Question:system of PDEs for a 2 wall heat diffusion problem

Question:system of PDEs for a 2 wall heat diffusion problem

Maple

Hi,

I need to solve the heat diffusion equation in a system with two enclosed walls.  The first wall (made of a material called fr4) goes from x=0 o x=d1, and the second wall (made of copper, that is cu)from x=d1 to x=d2.

u(x,t) is my unknown (change of temperature with time and space) for the first wall, and v(x,t) the unknown for the second wall.

So the system of PDEs to solve is:

>PDE:=[ diff(u(x,t),t) = afr4*diff(diff(u(x,t),t) ,

diff(diff(v(x,t),t) = acu*diff(diff(v(x,t),t) ]

As for initial conditions (time), I have that every point of the system was at ambient temperature at t=0:

u(x,0)=Tamb, and  v(x,0)=Tamb

As for the boundary contitions (space) I know that the front surface of the first wall (x=0) suffers from irradiation with intensity Is, and natural convection at the same time, so:

D[1] (u) (0,t) =kconv * ( u(0,t)-Tamb )                            - kirr * Is

The back surface of the second wall (x=d2) suffers only from natural convection, so the boundary would be:

D[1] (v) (d2,t) = - kconv * ( v(d2,t)- Tamb )

As our unknowns u(x,t) and v(x,t) are second derivative in x, so I know I need two boundary conditions in x.  I already gave above one boundary condition for u(x,t) at x=0, and another for v(x,t) at x=d2

It happens that the two walls are enclosed so they share a boundary at x=d1, where the heat flux just goes from one wall to another.  So the equation ruling at that boundary is:

-kfr4 * D[1] (u) (d1,t) = -kcu * D[1] (v) (d1,t)

It thought  with this equation I was given the extra needed boundary condition for u(x,t) and v(x,t)... but it seems that I need someting more... or I do not know what it is wrong

I defined as explained:

> IBC:={ u(x,0)=Tamb,   v(x,0)=Tamb ,

D[1] (u) (0,t) =kconv * ( u(0,t)-Tamb )                            - kirr * Is  ,

D[1] (v) (d2,t) = - kconv * ( v(d2,t)- Tamb ) ,

-kfr4 * D[1](u)(d1,t) = -kcu * D[1] (v)(d1,t)  }

And then execute:

> pds:=pdsolve(PDE,IBC,numeric);

And got the error:

> Error, in (pdsolve/numeric) initial/boundary conditions must be defined at one or two points for each independent variable.

I would appreciate very much if someone could help me to solve this problem. Thank you in advance.

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