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: Understanding the Concept:
Each operator here has the form \(L = A(x)\dfrac{d^2}{dx^2} + B(x)\dfrac{d}{dx} + C(x)\). Such an operator can always be written in the divergence, or Sturm-Liouville, form \(L = \dfrac{d}{dx}\Big(p(x)\dfrac{d}{dx}\Big) + q(x)\) exactly when \(B(x)\) equals the derivative of \(A(x)\). An operator in this divergence form is formally self-adjoint under the usual real inner product, because integrating by parts twice to move both derivatives off a test function \(u\) and onto another test function \(w\) brings back the same operator with no leftover first-derivative term. So the whole question reduces to a simple check on each option.

Step 2: Key Formula or Approach:
For \(L = A(x)\dfrac{d^2}{dx^2} + B(x)\dfrac{d}{dx} + C(x)\):
\[ L \text{ is formally self-adjoint} \iff B(x) = A'(x) \]
We check this condition for each of the four operators.

Step 3: Check option (A).
Here \(A(x)=x^2\), so \(A'(x)=2x\), while \(B(x)=3x\). Since \(3x \neq 2x\) in general, the condition fails, so (A) is NOT self-adjoint.

Step 4: Check option (B).
Here \(A(x)=1-x^2\), so \(A'(x)=-2x\), and \(B(x)=-2x\). These match exactly (\(-2x = -2x\)), so (B) IS self-adjoint. This is in fact the Legendre operator, a textbook example of a self-adjoint operator.

Step 5: Check option (C).
Here \(A(x)=3x-4x^3\), so \(A'(x)=3-12x^2\), and \(B(x)=3-12x^2\). These also match exactly, so (C) IS self-adjoint.

Step 6: Check option (D).
Here \(A(x)=x\), so \(A'(x)=1\), while \(B(x)=x^2\). Since \(x^2 \neq 1\) in general, the condition fails, so (D) is NOT self-adjoint.

Step 7: Final Answer:
Only the operators in (B) and (C) satisfy \(B(x)=A'(x)\) and are self-adjoint.\[ \boxed{\text{(B), (C)}} \]