2.8: Approximating solution using Method of Successive Approximation

It should be noted that this method is also referred to as Picard's iteration method. For the Initial value problem \(y^{\prime}=f(t, y), y\left(t_0\right)=y_0\), We can always translate to have the initial value to the origin and translate back after solving. Hence for simplicity in section \(\mathbf{2 . 8}\), we will assume initial value is at the origin: \(y^{\prime}=f(t, y), y(0)=0\). Recall the following Theorem from earlier

Theorem \(\PageIndex{1}\)

Suppose the functions \(z=f(t, y)\) and \(z=\frac{\partial f}{\partial y}(t, y)\) are continuous on \((a, b) \times(c, d)\) and the point \(\left(t_0, y_0\right) \in(a, b) \times(c, d)\), then there exists an interval \(\left(t_0-h, t_0+h\right) \subset(a, b)\) such that there exists a unique function \(y=\phi(t)\) defined on \(\left(t_0-h, t_0+h\right)\) that satisfies the following initial value problem:
\[  \notag
y^{\prime}=f(t, y), \quad y\left(t_0\right)=y_0 .
\]

Thm 2.8.1 is translated to origin version of Thm 2.4.2:

The next theorem is the previous theorem but using the assumption that we can move the initial condition.

Theorem \(\PageIndex{1}\)

Suppose the functions \(z=f(t, y)\) and \(z=\frac{\partial f}{\partial y}(t, y)\) are continuous for all \(t\) in \((-a, a) \times(-c, c)\), then there exists an interval \((-h, h) \subset(-a, a)\) such that there exists a unique function \(y=\phi(t)\) defined on \((-h, h)\) that satisfies the following initial value problem:
\[  \notag
y^{\prime}=f(t, y), \quad y(0)=0 .
\]

 This is a constructive proof and so it can be very useful.

Proof Outline

Given: \(y^{\prime}=f(t, y), y(0)=0 \quad\) Eqn \((*)\) \(f, \partial f / \partial y\) continuous \(\forall(t, y) \in(-a, a) \times(-b, b)\). Then \(y=\phi(t)\) is a solution to \(\left(^*\right)\) iff \(\phi^{\prime}(t)=f(t, \phi(t)), \quad \phi(0)=0\) iff \(\int_0^t \phi^{\prime}(s) d s=\int_0^t f(s, \phi(s)) d s, \quad \phi(0)=0\) iff \(\phi(t)=\phi(t)-\phi(0)=\int_0^t f(s, \phi(s)) d s\)

Thus \(y=\phi(t)\) is a solution to \((*)\) iff \(\phi(t)=\int_0^t f(s, \phi(s)) d s\)

Construct \(\phi\) using method of successive approximation - also called Picard's iteration method.

Let \(\phi_0(t)=0\) (or the function of your choice)
Let \(\phi_1(t)=\int_0^t f\left(s, \phi_0(s)\right) d s\)
Let \(\phi_2(t)=\int_0^t f\left(s, \phi_1(s)\right) d s\)

Let \(\phi_{n+1}(t)=\int_0^t f\left(s, \phi_n(s)\right) d s\)
Let \(\phi(t)=\lim _{n \rightarrow \infty} \phi_n(t)\)

To finish the proof, need to answer the following questions (see book or more advanced class):
1.) Does \(\phi_n(t)\) exist for all \(n\) ?
2.) Does sequence \(\phi_n\) converge?
3.) Is \(\phi(t)=\lim _{n \rightarrow \infty} \phi_n(t)\) a solution to \((*)\).
4.) Is the solution unique.

 

Example \(\PageIndex{1}\)

\(y^{\prime}=t+2 y . \quad\) That is \(f(t, y)=t+2 y\)

Solution

Let \(\phi_0(t)=0\)
Let \(\phi_1(t)=\int_0^t f(s, 0) d s=\int_0^t(s+2(0)) d s\)
\[\notag
=\int_0^t s d s=\left.\frac{s^2}{2}\right|_0 ^t=\frac{t^2}{2}
\]

Let \(\phi_2(t)=\int_0^t f\left(s, \phi_1(s)\right) d s=\int_0^t f\left(s, \frac{s^2}{2}\right) d s\)
\[ \notag
=\int_0^t\left(s+2\left(\frac{s^2}{2}\right)\right) d s=\frac{t^2}{2}+\frac{t^3}{3}
\]

Let \(\phi_3(t)=\int_0^t f\left(s, \phi_2(s)\right) d s=\int_0^t f\left(s, \frac{s^2}{2}+\frac{s^3}{3}\right) d s\)
\[ \notag
=\int_0^t\left(s+2\left(\frac{s^2}{2}+\frac{s^3}{3}\right)\right) d s=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6}
\]

Let \(\begin{aligned} \phi_4(t) & =\int_0^t f\left(s, \phi_3(s)\right) d s \\ & =\int_0^t f\left(s, \frac{s^2}{2}+\frac{s^3}{3}+\frac{s^4}{6}\right) d s \\ & =\int_0^t\left(s+2\left(\frac{s^2}{2}+\frac{s^3}{3}+\frac{s^4}{6}\right)\right) d s \\ & =\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6}+\frac{t^5}{15}\end{aligned}\)

# Importing libraries 
import matplotlib.pyplot as plt 
import numpy as np 
import math 

# Using Numpy to create an array X 
X = np.arange(-0.3, 0.3, 0.01) 

# Assign variables to the y axis part of the curve 
A1 = X**2/2 
A2 = X**2/2 + X**3/3 
A3 = X**2/2 + X**3/3 + X**4/6 

# Plotting both the curves simultaneously 
plt.plot(X, A1, color='r', label='$\\phi_1$') 
plt.plot(X, A2, color='g', label='$\\phi_2$') 
plt.plot(X, A3, color='b', label='$\\phi_3$') 

# Naming the x-axis, y-axis and the whole graph 
#plt.xlabel(" ") 
# plt.ylabel(" ") 
# plt.title(" ") 

# Adding legend, which helps us recognize the curve according to it's color 
plt.legend() 

# To load the display window 
plt.show() 

Determine formula for \(\phi_n\) :
Note patterns:
\[ \notag
\begin{array}{l}
\int_0^t s d s=\frac{t^2}{2}= \\
\int_0^t \frac{s^2}{2} d s=\frac{t^3}{3 \cdot 2}= \\
\int_0^t \frac{s^3}{3 \cdot 2} d s=\frac{t^4}{4 \cdot 3 \cdot 2}= \\
\int_0^t \frac{s^4}{4 \cdot 3 \cdot 2} d s=\frac{t^5}{5 \cdot 4 \cdot 3 \cdot 2}=
\end{array}
\]

Thus look for factorials.
\[ \notag
\begin{array}{l}
\phi_0(t)=0 \\
\phi_1(t)=\frac{t^2}{2} \\
\phi_2(t)=\frac{t^2}{2}+\frac{t^3}{3} \\
\phi_3(t)=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6} \\
\phi_4(t)=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6}+\frac{t^5}{15}=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{3 \cdot 2}+\frac{t^5}{5 \cdot 3}
\end{array}
\]

Thus \( \phi_n(t) =\)

Definition:

\[ \notag
\sum_{k=0}^{\infty} a_k x^k=  \lim _{n \rightarrow \infty} \sum_{k=0}^n a_k x^k  =a_0+a_1 x+a_2 x^2+a_3 x^3+\ldots
\]

Taylor's Theorem

\[ \notag
f(x)  =\sum_{k=0}^{\infty} \frac{f^{(k)}(0)}{k!} x^k =f(0)+f^{\prime}(0) x+\frac{f^{\prime \prime}(0)}{2} x^2+\frac{f^{\prime \prime \prime}(0)}{6} x^3+\ldots
\]

Example \(\PageIndex{2}\)

Approximate \(e^{bt}\) near 0.

Solution

\(e^t=\sum_{k=0}^{\infty} \frac{t^k}{k!}\) and thus \(e^{b t}=\sum_{k=0}^{\infty} \frac{b^k t^k}{k!}\) for \(t\) near 0 .
\[ \notag
\phi_n(t)=\sum_{k=2}^n \frac{2^{k-2}}{k!} t^k
\]

Thus \(\phi(t)=\lim _{n \rightarrow \infty} \phi_n(t)=\sum_{k=2}^{\infty} \frac{2^{k-2}}{k!} t^k=\frac{1}{4} \sum_{k=2}^{\infty} \frac{2^k}{k!} t^k\)
\[ \notag
=\frac{1}{4} \left(\sum_{k=0}^{\infty} \frac{2^{k}}{k!} t^k \quad -\quad \frac{2}{1!} t^1 \quad -\quad \frac{2}{2!} t^2 \right)
\] 

2.8: Approximating soln to IVP using seq of fns.
\[ \notag
\begin{array}{l}
\phi_0(t)=0, \quad \phi_1(t)=\frac{t^2}{2}, \quad \phi_2(t)=\frac{t^2}{2}+\frac{t^3}{3} \\
\phi_3(t)=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6}, \quad \phi_4(t)=\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^4}{6}+\frac{t^5}{15}
\end{array}
\]

2.7: Approximating soln to IVP using multiple tangent lines.
\[ \notag
y(t)=\left\{\begin{array}{ll}
0 & 0 \leq t \leq 0.1 \\
0.1 t-0.01 & 0.1 \leq t \leq 0.2 \\
0.22 t-0.034 & 0.2 \leq t \leq 0.3 \\
0.364 t-0.0772 & 0.3 \leq t \leq 0.4 \\
0.5328 t-0.14672 & 0.4 \leq t \leq 0.5
\end{array}\right.
\]