Past exam of the differential equations and applications course of the University of Oxford / 2009 / B / 6 / d / Solution

The forcing contains $x/y$, so matching it with a single similarity variable requires $\beta =1$. For $u=y^{\alpha }f(\eta )$, $\eta =x/y$, substitution givesMatching the source $y^{-1}\eta e^{-\eta }$ selects $\boxed{\alpha =0,\beta =1}$ and gives $f^{\prime \prime }+f^{\prime }=e^{-\eta }$. Multiplying the first-order equation for $f^{\prime }$ by $e^{\eta }$ yields $f^{\prime }=(\eta +C)e^{-\eta }$, so $f=-(\eta +1+C)e^{-\eta }+D$. The limits at zero and infinity give $D=0$, $C=-1$. HenceFor fixed positive $x$, the initial limit is also zero because the exponential dominates $x/y$. The function $-\eta e^{-\eta }$ decreases to $-1/e$ at $\eta =1$ and increases thereafter. ThereforeThe axis limits are zero, but there is no joint limit at $(0,0)$: approaching along $x=y$ gives $-1/e$. This does not violate any of the printed fixed-variable limits. It is this bounded similarity solution whose minimum was calculated; the unbounded corner-source additions in (b) need not have the same minimum.