# Error message when trying to solve 2 coupled PDEs using NDSolve

I'm currently facing a problem when trying to solve 2 coupled PDEs. I get the warning messages:

NDSolve::fembdcc: Cross-coupling of dependent variables in DirichletCondition[IPF==IPB/10,z==5.] is not supported in this version.

NDSolve::fembdcc: Cross-coupling of dependent variables in DirichletCondition[IPF==IPB/10,z==5.] is not supported in this version.

Here's the code I'm using:

 FirEqn = -D[IPF[z, t], z] - D[IPF[z, t], t]  - IPF[z, t]

SecEqn = -D[IPB[z, t], z] - D[IPB[z, t], t]  - IPB[z, t]

solIntEqn =
NDSolve[{FirEqn == 0, SecEqn == 0,
IPF[0, t] == 4 Exp[-(t - 1)^2],
IPB[0, t] == 4 Exp[-(t - 1)^2]/RICP,
IPF[5, t] == IPB[5, t]/ROCP}, {IPF, IPB}, {z, 0, 5}, {t, 0, 10}];


I know the issue is to do when I specify the boundary condition for z == 5, but that's one of the requirements I need. All help is appreciated! Cheers.

BTW, I'm using version 11.2

Also, ROCP = RICP = 1

• What are RICP and ROCP? It's always good to give all details for the equations as otherwise it's going to be guesswork. – user21 Jan 17 '20 at 6:59
• Sorry I forgot to mention. They are just variables that I’ve set values for. – Ibis_prod1gy Jan 17 '20 at 7:01
• Edit you post to include actual values for those variables. Be specific. That will make your question a much better post and you will get upvotes for it. – user21 Jan 17 '20 at 7:02
• Thank you for the feedback. Sorry for not including the values from before, I will be sure to be more thoughtful in the future. Cheers! – Ibis_prod1gy Jan 17 '20 at 7:09
• no worries. It's always good to tell new users how to write good posts, then this site can be very useful for you. Happy NDSolve-ing. – user21 Jan 17 '20 at 7:13

You need to specify initial conditions:

FirEqn = -D[IPF[z, t], z] - D[IPF[z, t], t] - IPF[z, t]

SecEqn = -D[IPB[z, t], z] - D[IPB[z, t], t] - IPB[z, t]

(* a wild guess *)
ROCP = 1;
RICP = 1;

solIntEqn = NDSolve[{
FirEqn == 0, SecEqn == 0,
IPF[0, t] == 4 Exp[-(t - 1)^2],
IPF[5, t] == IPB[5, t]/ROCP,
IPB[0, t] == 4 Exp[-(t - 1)^2]/RICP,
IPB[z, 0] == 4/(E*RICP),
IPF[z, 0] == 4/E
}, {IPF, IPB}, {z, 0, 5}, {t, 0, 10}]

• Thank you very much. I just tested it and it's exactly what I needed. – Ibis_prod1gy Jan 17 '20 at 9:27
• @Ibis_prod1gy, great. Not necessarily in this case, but it is generally a good idea to wait a little before you accept an answer. Not everyone has time immediately to look at post but still might have good ideas for a solution. Thanks for the accept though ;-) – user21 Jan 17 '20 at 10:09
• Ah, I’ll know better for the future. Cheers for the feedback! – Ibis_prod1gy Jan 17 '20 at 11:41