9 years, 8 days

## Clarification...

Thanks to everybody for your contribution.

For every Q, it is Q(r, x, tau).

For every U, it is U(x,tau)

There are cases where Q is estimated at r = 1. (the cases where U and Q are in the same equation)

If Maple cannot handle pdes with more than two independent variable, it means I will have to try another software.

The Systems of pdes is actually for a fixed-bed adsorption column filled with spherical adsorbents. The adsorbent is referred to as the particle while the fluid to be purified is the fluid phase.

The particle: Q(r, x, tau)

PDE:= diff(Q(r, tau), tau) = diff(Q(r, tau), \$(r, 2))+(2/r)*diff(Q(r, tau),r);

IBC:={Q(r, 0) = 0,(D[1](Q))(0, tau) = 0,(1/K)*(D[1](Q))(1, tau)=xi*(U-Q(1, tau)/K)};

Q is the dimensionless concentration of the particle

tau is the dimensionless time

x is the dimensionless axial distance along the column where the adsorbent in consideration is located

Fluid phase: U(x, tau)

PDE:= diff(U(x, tau),tau)+ psi*Theta*diff(U(x, tau),x)-(1/Pe)*psi*Theta*diff(U(x, tau),\$(x, 2))=-3*psi*xi*(U(x, tau)-Q/K);

IBC:={U(x, 0) = 0,U(0, tau) = 1+(1/Pe)*(D[1](U))(0, tau),(D[1](U))(1, tau)=0}

U is the dimensionless concentration of the fluid

tau is the dimensionless time

x is the dimensionless axial distance along the column through which the fluid is flowing.

Pe:=0.01:

psi:=6780:

Theta:=3.0:

xi:=10000:

NOTE:

1. the range of variable x is from 0 to 1.

2. the range of variable r is from 0 to 1

3. The Q in the first pde is evaluated at r = 1. That is why it can be subtracted from U(x,tau). (for compatibility sake)

4. The Q in the BC pde is evaluated at r = 1. That is why it can be subtracted from U(x,tau).  (for compatibility sake)

## Urgent Help...

Please, is there any one that can help with the solution of this problem?

I will really appreciate it. thanks in anticipation.

## Respond...

@Rouben Thanks. You are right. After a closer look, I discovered that Q is dependent on Q(r,x,tau) but U is dependent on U(x,tau).

1. The Q in the first pde is evaluated at r = 1. That is why it can be subtracted from U(x,tau).

2. The Q in the BC pde is evaluated at r = 1. That is why it can be subtracted from U(x,tau).

Therefor, the U is dependent on (x,tau) and Q is dependent on (r,x,tau).

Thanks in anticipation...