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Prove some facts about solutions of y′′ + y = 0

  1. Prove that there is exactly one solution of the second-order differential equation

        \[ y'' + y = 0 \]

    whose graph passes through the points (a_1, b_1) and (a_2, b_2) where a_1 - a_2 \neq n \pi, where n \in \mathbb{Z}.

  2. Is part (a) ever true if a_1 - a_2 = n \pi for n \in \mathbb{Z}?
  3. Generalize part (a) for the second-order differential equation

        \[ y'' + ky = 0. \]

    Include the case k = 0.


  1. Proof. The equation y'' + y = 0 is of the form

        \[ y'' + ay' + by = 0 \qquad \text{with} \qquad a = 0, b = 1. \]

    Therefore, using Theorem 8.7 (pages 326-327 of Apostol) we have d = a^2 - 4b = -4 and k = \frac{1}{2} \sqrt{-d} = 1. Since d < 0 we have solutions given by

        \[ y = c_1 \cos x + c_2 \sin x. \]

    The conditions in the problem tell us y(a_1) = b_1 and y(a_2) = b_2. From the first condition we have

        \[ c_1 \cos a_1 + c_2 \sin a_1 = b_1 \quad \implies \quad c_1 = \frac{b_1 - c_2 \sin a_1}{\cos a_1}. \]

    Using this expression for c_1 and the second condition we have

        \begin{align*}  &&c_1 \cos a_2 + c_2 \sin a_2 &= b_2 \\ \implies && \left( \frac{b_1 - c_2 \sin a_1}{\cos a_1} \right) \cos a_2 + c_2 \sin a_2 &= b_2 \\ \implies && b_1 \left( \frac{\cos a_2}{\cos a_1} \right) - c_2 \left( \frac{\sin (a_1 - a_2)}{\cos a_1} \right) &= b_2\\ \implies && \left( \frac{1}{\sin (a_1 - a_2)} \right) \left( b_1 \frac{\cos a_2}{\cos a_1} - b_2 \cos a_1 \right) &= c_2 \end{align*}

    where we use that a_1 - a_2 \neq n \pi so \sin (a_1 - a_2) \neq 0. Thus, c_1 and c_2 are uniquely determined, so the solution is unique. \qquad \blacksquare

  2. No, if a_1 - a_2 = n \pi, then \sin (a_1 - a_2) = 0 and we find the choice of c_2 is arbitrary.
  3. If k =0, then y = c_1 x + c_2 and so the first condition y(a_1) = b_1 implies

        \[ b_1 = c_1 a_1 + c_2 \quad \implies \quad c_2 = b_1 -c_1 a_1. \]

    Then, the second condition y(a_2) = b_2 implies

        \begin{align*}  && b_2 &= c_1 a_2 + c_2 \\  \implies && b_2 &= c_1 a_2 + b_1 - c_1 a_1 \\  \implies && b_2 &= c_1 (a_2 - a_1) + b_1 \\  \implies && c_1 &= \frac{b_2-b_1}{a_2-a_1} & (\text{if } a_2 - a_1 \neq 0) \\  \implies && c_2 &= b_1 - \frac{a_1(b_2-b_1)}{a_2-a_1}. \end{align*}

    Thus, y is uniquely determined.

    If k \neq 0, then y'' + k^2 y = 0 implies d = -4k^2 < 0. Therefore,

        \begin{align*}  && y&= c_1 \cos (kx) + c_2 \sin (kx) \\ \implies && b_1 &= c_1 \cos (ka_1) + c_2 \sin (ka_1) \\ \implies && c_1 &= \frac{b_1 - c_2 \sin (ka_1)}{\cos (ka_1)}, \end{align*}

    and

        \begin{align*}  && b_2 &= c_1 \cos (ka_2) + c_2 \sin (ka_2) \\ \implies && c_2 &= \frac{b_1 \cos (ka_2) - b_2 \cos (ka_1)}{\sin (k (a_1 - a_2))}.  \end{align*}

    Thus, c_1 and c_2 are uniquely determined as long as a_1 - a_2 \neq \frac{n \pi}{k}.

2 comments

  1. Kaitei says:

    Rori, I think this type of exercise is better solved with a matrix and linear equation systems solutions analysis. The solution is much more clean, and much more easy to visualize the necessary conditions to the existence of an unique solution.
    Just let A be the matrix with the coefficients of c1 and c2. If det A != 0, then there is an unique solution of the linear equation system.

    Also, thanks for your amazing work. It is really helpful :)

    • RoRi says:

      Yeah, I totally agree it is better, but at this point in the textbook we don’t know what a matrix is. I tried to limit myself to the tools we have available since that’s presumably what Apostol intended. I do think I mention in a few places that if you know linear algebra then using matrices for this sort of thing is faster/easier/better. In Chapters 6 & 7 of Volume 2 of Apostol (iirc) we’ll do a ton of solving these types of systems using linear algebra techniques.

      Thanks for reading!

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