Under what condition on the constants $c$ and $c'$ are the boundary conditions $$f(b)=cf(a)$$ and $$f'(b)=c'f'(a)$$ self-adjoint for the operator the operator $$\mathcal{L}f=\frac{d}{dx}\left(p_0(x)\frac{df}{dx}\right)+p_2(x)f$$ on $[a,b]$? ($p_0(x),p_2(x)$ are reals)

I don't know how to impose the boundary conditions to the problem, actually I don't know what to do to find that constants.

Using the Lagrange identity I found $$\left\langle f|\mathcal{L}|f\right\rangle=\langle f|\mathcal{L}^{\dagger}|f\rangle+\left[p_0(x)\left(\frac{df}{dx}f^{*}(x)-f(x)\frac{df^{*}}{dx}\right)\right]_{a}^{b}$$ and this condition requires that $$\left[p_0(x)\left(\frac{df}{dx}f^{*}(x)-f(x)\frac{df^{*}}{dx}\right)\right]_{a}^{b}=0$$ so $$p_0(b)\left(\frac{df}{dx}(b)f^{*}(b)-f(b)\frac{df^{*}}{dx}(b)\right)=p_0(a)\left(\frac{df}{dx}(a)f^{*}(a)-f(a)\frac{df^{*}}{dx}(a)\right)\,.$$

The answer is $${cc'}^{*}=\frac{p_0(a)}{p_0(b)}$$ where did I go wrong?

  • $\begingroup$ Is that the answer provided, or the one you came up with? $\endgroup$ – AlexanderJ93 Dec 6 '18 at 0:33
  • $\begingroup$ This is the provided answer $\endgroup$ – Marcos Pereira Dec 6 '18 at 1:41
  • $\begingroup$ Are you sure they did not assume $c,c'$ to be real? I would expect that to be the case. $\endgroup$ – DisintegratingByParts Dec 6 '18 at 17:42

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