ordinary differential equations
This is a HW question.
Derive the Taylor series for $\sin 2t$ by applying the Picard method to the
first-order system corresponding to the second-order initial value problem
$x" = −4x; x(0) = 0, x'(0) = 2.$
I know how to generate terms of a the Picard iteration if I have been given conditions of the form : $x = x + 2; x(0) = 2$ but not sure how to go about solving the above equation.
We are given the system:
$\tag 1 x'' = −x; x(0) = 1, x'(0) = 0.$
To linearize $(1)$, we let $x_1 = x$ and $x_2 = x'$, yielding:
$\tag 2 x'_1 = x' = x_2 ~~~ \text{and} ~~ x'_2 = x'' = -x_1.$
This gives us the matrix: $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}.$
Solving for the eigenvalues and eigenvectors of $A$, yields:
$\lambda_1 = -i, v_1 = (i, 1)$ and $\lambda_2 = i, v_2 = (-i, 1).$
Using the eigenvalues, eigenvectors and initial conditions, we arrive at:
$$\tag 3 x_1(t) = - \cos t ~~~~ \text{and} ~~~~ x_2(t) = \sin t.$$
You should verify that this satisfies the system in $(2)$ and derive it!
Note: Variations of solutions with varying signs are okay as long as they satisfy the linearized system (so, we can have $\cos t$ and $- \sin t$) as solutions too, but the Picard iteration will change accordingly.
We will want to compare the Taylor series for the solution above, so that would be:
$$\tag 4 \sin t = t- \frac{t^3}{6} + \frac{t^5}{120} + O(t^7) \\ \cos t = 1 - \frac{t^2}{2} + \frac{t^4}{24} - \frac{t^6}{720} + O(t^7)$$
Now, we want to form the Picard iterates on the linearized, first order system.
The Picard-Lindelof Iteration is given by:
$$\tag 5 \displaystyle x_0(t) = x_0, ~~x_{n+1}(t) = x_0 + \int^t_{t_0} f(s, x_n(s))ds$$
Of course, we are going to do this using both $x_1(t)$ and $x_2(t)$ as called for by the linearization that we did earlier (two solutions to consider).
The setup uses $x(0) = (x_1(0),x_2(0))^T = (-1, 0)^T$ and then does the Picard iterates using $(5)$, so we have:
$X_{0}= \begin{bmatrix} -1 \\0 \\ \end{bmatrix}$ and $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \\ \end{bmatrix}$, so
First Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 \\0 \\ \end{bmatrix} ds = \begin{bmatrix} -1 \\ t \\ \end{bmatrix}$
Second Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 \\ s \\ \end{bmatrix} ds = \begin{bmatrix} -1 + \frac{t^2}{2} \\ t \end{bmatrix}$
Third Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 +\frac{s^2}{2} \\ s \end{bmatrix} ds = \begin{bmatrix} -1 +\frac{t^2}{2} \\ t -\frac{t^3}{6} \end{bmatrix}$
Fourth Iteration:
$\displaystyle \begin{bmatrix} -1 \\0 \\ \end{bmatrix} + \int^{t}_{0} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} -1 + \frac{s^2}{2} \\ s -\frac{s^3}{6} \end{bmatrix} ds = \begin{bmatrix} -1 +\frac{t^2}{2} - \frac{t^4}{24} \\ t - \frac{t^3}{6} \end{bmatrix}$
Notice what this is converging to? Compare these to the series expansions we wrote above in $(4)$. The top term in your $- \cos t$ the bottom is $\sin t$, as expected.
Although we see that this is converging to the $x_1(t) = -\cos t$ and $x_2(t) = \sin t$ results, you should really prove this using an inductive argument.
For an inductive argument, you would write the vector we are deriving as a function of $n$ and then use induction that it is true and gives the $x_1(t)$ and $x_2(t)$.
A different yet very simple and easy derivation of Euler's method is to consider the following $y'=f(x,y)$ (I hope you don't mind me using $x$ and $y$, the diagram for the graphical interpretation is in terms of $x$ and $y$, sorry.) with $y(x_0)=y_0$
If you imagine the graph of $y=y(x)$ (the solution) then $y'=f(x,y)$ is the slope of the tangent to the graph at the point $(x,y)$. The graph also passes through the point $(x_0,y_0)$ from the initial conditions, combining this graphically we get:
As an approximation to $y_1=y(x_1)$, we take $Y_1$ where $(x_1,Y_1)$ is a point on the tangent, using the equation for a straight line $y-y_0=m(x-x_0)$ and substituting the point $(x_1,Y_1)$ gives
$$
Y_1=y_0+f(x_0,y_0)(x_1-x_0)
$$
Using $h=(x_1-x_0)$ we arrive at the iteration scheme for Euler's method
$$
Y_{n+1}=y_n+hf(x_n,y_n)
$$
Sorry that this method went off on a bit of a tangent (pardon the pun) to the original problem but I just think it is a great way to derive Euler's method with minimal calculus needed.
Best Answer
For your problem, we are given:
$\tag 1 \displaystyle x''= - 4x, x(0) = 0, x'(0) = 2 \rightarrow x(t) = \sin(2t).$
However, we are asked to apply the Picard Iteration method to the first order system.
So, we need to convert $(1)$ to a first order system, so let: $x_1 = x$ and $x_2 = x'$, thus:
$\tag 2 x'_1 = x' = x_2 ~~~ \text{and} ~~ x'_2 = x'' = -4x_1$.
This gives us the matrix: $A = \begin{bmatrix} 0 & 1 \\ -4 & 0 \\ \end{bmatrix} $
Solving for the eigenvalues and eigenvectors of $A$, yields:
$\lambda_1 = 2i, v_1 = (-\frac{i}{2}, 1)$
$\lambda_2 = -2i, v_2 = (\frac{i}{2}, 1)$
Using the eigenvalues, eigenvectors and initial conditions, we arrive at:
$x_1(t) = \sin(2t)$ and $x_2(t) = 2\cos(2t)$. You should verify that this satisfies the system in $(2)$ and derive it!
Now, we are going to write out the Taylor series for these two since we need it for comparison purposes:
$\tag 3 \displaystyle x_1(t) = \sin(2t) = 2 t-\frac{4 t^3}{3} + \frac{4 t^5}{15} + O(t^7)$
$\tag 4 \displaystyle x_2(t) = 2\cos(2t) = 2 - 4 t^2 + \frac{4 t^4}{3} - \frac{8 t^6}{45} + O(t^7)$
Now, you actually have to perform Picard Iteration on $x_1(t)$ and $x_2(t)$, but you said you know how to do this.
Of course, if everything works out, you should see these terms looking like $(3)$ and $(4)$.
Regards