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The initial data and $\ddot{u}=0$ give $u=\tau$. To first order in $ϵ$, replace $x$ and $y$ on the right-hand sides of the transverse equations by $x_0$ and $y_0$. Twice integrating with the initial rest conditions givesFor $\tau >U$, define $A_0={\int }_{-\infty }^{\infty }a(s)ds$ and $A_1={\int }_{-\infty }^{\infty }s,a(s)ds$. ThenHenceThese permanent changes are forms of displacement memory and velocity memory.

There is a discrepancy in the question's final instruction. The longitudinal equation actually givesand thereforeThus $z$ does not vanish to first order for a general allowed profile. It vanishes after the pulse under the additional hypothesis $A_0=0$ used in part (iv), but it need not vanish while that pulse is passing. The proper-time normalization independently gives the same relation $\dot{z}=xyA+O({ϵ}^2)$.

Now $A_0=0$, so every mass is again at rest after the pulse and $z=0+O({ϵ}^2)$. Put $k=ϵA_1>0$. The final transverse coordinates areThe unit vectors along $x=y$ and $x=-y$ are eigenvectors with eigenvalues $1-k$ and $1+k$, respectively. Consequently an initial circle becomes, to first order, an ellipse compressed along the line $x=y$ and stretched along $x=-y$. This persistent deformation is displacement memory.