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

For each source point $z$, construct the Green function from homogeneous solutions on either side:The endpoint conditions are then automatic. The equation $L_{x}G=\delta (x-z)$ requires continuity of $G$ and a unit jump in its first derivative: a discontinuity in $G$ would introduce an unwanted delta derivative, while integrating across the source fixes $G_x(z^{+},z)-G_x(z^{-},z)=1$. ThusSolving this two-by-two system, with the Wronskian $W=y_a{y}_b^{\prime }-y_b{y}_a^{\prime }\ne 0$, givesThis derivation also remains valid if one of the solution values at the matching point vanishes: only the Wronskian must be nonzero. For a regular nonsingular two-point problem it supplies the inverse kernel; at a singular endpoint, uniqueness requires separate checking, as part (d) illustrates.

The Green-function representation isThe delta identity gives $L{y}_p=f$ and the endpoint conditions follow from those of $G$. If a nonzero homogeneous solution also satisfies both endpoint conditions, it may be added; uniqueness is not automatic in that case.

For the Euler-Cauchy equation, substitute $y=x^n$ on an interval not containing zero. The coefficient of $x^{n-2}$ isThe distinct roots giveThese solutions are linearly independent since their Wronskian is $2x^3\ne 0$ away from zero. Both nevertheless approach zero as $x\to 0$, which will make the endpoint conditions in the next part insufficient for uniqueness.

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.