increment function for the implicit midpoint rule
$begingroup$
Assume that $h rightarrow phi(s; x; h; f ) $ is continuously diffrentiable in a neighborhood of 0. Then, $phi$ is consistent if and only if there is a continuous increment function $h rightarrow phi(s; y; h; f ) $ such that $phi (s; x; h; f ) = x + h phi(s; x; h; f) $; $phi(s; x; 0; f ) = f (s; x) $
For the explicit euler method $y_{n+1} = y_n + hf(t_n,y_n)$
So the increment function would be
$phi (s; x; h; f ) = f(s,x) $?
For the implicit euler method $y_{n+1} = y_n + hf(t_{n+1},y_{n+1})$
am i correct in saying the increment function would be
$phi (s; x; h; f ) = f(s+h,phi(s,x,h,f))$
How would I calculate the increment function for the implicit midpoint rule
$y_{n+1} = y_n + hf(t_n + h/2,frac{y_n + y_{n+1}}{2})$
would it be $phi (s; x; h; f ) = f(s+ h/2,frac{x+ phi(s,x,h,f)}{2})$
thank you
ordinary-differential-equations pde approximation
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
Assume that $h rightarrow phi(s; x; h; f ) $ is continuously diffrentiable in a neighborhood of 0. Then, $phi$ is consistent if and only if there is a continuous increment function $h rightarrow phi(s; y; h; f ) $ such that $phi (s; x; h; f ) = x + h phi(s; x; h; f) $; $phi(s; x; 0; f ) = f (s; x) $
For the explicit euler method $y_{n+1} = y_n + hf(t_n,y_n)$
So the increment function would be
$phi (s; x; h; f ) = f(s,x) $?
For the implicit euler method $y_{n+1} = y_n + hf(t_{n+1},y_{n+1})$
am i correct in saying the increment function would be
$phi (s; x; h; f ) = f(s+h,phi(s,x,h,f))$
How would I calculate the increment function for the implicit midpoint rule
$y_{n+1} = y_n + hf(t_n + h/2,frac{y_n + y_{n+1}}{2})$
would it be $phi (s; x; h; f ) = f(s+ h/2,frac{x+ phi(s,x,h,f)}{2})$
thank you
ordinary-differential-equations pde approximation
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
$endgroup$
add a comment |
$begingroup$
Assume that $h rightarrow phi(s; x; h; f ) $ is continuously diffrentiable in a neighborhood of 0. Then, $phi$ is consistent if and only if there is a continuous increment function $h rightarrow phi(s; y; h; f ) $ such that $phi (s; x; h; f ) = x + h phi(s; x; h; f) $; $phi(s; x; 0; f ) = f (s; x) $
For the explicit euler method $y_{n+1} = y_n + hf(t_n,y_n)$
So the increment function would be
$phi (s; x; h; f ) = f(s,x) $?
For the implicit euler method $y_{n+1} = y_n + hf(t_{n+1},y_{n+1})$
am i correct in saying the increment function would be
$phi (s; x; h; f ) = f(s+h,phi(s,x,h,f))$
How would I calculate the increment function for the implicit midpoint rule
$y_{n+1} = y_n + hf(t_n + h/2,frac{y_n + y_{n+1}}{2})$
would it be $phi (s; x; h; f ) = f(s+ h/2,frac{x+ phi(s,x,h,f)}{2})$
thank you
ordinary-differential-equations pde approximation
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
$endgroup$
Assume that $h rightarrow phi(s; x; h; f ) $ is continuously diffrentiable in a neighborhood of 0. Then, $phi$ is consistent if and only if there is a continuous increment function $h rightarrow phi(s; y; h; f ) $ such that $phi (s; x; h; f ) = x + h phi(s; x; h; f) $; $phi(s; x; 0; f ) = f (s; x) $
For the explicit euler method $y_{n+1} = y_n + hf(t_n,y_n)$
So the increment function would be
$phi (s; x; h; f ) = f(s,x) $?
For the implicit euler method $y_{n+1} = y_n + hf(t_{n+1},y_{n+1})$
am i correct in saying the increment function would be
$phi (s; x; h; f ) = f(s+h,phi(s,x,h,f))$
How would I calculate the increment function for the implicit midpoint rule
$y_{n+1} = y_n + hf(t_n + h/2,frac{y_n + y_{n+1}}{2})$
would it be $phi (s; x; h; f ) = f(s+ h/2,frac{x+ phi(s,x,h,f)}{2})$
thank you
ordinary-differential-equations pde approximation
ordinary-differential-equations pde approximation
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
edited Nov 27 '18 at 23:24
pablo_mathscobar
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
asked Nov 27 '18 at 23:16
pablo_mathscobarpablo_mathscobar
996
996
add a comment |
add a comment |
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