Identical E2 pages with different later differentials

For $t=0,1$, let $C_t^1=kx$, $C_t^2=ky$ and $dx=ty$, zero elsewhere. Let $x$ have filtration degree zero and $y$ degree two in the decreasing filtration. This means $F^p{C}_t^1=kx$ exactly for $p\le 0$ and $F^p{C}_t^2=ky$ exactly for $p\le 2$. Both filtrations are finite in each degree and preserved by $d$. Their graded terms are $kx$ at $(0,1)$ and $ky$ at $(2,0)$, zero elsewhere. As in F1, negate chain degree and filtration index to apply F2's homological formulas.

Through page two the numerator at each of these positions is the full displayed line and its denominator is zero: $dx$ already lies in $F^2{C}_t^2$, and no boundary from filtration zero enters the denominator at filtration two until page three. Thus $d_0=d_1=0$ and both $E_2$ pages have the same two lines. At $r=2$, F2's representative rule gives $d_{2}x=ty$. If $t=1$, the page-three cycle numerator at $x$ is zero because $dx\mathrm{\notin }{F}^3{C}_t^2=0$, and the page-three denominator at $y$ is all $ky=d(kx)$. Hence $E_3=0$. If $t=0$, both numerators remain full and both denominators remain zero on every page, so $E_3=E_{\mathrm{\infty }}=E_2$.

Direct total cohomology is zero for $t=1$ and is $kx$ in degree one and $ky$ in degree two for $t=0$. In the latter case the induced filtration has $F^0{H}^1=H^1,F^1{H}^1=0$ and $F^0{H}^2=F^1{H}^2=F^2{H}^2=H^2,F^3{H}^2=0$. These finite normalized filtrations have exactly the calculated stationary graded pieces; all later incident arrows are zero. Thus the difference persists in the actual filtered abutments, with strong convergence verified directly. No infinite choices or splitting assumptions occur.