Part BCSIR NET June 2023greens-function-dirichlet

Greens function dirichlet

For the unknown $y : [0,1] \to \mathbb{R}$ , consider the boundary value problem y″(x) + 2y(x) = 0 for x ∈ (0,1), y(0) = y(1) = 0. It is given that it corresponds to the integral equation $y(x) = 2\int_{0}^{1} K(x, t) y(t)$ dt. Which of the following is the kernel K(x, t)?

A.K(x,t) = t(1 − x) for t < x; x(1 − t) for t > x✓ Correct
B. $K(x,t) = t^{2}(1 - x)$ for $t < x; x^{2}(1 - t)$ for t > x
C. $K(x,t) = \sqrt{t}(1 - x)$ for $t < x; \sqrt{x}(1 - t)$ for t > x
D. $K(x,t) = \sqrt{t^{3}}(1 - x)$ for $t < x; \sqrt{x^{3}}(1 - t)$ for t > x

The idea, in plain words

Rewriting a boundary value problem as an integral equation means bundling the boundary conditions into a single kernel. That kernel is the response to a point poke at one location, and it takes a simple two-piece form depending on which side of the poke you are on.

Solution

The Green's function of −y″ with Dirichlet conditions on [0,1] is G(x,t) = t(1 − x) for t < x and x(1 − t) for t > x; then $y = 2\int{}G y$ .

Why each option is right or wrong

Checked against 2 symbolic computations

A.Correct
True.

K is the Green's function for −y″ with Dirichlet conditions on [0,1].

It must satisfy three things, and this one does all three: it vanishes at x = 0 and x = 1; it is continuous across t = x

since both branches give x(1 − x) there (SymPy: their difference is 0); and its t-derivative jumps by −1 across t = x, from (1 − x) below to −x above (SymPy: the jump is exactly −1).

The trigger: build a Green's function from solutions of the homogeneous equation satisfying each boundary condition, then normalise by the jump.
B.WrongExecution slip
False.

The branches $t^{2}(1 - x)$ and $x^{2}(1 - t)$ do not match at t = x — they give $x^{2}(1 - x)$ and $x^{2}(1 - x)$

which happen to agree

but the derivative jump is then 2x·… rather than the required −1, and the function no longer solves y″ = 0 away from t = x.

A Green's function for a second-order operator is piecewise linear here, not quadratic.
C.WrongHypothesis dropped
False.

$\sqrt{t}$ is not a solution of the homogeneous equation y″ = 0

so this kernel cannot reproduce the differential operator.

Each branch of a Green's function must itself solve the homogeneous problem on its side of the diagonal.
D.Wrong
False, for the same reason as $(C): \sqrt{t^{3}} =$ $t^{3/2}$ does not satisfy y″ = 0.

Only the linear branches in (A) do — which is a quick way to eliminate three of the four options without computing anything.

2 of these 3 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Greens function dirichlet

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