Question

The integral equation corresponding to the boundary value problem \[ \frac{d^2y}{dx^2} + \lambda y(x) = 0; \quad y(0) = 0; \quad y(1) = 0 \] is

where \[ k(x,t) = \begin{cases} t(1-x), & \text{if } t < x \\ x(1-t), & \text{if } t > x \end{cases} \]

• A. \( y(x) = \lambda \int_0^1 k(x,t) y(t) dt \)
• B. \( y(x) = 1 + \lambda \int_0^1 k(x,t) y(t) dt \)
• C. \( y(x) = 1 + \lambda^2 \int_0^1 k(x,t) y(t) dt \)
• D. \( y(x) = 1 + 3\lambda \int_0^1 k(x,t) y(t) dt \)

Hint:

The conversion of a Sturm-Liouville boundary value problem \( (p(x)y')' + q(x)y + \lambda r(x)y = 0 \) with homogeneous boundary conditions into an integral equation \( y(x) = \lambda \int_a^b G(x,t) r(t) y(t) dt \) is a standard procedure. The function \(G(x,t)\) is the Green's function for the operator \(L = (py')'+qy\). For the simple case \(y''+\lambda y=0\) with \(y(0)=y(1)=0\), the resulting integral equation is \(y(x) = \lambda \int_0^1 k(x,t) y(t) dt\), where \(k(x,t)\) is the specific kernel given.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to convert the second-order boundary value problem (BVP) into an equivalent Fredholm integral equation. The kernel of this equation is the Green’s function for the operator \(L = \tfrac{d^2}{dx^2}\) under the given boundary conditions.
Step 2: Key Formula or Approach:
The BVP is: \[ y''(x) + \lambda y(x) = 0, \quad y(0)=0, \; y(1)=0. \] Treating \(-\lambda y(x)\) as a forcing term \(f(x)\), we first solve \(y''(x) = f(x)\). After integrating twice and applying boundary conditions, we express \(y(x)\) in terms of an integral with a kernel \(G(x,t)\). Substituting back \(f(t) = -\lambda y(t)\), we obtain the Fredholm equation with kernel \(k(x,t) = -G(x,t)\).
Step 3: Detailed Explanation:
1. Solve \(y'' = f(x)\): \[ y(x) = \int_0^x (x-t)f(t)dt + C_1x + C_2. \] Using \(y(0)=0 \implies C_2=0\). 2. Apply \(y(1)=0\): \[ 0 = \int_0^1 (1-t)f(t)dt + C_1 \implies C_1 = -\int_0^1 (1-t)f(t)dt. \] 3. Substitution gives: \[ y(x) = \int_0^x (x-t)f(t)dt - x \int_0^1 (1-t)f(t)dt. \] 4. Rearranging yields: \[ y(x) = \int_0^1 G(x,t) f(t)\,dt, \] where \[ G(x,t) = \begin{cases} t(x-1), & t < x, \\ x(t-1), & t > x. \end{cases} \] 5. Substituting \(f(t) = -\lambda y(t)\): \[ y(x) = \lambda \int_0^1 k(x,t) y(t)\,dt, \] with \[ k(x,t) = \begin{cases} t(1-x), & t < x, \\ x(1-t), & t > x. \end{cases} \] Step 4: Final Answer:
The integral equation corresponding to the given BVP is: \[ y(x) = \lambda \int_0^1 k(x,t) \, y(t)\, dt, \] where \[ k(x,t) = \begin{cases} t(1-x), & t < x, \\ x(1-t), & t > x. \end{cases} \] This matches the required form.