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

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 $n \in \mathbb{Z}$ .
Is part (a) ever true if $a_1 - a_2 = n \pi$ for $n \in \mathbb{Z}$ ?
Generalize part (a) for the second-order differential equation
$y'' + ky = 0.$

Include the case $k = 0$ .

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 $. \qquad \blacksquare$
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.
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 $a_1 - a_2 \neq \frac{n \pi}{k}$ .

7 comments

June 10, 2020 at 9:25 am

Anonymous says:

In a) it seems you forgot to multiply $b \frac{\cos(a_2)}{\cos(a_1)$ by $cos(a_1)$ at the end.

- June 10, 2020 at 9:26 am
  
  Anonymous says:
  
  Yeah forgot the closing bracket. b\frac{\cos(a_2)}{\cos(a_1)}. Sorry for spam.
  
October 1, 2019 at 2:16 pm

Artem says:

I think there is a mistake in Apostol in (b). Actually, there are 2 distinct solutions. In case n = 1 mod 2, then we can write $sin(a_2) = sin(a_1 + n\pi) = -sin(a_1)$ , and we get a system where you simply need to set $b_1 = -b_2$ , in the case n = 1 mod 2. If you do the same with n = 2 mod 2, then setting $b_1 = b_2$ you get another solution.

- July 20, 2022 at 2:14 am
  
  S says:
  
  I suppose the question asks in case the bs of the points are arbitrary.
  
- July 20, 2022 at 3:11 am
  
  S says:
  
  Plus, as RoRi wrote, the solution, if exists in case bs satisfy the needed condition, is not unique.
  
April 18, 2016 at 7:00 pm

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 :)

- April 18, 2016 at 7:08 pm
  
  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!

Stumbling Robot

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

Related

7 comments

Point out an error, ask a question, offer an alternative solution (to use Latex type [latexpage] at the top of your comment): Cancel reply