Boundary Value Problem | Sahithyan's Notes

Also known as BVP. A differential equation with conditions specified at 2 distinct points, called boundary conditions.

Standard Form

y'' = f(x, y, y'), \quad a \leq x \leq b, \quad y(a) = \alpha,\ y(b) = \beta

Existence and Uniqueness

The BVP has a unique solution on $[a,b]$ iff

$f$ , $f_y$ , and $f_{y'}$ are continuous on $D = \{(x,y,y') \mid a \leq x \leq b,\ y, y' \in \mathbb{R}\}$
$f_y(x,y,y') > 0$ for all $(x,y,y') \in D$
A constant $M > 0$ exists with $|f_{y'}(x,y,y')| \leq M$ for all $(x,y,y') \in D$

Worked Example

Determine whether the BVP

y'' = y + \sin(y'), \quad 0 \leq x \leq 1, \quad y(0) = 0,\ y(1) = 0

has a unique solution.

Solution Reveal Hide

Write the equation as $y'' = f(x,y,y')$ with

f(x,y,y') = y + \sin(y')

The domain is

D = \{(x,y,y') \mid 0 \leq x \leq 1,\ y,y' \in \mathbb{R}\}

The first condition requires $f$ , $f_y$ , and $f_{y'}$ to be continuous on $D$ . Calculate

f_y(x,y,y') = 1, \qquad f_{y'}(x,y,y') = \cos(y')

The function $f(x,y,y') = y + \sin(y')$ is continuous because it is the sum of the continuous function $y$ and the continuous function $\sin(y')$ . The derivative $f_y = 1$ is constant, so it is continuous. The derivative $f_{y'} = \cos(y')$ is continuous. Therefore, the first condition holds.

The second condition requires $f_y(x,y,y') > 0$ throughout $D$ . Since

f_y(x,y,y') = 1 > 0

for every $(x,y,y') \in D$ , the second condition holds.

The third condition requires a constant $M > 0$ such that $|f_{y'}(x,y,y')| \leq M$ throughout $D$ . Since

|f_{y'}(x,y,y')| = |\cos(y')| \leq 1

for every $(x,y,y') \in D$ , choose $M = 1$ . The third condition holds.

All 3 conditions hold. Therefore, the BVP has a unique solution on $[0,1]$ .