I've been practicing on Homogenous Linear Equations with constant coefficients. The textbook I have has a problem that says:
Boundary Value Problems. When the values of a solution to a differential equation are specificed at two different points, these conditions are called boundary conditions. The purpose of this exercise is to show that for boundary value problems there is no existence-uniqueness theorem that is analogous to Theorem 1.
Given that every solution to
$$y''+y=0$$
is of the form
$$y(t)=c_1\cos t+c_2\sin t$$
Where $c_1$ and $c_2$ are arbitrary constants, show that
(a) There is a unique solution to the equation above that satisfies the boundary conditions $y(0)=2$ and $y\left(\dfrac{\pi}{2}\right)=0$ .
(b) There is no solution that satisfies $y(0)=2$ and $y(\pi)=0$ .
(c) There are infinitely many solutions that satisfy $y(0)=2$ and $y(\pi)=-2$ .
I just found the derivative of the given $y(t)$ and substituted the values. The questions that I have are.
What is the existence-uniqueness theory in simple words?
and for (c), if there are infinitely many solutions that satisfy the boundary conditions, does that mean that all solutions satisfy these boundary conditions?
Best Answer
(a):
$y(0)=2$ :
$c_1=2$
$\therefore y(t)=2\cos t+c_2\sin t$
$y\left(\dfrac{\pi}{2}\right)=0$ :
$c_2=0$
$\therefore y(t)=2\cos t$ , which has unique solution
(b):
$y(0)=2$ :
$c_1=2$
$\therefore y(t)=2\cos t+c_2\sin t$
$y(\pi)=0$ :
$-2=0$ , which is impossible
$\therefore$ There is no solution
(c):
$y(0)=2$ :
$c_1=2$
$\therefore y(t)=2\cos t+c_2\sin t$
$y(\pi)=-2$ :
$-2=-2$
$\therefore$ There are infinitely many solutions that $y(t)=2\cos t+c_2\sin t$
You will obviously discover that the cases of no solution or infinitely many solutions will only appear when the two positions of the conditions are both making at least one of the linear independent solutions become $0$ .