Eigenvalues and Eigenfunctions — Question 6

Question 6 – Solution

Strategy. Transform the differential equation and its weight together, not just its apparent frequency.

Step 1: Carry out the logarithmic substitution. Since $dt/dx=1/x$, $y'=u'/x$, $xy'=u'$ and $(xy')'=u''/x$. Multiplying the equation by $x$ gives $$\boxed{-u''=\lambda u,\qquad u(0)=u(2)=0.}$$ Also $dx/x=dt$, so the weighted squared norm is exactly $\int_0^2u^2dt$.

Step 2: Classify and normalize. At zero the affine function with two zero endpoints is zero. At a negative parameter, the left-zero hyperbolic sine cannot vanish at the right endpoint. For $\lambda=k^2>0$, $u=A\sin(kt)$ and $\sin(2k)=0$. Thus $$\boxed{\lambda_n=(n\pi/2)^2,\qquad \phi_n(x)=\sin\!\left(\frac{n\pi\log x}{2}\right),\quad n\geq 1.}$$ The eigenspace is the span of $\phi_n$. Its weighted squared norm is one, and $\phi_n'(1)=n\pi/2>0$, fixing both amplitude and sign.

Step 3: Read locations in the original coordinate. For $n=1$, the sine attains its unique maximum one when $t=1$, hence $\boxed{x=e}$, the geometric rather than arithmetic midpoint of the endpoints. For mode $n$, the interior zeros are $\boxed{x_j=e^{2j/n},\ j=1,\ldots,n-1}$. They form a geometric progression: logarithmic spacing becomes multiplicative spacing in $x$.

Step 4: Change the endpoint. For $[1,b]$ the transformed length is $\ell=\log b>0$. Consequently $$\boxed{\lambda_n=(n\pi/\log b)^2,\qquad \phi_n=\sqrt{\frac{2}{\log b}}\sin\!\left(\frac{n\pi\log x}{\log b}\right).}$$ The graph for $b=e^2$ uses the physical coordinate $x$, with the landmark $x=e$.