Eigenvalues and Eigenfunctions — Question 7

Question 7 – Solution

Strategy. Remove the first derivative with an invertible exponential change and keep track of the shifted spectral parameter.

Step 1: Transform the operator completely. The derivatives are $y'=e^{ax}(u'+au)$ and $y''=e^{ax}(u''+2au'+a^2u)$. Therefore $$-y''+2ay'=e^{ax}(-u''+a^2u),\qquad \boxed{-u''=(\lambda-a^2)u,\quad u(0)=u(L)=0.}$$ The exponential is never zero, so this is a bijection of solution spaces. The shifted parameter must be positive: zero or negative gives only the zero Dirichlet solution. All eigenvalues and eigenfunctions are thus $$\boxed{\lambda_n=a^2+(n\pi/L)^2,\quad E_n=\operatorname{span}\{e^{ax}\sin(n\pi x/L)\},\quad n\geq 1.}$$

Step 2: Normalize with the correct weight. The factors $e^{-2ax}$ and $e^{2ax}$ cancel in the squared norm. Hence $\boxed{\phi_n=\sqrt{2/L}\,e^{ax}\sin(n\pi x/L)}$ has unit weighted norm and positive initial derivative. The envelope changes heights and the locations of extrema, but its positivity leaves the zeros $jL/n$ unchanged.

Step 3: Locate the displaced maximum. For $y=e^x\sin x$, $y'=e^x(\sin x+\cos x)$ is positive up to $3\pi/4$ and negative afterwards on $(0,\pi)$. Thus $$\boxed{x_{\max}=3\pi/4,\qquad y_{\max}=e^{3\pi/4}/\sqrt 2.}$$ The sine itself peaks at $\pi/2$; multiplying by the increasing envelope shifts the peak without moving either zero.

Step 4: Solve the inverse spectral data. The gap $8-5=3$ equals $3\pi^2/L^2$, so $L=\pi$. Then $a^2+1=5$, giving $\boxed{a=2\text{ or }a=-2}$. Both choices give the entire same spectrum $4+n^2$, yet their exponential envelopes point in opposite directions. The eigenvalues alone cannot recover the sign of the drift parameter.