Separation of variables and well posedeness Let $k$ be a real constant. Prove that the equation
$$
u_t = u_{xx} +ku_x
$$
for $x\in [0,1]$ and $t\geq 0$ given with zero dirichlet boundary condition is well posed. The hint is to use separation of variables and Fourier expansion to show the norm of the solution is bounded in finite time horizon. 

I have only seen separation of variables for the wave equation but this seems more complicated to me because of the ODE found for $X$, namely
$$
X'' +kX'+\lambda X =0
$$
which I cannot solve easily given I do not know the discriminant's sign of the characteristic polynomial.
Any help or reference?

EDIT: I add the definition of well posedeness. Consider for all $t \geq 0$ the operator $E(t)$ that sends the initial condition $u_0$ to the function $x \rightarrow u(t,x)$. Then we say the PDE is well posed if for all $T&gt;0$ there exists a constant $C_T$ such that $||E(t)|| \leq C_T $ for $0\leq t \leq T.$ 

We are working on the function space $L^2$ and the norm of an operator is defined as usual by $||E(t)||= \sup_{u\in L^2, u\neq 0} \frac{||E(t)(u)||}{||u||}$ 

In this particular setting we want to show that if we fix $T&gt;0$ then there is a constant $C_T$ such that $||u(t,\cdot)|| \leq C_T||u_0||$ for $0 \leq t \leq T$. where $u_0$ is the intial condition.
Note that 

\begin{align*} \frac{\partial}{\partial t} \|u(\cdot, t)\|^2&amp;=  2 \int_0^1 u u_t \mathrm{d}x \\
&amp;=2 \int_0^1 u (u_{xx} + ku_x) \mathrm{d}x \\
&amp;= -2 \int_0^1 u_x^2 \mathrm{d}x + 2k \int_0^1 
 u u _x\mathrm{d}x \\
 &amp; =-2 \| u_x\|^2 + 2k \int_0^1 
 u u _x\mathrm{d}x\\
&amp;\le -2 \| u_x\|^2 + 2|k| \| u\| \cdot \| u_x\|\\
&amp;\le -2 \| u_x\|^2 + |k| (|k|^{-1} \| u_x\|^2 + |k|\| u\|^2) \\
&amp;= k^2\| u(\cdot , t)\|^2
\end{align*}

Divide $\| u\|^2$ on both sides of the inequality and integrate from $0$ to $t$, we have 

$$ \| u(\cdot, t)\|^2 \le e^{k^2t} \| u_0\|^2.$$
Note that 

\begin{align*} \frac{\partial}{\partial t} \|u(\cdot, t)\|^2&amp;=  2 \int_0^1 u u_t \mathrm{d}x \\
&amp;=2 \int_0^1 u (u_{xx} + ku_x) \mathrm{d}x \\
&amp;= -2 \int_0^1 u_x^2 \mathrm{d}x + 2k \int_0^1 
 u u _x\mathrm{d}x \\
 &amp; =-2 \| u_x\|^2 + 2k \int_0^1 
 u u _x\mathrm{d}x\\
&amp;\le -2 \| u_x\|^2 + 2|k| \| u\| \cdot \| u_x\|\\
&amp;\le -2 \| u_x\|^2 + |k| (|k|^{-1} \| u_x\|^2 + |k|\| u\|^2) \\
&amp;= k^2\| u(\cdot , t)\|^2
\end{align*}

Divide $\| u\|^2$ on both sides of the inequality and integrate from $0$ to $t$, we have 

$$ \| u(\cdot, t)\|^2 \le e^{k^2t} \| u_0\|^2.$$