Mixing
Runge-Kutta method for higher-order differential equations
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I am studying Numerical Analysis with the book of Richard L.Burden.
A question which I'm struggling with right now is following.
Transform the second-order initial-value problem
$y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$
into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.
Then,
$$u_1(t) = y(t), u_2(t) = y'(t)$$
$$u_1'(t) = u_2(t)$$
$$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
$$u_1(0) = -0.4, u_2(0) = -0.6$$
This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$
I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$
$f_1 = u_1'= u_2(t)$,
So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)
However, I can't understand the following.
$$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$
Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.
runge-kutta-methods
asked 6 hours ago
James
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I am studying Numerical Analysis with the book of Richard L.Burden.
A question which I'm struggling with right now is following.
Transform the second-order initial-value problem
$y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$
into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.
Then,
$$u_1(t) = y(t), u_2(t) = y'(t)$$
$$u_1'(t) = u_2(t)$$
$$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
$$u_1(0) = -0.4, u_2(0) = -0.6$$
This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$
I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$
$f_1 = u_1'= u_2(t)$,
So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)
However, I can't understand the following.
$$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$
Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.
runge-kutta-methods
asked 6 hours ago
James
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am studying Numerical Analysis with the book of Richard L.Burden.
A question which I'm struggling with right now is following.
Transform the second-order initial-value problem
$y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$
into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.
Then,
$$u_1(t) = y(t), u_2(t) = y'(t)$$
$$u_1'(t) = u_2(t)$$
$$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
$$u_1(0) = -0.4, u_2(0) = -0.6$$
This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$
I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$
$f_1 = u_1'= u_2(t)$,
So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)
However, I can't understand the following.
$$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$
Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.
runge-kutta-methods
asked 6 hours ago
James
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am studying Numerical Analysis with the book of Richard L.Burden.
A question which I'm struggling with right now is following.
Transform the second-order initial-value problem
$y'' - 2y' + 2y = e^{2t}sint$ for $0 leq t leq 1, $ with $y(0) = -0.4, y'(0) = -0.6, h=0.1$
into a system of ifrst order initial-value problems, and use the Runge-Kutta method ith h=0.1 to approximate the solution.
Then,
$$u_1(t) = y(t), u_2(t) = y'(t)$$
$$u_1'(t) = u_2(t)$$
$$u_2'(t) = e^{2t}sint - 2u_1(t) + u_2(t)$$
$$u_1(0) = -0.4, u_2(0) = -0.6$$
This initial conditions give $w_{1,0} = -0.4, w_{2,0}=-0.6$
I can understand that $k_{1,1} = hf_1(t_0, w_{1,0}, w_{2,0}) = hw_{2,0}$
$f_1 = u_1'= u_2(t)$,
So $f_1(t_0, w_{1,0}, w_{2,0}) = u_2(t_0, w_{1,0}, w_{2,0}) = w_{2,0}$ (By definition of $w_{i,j}$)
However, I can't understand the following.
$$k_{2,1} = hf_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2}) = hleft[w_{2,0} + frac{1}{2}k_{1,2}right]$$
Why does $f_1(t_0 + frac{h}{2}, w_{1,0} + frac{1}{2}k_{1,1}, w_{2,0} + frac{1}{2}k_{1,2})$ equal to $w_{2,0} + frac{1}{2}k_{1,2}$? It seems that third argument in the function comes out, but there is no detailed explanation in this book.
runge-kutta-methods
runge-kutta-methods
asked 6 hours ago
James
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
James
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
James
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
James
1
asked 6 hours ago
James
1
1
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
James is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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James is a new contributor. Be nice, and check out our Code of Conduct.
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