Past exam of the differential equations 2 course of the University of Oxford / 2017 / 3 / b

The leading regular perturbation equation has $c=0$. Its homogeneous solutions are $1$ and $(x+2)e^{-x}$, soThe boundary data must also be expanded: $2e^{ϵ}=2+2ϵ+\cdots$, while the right endpoint value is independent of $ϵ$. Thus $y_0(0)=2$ and $y_0(1)=3/e$, givingSince $2-3/e\ne 0$, the unique values are $A=0$, $B=1$. ThereforeIts derivatives are $y_0^{\prime }=-(x+1)e^{-x}$ and $y_0^{\prime \prime }=x{e}^{-x}$, so the leading differential equation and both leading endpoint values can be checked directly.

Let $L_{0}u=(x+1)u^{\prime \prime }+x{u}^{\prime }$. Matching powers in the regular perturbation expansion and in the Taylor series of the left endpoint value gives the complete recursive boundary value problemThe higher-order left endpoint values are not zero. The parameter-dependent boundary data in a regular perturbation hierarchy rule applies here.

First prove uniqueness for any one of these forced problems. The difference $v$ of two solutions obeys $L_{0}v=0$ and homogeneous Dirichlet boundary conditions. The positive integrating factor $r_0=e^x/(x+1{)}^2$ givesMultiplication by $\overline{v}$ and integration by parts yieldHence $v^{\prime }=0$, and the endpoint values force $v=0$. The kernel is trivial: zero is not an eigenvalue of this homogeneous Dirichlet boundary value problem.

Existence is also explicit. For arbitrary smooth forcing $g$ and endpoint values $u(0)=a$, $u(1)=b$, two integrations of $(p{u}^{\prime }{)}^{\prime }=r_{0}g$ givewhereSince $H(1)=2-3/e>0$, the unique choicealways meets the second endpoint value. Starting from the smooth $y_0$, induction with $g=-y_{n-1}$, $a=2/n!$, $b=0$ proves existence and uniqueness at every order.

Equivalently, the Fredholm alternative has no compatibility obstruction because the homogeneous kernel vanishes; the corresponding adjoint operator has no zero eigenvalue either. The direct construction above proves this without computing any particular $y_n$ beyond the leading term.