 Solution

Fault slip must taper toward zero at the tips of an isolated finite fault, because its two sides are still joined beyond the tips. Accumulating more central displacement at a fixed short length steepens the displacement gradient and loads the surrounding rock, especially near the tips. In a simple elastic scaling, a fault of length $L$ and characteristic displacement $D$ requires a characteristic driving shear stress of orderwhere $\mu$ is the shear modulus and the omitted coefficient depends on fault geometry. The host's finite strength prevents this gradient increasing indefinitely. Tip damage, new cracking and linkage with neighboring segments allow the fault system to lengthen and accommodate more displacement without unbounded tip loading. Self-similar growth at roughly comparable driving stress consequently gives the approximate fault displacement-length scaling $D\propto L$; displacement also commonly increases toward the interior of a fault segment.^[ref]

The word “must” describes this mechanical limitation over sustained growth. It is not a rule that every earthquake or every increment of accumulated fault slip lengthens the mapped trace: strong barriers, interactions and early establishment of fault length can permit substantial later displacement at nearly constant length.