Existence of solutions to Differential Equations

Contents

The simplest case

If the right hand side of a differential equation does not contain the unknown function then we can solve it by integrating: the solution of \begin{equation} \frac{dx}{dt} = f(t), \quad x(t_0) = x_0 \label{eq:simplest-ODE} \end{equation} is \[ x(t) = x_0 + \int_{t_0}^t f(s) ds. \] This is just the fundamental theorem of calculus.

In calculus we learn a number of tricks that allow us to solve differential equations with more complicated right hand sides. These tricks all involve some manipulation of the differential equation that turns it into the simplest case \eqref{eq:simplest-ODE}, so we can solve it by integrating. The most important cases are separable differential equations and first order linear differential equations. Click here for a review.

The existence and uniqueness theorems deal with the equations that we cannot solve using calculus.

The general case

Consider the differential equation \begin{equation} \frac{dx}{dt} = f(t, x), \quad x(t_0) = x_0 \label{eq:ODE} \end{equation} Assume the right hand side $f(t, x)$ in the equation \eqref{eq:ODE} is a function of two variables that is defined in the rectangle $R$ consisting of all $(x, t)$ with \[ x_0-\delta \leq x \leq x_0+\delta, \quad t_- \leq t \leq t_+~~. \]
The existence theorem. Suppose the right hand side $f$ is bounded, i.e. assume that there is a constant $M$ such that \[ |f(t, x)|\leq M \] holds for all $x$ and $t$. Suppose also that $f$ satisfies the Lipschitz condition, i.e. assume that there is a constant $L$ such that \[ |f(t, x) - f(t, y)| \leq L |x-y| \] holds for all $x, y$, and $t$. Then the differential equation \eqref{eq:ODE} has a solution $x(t)$, which is defined on some interval $t_0-\epsilon \lt t \lt t_0+\epsilon$ for some number $\epsilon\gt0$.

Proof by Picard iteration of the Existence Theorem

There is a technique for proving that a solution exists, which goes back to Émile Picard (1856—1941). Here is a simplified version of his proof. The (important) details follow below.

Not knowing any solution to the ODE, we begin with a first guess, namely $x_0(t) = x_0$. We try to improve on this guess by solving \[ \frac{dx_1}{dt} = f(t, x_0), \quad x_1(t_0) = x_0, \] which gives us a new function $x_1(t)$. The right hand side in this differential equation only contains $x_0$, so it is known and we can find $x_1(t)$ by integrating. The function $x_1(t)$ that we get is still not a solution of the ODE \eqref{eq:ODE} that we want to solve, so we try to improve it again by solving \[ \frac{dx_2}{dt} = f(t, x_1(t)), \quad x_1(t_0) = x_0, \] which gives us another new function $x_2(t)$. Picard's idea was to repeat this process infinitely often, which results in a sequence of functions $x_0(t)$, $x_1(t)$, $x_2(t)$, $\cdots$, in which the $n$th function is defined by solving \begin{equation} \frac{dx_n}{dt} = f(t, x_{n-1}(t)), \quad x_{n-1}(t_0) = x_0. \label{eq:picard-iteration} \end{equation} The key part of the whole construction is to prove that the functions $x_k(t)$ converge, as $k\to\infty$, and that the limit \[ x(t) \stackrel{\rm def}= \lim_{k\to\infty} x_k(t) \] is a solution of the ODE.

This procedure of generating a sequence of functions which approximate the solution whose existence we are trying to establish, is called Picard iteration.

Details of Picard’s proof

Choice of $\epsilon$

We will choose \begin{equation} \epsilon = \min \Bigl\{ \frac1{2L}, \frac{\delta}{M} \Bigr\} \label{eq:epsilon-choice} \end{equation} The reason for this choice will not become clear until later in the proof, so for now we will simply write $\epsilon$ and not worry about its specific value.

The $x_k(t)$ are well–defined

By definition we get $x_k(t)$ from $x_{k-1}(t)$ by solving \eqref{eq:picard-iteration}, i.e. by integrating \[ x_k(t) = x_0 + \int_{t_0}^t f(s, x_{k-1}(s))\; ds. \] Here we have to substitute $x_{k-1}(t)$ in the function $f(t, x)$. Since this function is only defined for $x_0-\delta\leq x\leq x_0+\delta$ (as far as we know), we have to be sure that $x_0-\delta\leq x_{k-1}(t) \leq x_0+\delta$ holds for all $t$ with $t_0-\epsilon \le t \le t_0+\epsilon$.

For the first approximation $x_0(t)$ this is automatically true since $x_0(t) = x_0$ is constant.

The second approximation $x_1(t)$ is a function whose derivative is bounded by \[ |x_1'(t)| = \bigl|f(t, x_0(t))\bigr| \leq M, \] as we have assumed in the theorem. Since $x_1$ is a function whose derivative is never more than $M$, and since $x_1(0) = x_0$, it follows that for $t_0-\epsilon \le t \le t_0+\epsilon$ one has \[ \bigl| x_1(t) - x_0 \bigr| \leq M |t-t_0| \leq M\epsilon. \] Our choice of $\epsilon$ implies that \begin{equation} |x_1(t)-x_0|\leq \delta, \text{ i.e. } x_0-\delta \le x_1(t) \le x_0+\delta \text { for }t_0-\epsilon \le t \le t_0+\epsilon. \label{eq:x1bound} \end{equation} Therefore $f(t, x_1(t))$ is well–defined, and we can repeat the same arguments: since $x_2'(t) = f(t, x_1(t))$ we have $|x_2'(t)|\leq M$, so that $x_2(t)$ cannot change more than $M\cdot \epsilon$ during a time interval of length $\epsilon$. Hence $x_2(t)$ also satisfies $x_0-\delta \le x_2(t) \le x_0+\delta$ for as long as $t_0-\epsilon \le t \le t_0+\epsilon$.

The difference $y_k$ between two successive approximations

The key to proving that the $x_k(t)$ converge is to estimate the difference between two successive $x_k$’s. We define \[ y_k(t) = x_k(t)-x_{k-1}(t). \] If we know something about the $y_k(t)$ then we can go back to the functions $x_k(t)$ by adding: \[ x_k(t) = x_0 + y_1(t) + y_2(t) + \cdots + y_k(t). \] We will measure the size of the terms $y_k(t)$ by means of \[ m_k \stackrel{\rm def}= \max _{t_0-\epsilon \le t \le t_0+\epsilon} \left| y_k(t) \right|. \]

An estimate for $m_1$

For $k=1$ we have $y_1(t) = x_1(t)-x_0$, and it follows from \eqref{eq:x1bound} that $|y_1(t)|\leq \delta$. Hence \begin{equation} m_1\leq \delta. \label{eq:m1bound} \end{equation}

Estimating $m_k$ for $k>1$

Since both $x_k$ and $x_{k-1}$ satisfy the same initial condition $x(t_0) = x_0$, we have \[ y_k(t_0) = x_k(t_0) - x_{k-1}(t_0) = x_0 - x_0 = 0. \] For the derivative of $y_k$ we have \[ \frac{dy_k}{dt} = \frac{dx_k}{dt} - \frac{dx_{k-1}}{dt} = f\bigl(t, x_{k-1}(t)\bigr) - f\bigl(t, x_{k-2}(t)\bigr) \] so that \[ \left| \frac{dy_k}{dt} \right| \leq L |x_{k-1}(t) - x_{k-2}(t)| = L |y_{k-1}(t)| \le Lm_{k-1}. \] Since $y_{k}(t_0) = 0$, this implies that for all $t$ in the interval $(t_0-\epsilon, t_0+\epsilon)$ we get \[ |y_k(t)|\leq L\epsilon m_{k-1}. \] The maximal value of $|y_k(t)|$ therefore also satisfies this inequality, so that we have \[ m_k \leq L\epsilon\cdot m_{k-1}. \] We now use \eqref{eq:epsilon-choice} again, where we chose $\epsilon \leq 1/(2L)$. This choice implies that $L\epsilon \leq \frac12$, so that \[ m_k\leq \tfrac12 m_{k-1} \leq \bigl(\tfrac12\bigr)^2 m_{k-2} \leq \bigl(\tfrac12\bigr)^3 m_{k-3} \leq \cdots \leq \bigl(\tfrac12\bigr)^{k-1} m_{1} \leq \bigl(\tfrac12\bigr)^{k-1} \delta. \]

The series $\sum y_k(t)$ converges; the sequence $x_k(t)$ converges

Since \[ |y_k(t)|\leq m_k \leq \bigl(\tfrac12\bigr)^{k-1} \delta, \] we have \[ \sum_{k=1}^\infty |y_k(t)| \leq \sum_{k=1}^\infty m_k \leq \sum_{k=1}^\infty \bigl(\tfrac12\bigr)^{k-1} \delta = 2\delta, \] by the geometric series.

Since the series $\sum |y_k(t)|$ converges, the series $\sum_k y_k(t)$ also converges. Hence \[ x(t) \stackrel{\rm def}= \lim_{k\to\infty} x_k(t)= \lim_{k\to\infty} x_0+y_1(t)+\cdots+y_k(t)= x_0+\sum_{k=1}^\infty y_k(t) \] exists—the sequence of functions $x_k(t)$ converges.

How fast does the sequence converge?

The difference between the $k$th approximation $x_k(t)$ and the limit $x(t)$ is \[ x(t) - x_k(t) = y_{k+1}(t) + y_{k+2}(t) + \cdots = \sum_{i=k+1}^\infty y_i(t). \] We can estimate it by \[ |x(t) - x_k(t)| \leq \sum_{i=k+1}^\infty |y_i(t)| \leq \sum_{i=k+1}^\infty \delta\bigl(\tfrac12\bigr)^{i-1} =\delta\bigl(\tfrac12\bigr)^{k-1}. \] So every approximation $x_k$ should be about twice as accurate as the previous one.

Proving that the limit is a solution

We have shown that the functions $x_k(t)$ have a limit $x(t)$, but we still have to show that this function satisfies the differential equation. We reason indirectly by showing that the limit $x(t)$ satisfies an integral equation, and then differentiating both sides of that equation. Here is the computation: \begin{align*} x(t) &= \lim_{k\to \infty} x_{k+1}(t) \\ &= \lim_{k\to\infty} x_0 +\int_0^t f(s, x_k(s)) ds \\ &= x_0 +\lim_{k\to\infty} \int_{t_0}^t f(s, x_k(s)) ds\\ &= x_0 +\int_{t_0}^t \lim_{k\to\infty} f(s, x_k(s)) ds\\ &= x_0 + \int_{t_0}^t f(s, x(s)) ds. \end{align*} Therefore $x(t)$ is differentiable, and its derivative is \[ \frac{dx}{dt} = f(t, x(t)). \]

Sticky points

Picard's construction involves taking the limit of a sequence of functions. In the proof we used the plausible looking “fact” \[ \lim_{n\to\infty} \int_a^b f_n(t) dt = \int_a^b \lim_{n\to\infty} f_n(t) dt, \] We also assumed that the limit function $x(t)$ is such that the integral $\int_{t_0}^t f(s, x(s))\, ds$ is well defined as a Riemann integral. This kind of question can be answered with some real analysis, in particular, by using the concept of “uniform convergence” for a sequence of functions. We will not address these points in this course.