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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!

2 Answers 2

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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 $$ \lambda/a = b^2/4a+\mu^2 \\ \lambda = ab^2/4 + a\mu^2. $$

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    Thank TrialAndError , but your transcendtal equation may be $tan(\mu L)=\mu (2a/b)$ – 2017-04-25
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    @CanrongTian : Yes, I see I dropped the $2a/b$ factor and the $L$. I'll add those back in. – 2017-04-25
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    Thank TrialAndError , I get it. – 2017-04-27
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@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}