PDE solving with seperating of variables and SL-problem I am stuck at the beginning of an excercise of PDE. The question is the following:
A thin bar with length L, so that x=0 --> x=L. The bar is totally insulated and has a temperature of $100°C$. The temperature of the environment is $0°C$. At t=0 the insulation at the point $x=L$ is taken away. At this moment, the heat exchange with the environment is happening according to the following equation:
$\frac{\delta u}{\delta x}(L,t)=-hu(L,t)$
Now I started out with the 1 dimensional heat transfer equation. And I wrote the solution as:
$u(x,t)=X(x)T(t)$
$T'(t)=a^2 \lambda T(t)$
$X''(x)=\lambda X(x)$
I sort of tried to find good boundary conditions to spot the Sturm-Liouville problem and start to solve this.
$u'(0,t) = 0$
$u(L,0) = 100$
$u(0,0) = 100$ 
Now I am a little bit stuck, and it has something to do with the law of heat transfer that is given at the beginning.
I want to solve it myself, but could someone give me some starting hints, or a beginning where I can build on?
Thank you very much
Thomas
EDIT:

 A: I think you want the heat equation with the following BC's:
$$u_{xx} = a u_t$$
$$u_x(0,t) = 0$$
$$u_x(L,t)+h u(L,t) = 0$$
$$u(x,0) = u_0=100$$
Separate variables as $u = X(x)T(t)$ to get
$$X'' +\lambda^2 X = 0$$
$$a T' + \lambda^2 T = 0$$
where $-\lambda^2$ is the separation constant.  The $x$ equation yields
$$X(x) = A \cos{\lambda x} + B \sin{\lambda x} $$
$$X'(x) = -A \lambda \sin{\lambda x} + B \lambda \cos{\lambda x}$$
$$X'(0) = 0 \implies B = 0$$
$$X'(L) + h X(L) = 0 \implies \lambda \tan{\lambda L} = h $$
To this last equation, there will be an infinite sequence of solutions $\lambda = \lambda_n$.  Thus, there will be an infinite sequence of solutions $X_n(x) = A_n \cos{\lambda_n x} $,  with $A_n$ yet to be determined.  
The $t$ equation yields
$$T_n(t) = C_n e^{-\lambda_n^2 t/a}$$
with $C_n$ yet to be determined.  Thus, the full solution is a linear combination of these solutions, i.e., a sum over $n$:
$$u(x,t) = \sum_{k=1}^{\infty} D_n e^{-\lambda_n^2 t/a} \cos{\lambda_n x} $$
where $D_n$ is given by
$$\frac{\displaystyle \int_0^L dx\,  u(x,0) \cos{\lambda_n x}}{\displaystyle \int_0^L dx\,  \cos^2{\lambda_n x}} =  \frac{2 u_0 \sin{\lambda_n L}}{\lambda_n L + \sin{\lambda_n L} \cos{\lambda_n L}}$$
ADDENDUM
Here is a plot of the solution for $t \in \{0,0.01,0.1,0.2,0.5,1.0\}$ for $a=1$, $L=1$, and $h=1$.  The plot was generated in Mathematica.  I used $100$ terms in the sum.

