Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2021 / Paper 1 / 3

For a fixed source point $\xi \in (a,b)$, the Green function is defined bywhere $\delta$ is the Dirac delta distribution and the operator acts on $x$. The corresponding solution isApplying $L_x$ under the integral gives $f(x)$ by the defining delta identity, while the endpoint values vanish. With the stated continuous coefficients, one may define the normalized distributional equation $G_{xx}+(\beta /\alpha )G_x+(\gamma /\alpha )G=\delta (x-\xi )/\alpha (\xi )$; this avoids multiplying a general distributional second derivative by a merely continuous coefficient.

Existence of this inverse Green function additionally requires that the homogeneous two-point boundary value problem have only the zero solution. The assumptions printed in the question alone do not ensure this: $y^{\prime \prime }+y$ on $[0,\pi ]$ has the nonzero zero-endpoint solution $\sin x$. At such resonance there are solvability conditions and no ordinary inverse kernel for arbitrary $f$. The construction below is nonresonant.

Suppose the piecewise smooth Green function has jump $[G{]}_{\xi }=G({\xi }^{+};\xi )-G({\xi }^{-};\xi )$. Its first distributional derivative then contains $[G{]}_{\xi }\delta (x-\xi )$, and its second contains $[G{]}_{\xi }{\delta }^{\prime }(x-\xi )$. In the normalized equation from part (a), the first-derivative term can supply a delta but no delta derivative. The right side has only a delta. Thus its delta-derivative coefficient must vanish:Equivalently, the two homogeneous branches must join continuously. This is the continuity part of the jump condition for a second-order Green function.

Now $G$ is continuous, so $G_x$ has no delta contribution. Integrate its normalized differential equation over $(\xi -ϵ,\xi +ϵ)$. The bounded lower-order terms have integrals tending to zero, while the integral of $G_{xx}$ is the difference of the two one-sided first derivatives. Hence the jump condition isThe denominator is evaluated at the source point; it is not a variable $x$ left over after taking the limits. The positivity or negativity of this jump follows the sign convention used for the leading coefficient of $L$.

The homogeneous characteristic equation is $r^2-2r+2=0$, with roots $1\pm i$. Choose the endpoint-adapted homogeneous solutionsso that $u(0)=0$ and $v(\pi /2)=0$. Their Wronskian iswhich never vanishes. A continuous Green function with the required endpoints has the form $C(\xi )u(x_{<})v(x_{>})$, where $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. Its derivative jump is $C(\xi )W(\xi )$. Because the leading coefficient is one, part (b) requires this to equal one. ThereforeIn particular, the branches are $-e^{x-\xi }\sin x\cos \xi$ for $x\le \xi$ and $-e^{x-\xi }\sin \xi \cos x$ for $x\ge \xi$. This is the Dirichlet Green function for a conjugated harmonic oscillator with interval length $\pi /2$, drift parameter one and oscillator wave number one.

Insert the source into the Green-function representation and split the integral at its source discontinuity in derivative:Using ${\sin }^{2}x+{\cos }^{2}x=1$, the answer isBoth endpoint values are zero. As a direct check, conjugating by $e^x$ sends $y^{\prime \prime }-2y^{\prime }+2y$ to $e^x(v^{\prime \prime }+v)$, and $v=1-\cos x-\sin x$ gives $v^{\prime \prime }+v=1$. Thus the differential equation and both boundary conditions are satisfied.