Runge-Kutta fourth order with negative stepsize












0














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







share|cite|improve this question
























  • 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 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
















0














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







share|cite|improve this question
























  • 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 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














0












0








0







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







share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 12:20









LutzL

55.8k42054




55.8k42054










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 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


















  • 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 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
















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










1 Answer
1






active

oldest

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2














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.






share|cite|improve this answer





















  • 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











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









2














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.






share|cite|improve this answer





















  • 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
















2














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.






share|cite|improve this answer





















  • 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














2












2








2






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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 13:50









LutzL

55.8k42054




55.8k42054












  • 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


















  • 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
















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


















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