Initial Value Problem (IVP) - Existence and Uniqueness Proof
To solve the Initial Value Problem:
$$ X'(t) = f(X(t), t), \quad X(t_0) = X_0 $$
Assume that the solution exists, i.e.,
$$ X'(t) = f(X(t), t) $$
Take integral on both sides:
$$ \int_{t_0}^{t} X'(s) \, ds = \int_{t_0}^{t} f(X(s), s) \, ds $$
By Fundamental Theorem of Calculus, the above becomes:
$$ X(t) - X_0 = \int_{t_0}^{t} f(X(s), s) \, ds $$
So:
$$ X(t) = X_0 + \int_{t_0}^{t} f(X(s), s) \, ds \tag{ii} $$
Hence, \(X(t)\) is a solution of IVP (i) iff it is a solution of (ii).
The rest of (ii) again using Leibniz's rule:
$$ F'(t) = \frac{d}{dt} \left( \int_{u(t)}^{v(t)} f(x,t) \, dx \right) \Rightarrow f(x,t) = f(x,t) \cdot \frac{dx}{dt} + \int_{u(t)}^{v(t)} \frac{\partial f(x,t)}{\partial t} \, dt $$
Given:
$$ X(t) = X_0 + \int_{t_0}^{t} f(X(s), s) \, ds $$
Then:
$$ X'(t) = \frac{d}{dt} \left( X_0 + \int_{t_0}^{t} f(X(s), s) \, ds \right) $$
Let:
$$ \frac{\partial}{\partial t} X(t) = 0 \tag{iii} $$
Now solving the integral part:
$$ F'(t) = \frac{d}{dt} \int_{t_0}^{t} f(X(s), s) \, ds $$
$$ F'(x) = f(x, h(x)) L'(x) + f(u, g(x)) g'(x) $$
Let:
\( h(x) = t \Rightarrow h'(x) = 1 \), \( g(x) = t_0 \Rightarrow g'(x) = 0 \)
So:
$$ F'(x) = f(x,t) \cdot 1 - f(x,t_0) \cdot 0 = f(x,t) $$
$$ f'(x) = f(x,t) \tag{iv} $$
And from (iii) and (iv):
$$ X'(t) = 0 + f(X(t), t) \Rightarrow X'(t) = f(X(t), t) \tag{v} $$
Also,
$$ X(t_0) = X_0 + \int_{t_0}^{t_0} f(X(s), s) \, ds = X_0 $$
Therefore, equation (ii) implies (i). This completes the proof.
