Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 3 / d

Choose $y_a=x$ and $y_b=x^3-x$, satisfying the left and right conditions respectively. Their Wronskian isPart (a) therefore supplies the valid kernelIt is continuous at the source, its first derivative has jump one, and it vanishes at both endpoints. However, it is not unique: $x^3-x$ is a nonzero homogeneous solution satisfying both stated endpoint conditions. Thus $G_0+(x^3-x)h(z)$ has the same defining properties for any suitable $h$.

This is a nonunique Green kernel at a singular endpoint with two vanishing modes. Choosing $y_a=x^3$ instead also gives a valid kernel, corresponding to the stronger selection $y(x)/x\to 0$ at the left endpoint. The original problem states no such selection, so the family must be acknowledged.

The original PDF shows $x^2$ in the nonzero source branch. Inserting that source into $G_0$ gives, for $x<1/2$,For $x\ge 1/2$, split the integral at $z=x$:Because of the homogeneous freedom established in part (i), the full answer to the stated endpoint problem isAt the interface, both branches have value $-5/96$ and derivative $-5/48$, so no extra delta or delta derivative occurs there. Direct substitution gives zero source on the left and $x^2$ on the right; the endpoint values are zero for every $C$. If the extra condition $y/x\to 0$ is imposed, it selects $C=-5/48$, but that is additional information absent from the printed question.