Question

Which of the following operators is/are self-adjoint?

• A. \(x^2\dfrac{d^2}{dx^2} + 3x\dfrac{d}{dx} + x^2\)
• B. \((1-x^2)\dfrac{d^2}{dx^2} - 2x\dfrac{d}{dx} + 3x\)
• C. \((3x-4x^3)\dfrac{d^2}{dx^2} + (3-12x^2)\dfrac{d}{dx} + 12\)
• D. \(x\dfrac{d^2}{dx^2} + x^2\dfrac{d}{dx} + \dfrac{5x}{3}\)

Hint:

For \(L = A(x)D^2 + B(x)D + C(x)\), the operator is formally self-adjoint exactly when \(B(x) = A'(x)\), which lets it be written as \(\dfrac{d}{dx}\big(A(x)\dfrac{d}{dx}\big) + C(x)\). Check this coefficient match for each option.

Answer 1

The Correct Option is B, C

Step 1: Derive the self-adjointness condition from scratch using integration by parts.
For $L = A(x)D^2 + B(x)D + C(x)$ (with $D=d/dx$) acting on functions vanishing at the boundary points, the formal adjoint is found by moving derivatives off one function $u$ onto another $w$ in $\int w\,(Lu)\,dx$. Integrating $\int w A u''\,dx$ by parts twice gives $\int (Aw)'' u\,dx$ (boundary terms drop), and integrating $\int wBu'\,dx$ by parts once gives $-\int (Bw)' u\,dx$. Collecting terms, the adjoint operator turns out to be $L^\dagger w = (Aw)'' - (Bw)' + Cw = Aw'' + (2A'-B)w' + (A''-B'+C)w$.

Step 2: Compare $L^\dagger$ with $L$ to get the condition.
$L$ itself is $Aw'' + Bw' + Cw$. For $L^\dagger = L$, the coefficients of $w'$ must match: $2A'-B = B$, which gives $B = A'$. (The coefficients of $w''$ already match automatically since both are $A$.) So self-adjointness is equivalent to $B(x)=A'(x)$, the same condition used in the other method, but here it comes directly out of the integration-by-parts bookkeeping instead of being quoted.

Step 3: Apply the condition option by option, starting from (D).
For (D): $A=x \Rightarrow A'=1$, but $B=x^2$. Since $x^2 \ne 1$ except at isolated points, this fails everywhere except a single value of $x$, so (D) is NOT self-adjoint.

Step 4: Check (A).
$A = x^2 \Rightarrow A' = 2x$, while $B = 3x$. The ratio $B/A' = 3/2 \ne 1$ for all $x\ne 0$, so (A) is NOT self-adjoint.

Step 5: Check (C).
$A = 3x - 4x^3 \Rightarrow A' = 3 - 12x^2$, and directly reading off $B = 3-12x^2$ from the given operator shows $B$ and $A'$ are the identical polynomial, so (C) IS self-adjoint.

Step 6: Check (B).
$A = 1-x^2 \Rightarrow A' = -2x$, and the operator gives $B = -2x$ directly, so $B=A'$ again holds exactly, and (B) IS self-adjoint. This is the well-known Legendre differential operator.

Final Answer:
Working the adjoint out from first principles confirms the same pair: (B) and (C) are self-adjoint, while (A) and (D) are not.\[ \boxed{\text{(B), (C)}} \]