The Method of Successive Approximations Examples 2

The Method of Successive Approximations for First Order Differential Equations Examples 2

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)

\begin{align} \quad \phi_{n+1}(t) = \int_0^t f(s, \phi_n(s)) \: ds \end{align}

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 another example of applying the method of successive approximations to solve first order initial value problems.

Example 1

Find the functions $\phi_1$, $\phi_2$, and $\phi_3$ using the Method of Successive Approximations for the differential equation $\frac{dy}{dt} = t^2 y - t$ with the initial condition $y(0) = 0$.

Let $f(t, y) = t^2 y - t$. Note that $f$ is continuous on all of $\mathbb{R}^2$ and $\frac{df}{dt} = 2ty - 1$ is continuous on all of $\mathbb{R}^2$ so a unique solution exists. Define $\phi_0(t) = 0$. We will now compute the first three approximation functions:

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\begin{align} \quad \phi_1(t) = \int_0^t f(s, \phi_0(s)) \: ds \\ \quad \phi_1(t) = \int_0^t s^2(0) - s \: ds \\ \quad \phi_1(t) = \int_0^t -s \: ds \\ \quad \phi_1(t) = -\frac{t^2}{2} \end{align}

(3)

\begin{align} \quad \phi_2(t) = \int_0^t f(s, \phi_1(s)) \: ds \\ \quad \phi_2(t) = \int_0^t \left ( s^2\left ( -\frac{s^2}{2} \right ) - s \right ) \: ds \\ \quad \phi_2(t) = \int_0^t \left ( -\frac{s^4}{2} - s \right ) \: ds \\ \quad \phi_2(t) = -\frac{s^5}{2 \cdot 5} - \frac{s^2}{2} \end{align}

(4)

\begin{align} \quad \phi_3(t) = \int_0^t f(s, \phi_2(t)) \: ds \\ \quad \phi_3(t) = \int_0^t \left ( s^2 \left ( -\frac{s^5}{2 \cdot 5} - \frac{s^2}{2} \right ) - s \right ) \: ds \\ \quad \phi_3(t) = \int_0^t \left ( - \frac{s^7}{2 \cdot 5} - \frac{s^4}{2} - s \right ) \: ds \\ \quad \phi_3(t) = -\frac{t^8}{2 \cdot 5 \cdot 8} - \frac{t^5}{2 \cdot 5} - \frac{t^2}{2} \end{align}

Example 2

Find the functions $\phi_1$, $\phi_2$, and $\phi_3$ using the Method of Successive Approximations for the differential equation $\frac{dy}{dt} = y - t + 1$ with the initial condition $y(0) = 0$. Then find the exact solution to this initial value problem by taking the limit of the sequence of approximations $\{ \phi_0, \phi_1, \phi_2, ... \}$ as $n \to \infty$.

Let $f(t, y) = y - t + 1$. Clearly this function is continuous on all of $\mathbb{R}^2$ and $\frac{\partial f}{\partial t} = -1$ is continuous on all of $\mathbb{R}^2$ as well, and so there exists a unique solution $\phi(t)$ to this differential equation. Define $\phi_0(t) = 0$.

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\begin{align} \quad \phi_1(t) = \int_0^t f(s, \phi_0(s)) \: ds \\ \quad \phi_1(t) = \int_0^t \left ( -s + 1 \right ) \: ds \\ \quad \phi_1(t) = -\frac{t^2}{2} + t \end{align}

(6)

\begin{align} \quad \phi_2(t) = \int_0^t f(s, \phi_1(s)) \: ds \\ \quad \phi_2(t) = \int_0^t \left ( -\frac{s^2}{2} + s - s + 1 \right ) \: ds \\ \quad \phi_2(t) = \int_0^t \left ( -\frac{s^2}{2} + 1 \right ) \: ds \\ \quad \phi_2(t) = -\frac{t^3}{3 \cdot 2} + t \end{align}

(7)

\begin{align} \quad \phi_3(t) = \int_0^t f(s, \phi_2(s)) \: ds \\ \quad \phi_3(t) = \int_0^t \left ( -\frac{s^3}{3 \cdot 2} + s - s + 1 \right ) \: ds \\ \quad \phi_3(t) = \int_0^t \left ( - \frac{s^3}{3 \cdot 2} + 1 \right ) \: ds \\ \quad \phi_3(t) = -\frac{t^4}{4!} + t \end{align}

It's not hard to see that for $n \in \mathbb{N}$ we have that $\phi_n(t) = -\frac{t^{n+1}}{(n+1)!} + t$. Taking the limit as $n \to \infty$ gives us that:

(8)

\begin{align} \quad \lim_{n \to \infty} \phi_n(t) = \lim_{n \to \infty} \left ( -\frac{t^{n+1}}{(n+1)!} + t \right ) \\ \quad \lim_{n \to \infty} \phi_n(t) = t \end{align}

Therefore $\phi(t) = t$ is the unique solution to this initial value problem. Clearly $\phi(t) = t$ satisfies the initial condition of $\phi(0) = 0$. Furthermore, a quick substitution of this function into our differential equation shows that indeed $\phi(t) = t$ is a solution.