Find the exact solution to a boundary-value problem involving a differential equation with a Dirac delta distribution | Step-by-Step Solution

Hello! This is a fascinating problem that combines differential equations with distribution theory.

1. What We're Solving:

We need to find a function u(x) that satisfies the differential equation -u''(x) = δ(x - 1/2) on the interval (0,1), with the boundary conditions u(0) = u(1) = 0. The δ(x - 1/2) is a Dirac delta function concentrated at x = 1/2.

2. The Approach:

The key insight is that the Dirac delta function creates a "jump" in the derivative of u(x) at x = 1/2, even though u(x) itself remains continuous. We'll solve this by:

• Finding the general solution in each region (before and after x = 1/2)
• Applying continuity and jump conditions
• Using boundary conditions to determine constants

3. Step-by-Step Solution:

Step 1: Understand what the delta function means The equation -u''(x) = δ(x - 1/2) tells us that u''(x) = 0 everywhere except at x = 1/2, where there's an impulse.

Step 2: Solve in each region For 0 < x < 1/2 and 1/2 < x < 1, we have u''(x) = 0, so:

• For 0 < x < 1/2: u(x) = A₁x + B₁
• For 1/2 < x < 1: u(x) = A₂x + B₂

Step 3: Apply boundary conditions

• u(0) = 0 gives us B₁ = 0
• u(1) = 0 gives us A₂ + B₂ = 0, so B₂ = -A₂

So far: u(x) = A₁x for x < 1/2, and u(x) = A₂(x - 1) for x > 1/2

Step 4: Apply continuity at x = 1/2 u(1/2⁻) = u(1/2⁺) means: A₁(1/2) = A₂(1/2 - 1) = -A₂/2 This gives us: A₁/2 = -A₂/2, so A₁ = -A₂

Step 5: Apply the jump condition The delta function creates a jump in u'(x). Integrating -u''(x) = δ(x - 1/2) from just before to just after x = 1/2: -[u'(1/2⁺) - u'(1/2⁻)] = 1

Since u'(x) = A₁ for x < 1/2 and u'(x) = A₂ for x > 1/2: -(A₂ - A₁) = 1, so A₁ - A₂ = 1

Step 6: Solve for the constants We have: A₁ = -A₂ and A₁ - A₂ = 1 Substituting: -A₂ - A₂ = 1, so -2A₂ = 1, giving A₂ = -1/2 and A₁ = 1/2

4. The Answer:

The solution is: $$u(x) = \begin{cases} \frac{x}{2} & \text{for } 0 \leq x \leq \frac{1}{2} \\ \frac{1-x}{2} & \text{for } \frac{1}{2} \leq x \leq 1 \end{cases}$$

This creates a "tent" or "hat" function that peaks at x = 1/2 with u(1/2) = 1/4.

5. Memory Tip:

Think of the Dirac delta as creating a "kink" in your solution - the function stays continuous, but its derivative jumps by exactly the strength of the delta impulse. The solution looks like a tent pitched right at the location of the delta function!

Great work tackling this advanced topic! The combination of classical differential equations with distribution theory opens up many applications in physics and engineering.