Existence-Uniqueness Theorem for First-Order Differential Equation
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Theorem
Let $P$ and $Q$ be continuous real functions on some open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
There exists a unique function $\map f x = y$ on $I$ that satisfies the linear first order ordinary differential equation:
- $(1): \quad y' + \map P x y = \map Q x$
along with the initial condition:
- $\map f a = b$
This function is:
- $\ds \map f x = be^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$
where $\ds \map A x = \int_a^x \map P t \rd t$
Proof
Existence
Because $P$ and $Q$ are continuous, they are integrable.
Hence we may use the Fundamental Theorem of Calculus.
\(\ds \map {f'} x\) | \(=\) | \(\ds -b \map P x e^{-\map A x} - \map P x e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t + e^{-\map A x} \map Q x e^{\map A x}\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds -\map P x \paren {b e^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t} + \map Q x\) | by factoring | |||||||||||
\(\ds \) | \(=\) | \(\ds -\map P x \map f x + \map Q x\) | by substitution |
Therefore:
- $\map {f'} x + \map P x \map f x = \map Q x$
For the initial condition:
\(\ds \map f a\) | \(=\) | \(\ds b e^{-\map A a} + e^{-\map A a} \int_a^a \map Q t e^{\map A t}dt\) | substituting of $a$ into $\map f x$ | |||||||||||
\(\ds \) | \(=\) | \(\ds b e^0 + e^0 \cdot 0\) | Integral on Zero Interval | |||||||||||
\(\ds \) | \(=\) | \(\ds b\) | simplification |
$\Box$
Uniqueness
Let $f$ be a solution to $(1)$ which satisfies the initial condition.
Let $\map g x = \map f x e^{\map A x}$.
\(\ds \map {g'} x\) | \(=\) | \(\ds \map {f'} x e^{\map A x} + \map P x \map f x e^{\map A x}\) | Product Rule for Derivatives | |||||||||||
\(\ds \) | \(=\) | \(\ds e^{\map A x} \map Q x\) | by Substitution |
Moreover:
- $\map g a = b$
By the Fundamental Theorem of Calculus:
- $\ds \map g x = \int_a^x \map Q t e^{\map A t} \rd t + b$
Furthermore:
- $\map f x = \map g x e^{-\map A x}$
Therefore, we can conclude that:
- $\ds \map f x = b e^{-\map A x} + e^{-\map A x} \int_a^x \map Q t e^{\map A t} \rd t$
$\blacksquare$
Sources
- 1967: Tom M. Apostol: Calculus Volume 1: $\S 8.3$