The Method of Successive Approximations Diff. Eqs. Examples 1

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)

\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 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.

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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 f(s, 0) \: ds \\ \quad \phi_1(t) = \int_0^t -1 \: ds \\ \quad \phi_1(t) = -t \\ \end{align}

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\begin{align} \quad \phi_2(t) = \int_0^t f(s, \phi_1(s)) \: ds \\ \quad \phi_2(t) = \int_0^t f(s, -s) \: ds \\ \quad \phi_2(t) = \int_0^t [s - 1] \: ds \\ \quad \phi_2(t) = \frac{t^2}{2} - t \\ \end{align}

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\begin{align} \quad \phi_3(t) = \int_0^t f(s, \phi_2(s)) \: ds \\ \quad \phi_3(t) = \int_0^t f \left (s, \frac{s^2}{2} - s \right ) \: ds \\ \quad \phi_3(t) = \int_0^t \left [ -\frac{s^2}{2} + s - 1 \right ] \: ds \\ \quad \phi_3(t) = - \frac{t^3}{6} + \frac{t^2}{2} - t \end{align}

It's not hard to prove by mathematical induction that:

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\begin{align} \quad \phi_n(t) = -t + \frac{t^2}{2} - \frac{t^3}{3} + ... + \frac{(-1)^n t^n}{n!} = \sum_{k=1}^{n} \frac{(-1)^k t^k}{k!} \end{align}

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:

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\begin{align} \quad \lim_{k \to \infty} \biggr \rvert \frac{(-1)^{k+1} t^{k+1}}{(k+1)!} \frac{k!}{(-1)^k t^{k}}\biggr \rvert = \lim_{k \to \infty} \biggr \rvert \frac{-t}{k} \biggr \rvert = \lim_{k \to \infty} \frac{t}{k} = 0 < 1 \end{align}

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.

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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 f(s, 0) \: ds \\ \quad \phi_1(t) = \int_0^t s^2 \: ds \\ \quad \phi_1(t) = \frac{t^3}{3} \end{align}

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\begin{align} \quad \phi_2(t) = \int_0^t f(s, \phi_1(s)) \: ds \\ \quad \phi_2(t) = \int_0^t f\left ( s, \frac{s^3}{3} \right ) \: ds \\ \quad \phi_2(t) = \int_0^t \left [ s^2 + \left ( \frac{s^3}{3} \right )^2 \right ] \: ds \\ \quad \phi_2(t) = \int_0^t \left [ s^2 + \frac{s^6}{9} \right ] \: ds \\ \quad \phi_2(t) = \frac{t^3}{3} + \frac{t^7}{ 7 \cdot 9 } \end{align}

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\begin{align} \quad \phi_3(t) = \int_0^t f(s, \phi_2(s)) \: ds \\ \quad \phi_3(t) = \int_0^t f \left ( s, \frac{s^3}{3} + \frac{s^7}{7 \cdot 9} \right ) \: ds \\ \quad \phi_3(t) = \int_0^t \left [ s^2 + \left ( \frac{s^3}{3} + \frac{s^7}{7 \cdot 9} \right )^2 \right ] \: ds \\ \quad \phi_3(t) = \int_0^t \left [ s^2 + \frac{s^6}{9} + \frac{2s^{10}}{3 \cdot 7 \cdot 9} + \frac{s^{14}}{49 \cdot 81} \right ] \: ds \\ \quad \phi_3(t) = \frac{t^3}{3} + \frac{t^7}{7 \cdot 9} + \frac{2t^{11}}{3 \cdot 7 \cdot 9 \cdot 11} + \frac{t^{15}}{15 \cdot 49 \cdot 81} \end{align}