Past exam of the differential equations 2 course of the University of Oxford / 2023 / 3 / b

For the positive coefficient of $y^{\prime \prime }$, setting $ϵ=0$ gives the reduced equation $y_0^{\prime }+x{y}_0=0$, with general solution $y_0=C{e}^{-x^2/2}$. A fast decaying homogeneous solution of $ϵy^{\prime \prime }+y^{\prime }=0$ is $e^{-x/ϵ}$, placing the boundary layer at the left endpoint. Thus use the right boundary condition for the outer solution: $y_0(1)=0$ impliesThe dominant balance $ϵ/{\delta }^2\sim 1/\delta$ gives width $\delta =ϵ$. With the inner variable $X=x/ϵ$ and $Y(X)=y(ϵX)$, the exact rescaled equation isAt leading order, $Y_{0,XX}+Y_{0,X}=0$, so $Y_0=A+B{e}^{-X}$. The asymptotic matching condition $Y_0(\infty )=y_{\mathrm{out},0}(0)=0$ sets $A=0$, and $Y_0(0)=1$ sets $B=1$. ThereforeThe additive composite expansion has zero overlap to subtract. Its graph starts at one, decreases steeply over $x=O(ϵ)$, and then stays close to zero; its initial slope is $-1/ϵ$. The right endpoint discrepancy $e^{-1/ϵ}$ is exponentially small. The left panel of the figure below sketches this profile. A right-end inner expansion would instead contain a growing $e^X$ term and could not repair the left endpoint mismatch.

With the negative coefficient of $y^{\prime \prime }$, the reduced equation is unchanged, but the fast mode now decays away from the right endpoint. Apply the left boundary condition to the outer solution, givingThe boundary layer still has width $O(ϵ)$. Put $X=(1-x)/ϵ$ and $Y(X)=y(1-ϵX)$. The rescaled equation isIts leading inner expansion has $Y_0=A+B{e}^{-X}$. The asymptotic matching condition is $Y_0(\infty )=e^{-1/2}$, while $Y_0(0)=0$. ThusSubtracting the overlap $e^{-1/2}$ in the additive composite expansion givesIts graph starts exponentially close to one, follows the gently decreasing outer solution, and drops from near $e^{-1/2}$ to zero in a right boundary layer of width $O(ϵ)$. The right endpoint value is exactly zero; the left endpoint error is $-e^{-1/2-1/ϵ}$. Its right endpoint slope is $-e^{-1/2}(1+1/ϵ)$, showing the rapid final descent.