Existence-Uniqueness Theorem for Homogeneous First-Order Differential Equation

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Theorem

Let $P \left({x}\right)$ be a continuous function on an open interval $I \subseteq \R$.

Let $a \in I$.

Let $b \in \R$.

Let $f \left({x}\right) = y$ be a function satisfying the differential equation:

$y' + P \left({x}\right) y = 0$

and the initial condition:

$f \left({a}\right) = b$


Then there exists a unique function satisfying these initial conditions on the interval $I$.

That function takes the form:

$f \left({x}\right) = b e^{-A \left({x}\right)}$

where:

$\displaystyle A \left({x}\right) = -\int_a^x P \left({t}\right) \mathrm d t$


Proof

Existence

Differentiating $f \left({x}\right) = b e^{-A \left({x}\right)}$ with respect to $x$:

\(\displaystyle f' \left({x}\right)\) \(=\) \(\displaystyle b e^{-A \left({x}\right)} \cdot \left({-A' \left({x}\right)}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle -b P \left({x}\right) e^{-A \left({x}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle -P \left({x}\right) f \left({x}\right)\)


So thedifferential equation becomes:

$f' \left({x}\right) + P \left({x}\right) f \left({x}\right) = -P \left({x}\right) f \left({x}\right) + P \left({x}\right) f \left({x}\right) = 0$


For the initial condition:

\(\displaystyle f \left({a}\right)\) \(=\) \(\displaystyle b e^{-A \left({a}\right)}\)
\(\displaystyle \) \(=\) \(\displaystyle b e^{-\int_a^a P \left({x}\right) \ \mathrm d x}\)
\(\displaystyle \) \(=\) \(\displaystyle b e^0\)
\(\displaystyle \) \(=\) \(\displaystyle b\)


Thus such a function exists satisfying the conditions.

$\Box$


Uniqueness

Suppose that $f$ is a function satisfying the initial conditions.

Let $g \left({x}\right) = f \left({x}\right) e^{A \left({x}\right)}$.


By the Product Rule:

\(\displaystyle g' \left({x}\right)\) \(=\) \(\displaystyle f' \left({x}\right) e^{A \left({x}\right)} + f \left({x}\right) e^{A \left({x}\right)}\cdot A' \left({x}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle e^{A \left({x}\right)} (f' \left({x}\right) + P \left({x}\right) f \left({x}\right))\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)


So $g \left({x}\right)$ must be constant.

Therefore:

$g \left({x}\right) = g \left({a}\right) = f \left({a}\right) e^{A \left({a}\right)} = f \left({a}\right) = b$


From this, we conclude that:

$f \left({x}\right) = g \left({x}\right) e^{-A \left({x}\right)} = b e^{-A \left({x}\right)}$

$\blacksquare$


Sources