Question: PDE system solution boundary condition problem

i'm trying to solve a sysem of pde, but i get this error 

Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {x(0, z, r), x(t, z, 0.5e-2), xeq(t, z, 0.5e-2), y(0, z, r), y(t, 0, r)} in the given boundary conditions {x(0, z, r) = 0, x(t, z, 0.5e-2) = xeq(t, z, 0.5e-2), y(0, z, r) = 0, y(t, 0, r) = 0}
> 

> restart; with(PDEtools);
[CanonicalCoordinates, ChangeSymmetry, CharacteristicQ, 

  CharacteristicQInvariants, ConservedCurrentTest, 

  ConservedCurrents, ConsistencyTest, D_Dx, DeterminingPDE, 

  Eta_k, Euler, FromJet, FunctionFieldSolutions, 

  InfinitesimalGenerator, Infinitesimals, IntegratingFactorTest, 

  IntegratingFactors, InvariantEquation, InvariantSolutions, 

  InvariantTransformation, Invariants, Laplace, Library, PDEplot, 

  PolynomialSolutions, ReducedForm, SimilaritySolutions, 

  SimilarityTransformation, Solve, SymmetryCommutator, 

  SymmetryGauge, SymmetrySolutions, SymmetryTest, 

  SymmetryTransformation, TWSolutions, ToJet, build, casesplit, 

  charstrip, dchange, dcoeffs, declare, diff_table, difforder, 

  dpolyform, dsubs, mapde, separability, splitstrip, splitsys, 

  undeclare]
> Y := diff_table(y(t, z, r));

> X := diff_table(x(t, z, r));

> Xeq := diff_table(xeq(t, z, r));

> Yeq := diff_table(yeq(t, z, r));

> pde[1] := Y[t]-3*(Yeq[]-Y[])+Y[z] = 0;
        / d            \                                
        |--- y(t, z, r)| - 3 yeq(t, z, r) + 3 y(t, z, r)
        \ dt           /                                

             / d            \    
           + |--- y(t, z, r)| = 0
             \ dz           /    
> pde[2] := X[t]-0.2e-1*X[r, r]-0.2e-1*X[r]/r = 0;
         / d            \        / d  / d            \\
         |--- x(t, z, r)| - 0.02 |--- |--- x(t, z, r)||
         \ dt           /        \ dr \ dr           //

                   / d            \    
              0.02 |--- x(t, z, r)|    
                   \ dr           /    
            - --------------------- = 0
                        r              
> pde[3] := 5*(Yeq[]-Y[]) = 0.1e-2*X[r];
                                           / d            \
     5 yeq(t, z, r) - 5 y(t, z, r) = 0.001 |--- x(t, z, r)|
                                           \ dr           /
> pde[4] := Xeq[] = 100*Yeq[]/(1+1000*Yeq[]);
                               100 yeq(t, z, r)   
              xeq(t, z, r) = ---------------------
                             1 + 1000 yeq(t, z, r)
> bc[1] := eval(X[], t = 0) = 0;
                         x(0, z, r) = 0
> bc[2] := eval(Y[], t = 0) = 0;
                         y(0, z, r) = 0
> bc[3] := eval(Y[], z = 0) = 0;
                         y(t, 0, r) = 0
> bc[4] := eval(X[r], r = 0.16e-1) = 0;
                 / d                         \    
             eval|--- x(t, z, r), {r = 0.016}| = 0
                 \ dr                        /    
> bc[5] := eval(X[], r = 0.5e-2) = eval(Xeq[], r = 0.5e-2);
               x(t, z, 0.005) = xeq(t, z, 0.005)
> sys := {bc[1], bc[2], bc[3], bc[4], bc[5], pde[1], pde[2], pde[3], pde[4]};
 /                                                               
 |                                                               
<                                       / d            \  / d    
 |5 yeq(t, z, r) - 5 y(t, z, r) = 0.001 |--- x(t, z, r)|, |--- x(
 \                                      \ dr           /  \ dt   

                                                 / d            \   
                                            0.02 |--- x(t, z, r)|   
          \        / d  / d            \\        \ dr           /   
  t, z, r)| - 0.02 |--- |--- x(t, z, r)|| - --------------------- = 
          /        \ dr \ dr           //             r             

     / d            \                                
  0, |--- y(t, z, r)| - 3 yeq(t, z, r) + 3 y(t, z, r)
     \ dt           /                                

     / d            \                      
   + |--- y(t, z, r)| = 0, x(0, z, r) = 0, 
     \ dz           /                      

  x(t, z, 0.005) = xeq(t, z, 0.005), 

                   100 yeq(t, z, r)                     
  xeq(t, z, r) = ---------------------, y(0, z, r) = 0, 
                 1 + 1000 yeq(t, z, r)                  

                                                       \ 
                                                       | 
                      / d                         \     >
  y(t, 0, r) = 0, eval|--- x(t, z, r), {r = 0.016}| = 0| 
                      \ dr                        /    / 
> cond := {bc[1], bc[2], bc[3], bc[4], bc[5]};
      /                                                   
     { x(0, z, r) = 0, x(t, z, 0.005) = xeq(t, z, 0.005), 
      \                                                   

       y(0, z, r) = 0, y(t, 0, r) = 0, 

           / d                         \    \ 
       eval|--- x(t, z, r), {r = 0.016}| = 0 }
           \ dr                        /    / 

> soluzione := pdsolve(sys, [y(t, z, r), yeq(t, z, r), x(t, z, r), xeq(t, z, r)]);
Error, (in PDEtools:-Library:-NormalizeBoundaryConditions) unable to isolate the functions {x(0, z, r), x(t, z, 0.5e-2), xeq(t, z, 0.5e-2), y(0, z, r), y(t, 0, r)} in the given boundary conditions {x(0, z, r) = 0, x(t, z, 0.5e-2) = xeq(t, z, 0.5e-2), y(0, z, r) = 0, y(t, 0, r) = 0}
> 
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