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Boundary Value Problems (BVP)
Definition A Boundary Value Problem (BVP) is a differential equation together with conditions specified at two or more points. Formally:
L[y] = f (x), a ≤ x ≤ b with boundary conditions:
y(a) = α, y(b) = β (Dirichlet type) or
y′(a) = α, y′(b) = β (Neumann type)
Important Point
Key Point: Unlike initial value problems (IVPs), BVPs require conditions at two different points, which makes them trickier.
Types of BVP
Definition Linear vs Nonlinear:
y′′^ + p(x)y′^ + q(x)y = f (x) (Linear), y′′^ + y^2 = 0 (Nonlinear)
Self-adjoint vs Non-self-adjoint:
d dx
[p(x)y′] + q(x)y = f (x)
Dirichlet, Neumann, Robin conditions:
y(a) = α, y(b) = β (Dirichlet)
y′(a) = α, y′(b) = β (Neumann) ay(a) + by′(a) = α (Robin)
General Method to Solve Linear BVP
Formula For a second-order linear BVP:
y′′^ + p(x)y′^ + q(x)y = f (x), y(a) = α, y(b) = β
- Solve homogeneous equation:
y h′′ + p(x)y h′ + q(x)yh = 0, yh = C 1 y 1 (x) + C 2 y 2 (x)
- Find particular solution:
y′′ p + p(x)y′ p + q(x)yp = f (x)
- General solution: y(x) = yh + yp
- Apply boundary conditions to solve for constants C 1 , C 2
Example
Example: Solve
y′′^ − y = ex, 0 ≤ x ≤ 1 , y(0) = 0, y(1) = 0 Step 1: Homogeneous solution:
y h′′ − yh = 0 =⇒ r^2 − 1 = 0 =⇒ r = ± 1
yh = C 1 ex^ + C 2 e−x Step 2: Particular solution:
yp = Axex^ =⇒ A =
, yp =
xex
Step 3: General solution:
y(x) = C 1 ex^ + C 2 e−x^ +
xex
Step 4: Apply BCs:
y(0) = 0 =⇒ C 1 + C 2 = 0 =⇒ C 2 = −C 1
y(1) = 0 =⇒ C 1 (e − e−^1 ) +
e = 0 =⇒ C 1 = −
e 2(e − e−^1 )
y(x) =
−e 2(e − e−^1 )
(ex^ − e−x) +
xex
Formula
Homogeneous self-adjoint equation:
d dx
h p(x)
dy dx
i
Definition
Converting a general second-order ODE to self-adjoint form: Given a general ODE: y′′^ + P (x)y′^ + Q(x)y = f (x)
- Find the integrating factor:
μ(x) = e
R (^) P (x)dx
- Multiply the ODE by μ(x):
μ(x)y′′^ + μ(x)P (x)y′^ + μ(x)Q(x)y = μ(x)f (x)
- Combine first two terms as a derivative:
d dx
[μ(x)y′] + μ(x)Q(x)y = μ(x)f (x)
Important Point
Properties of Self-Adjoint 2nd Order ODEs:
- The differential operator L[y] = (^) dxd [p(x)y′] + q(x)y is symmetric: Z (^) b
a
uL[v]dx =
Z (^) b
a
vL[u]dx
- Green’s function exists and is symmetric: G(x, ξ) = G(ξ, x)
- Frequently appears in eigenvalue problems:
d dx
[p(x)y′] + q(x)y + λw(x)y = 0
- Corresponds to energy minimization problems in physics.
Example
Example: Convert y′′^ + (^2) x y′^ − (^) x^22 y = 0 to self-adjoint form. Step 1: Identify coefficients: P (x) = (^2) x , Q(x) = − (^) x^22 Step 2: Integrating factor:
μ(x) = e
R (^) P (x)dx = e
R (^2) x dx^ = e2 ln^ x^ = x^2
Step 3: Multiply ODE by μ(x) = x^2 :
x^2 y′′^ + 2xy′^ − 2 y = 0
Step 4: Combine first two terms as derivative:
d dx
[x^2 y′] − 2 y = 0
Self-adjoint form: d dx
[x^2 y′] − 2 y = 0
Sturm Separation Theorem
Definition Consider the second-order linear homogeneous differential equation:
y′′^ + p(x)y′^ + q(x)y = 0 where p(x) and q(x) are continuous on an interval I. Let y 1 (x) and y 2 (x) be two linearly independent solutions. Sturm Separation Theorem:
Between any two consecutive zeros of y 1 (x), there exists exactly one zero of y 2 (x).
Note: The zeros of linearly independent solutions interlace and do not coincide.
Sturm Comparison Theorem
Definition Consider two second-order linear differential equations on the same interval I:
y′′^ + p 1 (x)y = 0, z′′^ + p 2 (x)z = 0 where p 1 (x), p 2 (x) are continuous on I, and let y(x) and z(x) be nontrivial solutions. Sturm Comparison Theorem: If p 1 (x) ≥ p 2 (x) for all x ∈ I, then between any two consecutive zeros of y(x), there exists at least one zero of z(x).
Formula Properties:
- Solutions of the equation with larger p(x) oscillate at least as much as the equation with smaller p(x).
- If p 1 (x) > p 2 (x) strictly, z(x) has at least one zero in every interval between consecutive zeros of y(x).
- If y(x) has infinitely many zeros, so does z(x).
- Linearly independent solutions cannot have a zero at the same point.
- Useful for analyzing the nature of solutions (oscillatory or non-oscillatory) and eigenvalue problems.
Example
Example 1: Compare y′′^ + y = 0 and z′′^ + 4z = 0.
- p 1 = 1, p 2 = 4 (p 1 < p 2 )
- Solutions: y = sin x, z = sin 2x
- Zeros: y(x) = 0 =⇒ x = nπ, z(x) = 0 =⇒ x = nπ/ 2
- Observation: Between consecutive zeros of y, z has exactly one zero.
- Conclusion: z(x) oscillates faster, consistent with Sturm comparison.
Example
Example 2: Compare y′′^ − y = 0 and z′′^ + z = 0.
- p 1 = − 1 , p 2 = 1 (p 1 < p 2 )
- Solutions: y = ex, e−x^ (no zeros), z = sin x, cos x (infinitely many zeros)
- Observation: Non-oscillatory solution y compared with oscillatory solution z
- Conclusion: Solution with larger p(x) oscillates more, as predicted.
Example
Example 3: Compare y′′^ + 9y = 0 and z′′^ + 4z = 0.
- Solutions: y = sin 3x, z = sin 2x
- Zeros: y(x) = 0 =⇒ x = nπ/3, z(x) = 0 =⇒ x = nπ/ 2
- Observation: Between consecutive zeros of z at 0 and π/2, y has a zero at π/ 3
- Conclusion: Higher p(x) gives faster oscillation; Sturm comparison verified.
Nature of Solutions of Second-Order Differential Equa-
tions
Definition Oscillatory Solution: A solution y(x) of
y′′^ + p(x)y′^ + q(x)y = 0
is called oscillatory if it has infinitely many zeros in the interval. Non-oscillatory Solution: A solution is non-oscillatory if it has at most finitely many zeros.
Example
Example 3: At Most One / Infinitely Many Solutions (BVP) Equation: y′′^ + π^2 y = 0, y(0) = 0, y(1) = 0 General solution: y = C 1 cos πx + C 2 sin πx Apply BC: y(0) = C 1 = 0, y(1) = C 2 sin π = 0 =⇒ C 2 arbitrary Conclusion: Nontrivial solutions exist → BVP has infinitely many solutions.
Extremely Hard Conceptual CSIR-NET Style MSQs
(with Solutions)
Question 1
Consider the Sturm-Liouville problem
y′′^ + λy = 0, y(0) = 0, y(L) = 0.
Let λn denote the n-th eigenvalue and yn(x) the corresponding eigenfunction. De- termine which statements are correct.
- {λn} is countably infinite, λn → ∞ as n → ∞.
- Each yn(x) has exactly n − 1 zeros in (0, L).
- For any f ∈ L^2 (0, L), the series
P∞
n=1⟨f, yn⟩yn^ converges uniformly to^ f^ (x).
- The eigenfunctions {yn} are orthogonal but not necessarily complete in L^2 (0, L).
Solution: - Characteristic equation: y = A sin(
λx), BC y(L) = 0 =⇒
λnL = nπ, n = 1, 2 ,.... - λn = (nπ/L)^2 → ∞ ⇒ statement 1 correct. - Eigenfunction yn = sin(nπx/L) has exactly n − 1 zeros in (0, L) ⇒ statement 2 correct. - Com- pleteness in L^2 ensures Fourier sine expansion converges in L^2 norm, but uniform convergence requires f ∈ C^1 or Lipschitz ⇒ statement 3 false in general. - Or- thogonal and complete in L^2 ⇒ statement 4 false. Correct: 1,
Question 2
Let y 1 (x) be a nontrivial solution of the Airy-type equation
y′′^ + xy = 0, x > 0.
Also, let y 2 (x) solve y 2 ′′ +y 2 = x^2 +2, y 2 (0) = y 2 ′(0) = 0. Determine which statements are correct.
- y 1 (x) has infinitely many zeros accumulating at infinity.
- y 1 (x) is bounded for all x > 0.
- y 2 (x) has exactly one zero at x = 0.
- y 2 (x) has infinitely many zeros.
Solution: - y 1 ′′ + xy 1 = 0 is Airy’s equation: oscillatory for large x ⇒ infinitely many zeros, unbounded ⇒ statements 1 true, 2 false. - y 2 ′′ + y 2 = x^2 + 2; solution y 2 (x) = x^2 satisfies IC ⇒ only zero at x = 0 ⇒ statement 3 true, 4 false. Correct: 1,
Question 3
Consider the first-order nonlinear ODE
y′^ = x^2 + y^2 , y(0) = 1, x ∈ [0, 1].
Which statements are correct?
- A unique local solution exists.
- The solution remains bounded in [0, 1].
- The solution exhibits blow-up at finite x < 1.
- The solution can become negative for some x > 0.
Solution: - f (x, y) = x^2 + y^2 is continuous and Lipschitz in y ⇒ Picard-Lindel¨of theorem ⇒ unique local solution ⇒ 1 true. - Compare y′^ ≥ y^2 with y′^ = y^2 , y(0) = 1 =⇒ y = 1/(1 − x) blows up at x = 1 ⇒ 2 false, 3 true. - y′^ > 0 for all x, y(0) > 0 ⇒ solution remains positive ⇒ 4 false. Correct: 1,
with boundary conditions:
α 1 y(a) + α 2 y′(a) = 0, β 1 y(b) + β 2 y′(b) = 0
where:
- p(x), p′(x), q(x), w(x) are continuous on [a, b]
- p(x) > 0, w(x) > 0 (weight function)
- λ is the eigenvalue and y(x) the corresponding eigenfunction
—
2. Standard Form
y′′^ + [λw(x) − q(x)]y = 0, if p(x) = 1 Self-adjoint form: (p(x)y′)′^ + [λw(x) − q(x)]y = 0 —
3. Properties of Sturm-Liouville Problems
- Eigenvalues are real: λn ∈ R.
- Orthogonality: Eigenfunctions corresponding to distinct eigenvalues are orthog- onal w.r.t weight function w(x): Z (^) b
a
ym(x)yn(x)w(x)dx = 0, m ̸= n
- Eigenvalues ordering: λ 1 < λ 2 <... , with λn → ∞
- Completeness: Eigenfunctions form a complete set in L^2 w(a, b):
f (x) =
X^ ∞
n=
cnyn(x), cn =
R (^) b a (^) Rf (x)yn(x)w(x)dx b a y n^2 (x)w(x)dx
- Oscillation theorem: yn(x) corresponding to λn has exactly n − 1 zeros in (a, b).
- Green’s function: The associated inhomogeneous problem
(p(x)y′)′^ + [λw(x) − q(x)]y = f (x)
can be solved using
y(x) =
Z (^) b
a
G(x, t)f (t)dt
- Self-adjointness:
R (^) b a y^1 Ly^2 dx^ =^
R (^) b a y^2 Ly^1 dx,^ L[y] = (py
′)′ (^) + qy
4. Common Formulas
- Eigenvalues for Dirichlet problem on [0, L]: y′′^ + λy = 0, y(0) = y(L) = 0
λn =
nπ L
, yn(x) = sin
nπx L
, n = 1, 2 ,...
- Eigenvalues for Neumann problem: y′(0) = y′(L) = 0
λn =
nπ
L
, yn(x) = cos
nπx L
, n = 0, 1 , 2 ,...
0
sin
mπx L
sin
nπx L
dx = 0, m ̸= n
- Eigenfunction zero counting: yn(x) has n − 1 zeros in (0, L)
- Expansion of function in eigenfunctions:
f (x) =
X^ ∞
n=
R L
(^0) Rf (x)yn(x)dx L 0 y (^2) n(x)dx yn(x)
5. Solved MSQs
MSQ 1
Consider S-L problem:
y′′^ + λy = 0, y(0) = 0, y(π) = 0
Which are correct?
- λn = n^2
- yn = sin(nx) has n − 1 zeros
- Eigenfunctions corresponding to distinct eigenvalues are orthogonal
- Series
P
cnyn converges uniformly to any continuous f (x)
Solution: 1,2,3 are correct; uniform convergence requires additional smoothness. Correct: 1,2,
MSQ 5
Nonhomogeneous S-L: (py′)′^ + (λw − q)y = f (x), Dirichlet BC. Which statements are true?
- Solution can be written using Green’s function
- Solution is unique for non-eigenvalue λ
- Eigenfunctions form complete set in L^2 w
- Eigenvalues can be complex
Solution: Green’s function exists, solution unique if λ ̸= λn, eigenfunctions com- plete, eigenvalues real Correct: 1,2,
10 Tricky Conceptual MSQs on Sturm-Liouville and
Sturm Theorems
Question 1
Consider the Sturm-Liouville problem:
y′′^ + λy = 0, y(0) = 0, y(π) = 0
Let yn(x) be the eigenfunction corresponding to λn. Which statements are correct?
- λn = n^2 , n = 1, 2 ,...
- yn(x) has exactly n − 1 zeros in (0, π)
- yn(x) is orthogonal to ym(x) for m ̸= n
- The eigenfunctions form a complete set in C[0, π] for uniform convergence
Solution: 1,2,3 are correct. Uniform convergence requires extra smoothness. Correct: 1,2, —
Question 2
Let y′′ 1 + q 1 (x)y 1 = 0 and y′′ 2 + q 2 (x)y 2 = 0, x ∈ (0, 1) with q 1 (x) < q 2 (x). According to Sturm Comparison Theorem:
- Between two zeros of y 1 , there is at least one zero of y 2
- y 1 has more zeros than y 2
- Zeros of y 2 interlace with zeros of y 1
- y 2 can have a zero outside (0, 1) while y 1 does not
Solution: Sturm Comparison Theorem: q 1 < q 2 =⇒ between any two zeros of y 1 , y 2 has at least one zero, interlacing occurs. Correct: 1, —
Question 3
Consider y′′^ + xy = 0 on (0, ∞). Let y(x) be any nontrivial solution.
- y(x) has finitely many zeros
- y(x) oscillates infinitely often as x → ∞
- y(x) is monotone eventually
- y(x) has a zero at x = 0
Solution: This is Airy’s equation. Solutions oscillate infinitely often as x → ∞; zeros accumulate. Not monotone, zero at x = 0 depends on IC. Correct: 2 —
Question 4
Let y 1 and y 2 be two solutions of y′′^ + q(x)y = 0 with y 1 (0) = 0, y 1 ′(0) > 0, y 2 (0) = 1, y′ 2 (0) = 0. Which are true about the zeros of y 1 , y 2?
- Zeros of y 1 and y 2 interlace
- y 2 has at most one zero in (0, ∞)
- y 1 has infinitely many zeros if q(x) > 0 for large x
- y 2 never crosses zero
Solution: Interlacing of zeros holds for linearly independent solutions; if q(x) > 0 even- tually, y 1 oscillates infinitely. y 2 may cross zero once. Correct: 1, —
Question 5
For the BVP y′′^ + λy = 0, y(0) = y(π/2) = 0, which statements are correct?
- Eigenvalues: λn = 4n^2
- Eigenfunctions: yn = sin(2nx)
- yn has n zeros in (0, π/2)
- Eigenfunctions are orthogonal
Solution: Solve sin(
λπ/2) = 0 =⇒
λπ/2 = nπ =⇒ λn = 4n^2 , eigenfunctions sin(2nx), n zeros. Orthogonal. Correct: 1,2, —
Question 9
Sturm Separation Theorem: y′′ 1 +q(x)y 1 = 0, y′′ 2 +q(x)y 2 = 0 linearly independent. Which is true?
- Between two consecutive zeros of y 1 , y 2 has exactly one zero
- y 1 , y 2 may share a zero
- Zeros of y 2 lie outside intervals defined by y 1
- y 2 has at most two zeros between consecutive zeros of y 1
Solution: Separation theorem: exactly one zero of y 2 between consecutive zeros of y 1. Correct: 1 —
Question 10
Periodic S-L: y′′^ + λ^2 y = 0, y(0) = y(T ), y′(0) = y′(T ). Which statements are true?
- Positive eigenvalues λn = 2nπ/T
- Eigenfunctions: sin(2nπx/T ), cos(2nπx/T )
- Zeros of sin eigenfunctions interlace with cos eigenfunctions
- Eigenvalues can be negative
Solution: Positive eigenvalues λn = 2nπ/T , eigenfunctions sin, cos, zeros interlace. Eigen- values real, non-negative. Correct: 1,2,
Extremely Complicated Conceptual Question
Consider the following boundary value problem defined on [0, ∞):
y′′(x) + q(x)y(x) = f (x), y(0) = 0, lim x→∞ y(x) bounded
where q(x) and f (x) are continuous on [0, ∞) and satisfy:
q(x) =
x^2 + sin x, 0 ≤ x ≤ π 1 + cos(x)/x, x > π
, f (x) = e−x^ sin(x)
Let y 1 (x) be a solution of the corresponding homogeneous equation y′′^ + q(x)y = 0 with y 1 (0) = 0, y′ 1 (0) = 1, and let y 2 (x) be another linearly independent solution of the homogeneous problem with y 2 (0) = 1, y′ 2 (0) = 0. Which of the following statements are necessarily true? (MSQ type, select all that apply)
- Between any two consecutive zeros of y 1 (x) in (0, π), y 2 (x) has exactly one zero.
- y 1 (x) has infinitely many zeros as x → ∞ because q(x) > 0 eventually.
- The solution y(x) of the non-homogeneous problem can be represented using a Green’s function, and it is bounded on [0, ∞).
- y 1 (x) is concave wherever y 1 (x) > 0 in (π, ∞) and q(x) > 0.
- The zeros of y(x) for the non-homogeneous equation coincide exactly with zeros of y 1 (x).
- Any linear combination c 1 y 1 + c 2 y 2 cannot be monotone on (π, ∞).
Detailed Solution:
- True: By Sturm Separation Theorem, zeros of two linearly independent solutions interlace in any interval where q(x) is continuous.
- True: For large x, q(x) ≈ 1 + cos(x)/x > 0, so by Sturm Oscillation Theorem, y 1 (x) oscillates infinitely often.
- True: The non-homogeneous problem can always be represented via Green’s function for self-adjoint operator, and boundedness is ensured by the exponential decay of f (x) and positive q(x) at infinity.
- True: y′′ 1 = −q(x)y 1 < 0 if y 1 > 0 and q(x) > 0, so the solution is concave.
- False: Zeros of non-homogeneous solution are generally shifted relative to homogeneous solution zeros; they do not coincide exactly.
- True: Any nontrivial linear combination of two linearly independent oscilla- tory solutions will have multiple zeros; it cannot be monotone eventually. Correct options: 1,2,3,4,
