📂Odinary Differential Equations

Existence and Uniqueness of Solutions for Initial Value Problems of First-Order Ordinary Differential Equations

Theorem¹

Let $E$ be open in $\mathbb{R}^{n}$ and given the following initial value problem concerning $f \in C^{1} (E)$ and $\phi_{0} \in E$.

$$\begin{cases} \dot{ \phi } = \mathbf{f} ( \phi ) \\ \phi (0) = \phi_{0} \end{cases}$$

Then, there exists a unique solution $\phi (t)$ to the given initial value problem in some interval $[-h,h] \subset \mathbb{R}$.

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• $C^{1}$ is a set of functions with continuous derivatives.

Proof

Strategy: Show existence first, then uniqueness. The ball in Euclidean space $\mathbb{R}^{n}$ is denoted by $$\begin{align*} B \left( \mathbf{x}_{0} ; d \right) :=& \left\{ \mathbf{x} \in \mathbb{R}^{n} : | \mathbf{x}_{0} - \mathbf{x} | < d \right\} \\ B \left[ \mathbf{x}_{0} ; d \right] :=& \left\{ \mathbf{x} \in \mathbb{R}^{n} : | \mathbf{x}_{0} - \mathbf{x} | \le d \right\} \end{align*}$$ It is obviously permissible to use the Lipschitz condition in place of $f \in C^{1} (E)$ for a stronger condition.

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Part 1. $f$ is locally Lipschitz

Locally Lipschitz condition: If $f \in C^{1}(E)$, then $f$ is locally Lipschitz in $E$.

Assuming $\dfrac{\partial f}{\partial y}$ is continuous, by the theorem, $f$ is locally Lipschitz. Hence, according to the definition of locally Lipschitz, for all $\mathbf{x} , \mathbf{y} \in B \left( \phi_{0} ; \varepsilon \right) \subset E$, there exists a $\varepsilon, K > 0$ that satisfies the following equation.

$$\left| f(t,y_{1}) - f(t,y_{2}) \right| \le K \left| y_{1} - y_{2} \right|$$

Relation between continuity and compactness: Let $X$ be a compact metric space, $Y$ a metric space, and $f:X\to Y$ continuous. Then, $f(X)$ is compact.

Since $f$ is continuous, it is bounded on the compact set $B : = B \left[ \phi_{0} ; {{ \varepsilon } \over {2}} \right]$, and $\displaystyle M : = \sup_{ \mathbf{x} \in B } | f ( \mathbf{x} ) |$ can be obtained.

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Part 2. Picard Method

Given an initial value problem with $E$ open in $\mathbb{R}^{n}$ and concerning $f \in C^{1} (E)$

$$\begin{cases} \dot{ \phi } = f ( \phi ) \\ \phi (0) = \phi_{0} \end{cases}$$ Defining the sequence of functions $\left\{ \mathbf{u}_{k} (t) \right\} _{ k =0}^{ \infty }$ as

$$\begin{cases} \mathbf{u}_{0} (t) = \phi_{0} \\ \displaystyle \mathbf{u}_{k+1} (t) = \phi_{0} + \int_{0}^{t} f \left( \mathbf{u}_{k} (s) \right) ds \end{cases}$$

then the continuous function $\displaystyle \mathbf{u} (t) := \lim_{k \to \infty} \mathbf{u}_{k} (t)$ is the solution to the given initial value problem.

Assume that the continuous function $\mathbf{u}_{k} (t)$ is defined as

$$\begin{cases} \mathbf{u}_{0} (t) = \phi_{0} \\ \displaystyle \mathbf{u}_{k+1} (t) = \phi_{0} + \int_{0}^{t} f \left( \mathbf{u}_{k} (s) \right) ds \end{cases}$$

$\mathbf{u}_{k}$ and $f$ are continuous in $[-h,h]$, so $( f \circ \mathbf{u}_{k} )$ is also continuous.

Fundamental Theorem of Calculus: If the function $f$ is continuous on the closed interval $[a,b]$, then the function $\displaystyle F(x) = \int_{a}^{x} f(t) dt$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $${{dF(x)} \over {dx}} = f(x)$$

Then, by the fundamental theorem of calculus, $\displaystyle \mathbf{u}_{k+1} (t) = \phi_{0} + \int_{0}^{t} f \left( \mathbf{u}_{k} (s) \right) ds$ is also continuous in $[-h, h]$. With a slight rearrangement and setting up an inequality, for all $t \in [-h, h]$,

$$\begin{align*} | \mathbf{u}_{k+1} (t) - \phi_{0} | \le & \int_{0}^{t} \left| f \left( \mathbf{u}_{k} (s) \right) \right| ds \\ \le & \int_{0}^{h} \left| f \left( \mathbf{u}_{k} (s) \right) \right| ds \\ \le & \int_{0}^{h} M ds \\ \le & Mh \end{align*}$$

In other words, by choosing $\displaystyle h \in \left( 0 , {{ \varepsilon } \over { 2M }} \right]$, $\mathbf{u}_{k} (t)$ can be defined as a continuous function for all $t \in [-h,h]$ and $k = 1,2,3 \cdots$.

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Part 3. Cauchy Sequence $\left\{ \mathbf{u}_{k} \right\}_{k=0}^{\infty}$

We will calculate the supremum for $t \in [-h , h ]$ over $| \mathbb{ u }_{ j + 1 } - \mathbb{ u }_{ j } |$.

• Case 1. $j = 1$
  $$\begin{align*} | \mathbb{ u }_{ 2 } ( t ) - \mathbb{ u }_{ 1 } ( t ) | \le & \int_{0}^{t} \left| f \left( \mathbf{u}_{1} (s) \right) - f \left( \mathbf{u}_{0} (s) \right) \right| ds \\ \le & K \int_{0}^{t} | \mathbb{ u }_{ 1 } ( s ) - \mathbb{ u }_{ 0 } ( s ) | ds \\ \le & K h \sup_{ t \in [ - h , h ] } | \mathbb{ u }_{ 1 } ( t ) - \phi_{ 0 } | \\ \le & {{ K h \varepsilon } \over { 2 }} \end{align*}$$
• Case 2. $j > 1$
  $$\begin{align*} | \mathbb{ u }_{ j+1 } ( t ) - \mathbb{ u }_{ j } ( t ) | \le & \int_{0}^{t} \left| f \left( \mathbf{u}_{ j } (s) \right) - f \left( \mathbf{u}_{ j - 1 } (s) \right) \right| ds \\ \le & K \int_{0}^{t} | \mathbb{ u }_{ j } ( s ) - \mathbb{ u }_{ j - 1 } ( s ) | ds \\ \le & K h \sup_{ t \in [ - h , h ] } | \mathbb{ u }_{ j } ( t ) - \mathbf{u}_{j-1} ( t ) \end{align*}$$

Recursively, by Case 1.

$$| \mathbb{ u }_{ j+1 } ( t ) - \mathbb{ u }_{ j } ( t ) | \le {{ (Kh)^{j} \varepsilon } \over {2}}$$

Let $m > k > N$ and $\displaystyle h \in \left( 0 , {{1} \over { K }} \right)$, and assume $c := K h$, then

$$\begin{align*} | \mathbb{ u }_{ m } ( t ) - \mathbb{ u }_{ k } ( t ) | \le & \sum_{j = k}^{m-1} | \mathbf{u}_{j+1} (t) - \mathbf{u}_{j} (t) | \\ \le & \sum_{j = N}^{ \infty } | \mathbf{u}_{j+1} (t) - \mathbf{u}_{j} (t) | \\ \le & \sum_{j = N}^{ \infty } {{ ( K h )^{ j } \varepsilon } \over {2}} = {{ c^{N} } \over { 1 - c}} {{ \varepsilon } \over {2}} \end{align*}$$

$|c| < 1$, so when $N \to \infty$, $\displaystyle {{ c^{N} } \over { 1 - c}} {{ \varepsilon } \over {2}}$ converges to $0$. In other words, for all $\varepsilon > 0$,

$$m , k \ge N \implies \| \mathbf{u}_{m} - \mathbf{u}_{k} \| = \sup_{ t \in [-h , h ] } | \mathbf{u}_{m} ( t ) - \mathbf{u}_{k} ( t ) | < \varepsilon$$

there exists a $N$ that satisfies it. This means $\left\{ \mathbf{u}_{k} \right\}_{k=0}^{\infty}$ is a Cauchy sequence in $C [ - h , h ]$. (Of course, the selected $h$ must satisfy $\displaystyle 0 < h < \min \left( {{b} \over {M}} , {{1} \over {K}} \right)$.)

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Part 4. Banach Space

Since $C [ -h , h ]$ is a Banach space, $\displaystyle \mathbf{u} (t) := \lim_{k \to \infty} \mathbf{u}_{k} (t)$ is a continuous function. As $\mathbf{u}$ is continuous, so is $( f \circ \mathbf{u} )$, and by the fundamental theorem of calculus,

$$\dot{\mathbf{u} } (t) = \left( \phi_{0} + \int_{0}^{t} f \left( u(s) \right) ds \right)' = f \left( \mathbf{u} (t) \right)$$

Also, if $t = 0$ then $\displaystyle \mathbf{u} (0) = \phi_{0} + \int_{0}^{0} f \left( u(s) \right) ds = \phi_{0} + 0 = \phi_{0}$ Hence, $\mathbf{u}$ exists as a solution to the given initial value problem for all $t \in [ - h , h ]$.

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Part 5. Uniqueness

Assuming $\mathbf{u}$ and $\mathbf{v}$ are solutions to the given initial value problem. If $| \mathbf{u} (t) - \mathbf{v} (t) |$ maximizes the value for some $t$, let it be $t_{0} \in [-h , h]$,

$$\begin{align*} \| \mathbf{u} - \mathbf{v} \| =& \sup_{t \in [-h,h ] } | \mathbf{u} (t) - \mathbf{v} (t) | \\ =& \left| \int_{0}^{t_{0} } \left[ f \left( \mathbf{u} (s) \right) - f \left( \mathbf{v} (s) \right) \right] ds \right| \\ \le & \int_{0}^{t_{0} } \left| f \left( \mathbf{u} (s) \right) - f \left( \mathbf{v} (s) \right) \right| ds \\ \le & K \int_{0}^{t_{0} } \left|\mathbf{u} (s) - \mathbf{v} (s) \right| ds \\ \le & K h \sup_{s \in [-h, h ] } \left| \mathbf{u} (s) - \mathbf{v} (s) \right| \\ \le & K h \| \mathbf{u} - \mathbf{v} \| \end{align*}$$

In summary, since $\| \mathbf{u} - \mathbf{v} \| \le K h \| \mathbf{u} - \mathbf{v} \|$ and because of $Kh < 1$, it must be $\| \mathbf{u} - \mathbf{v} \| = 0$. Therefore, in $[-h, h ]$, it must be $\| \mathbf{u} \| = \| \mathbf{v} \|$.

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