Steady State Temperature Profile of a Centrally Heated Pipe (or rod)

This question is inspired by a question from 'Robomaze'. The OP now seems to have lost interest in it but I still think it's an interesting issue.

I've cleaned up the original question and made it more concise.

A pipe (or rod) of total length $$2L_2$$ has a central section of length $$2L_1$$ that is heated constantly and uniformly at a rate of $$q\,\mathrm{Wm^{-1}}$$. The non-heated part loses heat through convection (but not radiation)

What is the temperature profile ($$T(x)$$) of the pipe at steady state?

1. Heated section:

The go-to equation here is Fourier's Heat Equation (with thermal load $$q$$):

$$\frac{\partial T}{\partial t}=\alpha \frac{\partial^2 T}{\partial x^2}+q$$ Or: $$T_t=\alpha T_{xx}+q\tag{1}$$ 2. Non-heated sections:

These can be treated as convective heat losing fins, the derivation of which can be found here. The resulting differential equation is (also steady state): $$T''-\frac{Ph}{Ak}(T-T_{\infty})=0\tag{2}$$ Let's make a small transformation of dependent variable: $$T-T_{\infty}=u$$

Because we're looking for steady state, so $$u_t=0$$, $$(1)$$ becomes: $$\alpha u''(x)+q=0\tag{3}$$

$$(3)$$ integrates easily: $$u'=-\frac{q}{\alpha}x+c_1$$ $$u=-\frac{q}{2\alpha}x^2+c_1 x+ c_2$$ Also, because the system's geometry is symmetric about $$x=0$$, that means the maximum of $$u$$ must also be at $$x=0$$, so: $$u'(0)=0 \Rightarrow c_1=0$$ $$\Rightarrow u(x)=-\frac{q}{2\alpha}x^2+ c_2$$ $$(2)$$, after the transformation of $$T$$ becomes:

$$u''-m^2u=0\tag{4}$$ $$\text{where }m^2=\frac{Ph}{Ak}$$ (For the definition of symbols please consult the link above)

$$(4)$$ solves to: $$u(x)=c_3e^{mx}+c_4e^{-mx}$$ As a boundary condition, I choose 'no heat loss at the end-points', so: $$u'(L_2)=0$$

Thus: $$mc_3e^{mL_2}-mc_4e^{-mL_2}=0\tag{5}$$

The temperature profile must be differentiable so that the derivatives in $$L_1$$ must be equal:

$$mc_3e^{mL_1}-mc_4e^{-mL_1}=-\frac{q}{\alpha}L_1\tag{6}$$

$$(5)$$ and $$(6)$$ allow to determine $$c_3$$ and $$c_4$$.

Finally there can only be one value of $$T_1$$, so:

$$-\frac{q}{2\alpha}L_1^2+ c_2=c_3e^{mL_1}+c_4e^{-mL_1}\tag{7}$$

With knowledge of $$c_3$$ and $$c_4$$, $$c_2$$ can then be determined.