The Method of Successive Approximations for First Order Differential Equations Examples 1
Recall from The Method of Successive Approximations page that by The Method of Successive Approximations (Picard's Iterative Method), if $\frac{dy}{dt} = f(t, y)$ is a first order differential equation and with the initial condition $y(0) = 0$ (if the initial condition is not $y(0) = 0$ then we can apply a substitution to translate the differential equation so that $y(0) = 0$ becomes the initial condition) and if both $f$ and $\frac{\partial f}{\partial y}$ are both continuous on some rectangle $R$ for which $-a ≤ t ≤ a$ and $-b ≤ y ≤ b$ then $\lim_{n \to \infty} \phi_n(t) = \lim_{n \to \infty} \int_0^t f(s, \phi_{n-1}(s)) \: ds = \phi(t)$ where $y = \phi(t)$ is the unique solution to this initial value problem.
Furthermore, recall that the functions $\{ \phi_0, \phi_1, \phi_2, ..., \phi_n, ... \}$ are successively better approximations of the unique solution $y = \phi(t)$. We start with $\phi_0(t) = 0$ and the rest of the functions, $\phi_1, \phi_2, ..., \phi_n, ...$ can be obtained with the following recursive formula:
(1)We also noted that if $\phi_k(t) = \phi_{k+1}(t)$ for some $k$, then we have that $y = \phi_k(t)$ is the unique solution we're looking for.
We will now look at some examples of applying the method of successive approximations to solve first order initial value problems.
Example 1
Solve the initial value problem $\frac{dy}{dt} = -y - 1$ with the initial condition $y(0) = 0$ using the Method of Successive Approximations.
Let $f(t, y) = -y - 1$. Clearly $f$ is continuous on all of $\mathbb{R}^2$ and also $\frac{\partial f}{\partial y} = -1$ is continuous on all of $\mathbb{R}^2$ and so a unique solution exists. We can express our first order differential equation as $\frac{dy}{dt} = f(t, y)$. Define $\phi_0(t) = 0$. We will now compute some of the approximation functions until we see a pattern emerging.
(2)It's not hard to prove by mathematical induction that:
(5)Therefore $\phi(t) = \sum_{k=1}^{\infty} \frac{(-1)^k t^k}{k!}$ provided that this series converges. To show that it converges, we can apply the ratio test:
(6)By the ratio test, we have that the series $\sum_{k=1}^{\infty} \frac{(-1)^k t^k}{k!}$ converges and thus it converges to the unique solution $\phi(t) = \sum_{k=1}^{\infty} \frac{(-1)^k t^k}{k!}$.
Example 2
Find the functions $\phi_1$, $\phi_2$, and $\phi_3$ using the Method of Successive Approximations for the differential equation $\frac{dy}{dt} = t^2 + y^2$ with the initial condition $y(0) = 0$.
Let $f(t, y) = t^2 + y^2$. Then $f$ is continuous on all of $\mathbb{R}^2$ and $\frac{\partial f}{\partial y} = 2y$ is continuous on all of $\mathbb{R}^2$ so a unique solution exists. Define $\phi_0(t) = 0$. We will compute the first three approximation functions.
(7)