Heat equation with convection loss - Steady State confusion! The Fourier PDE for a uniform $1D$ rod with convection loss all along the length $L$ is given by:
$$u_t=\alpha u_{xx}-\beta u$$
Where both $\alpha$ and $\beta$ are Real and positive numbers.
Domain: $[0,L]$
Boundary conditions (BCs):
$$u(0,t)=0\text{ and }u_x(L,t)=0$$
Initial condition:
$$u(x,0)=T_0$$

*

*Steady State solution ($u_t=0$):
$$u''(x)-\frac{\beta}{\alpha}u(x)=0$$
If:
$$\lambda^2=\frac{\beta}{\alpha}$$
Then:
$$u(x)=c_1e^{\lambda x}+c_2e^{-\lambda x}$$
The integration constants $c_1$ and $c_2$ are obtained by applying the BCs. The obtained solution is in agreement with published solutions for cooling fins (for example).


*Separation of values:
Ansatz:
$$u(x,t)=X(x)\Theta(t)$$
Separation:
$$X\Theta'=\alpha \Theta X''+\beta X\Theta$$
$$\frac{\Theta'}{\Theta}=\alpha \frac{X''}{X}
+\beta$$
$$\frac{\Theta'}{\Theta}-\beta=\alpha \frac{X''}{X}$$
$$\frac{\Theta'}{\alpha \Theta}-\frac{\beta}{\alpha}= \frac{X''}{X}=-m^2\tag{1}$$
$$\frac{\Theta'}{\alpha \Theta}-\frac{\beta}{\alpha}=-m^2$$
$$\frac{\Theta'}{ \Theta}-\beta=-m^2\alpha$$
$$\frac{\Theta'}{ \Theta}=-m^2\alpha+\beta$$
$$\Theta=\exp[(-m^2\alpha+\beta)t]$$
As the rod's overall temperature decreases with time ($T_0>0$):
$$-m^2\alpha+\beta<0 \Rightarrow -m^2<0$$
From $(1)$:
$$X''(x)+m^2X(x)=0$$
With the BCs:
$$X_n(x)=A_n\sin mx$$
$$m=\frac{n\pi}{2L}\text{ with }n=1,3,5,...$$
So that:
$$u_n(x,t)=A_n\exp[(-m^2\alpha+\beta)t]\sin mx$$
So the steady state solution is:
$$u_n(x,+\infty)=A_n\sin mx$$
which is  very different from the solution found under $1.$
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{k_{n} \equiv {\pars{2n + 1}\pi \over 2L}}$, lets $\ds{\mrm{u}\pars{x,t} =
\sum_{n = 0}^{\infty}a_{n}\pars{t}\sin\pars{k_{n}x}}$ which already satisfies the boundary condition $\ds{\mrm{u}\pars{0,t} = \mrm{u}_{x}\pars{L,t} = 0}$. However, $\ds{\mrm{u}\pars{x,t}}$ must satisfies the above differential equation:
\begin{align}
&\sum_{n = }^{\infty}\dot{a}_{n}\pars{t}\sin\pars{k_{n}x} =
\alpha\sum_{n = 0}^{\infty}a_{n}\pars{t}\pars{-k_{n}^{2}}\sin\pars{k_{n}x} -
\beta\sum_{n = 0}^{\infty}a_{n}\pars{t}\sin\pars{k_{n}x}
\end{align}
Multiply both members by $\ds{\pars{2/L}\sin\pars{k_{n}x}}$ and integrate over $\ds{x \in \pars{0,L}}$ which leads to
\begin{align}
&\dot{a}_{n}\pars{t} + \pars{\alpha k_{n}^{2} + \beta}a_{n}\pars{t} = 0
\implies
a_{n}\pars{t} = a_{n}\pars{0}\expo{-\pars{\alpha k_{n}^{2} + \beta}t}
\end{align}
The solution is reduced to
\begin{align}
\mrm{u}\pars{x,t} & =
\expo{-\beta t}\sum_{n = 0}a_{n}\pars{0}\exp\pars{-\alpha k_{n}^{2}t}
\sin\pars{k_{n}x}
\end{align}
However,
\begin{align}
& T_{0} = \mrm{u}\pars{x,0} =
\sum_{n = 0}a_{n}\pars{0}\sin\pars{k_{n}x}
\end{align}
Multiply both members by $\ds{\pars{2/L}\sin\pars{k_{n}x}}$ and integrate over $\ds{x \in \pars{0,L}}$:
\begin{align}
& {1 \over 2n + 1}\,{4 \over \pi}\,T_{0} =  a_{n}\pars{0}
\end{align}
\begin{align}
\mrm{u}\pars{x,t} & =
\bbx{{4T_{0} \over \pi}\expo{-\beta t}
\sum_{n = 0}{\exp\pars{-\alpha k_{n}^{2}t}\sin\pars{k_{n}x} \over
2n + 1}}
\,,\qquad k_{n} \equiv {\pars{2n + 1}\pi \over 2L} \\ &
\end{align}
A: Consider your initial steady state solution $u(x)=c_1 e^{\lambda x} + c_2 e^{-\lambda x}$.
Substitute $\lambda=im$ to find:
$$u(x)=c_1 e^{\lambda x} + c_2 e^{-\lambda x}=c_1 e^{imx} + c_2 e^{-im x} \\
= c_1(\cos(mx)+i\sin(mx)) + c_2(\cos(-mx) +i\sin(-mx))\\
 = A_m \sin(mx) + B_m\cos(mx)$$
Apply the first boundary condition to find $u(x)=A_m \sin(mx)$, which agrees with the steady state solution you found in 2.
