How to compute the first eigenvalue of Laplace-convection operator in a line segment? The eigenfunction equation to be solved is 
\begin{cases} -au_{xx}+bu_x=\lambda u, &\text{in } [0, L], \\ u_x(0)=u(L)=0, & \end{cases}
here $a, b, L$ are positive constants, and $\lambda$ is the eigenvalue.
I have tried the eigenfunction $coskx$, or $sin kx$, but not true. So I'd appreciate it if someone could enlighten me! Thanks in advance! 
 A: The solutions come out of looking at the equation
$$
        ar^2-br+\lambda = 0 \\
         r^2-(b/a)r +(\lambda/a) = 0 \\
         (r-b/2a)^2+(\lambda/a-b^2/4a^2)=0 \\
          r = b/2a\pm i\sqrt{\lambda/a-b^2/4a}.
$$
So long as $\lambda \ne b^2/4a$, the roots are distinct, leading to solutions
$$
           u = Ae^{bx/2a}\cos(\sqrt{\lambda/a-b^2/4a}\,x)+Be^{bx/4a}\sin(\sqrt{\lambda/a-b^2/4a}\,x),
$$
where $A$ and $B$ are constants. Then $u(L)=0$ forces
$$
          u = Ce^{bx/2a}\sin(\sqrt{\lambda/a-b^2/4a}(x-L)),
$$
where $C$ is a constant. The condition $u_{x}(0)=0$ gives and equation for $\lambda$:
\begin{align}
         0= u_x(L) =& -Ce^{bL/2a}\cos(\sqrt{\lambda/a-b^2/4a}L)\sqrt{\lambda/a-b^2/4a} \\
     & +C\frac{b}{2a}e^{bL/2a}\sin(\sqrt{\lambda/a-b^2/4a}L)
\end{align}
The resulting transcendtal equation has the form
$$
    \tan(\mu L) = \frac{2a}{b}\mu,\;\;\; \mu = \sqrt{\lambda/a-b^2/4a}
$$
Knowing the solutions $\mu$, the solutions $\lambda$ are determined from
$$
          \lambda/a = b^2/4a+\mu^2 \\
            \lambda = ab^2/4 + a\mu^2.
$$
A: @DisintegratingByParts: As for the other problem, whether my solution is true?
\begin{cases} -au_{xx}+bu_x=\lambda u, &\text{in } [-L, L], \\ u(-L)=u(L)=0, & \end{cases}
By using your approach, we find \begin{equation} r=\frac{b}{2a}\pm i\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}.  \end{equation}
As long as $\lambda>\frac{b^2}{4a}$, the solution can be written in the form:
$$u = Ae^{bx/2a}\cos(\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,x)+Be^{bx/2a}\sin(\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,x).$$
The boundary conditions imply that $$u = e^{bx/2a}\sin(\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,x).$$
Substitution $u(L)=0$ into the above equation yields
\begin{align}
         0= u(L) =&  e^{bL/2a}\sin(\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,L) \\
    0= u(L) = & \sin(\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,L) \\
{n\pi}=&\sqrt{\frac{\lambda}{a}-\frac{b^2}{4a^2}}\,L\\
(\frac{n\pi}{L})^2=&\frac{\lambda}{a}-\frac{b^2}{4a^2}\\
\lambda=&\frac{b^2}{4a}+a(\frac{n\pi}{L})^2, \mbox{ here }n=1,2\cdots
\end{align}
