... e−i/2tx2R2∈ I7 /4+ 1,1 /4 (φL). (4. 20)Inductively, we look for ul∈ I3 /4, 1 /4 (φL) such thatt2Dt− txDx+ t2H + itn +22− txqul+ e−i/2tx2R2∈ I7 /4+ l+1,1 /4 (φL). (4. 21) 4 The power ... termin (4. 1), and with R3(z,w,t) supported in x 4 . Finally, we solveP U 4 = −R3+ E 4 , (4. 6)with zero initial conditions, where E 4 satisfies the condition for the error termin (4. 1). Clearly ... i12+3 4 −2n +1 4 + i + x˜qσ3 /4 (u0)=σ7 /4 (e−i/2tx2R2). (4. 19)This gives an equation of the formLx∂x+xW+n2+ x˜qσ3 /4 (u0)=−iσ7 /4 (e−i/2tx2R2),so by (2.6), σ3 /4 (u0)...