## Heat equation initial/boundary conditions...

What would be the IBC conditions for the 1D heat equation describing the temperature distribution along a circular rod with a heat source at its base?

## Why does pdsolve behave this way?...

Consider a standard initial/boundary value problem for the heat equation on the interval x ∈ [0,1]:

restart;
pde := diff(u(x,t),t) = diff(u(x,t),x,x);
ic := u(x,0) = f(x);
bc := u(0,t)=0,  u(1,t)=0;

Then
pdsolve({pde, ic, bc});
produces the expected Fourier series solution.

However, if we change the interval to x ∈ [-1,1], as in:
bc := u(-1,t)=0,  u(1,t)=0;
pdsolve({pde, ic, bc});

then Maple fails to return a solution.  Why?

## finite difference and 2 d plot ...

Hello

I would like to solve the two dimensional heat equation in the square [-1,1]^2  using finite difference.

The following code  does not gives me the right answer.

I appreciate any help

heat2dequation.mw

## on pdsolve 2017, heat PDE, homogenous neumann B.C...

I am trying Maple 2017 pdsolve for heat PDE in 1D. It seems Maple can solve now  heat PDE with homogeneous dirichlet boundary conditions (good). But when I set the boundary conditions to homogeneous neumann B.C. instead, I get an answer when this B.C. is prescribed to the left side. When this B.C. is on the right side, I get an error. Which is strange.

I am newbie in Maple, so may be I am doing something wrong in the syntax?  In addition, the answer I get when homogeneous neumann B.C. is on the left side, does not match my hand solution, which I know is correct. I'll show this below.

First, here is the case where it works. homogeneous dirichlet boundary conditions on both sides:

```restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x\$2);
bc:=u(0,t)=0,u(L,t)=0;
sol:=pdsolve([pde,bc]) assuming 0<L:
```

This answer is correct. Now when setting the right side to homogeneous neumann B.C. I get an error

```restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x\$2);
bc:=u(0,t)=0,D[1](u)(L,t)=0;
pdsolve([pde,bc]) assuming 0<L;
```

I think may be it does not like `L` there in the B.C. But how else to tell it this B.C.? The above is the only syntax I know. And finally, when using homogeneous neumann B.C. on the left side, I get this result

```restart;
pde:=diff(u(x,t),t)=k*diff(u(x,t),x\$2);
bc:=D[1](u)(0,t)=0,u(L,t)=0;
pdsolve([pde,bc]) assuming 0<L;
```

The correct answer for this B.C. is

The answer should be series solution as well with eigenvalues. I think if I expand Maple solution in series may be I will get it to match my hand solution. I need to look at this more later.

my question is: Why do I get an error when homogeneous neumann B.C. is on the right side but not on the left side?

I suspect I am not entering the B.C. correctly? If so, How does one enter homogeneous neumann B.C. for this 1D heat PDE?

## How to solve this PDE problem?...

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hai All,

i have problem to soilve pde in the maple

can anyone suggest / idea to solve in maple?

## Mathematical model of time-variant temperature dis...

My task is to develop a mathematical model of time-variant temperature
distribution in a bar with Maple.
The bar is made of aluminum. The length is 202 mm and the diameter
is 8mm.
Heat is supplied from the kitchen lighter where the flame burns on butane.
The height of the flame is about 4 mm.
Your model should be able to answer how long the time it takes to reach
a certain value of temperature at the distance of 10 cm from the heat source.

anyone can help to answer this? im totally new to maple.. hopefully some1 can help me to answer this..

## pdsolve not returning anything...

Trying to solve the 1-dimensional heat equation with maple with constant boundary temperatures:

restart;

with(PDETools):
U := diff_table(u(x,t)):
pde := U[t]=U[x,x];
bc := u(0, t) =0, u(1, t) = 1, u(x,0)=x;
pdsolve([pde,bc]);

The solution of this equation is u(x,t)=x , but pdsolve(...) does not return anything at all! What is going wrong? Is it too hard PDE for maple? And if it is too hard, where can be found the types of equations, which are too hard and not too hard? Thank you.

## Heat Equation with Robin Boundary Condition and In...

I am trying to get a solution to the heat equation with multiple boundary conditions.

Most of them work but I am having trouble with two things: a Robin boundary condition and initial conditions.

First, here are my equations that work:

returns a solution (actually two including u(x,y,z,t)=0).

However, when I try to add:

or

I no longer get a solution.

Any guidance would be appreciated.

Regards.

I have uploaded a worksheet with the equations...

## boundary conditions are not been utilized need hel...

hello friends am working on 2d heat equation and i have writen maple code for it but its not running because my scheme is having fictitious points which are removed by using boundary conditions but in this code which i have written the line in bold is not using boundary conditions due to which my fictitious points are still in programe and its not working am attaching my worksheet

trytimg.mw

please let me know where am wrong thank you...

## Heat transfer with phase change and non homogeneou...

Hi,

I'm currently working on the modelling of a thermodynamic process.

Briefly, I cool down a solution (water + polymer) from -5°C to -15°C to induce a phase separation. At the end (and after removing of the water by lyophilisation) I obtain a porous sponge like material.

The process uses a home made cooling system which can be described like this:

- A Peltier module

- An aluminium layer recovered by teflon (And also a layer of ethanol)

## Heat equation with time dependant boudary conditio...

Hi,

I'm currently working on chemical process thermal exchange and particularly on the solving of the heat equation using a time dependant boundary condition.

Briefly, the process consists in two layers of different materials (M1 and M2, thickness L1 and L2). The bottom part of the material M1 (z=0) is cooled down from Ti to Tf with the function T(0,t)=Ti-R*t (R is the cooling rate in °C.min-1) until T(0,t)=Tf. Here the equilibrium is reached in t=(Tf-Ti...

## heat equation doesn't like piecewise condition...

Heat equation using piecewise conditions in Maple produce (unwanted?) oscillations at the transition.  Not sure if this is normal or a side effect of using a piecewise boundary condition in the equation?

I came across a slide presentation for the heat equation having Neuman conditions using a piecewise boundary condition and I thought I would apply the example to Maple.  The piecewise nature of the boundary condition in Maple causes oscillations at the transition...

## Modelling the heat equation...

I would like if someone could help me with an example of the heat equation, this is lengthy so please bare with me.  When trying to recall your studies it is quite difficult to re-absorb all the different nomenclatures used by different people and put them into something you understand.  Some people use "c" for specific heat, some people use "s", some people use "a" for thermal diffusivity and some people use "k" which is also used for thermal conductivity. ...

## Heat equation understanding...

Okay what am I doing wrong here?  It's an examination of the simple 1-d heat diffusion equation in a rod with homogenous boundary conditions.  I'm looking for the temperature distribution over a thin rod with unit length 1.  Both ends are held at 0.  And the initial temperature distribution across the rod I have set equal to x, with the diffusivity k=1/10.

So first set up the 1-d homogeneous heat equation
he:=diff(u(x,t),t)=k*diff(u(x,t),t,t):