Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation
Theorem
Let $\map P x$ be a continuous function on an open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
Let $\map f x = y$ be a function satisfying the differential equation:
- $y' + \map P x y = 0$
and the initial condition:
- $\map f a = b$
Then there exists a unique function satisfying these initial conditions on the interval $I$.
That function takes the form:
- $\map f x = b e^{-\map A x}$
where:
- $\ds \map A x = -\int_a^x \map P t \rd t$
Proof
Existence
Differentiating $\map f x = b e^{-\map A x}$ with respect to $x$:
\(\ds \map {f'} x\) | \(=\) | \(\ds b e^{-\map A x} \cdot \paren {-\map {A'} x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -b \map P x e^{-\map A x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\map P x \map f x\) |
So thedifferential equation becomes:
- $\map {f'} x + \map P x \map f x = -\map P x \map f x + \map P x \map f x = 0$
For the initial condition:
\(\ds \map f a\) | \(=\) | \(\ds b e^{-\map A a}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b e^{-\int_a^a \map P x \rd x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b e^0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b\) |
Thus such a function exists satisfying the conditions.
$\Box$
Uniqueness
Suppose that $f$ is a function satisfying the initial conditions.
Let $\map g x = \map f x e^{\map A x}$.
By Product Rule for Derivatives:
\(\ds \map {g'} x\) | \(=\) | \(\ds \map {f'} x e^{\map A x} + \map f x e^{\map A x} \cdot \map {A'} x\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds e^{\map A x} \paren {\map {f'} x + \map P x \map f x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
So $\map g x$ must be constant.
Therefore:
- $\map g x = \map g a = \map f a e^{\map A a} = \map f a = b$
From this, we conclude that:
- $\map f x = \map g x e^{-\map A x} = b e^{-\map A x}$
$\blacksquare$
Sources
- 1967: Tom M. Apostol: Calculus Volume 1: $\S 8.2$