Existence-Uniqueness Theorem for First-Order Differential Equation
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Contents
Theorem
Let $P$ and $Q$ be continuous functions on some open interval $I \subseteq \R$.
Let $a \in I$.
Let $b \in \R$.
There is a unique function $f(x)=y$ on $I$ that satisfies the differential equation
- $y' + P(x)y = Q(x)$ along with the initial condition
- $f(a)=b$
This function is
- $\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$ where $\displaystyle A(x) = \int_a^x P(t)dt$
Proof
Existence
If $P$ and $Q$ are continuous, then they are integrable, and we may use the Fundamental Theorem of Calculus on definite integrals involving these functions.
| \(\displaystyle f'(x)\) | \(=\) | \(\displaystyle -bP(x)e^{-A(x)} - P(x)e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt + e^{-A(x)} Q(x) e^{A(x)}\) | by the Product Rule | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle -P(x) (be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)}dt) + Q(x)\) | by factoring | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle -P(x) f(x) + Q(x)\) | by Substitution |
Therefore, $f'(x) + P(x) f(x) = Q(x)$. For the initial condition,
| \(\displaystyle f(a)\) | \(=\) | \(\displaystyle be^{-A(a)} + e^{-A(a)} \int_a^a Q(t) e^{A(t)}dt\) | by Substitution of $a$ into $f(x)$ | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle be^0 + e^0 \cdot 0\) | by Integral on Zero Interval | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle b\) | Simplification using arithmetic |
Uniqueness
Suppose $f$ is a solution to the differential equation and satisfies the initial condition. Let $g(x) = f(x) e^{A(x)}$.
| \(\displaystyle g'(x)\) | \(=\) | \(\displaystyle f'(x) e^{A(x)} + P(x) f(x) e^{A(x)}\) | by the Product Rule | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle e^{A(x)} Q(x)\) | by Substitution |
Moreover, $g(a) = b$. By the Second Fundamental Theorem of Calculus,
- $\displaystyle g(x) = \int_a^x Q(t) e^{A(t)} dt + b$
Furthermore, $\displaystyle f(x) = g(x) e^{-A(x)}$. Therefore, we can conclude that
- $\displaystyle f(x) = be^{-A(x)} + e^{-A(x)}\int_a^x Q(t) e^{A(t)} dt$
$\blacksquare$
Sources
- 1967: Tom M. Apostol: Calculus Volume 1: $\S 8.3$
