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    Entire functions satisfying harmonic-oscillator and double-angle identities

    P-BKF07-2A

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    2007
    Status
    draft

    Let \(f,g\) be entire functions satisfying, for every \(z\in\mathbb C\), \[ f'(z)=g(z),\qquad g'(z)=-f(z),\qquad f(2z)=2f(z)g(z). \] Find all possibilities for \(f\).

    Solution

    1. Every solution of the first two identities has the form\[ f(z)=ae^{iz}+be^{-iz}, \qquad g(z)=aie^{iz}-bie^{-iz} \]for some \(a,b\in\CC\).

    Differentiating \(f'=g\) and using \(g'=-f\) gives \[ f''=-f. \] The entire solutions of this constant-coefficient differential equation are \[ f(z)=ae^{iz}+be^{-iz}. \] Since \(g=f'\), the displayed formula for \(g\) follows.

    2. The identity\[ f(2z)=2f(z)g(z) \]is equivalent to\[ (a-2ia^2)e^{2iz}+(b+2ib^2)e^{-2iz}=0 \]for every \(z\in\CC\).

    Using step 1, \[ f(2z)=ae^{2iz}+be^{-2iz}, \] while \[ \begin{aligned} 2f(z)g(z) &= 2(ae^{iz}+be^{-iz})(aie^{iz}-bie^{-iz}) \\ &= 2ia^2e^{2iz}-2ib^2e^{-2iz}, \end{aligned} \] because the two mixed terms cancel. Moving the right-hand side to the left gives the claim.

    3. The identity in step 2 holds for all \(z\) if and only if\[ a-2ia^2=0 \qquad\text{and}\qquad b+2ib^2=0. \]

    Multiplying the identity in step 2 by \(e^{2iz}\) gives \[ (a-2ia^2)e^{4iz}+(b+2ib^2)=0 \] for every \(z\). Since \(e^{4iz}\) is nonconstant, an identity of the form \(Ce^{4iz}+D=0\) for all \(z\) forces \(C=D=0\). The converse is immediate.

    4. The possible coefficients are\[ a\in\left\{0,-\frac{i}{2}\right\}, \qquad b\in\left\{0,\frac{i}{2}\right\}. \]

    The equations from step 3 factor as \[ a(1-2ia)=0, \qquad b(1+2ib)=0. \] Solving them gives exactly the displayed possibilities.

    5. Hence the complete list of possibilities for \(f\) is\[ \boxed{ 0,\qquad -\frac{i}{2}e^{iz},\qquad \frac{i}{2}e^{-iz},\qquad \sin z }. \]

    Substituting the four pairs \((a,b)\) from step 4 into the formula for \(f\) in step 1 gives the first three functions and \[ -\frac{i}{2}e^{iz}+\frac{i}{2}e^{-iz} = \frac{e^{iz}-e^{-iz}}{2i} = \sin z. \] Conversely, step 3 shows that every such pair satisfies the double-angle identity, while step 1 already gives the first two differential identities.

    6. Q.E.D.

    Step 5 gives all and only the possibilities.