Example 19.2-1 Boundary-Layer Theory applied to Unsteady Evaporation

>  restart;
>  IxA:=t->int(xA(t,z),z=0...delta(t)); Since Xa is assumed to 
be zero outside the boundary layer, this is the same as the integral 
from zero to infinity.
>  eq:=diff(IxA(t),t)=-DAB*D[2](xA)(t,0)/(1-xA0); eq. 19.2-3
>  xA:=(t,z)->xA0*f(z/delta(t)); eq. 19.2-4
>  eq;
>  f(1):=0;
>  simplify(eq);
>  Dif:=diff(xA(t,z),z);Dif:=unapply(Dif,z);
>  eq2:=-xA0*D(delta)(t)*int(zeta*diff(f(zeta),zeta),
zeta=0...1)=-DAB*Dif(0)/(1-xA0);
>  s:=dsolve({eq2,delta(0)=0},delta(t));
>  assign(s[1]);delta:=simplify(unapply(delta(t),t));
>  f(1):=0; simplify(delta(t)); I thought Maple knew integration 
by parts and could simplify this considerably. The integral should 
be simply minus the integral of f from 0 to 1.
>  f:=zeta->(1-zeta)^2;
>  delta:=value(delta(t)); The use of value seems to be the only 
way to get Maple to evaluate the integral.
>  delta:=unapply(delta,t); Now we need to make delta a function 
again.
>  NA0:=t->c*int(diff(xA(t,z),t),z=0...delta(t)); Using 19.2-1 
again.
>  NA0(t)/c;
>  dVAdt:=simplify(%,assume=positive);
>  VA:=int(dVAdt,t=0...t1);
>  VAFick:=xA0*sqrt(4*DAB*t1/Pi);
>  VA/VAFick;
>  psi2:=simplify(%,assume=positive); This agrees with the value 
found in the text after eq.19.2-10 for the assumed function: 
f(zeta) = (1-zeta)^2..

>