Runge-Kutta fourth order with negative stepsize
I am solving an ODE using the Runge-Kutta method 4th order and the integration is backward i.e step size ($h$) is negative. All the references that I have seen consider the positive step size.
Is it ok to change $h$ to $-h$ in Runge-Kutta method or is there any change in sign in $K$ values?
differential-equations numerical-methods runge-kutta-methods
edited Nov 24 at 12:20
LutzL
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
add a comment |
I am solving an ODE using the Runge-Kutta method 4th order and the integration is backward i.e step size ($h$) is negative. All the references that I have seen consider the positive step size.
Is it ok to change $h$ to $-h$ in Runge-Kutta method or is there any change in sign in $K$ values?
differential-equations numerical-methods runge-kutta-methods
edited Nov 24 at 12:20
LutzL
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedurey_next = rk4step(f,t,y,h)that works correctly for positiveh, then it works also correctly for negativeh. Remember that the time step remainst_next = t+h.
– LutzL
Nov 24 at 13:42
add a comment |
I am solving an ODE using the Runge-Kutta method 4th order and the integration is backward i.e step size ($h$) is negative. All the references that I have seen consider the positive step size.
Is it ok to change $h$ to $-h$ in Runge-Kutta method or is there any change in sign in $K$ values?
differential-equations numerical-methods runge-kutta-methods
edited Nov 24 at 12:20
LutzL
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
I am solving an ODE using the Runge-Kutta method 4th order and the integration is backward i.e step size ($h$) is negative. All the references that I have seen consider the positive step size.
Is it ok to change $h$ to $-h$ in Runge-Kutta method or is there any change in sign in $K$ values?
differential-equations numerical-methods runge-kutta-methods
differential-equations numerical-methods runge-kutta-methods
edited Nov 24 at 12:20
LutzL
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
edited Nov 24 at 12:20
LutzL
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
edited Nov 24 at 12:20
LutzL
55.4k42053
edited Nov 24 at 12:20
LutzL
55.4k42053
edited Nov 24 at 12:20
LutzL
55.4k42053
55.4k42053
asked Nov 24 at 11:41
mageshwaran T
91
asked Nov 24 at 11:41
mageshwaran T
91
asked Nov 24 at 11:41
mageshwaran T
91
91
No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedurey_next = rk4step(f,t,y,h)that works correctly for positiveh, then it works also correctly for negativeh. Remember that the time step remainst_next = t+h.
– LutzL
Nov 24 at 13:42
add a comment |
No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedurey_next = rk4step(f,t,y,h)that works correctly for positiveh, then it works also correctly for negativeh. Remember that the time step remainst_next = t+h.
– LutzL
Nov 24 at 13:42
No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedure
y_next = rk4step(f,t,y,h) that works correctly for positive h, then it works also correctly for negative h. Remember that the time step remains t_next = t+h.– LutzL
Nov 24 at 13:42
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedure
y_next = rk4step(f,t,y,h) that works correctly for positive h, then it works also correctly for negative h. Remember that the time step remains t_next = t+h.– LutzL
Nov 24 at 13:42
add a comment |
1 Answer
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If you have a procedure
y_next = rk4step(f,t,y,h)
that works correctly for positive h, then this same procedure also works correctly for negative h. Remember that the time step remains
t_next = t+h
The only problem that may arise is the control of the loop. If the sampling times are given as array, then the loop
for k in range(1,len(t)):
y[k] = rk4step(f,t[k-1],y[k-1], t[k]-t[k-1])
will work independent of the direction of the time sample points.
If the loop control is based on the end time, then while t < tf works for positive h, for negative h one has to switch the sign or include h as in while 0<(tf-t)*h.
answered Nov 24 at 13:50
LutzL
55.4k42053
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you usek.. = h*f(..)or shift the multiplication withhto the combinations of thek,y_next = y + h*(k...)does not matter, as long as you use only one variant of these.
– LutzL
Nov 24 at 14:37
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
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oldest
votes
If you have a procedure
y_next = rk4step(f,t,y,h)
that works correctly for positive h, then this same procedure also works correctly for negative h. Remember that the time step remains
t_next = t+h
The only problem that may arise is the control of the loop. If the sampling times are given as array, then the loop
for k in range(1,len(t)):
y[k] = rk4step(f,t[k-1],y[k-1], t[k]-t[k-1])
will work independent of the direction of the time sample points.
If the loop control is based on the end time, then while t < tf works for positive h, for negative h one has to switch the sign or include h as in while 0<(tf-t)*h.
answered Nov 24 at 13:50
LutzL
55.4k42053
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you usek.. = h*f(..)or shift the multiplication withhto the combinations of thek,y_next = y + h*(k...)does not matter, as long as you use only one variant of these.
– LutzL
Nov 24 at 14:37
add a comment |
If you have a procedure
y_next = rk4step(f,t,y,h)
that works correctly for positive h, then this same procedure also works correctly for negative h. Remember that the time step remains
t_next = t+h
The only problem that may arise is the control of the loop. If the sampling times are given as array, then the loop
for k in range(1,len(t)):
y[k] = rk4step(f,t[k-1],y[k-1], t[k]-t[k-1])
will work independent of the direction of the time sample points.
If the loop control is based on the end time, then while t < tf works for positive h, for negative h one has to switch the sign or include h as in while 0<(tf-t)*h.
answered Nov 24 at 13:50
LutzL
55.4k42053
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you usek.. = h*f(..)or shift the multiplication withhto the combinations of thek,y_next = y + h*(k...)does not matter, as long as you use only one variant of these.
– LutzL
Nov 24 at 14:37
add a comment |
If you have a procedure
y_next = rk4step(f,t,y,h)
that works correctly for positive h, then this same procedure also works correctly for negative h. Remember that the time step remains
t_next = t+h
The only problem that may arise is the control of the loop. If the sampling times are given as array, then the loop
for k in range(1,len(t)):
y[k] = rk4step(f,t[k-1],y[k-1], t[k]-t[k-1])
will work independent of the direction of the time sample points.
If the loop control is based on the end time, then while t < tf works for positive h, for negative h one has to switch the sign or include h as in while 0<(tf-t)*h.
answered Nov 24 at 13:50
LutzL
55.4k42053
If you have a procedure
y_next = rk4step(f,t,y,h)
that works correctly for positive h, then this same procedure also works correctly for negative h. Remember that the time step remains
t_next = t+h
The only problem that may arise is the control of the loop. If the sampling times are given as array, then the loop
for k in range(1,len(t)):
y[k] = rk4step(f,t[k-1],y[k-1], t[k]-t[k-1])
will work independent of the direction of the time sample points.
If the loop control is based on the end time, then while t < tf works for positive h, for negative h one has to switch the sign or include h as in while 0<(tf-t)*h.
answered Nov 24 at 13:50
LutzL
55.4k42053
answered Nov 24 at 13:50
LutzL
55.4k42053
answered Nov 24 at 13:50
LutzL
55.4k42053
answered Nov 24 at 13:50
LutzL
55.4k42053
55.4k42053
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you usek.. = h*f(..)or shift the multiplication withhto the combinations of thek,y_next = y + h*(k...)does not matter, as long as you use only one variant of these.
– LutzL
Nov 24 at 14:37
add a comment |
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you usek.. = h*f(..)or shift the multiplication withhto the combinations of thek,y_next = y + h*(k...)does not matter, as long as you use only one variant of these.
– LutzL
Nov 24 at 14:37
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
Thanks for the answer. My only worry is the signs within K's . Is h needs to be multiplied to all K's or just at the end while calculating y_next.
– mageshwaran T
Nov 24 at 14:14
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you use
k.. = h*f(..) or shift the multiplication with h to the combinations of the k, y_next = y + h*(k...) does not matter, as long as you use only one variant of these.– LutzL
Nov 24 at 14:37
As I said, you use the same sequence of computations as for positive $h$, the RK4 step does not depend on the sign of $h$. If you use
k.. = h*f(..) or shift the multiplication with h to the combinations of the k, y_next = y + h*(k...) does not matter, as long as you use only one variant of these.– LutzL
Nov 24 at 14:37
add a comment |
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No, if you do it consistently, then there should be no change. Also check your loop conditions, going backwards is not as intuitive as going forward.
– LutzL
Nov 24 at 12:18
You mean there is no change in equations and I can use -h instead of h. Can you give me any references.
– mageshwaran T
Nov 24 at 13:39
I do not know where this is explicitly stated, Hairer-Norsett-Wanner is always a good reference to look into. If you have a procedure
y_next = rk4step(f,t,y,h)that works correctly for positiveh, then it works also correctly for negativeh. Remember that the time step remainst_next = t+h.– LutzL
Nov 24 at 13:42