18.03 PS4 Extra Credit Solution—Spring 09 (10 pts = 4 + 4 + 2)Xa.(4 pts) Start withy0< z0. The property|fy| ≤Aimplies|f(x0, y0)-f(x0, z0)|=Zz0y0fy(x0, y)dy≤Zz0y0|fy(x0, y)|dy≤Zz0y0Ady=A(z0-y0)In particular,f(x0, z0)-f(x0, y0)≥ -A(z0-y0)We havey1=y0+hf(x0, y0) andz1=z0+hf(x0, z0). Hence, sinceAh <1,z1-y1= (z0-y0)+h[f(x0, z0)-f(x0, y0)]≥(z0-y0)-hA(z0-y0) = (1-Ah)(z0-y0)>0In other words, we have shown thaty0< z0=⇒y1< z1.The same reasoning withindices shifted by 1 shows thaty1< z1=⇒y2< z2, andy2< z2=⇒y3< z3, etc. Ingeneral, by mathematical induction,yn< znfor all positive integersn.Xb.(4 pts) The proof in part (a) really only depended on the lower boundfy≥ -A, orequivalentlyfy>-1/h. We repeat the argument here. Suppose thatfy(x0, y)>-1/hfory0≤y≤z0. ThenhZz0y0fy(x0, y)dy > hZz0y0-1hdy=-(z0-y0)=⇒z1-y1= (z0-y0) +hZz0y0fy(x0, y)dy >(z0-y0)-(z0-y0) = 0so the paths do not cross.In particular, whenfy≥0 everywhere, there will never becrossing paths no matter whathis. Thus there are no crossings whenf(x, y) = 0.5y+ 1,f(x, y) =y, andf(x, y) =y-x.
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