution.
1. Comparing the given equation with Equation (2.2) we see that p (x) = 1 and q (x) = 0, so L = −d2/dx2
− 1. A similar boundary
value problem with the same boundary conditions (Dirichlet type) was discussed in detail in Example 2.1. We leave it to the reader as a reading exercise to obtain the following eigenfunctions and eigen- values of the boundary value problem consisting of the equation Ly = λy and boundary conditions in Equation (2.87):
λn =n2π2− 1, yn(x) =
√
2 sinnπx, n = 1, 2, 3, . . . (2.88) Substituting Equations (2.88) into Equation (2.29), we have
G(x, x′) = 2 ∞ X n=1 sinnπx · sin nπx′ n2π2− 1 . (2.89)
2. The solution to the equationy′′+y = f (x), which is a particular case
of Equation (2.22) with f (x) = −x, is given by Equation (2.27); thus, y(x) = b Z a G(x, x′)(−x′)dx′= ∞ X n=1 2 sinnπx n2π2− 1 1 Z 0 (−x′) sinnπx′dx′ = 2 ∞ X n=1 sinnπx n2π2− 1 (−1)n nπ .
We see that this solution satisfies the given boundary conditions.
Example 2.8. Obtain the eigenfunction expansion (generalized Fourier se- ries expansion) of the functionf (x) = x2(1
− x) by using the eigenfunc-
tions of the Sturm-Liouville problem
y′′+λy = 0, 0≤ x ≤ π/2, (2.90)
y′(0) =y′(π/2) = 0. (2.91)
104 2. Sturm-Liouville Theory
Solution. First, prove as a reading exercise that the eigenvalues and eigen- functions of this boundary value problem are
λ = 4n2, yn(x) = cos 2nx, n = 0, 1, 2, . . . (2.92) A Fourier series expansion, given in Equation (2.13), of the functionf (x) using the eigenfunctions above is
x2(1− x) = ∞ X n=0 cnyn(x) = ∞ X n=0 cncos 2nx. (2.93) Sincef and f′ are continuous functions, this expansion will converge to
x2(1− x) for 0 < x < π/2, as was shown previously. In Equation (2.90), we see that the functionr(x) = 1, and thus the coefficients of this expan- sion obtained from Equation (2.14) are
c0 = π/2 R 0 x2(1− x)dx π/2 R 0 dx = π 2 4 1 3− π 8 , cn = π/2 R 0 x2(1− x) cos 2nx dx π/2 R 0 cos22nx dx = (−1) n n4 1− 3π 4 + 3 2πn2 − 3 2πn4 , n = 1, 2, 3, . . .
Figure 2.2 shows the partial sum (n = 10) of this series, compared with the original functionf (x) = x2(1− x).
Two important special functions, Legendre and Bessel functions, are discussed in detail in Chapters 9 and 8, respectively. In the following two examples, they serve simply as illustrations of Sturm-Liouville problems.
Example 2.9 (Fourier-Legendre Series). The Legendre equation is d
dx
1− x2 y′ + λy = 0 (2.94) forx on the closed interval [−1, 1]. There are no boundary conditions in a straight form becausep (x) = 1 − x2vanishes at the endpoints. However, we seek a finite solution, a condition that in this case acts as a boundary condition.
Figure 2.2. Graphs of the function f (x) = x2(1
− x) (dashed line) and partial
sum withn = 10 of the Fourier expansion of f (x) (solid line).
Solution. The Legendre polynomials,Pn(x), are the only solutions of Leg- endre’s equation that are bounded on the closed interval [−1, 1]. The set of functions{Pn(x)}, where n = 0, 1, 2, . . ., is orthogonal with respect to the weight function r(x) = 1 on the interval [−1, 1], in which case the orthogonality relation is
1
Z
−1
Pn(x)Pm(x)dx = 0 for m 6= n. (2.95)
The eigenfunctions for this problem are thusPn(x) with eigenvalues λ = n(n + 1) for n = 0, 1, 2, . . ., as we will see in Chapter 9.
Iff (x) is piecewise smooth on [−1, 1], the series
∞ X n=0 cnPn(x) (2.96) converges to 1 2f (x0+ 0) +f (x0− 0) (2.97) at any point x0 on (−1, 1). Because r(x) = 1 in Equation (2.14), the
coefficientscn are cn = 1 R −1 f (x)Pn(x)dx 1 R −1 P2 n(x)dx , (2.98)
106 2. Sturm-Liouville Theory
or written in terms of the scalar product, cn = f (x) · Pn
(x) Pn(x) · Pn(x)
. (2.99)
Example 2.10 (Fourier-Bessel Series). Solve the Sturm-Liouville problem
(xy′)′+ λx − ν 2 x y = 0, 0≤ x ≤ 1, (2.100)
with boundary conditions such thaty(0) is finite and y(1) = 0. Here ν is a constant.
Solution. The eigenvalues for this problem areλ = jn2 for n = 1, 2, . . ., wherej1, j2,j3, . . . are the positive zeros of the functionsJν(x), which
are Bessel functions of orderν. If f (x) is a piecewise smooth function on the interval [0, 1], then for 0 < x < 1 it can be resolved in the series
∞ X n=1 cnJν(jnx), (2.101) which converges to 1 2f (x0+ 0) +f (x0− 0) . (2.102) Since, in this Sturm-Liouville problem,r(x) = x, the coefficients cn are
cn = 1 R 0 xf (x)Jν(jnx)dx 1 R 0 xJν2(jnx)dx , (2.103)
or in terms of the scalar product,
cn = f (x) · Jν(jnx)
Jν(jnx) · Jν(jnx)
. (2.104)
Problems
2.1. Find eigenvalues and eigenfunctions of the Sturm-Liouville problem for the equationy′′(x) + λy(x) = 0 with the following boundary conditions:
1. y(0) = 0 and y′(l) = 0; 2. y′(0) = 0 andy(π) = 0; 3. y′(0) = 0 andy′(1) = 0; 4. y′(0) = 0, y′(1) +y(1) = 0; 5. y′(0) +y(0) = 0, y(1) = 0; 6. y′(0) = 0, y′(l) = 0;
7. y(−l) = y(l), y′(−l) = y′(l) (periodic boundary conditions). As was
noted above, if the boundary conditions are periodic then the eigenvalues can be degenerate. Show that in this problem two linearly independent eigenfunctions exist for each eigenvalue.
2.2.
1. Find the eigenvalues and eigenfunctions of the boundary value problem
y′′+y′+λy = 0, y(0) = 0, y(5) = 0.
2. Write this differential equation in self-adjoint form and determine an ex- pression for the orthogonality relation.
2.3.
1. Find the eigenvalues and eigenfunctions of the boundary value problem
x2y′′+xy′+λy = 0, 1≤ x ≤ 10, y(1) = 0, y(10) = 0.
2. Write this differential equation in self-adjoint form and determine an ex- pression (2.11) for the orthogonality relation.
2.4. The Hermite differential equationy′′− 2xy′+ 2ny = 0, n = 0, 1, 2, . . . has
polynomial solutionsHn(x). Write this differential equation in self-adjoint form
and determine an expression (2.11) for the orthogonality relation.
2.5. Write the Chebyshev equation in the form of a Sturm-Liouville equation and identifyp (x), q (x), and r(x). Write the orthogonality condition for Chebyshev
polynomials of the first and second kind.