Past exam of the differential equations 2 course of the University of Oxford / 2014 / 2

Work first on $0<x<1$. Divide by $x-1$ to put the equation in leading-coefficient-one form. Its first-derivative coefficient is $-x/(x-1)$, so the integrating factor for the self-adjoint differential operator isThe resulting coefficients areIndeed $p^{\prime }(x)=x{e}^{-x}/(1-x{)}^2$, and multiplying the original equation by $-e^{-x}/(1-x{)}^2$ gives $(p{y}^{\prime }{)}^{\prime }+qy=F$. These transformed coefficients are singular at the right endpoint; the boundary value there is understood as the endpoint limit.

Substituting $y=x$ and $y=e^x$ in the homogeneous original equation gives zero in both cases. Their Wronskian iswhich is nonzero for $0<x<1$. Thus they are linearly independent on the solution interval.

To construct a Dirichlet Green function, choose homogeneous solutions adapted to the two endpoints:Their Wronskian is again $(x-1)e^x$, and $pW=-1$. Therefore the Green function for $L=(pd/dx{)}^{\prime }+q$ with zero endpoint values isFor fixed $s\in (0,1)$ it is continuous in $x$, vanishes at both endpoints, and satisfies the unit-flux jump conditionConsequently $L_{x}G(x,s)=\delta (x-s)$ in the sense of distributions. This is a singular-endpoint Green function with a double-zero right solution.

The endpoint data are nonhomogeneous, so use the homogeneous lifting $2x$ and writeAlthough $F(s)$ is singular at one, the integral exists for each interior $x$: the right solution satisfies $v(s)\sim e(1-s{)}^2/2$, cancelling the $(1-s{)}^{-2}$ forcing when $s>x$.

Splitting the integral at $s=x$ giveswhere elementary antiderivatives yieldFor the second equality, an antiderivative is $(1-e^{1-s})/(1-s)$, whose endpoint limit at one is $-1$. Simplifying the Green representation givesIt has the prescribed endpoint values. Direct substitution gives $(x-1)y^{\prime \prime }-x{y}^{\prime }+y=1$, and at the degenerate endpoint it also obeys $-y^{\prime }(1)+y(1)=1$. The only homogeneous solution satisfying both zero endpoint conditions is zero, so this solution is unique in the corresponding classical class.

Let $X=(x,y)$ be the integration variable and $P=(\xi ,\eta )$ the interior source point. Green's second identity, obtained from Green's theorem by applying the divergence theorem to $u\nabla G-G\nabla u$, isThe outward normal is used throughout. To obtain the prescribed representation with its positive volume term, chooseIndeed, substituting $\Delta u=f$ makes the left side $u(P)-{\int }_{D}Gf$. On the Dirichlet portion, $u=g$ and $G=0$ leave only $g{\partial }_{n}G$. On the Neumann portion, both normal derivatives vanish. Rearranging gives exactly the desired boundary-plus-volume representation. The distributional identity can equivalently be justified by excluding a small circle around $P$ and using the logarithmic singularity there.

For the quarter-plane, the fundamental solution of the Laplace equation with this sign convention is $\Phi (X,P)=(2\pi {)}^{-1}\log |X-P|$, since $\Delta \Phi ={\delta }_P$. The method of images uses an equal-sign reflection across the Neumann boundary $x=0$ and an opposite-sign reflection across the Dirichlet boundary $y=0$. HenceEquivalently,Only $(\xi ,\eta )$ lies inside the domain, so the interior Laplacian has exactly the required unit point source. At $y=0$ numerator and denominator agree, giving $G=0$. The formula is even in $x$, giving ${\partial }_{x}G=0$ at $x=0$, and therefore zero outward normal derivative there.

On the Dirichlet boundary $y=0$, the outward normal is $-{\partial }_y$, and differentiation gives the useful checkThis is the mixed Dirichlet-Neumann Green function in a quadrant. The image solution tends to zero at spatial infinity; on the unbounded domain the representation uses data and solutions for which the boundary term at infinity in Green's second identity vanishes.