Solving Boundary Conditions for One-Dimensional Heat Flow Equation

[Tomsk]

I have done most of a question except for the most important part, putting in the boundary conditions, I can't really interpret them.

The question is:

The temperature T of a one dimensional bar whose sides are perfectly insulated obeys the heat flow equation

$$\frac{\partial T}{\partial t} = \kappa\frac{\partial^2 T}{\partial x^2}$$

where kappa is a positive constant.

I managed to solve this, with -c^2 as a separation constant, and I got:

$$T(x,t) = X(x)F(t) = (A_{1} \cos{\frac{cx}{\sqrt{\kappa}}} + A_{2} \sin{\frac{cx}{\sqrt{\kappa}}})e^{-c^2 t}$$

But then the question says,

The bar extends from x=0 to x=L and is perfectly insulated at x=L. At t<0 the temperature is 0 degC throughout the bar and at t=0 the uninsulated end is placed in contact with a heat bath at 100 degC. Show that the temperature of the bar at subsequent times is given by:

$$\frac{T}{100} = 1 - \sum_{n=0}^{\infty} \frac{4}{(2n + 1)\pi} \sin{\left(\frac{(2n+1)\pi x}{2L}\right)} exp{\left(-\kappa\left(\frac{(2n+1)\pi}{2L}\right)^{2} t \right)}$$

And I can't figure out how to get this. I got T(0,t) = 100, therefore A1 e^(-c^2 t) = 100, but that doesn't tell me much. I know I need to sum over c or n at some point, but am I right in thinking you can't sum over c yet because it's a real arbitrary constant, rather than an integer n? That would probably give you the 1 though from n=0. But the problem is with the insulated end x=L, surely if it's insulated it won't lose heat, so the temperature would just go up?

 

[SGT]

The maximum temperature at any point is 100 degrees. The temperature at the insulated end will only reach this limit at t = infinity. So you have T(L,inf) = 100.

 

[Tomsk]

Thanks very much, that looks like it'll sort it. It's funny how its always such a little thing missing... Maybe it's just me.

 

[Tomsk]

Thanks very much, that looks like it'll sort it. It's funny how its always such a little thing missing... Maybe it's just me.

 

[Tomsk]

Thanks very much, that looks like it'll sort it. It's funny how its always such a little thing missing... Maybe it's just me.