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    A decaying solution of $y^{(4)}+y=0$ with prescribed initial data

    P-BKF98-2

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    1998
    Status
    draft

    Find a function \(y(x)\) for \(x\ge0\) such that \[ y^{(4)}+y=0, \qquad y(0)=0, \qquad y'(0)=1, \] and \[ \lim_{x\to\infty}y(x)= \lim_{x\to\infty}y'(x)=0. \]

    Solution

    1. The characteristic equation is\[ r^4+1=0, \]with roots\[ e^{i\pi/4}, \quad e^{3i\pi/4}, \quad e^{5i\pi/4}, \quad e^{7i\pi/4}. \]

    The equation \(r^4=-1\) has the four solutions \[ r = \exp\left( \frac{(2k+1)\pi i}{4} \right), \qquad k=0,1,2,3. \]

    2. The roots with negative real part are\[ \frac{-1+i}{\sqrt2} \qquad\text{and}\qquad \frac{-1-i}{\sqrt2}. \]

    These are \[ e^{3i\pi/4} \qquad\text{and}\qquad e^{5i\pi/4}. \] Their real part is \(-1/\sqrt2\), while the other two roots have positive real part.

    3. Every real solution whose value and first derivative tend to zero as\(x\to\infty\) has the form\[ y(x) = e^{-x/\sqrt2} \left( A\cos\frac{x}{\sqrt2} +B\sin\frac{x}{\sqrt2} \right) \]for real constants \(A,B\).

    The two roots from step 2 give the real basis \[ e^{-x/\sqrt2}\cos\frac{x}{\sqrt2}, \qquad e^{-x/\sqrt2}\sin\frac{x}{\sqrt2}. \] Both functions and their first derivatives tend to zero. The other two characteristic roots have positive real part; any nonzero contribution from them grows exponentially along an unbounded sequence and cannot satisfy the two decay conditions. Thus precisely the displayed two-dimensional real space consists of decaying solutions.

    4. The condition\[ y(0)=0 \]forces\[ A=0. \]

    Substitution of \(x=0\) in step 3 gives \(y(0)=A\).

    5. With \(A=0\),\[ y'(0) = \frac{B}{\sqrt2}. \]

    Write \[ a=\frac1{\sqrt2}. \] Then \[ y(x)=B e^{-ax}\sin(ax), \] so \[ y'(x) = Ba e^{-ax} \bigl( \cos(ax)-\sin(ax) \bigr). \] At \(x=0\) this gives \(y'(0)=Ba=B/\sqrt2\).

    6. The condition\[ y'(0)=1 \]forces\[ B=\sqrt2. \]

    Apply step 5.

    7. The required function is\[ \boxed{ y(x) = \sqrt2\,e^{-x/\sqrt2} \sin\frac{x}{\sqrt2} }. \]

    Steps <1>3–<1>6 determine this solution. It solves the differential equation because it is built from characteristic roots of \(r^4+1\), and its value and first derivative tend to zero because they are bounded trigonometric factors times \(e^{-x/\sqrt2}\).

    8. Q.E.D.

    Step 7 verifies all requested conditions.