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On the positive half-line, the Green function for $-d^2/d{x}^2+1$ with a zero Dirichlet boundary condition at the origin and decay at infinity isIt is continuous, has derivative jump $G_x(\xi +,\xi )-G_x(\xi -,\xi )=-1$, and is nonnegative. It is the negative of the Dirichlet half-line Green function for d2 minus 1; the sign of the leading differential term matters.

Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

For $x\ne \xi$, the homogeneous solutions are $e^x,e^{-x}$. The left boundary condition selects $\sinh x$, while boundedness at infinity selects $e^{-x}$ on the right. Thus write $G=A(\xi )\sinh x$ for $x<\xi$ and $G=B(\xi )e^{-x}$ for $x>\xi$.

A unit Dirac delta distribution in the second derivative requires continuity of $G$ and a derivative jump of one:The second equation and the first give $-A{e}^{\xi }=1$, so the Dirichlet half-line Green function for d2 minus 1 isThe second expression is also a method of images construction. Applying the Green's function to the forcing givesFor bounded forcing this is bounded, vanishes at zero and solves the equation; a bounded homogeneous solution vanishing at zero is zero, proving uniqueness. More general forcing is allowed when the integrals converge and the resulting function has the required growth. Arbitrary unbounded forcing need not admit a bounded solution; this distinction is relevant in part (b).

For $r>0$, the substitution $y=ru$ removes the first derivative in the three-dimensional radial Laplacian, since $u^{\prime \prime }+2u^{\prime }/r=y^{\prime \prime }/r$. Bounded $u$ at zero implies $y(0)=0$. Bounded $u$ at infinity means $y=O(r)$; it does not require bounded $y$. For example, constant forcing $g=1$ has the bounded particular solution $u=-1$, whose reduced function is $y=-r$. A Dirichlet half-line Green function for d2 minus 1 can still represent this solution when its integral converges, but the resulting reduced function has the growth demanded by the forcing.