solving higher orders of boundary value problems numerically












0














I want to solve a sixthorder Boundary Value Problem Numerically.
(I have reduced the sixth order of this problem to a system of first order odes.) But I do not know that how to apply the boundary conditions on it. My fourth-order problem is:
$$frac{Ad^4w}{dzeta^4}+frac{Bd^2w}{dzeta^2}+{Cw}+frac{D}{(1-w)^2}+frac{E}{(1-w)^3}=0$$
Boundaryspace Conditions:
$$zeta=0 rightarrow
begin{cases}
w(0)=0 \[2ex]
frac{dw(0)}{dzeta}=0
end{cases}
,~~~~
zeta=1 rightarrow
begin{cases}
frac{d^2w(0)}{dzeta^2}=0 \[2ex]
frac{d^3w(0)}{dzeta^3}=0
end{cases}
$$



I want to solve this ode numerically and I have reduced the order and I generated a system of first order ode's. But I want to solve this problem by shooting method and I do not know how to apply the boundary conditions on it.




Also, how to apply this on the sixth-order problem $$frac{Ad^6w}{dzeta^6}+ frac{Bd^5w}{dzeta^5}+frac{Cd^4w}{dzeta^4}+frac{Dd^3w}{dzeta^3}+{Ew}+frac{F}{(1-w)^2}+frac{G}{(1-w)^3}=0$$
Boundaryspace Conditions:



$$zeta=0 rightarrow
begin{cases}
w(0)=0 \
frac{dw(0)}{dzeta}=0 \
frac{d^3w(0)}{dzeta^3}=0
end{cases}
,~~~zeta=1 rightarrow
begin{cases}
w(1)=0 \
frac{dw(1)}{dzeta}=0 \
frac{d^3w(1)}{dzeta^3}=0
end{cases}
$$



is it possible to consider $w(1)=0$ as $frac{d^2w(0)}{zeta^2}=a_0$, $w'(1)$ as $frac{Cd^4w(0)}{zeta^4}=b_0$, $w'''(0)$ as $frac{Cd^5w(0)}{zeta^5}=c_0$



I would be appreciating if somebody could help me that which methods are suitable for such problems.






differential-equations numerical-methods







share|cite|improve this question
























  • Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
    – LutzL
    Nov 22 '17 at 12:45










  • It is $frac{d^kw(zeta)}{dzeta^k}$
    – Nima
    Nov 22 '17 at 16:54


















0














I want to solve a sixthorder Boundary Value Problem Numerically.
(I have reduced the sixth order of this problem to a system of first order odes.) But I do not know that how to apply the boundary conditions on it. My fourth-order problem is:
$$frac{Ad^4w}{dzeta^4}+frac{Bd^2w}{dzeta^2}+{Cw}+frac{D}{(1-w)^2}+frac{E}{(1-w)^3}=0$$
Boundaryspace Conditions:
$$zeta=0 rightarrow
begin{cases}
w(0)=0 \[2ex]
frac{dw(0)}{dzeta}=0
end{cases}
,~~~~
zeta=1 rightarrow
begin{cases}
frac{d^2w(0)}{dzeta^2}=0 \[2ex]
frac{d^3w(0)}{dzeta^3}=0
end{cases}
$$



I want to solve this ode numerically and I have reduced the order and I generated a system of first order ode's. But I want to solve this problem by shooting method and I do not know how to apply the boundary conditions on it.




Also, how to apply this on the sixth-order problem $$frac{Ad^6w}{dzeta^6}+ frac{Bd^5w}{dzeta^5}+frac{Cd^4w}{dzeta^4}+frac{Dd^3w}{dzeta^3}+{Ew}+frac{F}{(1-w)^2}+frac{G}{(1-w)^3}=0$$
Boundaryspace Conditions:



$$zeta=0 rightarrow
begin{cases}
w(0)=0 \
frac{dw(0)}{dzeta}=0 \
frac{d^3w(0)}{dzeta^3}=0
end{cases}
,~~~zeta=1 rightarrow
begin{cases}
w(1)=0 \
frac{dw(1)}{dzeta}=0 \
frac{d^3w(1)}{dzeta^3}=0
end{cases}
$$



is it possible to consider $w(1)=0$ as $frac{d^2w(0)}{zeta^2}=a_0$, $w'(1)$ as $frac{Cd^4w(0)}{zeta^4}=b_0$, $w'''(0)$ as $frac{Cd^5w(0)}{zeta^5}=c_0$



I would be appreciating if somebody could help me that which methods are suitable for such problems.






differential-equations numerical-methods







share|cite|improve this question
























  • Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
    – LutzL
    Nov 22 '17 at 12:45










  • It is $frac{d^kw(zeta)}{dzeta^k}$
    – Nima
    Nov 22 '17 at 16:54
















0












0








0







I want to solve a sixthorder Boundary Value Problem Numerically.
(I have reduced the sixth order of this problem to a system of first order odes.) But I do not know that how to apply the boundary conditions on it. My fourth-order problem is:
$$frac{Ad^4w}{dzeta^4}+frac{Bd^2w}{dzeta^2}+{Cw}+frac{D}{(1-w)^2}+frac{E}{(1-w)^3}=0$$
Boundaryspace Conditions:
$$zeta=0 rightarrow
begin{cases}
w(0)=0 \[2ex]
frac{dw(0)}{dzeta}=0
end{cases}
,~~~~
zeta=1 rightarrow
begin{cases}
frac{d^2w(0)}{dzeta^2}=0 \[2ex]
frac{d^3w(0)}{dzeta^3}=0
end{cases}
$$



I want to solve this ode numerically and I have reduced the order and I generated a system of first order ode's. But I want to solve this problem by shooting method and I do not know how to apply the boundary conditions on it.




Also, how to apply this on the sixth-order problem $$frac{Ad^6w}{dzeta^6}+ frac{Bd^5w}{dzeta^5}+frac{Cd^4w}{dzeta^4}+frac{Dd^3w}{dzeta^3}+{Ew}+frac{F}{(1-w)^2}+frac{G}{(1-w)^3}=0$$
Boundaryspace Conditions:



$$zeta=0 rightarrow
begin{cases}
w(0)=0 \
frac{dw(0)}{dzeta}=0 \
frac{d^3w(0)}{dzeta^3}=0
end{cases}
,~~~zeta=1 rightarrow
begin{cases}
w(1)=0 \
frac{dw(1)}{dzeta}=0 \
frac{d^3w(1)}{dzeta^3}=0
end{cases}
$$



is it possible to consider $w(1)=0$ as $frac{d^2w(0)}{zeta^2}=a_0$, $w'(1)$ as $frac{Cd^4w(0)}{zeta^4}=b_0$, $w'''(0)$ as $frac{Cd^5w(0)}{zeta^5}=c_0$



I would be appreciating if somebody could help me that which methods are suitable for such problems.






differential-equations numerical-methods







share|cite|improve this question















I want to solve a sixthorder Boundary Value Problem Numerically.
(I have reduced the sixth order of this problem to a system of first order odes.) But I do not know that how to apply the boundary conditions on it. My fourth-order problem is:
$$frac{Ad^4w}{dzeta^4}+frac{Bd^2w}{dzeta^2}+{Cw}+frac{D}{(1-w)^2}+frac{E}{(1-w)^3}=0$$
Boundaryspace Conditions:
$$zeta=0 rightarrow
begin{cases}
w(0)=0 \[2ex]
frac{dw(0)}{dzeta}=0
end{cases}
,~~~~
zeta=1 rightarrow
begin{cases}
frac{d^2w(0)}{dzeta^2}=0 \[2ex]
frac{d^3w(0)}{dzeta^3}=0
end{cases}
$$



I want to solve this ode numerically and I have reduced the order and I generated a system of first order ode's. But I want to solve this problem by shooting method and I do not know how to apply the boundary conditions on it.




Also, how to apply this on the sixth-order problem $$frac{Ad^6w}{dzeta^6}+ frac{Bd^5w}{dzeta^5}+frac{Cd^4w}{dzeta^4}+frac{Dd^3w}{dzeta^3}+{Ew}+frac{F}{(1-w)^2}+frac{G}{(1-w)^3}=0$$
Boundaryspace Conditions:



$$zeta=0 rightarrow
begin{cases}
w(0)=0 \
frac{dw(0)}{dzeta}=0 \
frac{d^3w(0)}{dzeta^3}=0
end{cases}
,~~~zeta=1 rightarrow
begin{cases}
w(1)=0 \
frac{dw(1)}{dzeta}=0 \
frac{d^3w(1)}{dzeta^3}=0
end{cases}
$$



is it possible to consider $w(1)=0$ as $frac{d^2w(0)}{zeta^2}=a_0$, $w'(1)$ as $frac{Cd^4w(0)}{zeta^4}=b_0$, $w'''(0)$ as $frac{Cd^5w(0)}{zeta^5}=c_0$



I would be appreciating if somebody could help me that which methods are suitable for such problems.






differential-equations numerical-methods




differential-equations numerical-methods






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









LutzL

56.2k42054




56.2k42054










asked Nov 10 '17 at 9:12









Nima

154




154












  • Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
    – LutzL
    Nov 22 '17 at 12:45










  • It is $frac{d^kw(zeta)}{dzeta^k}$
    – Nima
    Nov 22 '17 at 16:54




















  • Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
    – LutzL
    Nov 22 '17 at 12:45










  • It is $frac{d^kw(zeta)}{dzeta^k}$
    – Nima
    Nov 22 '17 at 16:54


















Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
– LutzL
Nov 22 '17 at 12:45




Please ask a new question or add to the old question, do not overwrite a question that already has a valid answer. Is it $dfrac{d^kw(ζ)}{dζ^k}$ or do you mean $dfrac{w^{(k)}(ζ)}{ζ^k}$?
– LutzL
Nov 22 '17 at 12:45












It is $frac{d^kw(zeta)}{dzeta^k}$
– Nima
Nov 22 '17 at 16:54






It is $frac{d^kw(zeta)}{dzeta^k}$
– Nima
Nov 22 '17 at 16:54












1 Answer
1






active

oldest

votes


















1














You can apply the shooting method. So you will start with the initial conditions $w(0)=w'(0)=0$ and $w''(0)=a_0$ and $w'''(0)=b_0$. Then simulate the system and check the values $w''(1)$ and $w'''(1)$. I would create a grid of $a_0,b_0$ values and let multiple simulations run. Then look what is closest to your boundary conditions, then refine the grid points of $a_0,b_0$ and simulate the system again. Proceed until you have found a solution that is precise enough.



Just to summarize. You turn the boundary value problem into an initial value problem with two parameters. You tune the two parameters such that you get closer to the solution of the boundary value problem.






share|cite|improve this answer





















  • How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
    – Nima
    Nov 22 '17 at 9:13













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You can apply the shooting method. So you will start with the initial conditions $w(0)=w'(0)=0$ and $w''(0)=a_0$ and $w'''(0)=b_0$. Then simulate the system and check the values $w''(1)$ and $w'''(1)$. I would create a grid of $a_0,b_0$ values and let multiple simulations run. Then look what is closest to your boundary conditions, then refine the grid points of $a_0,b_0$ and simulate the system again. Proceed until you have found a solution that is precise enough.



Just to summarize. You turn the boundary value problem into an initial value problem with two parameters. You tune the two parameters such that you get closer to the solution of the boundary value problem.






share|cite|improve this answer





















  • How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
    – Nima
    Nov 22 '17 at 9:13


















1














You can apply the shooting method. So you will start with the initial conditions $w(0)=w'(0)=0$ and $w''(0)=a_0$ and $w'''(0)=b_0$. Then simulate the system and check the values $w''(1)$ and $w'''(1)$. I would create a grid of $a_0,b_0$ values and let multiple simulations run. Then look what is closest to your boundary conditions, then refine the grid points of $a_0,b_0$ and simulate the system again. Proceed until you have found a solution that is precise enough.



Just to summarize. You turn the boundary value problem into an initial value problem with two parameters. You tune the two parameters such that you get closer to the solution of the boundary value problem.






share|cite|improve this answer





















  • How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
    – Nima
    Nov 22 '17 at 9:13
















1












1








1






You can apply the shooting method. So you will start with the initial conditions $w(0)=w'(0)=0$ and $w''(0)=a_0$ and $w'''(0)=b_0$. Then simulate the system and check the values $w''(1)$ and $w'''(1)$. I would create a grid of $a_0,b_0$ values and let multiple simulations run. Then look what is closest to your boundary conditions, then refine the grid points of $a_0,b_0$ and simulate the system again. Proceed until you have found a solution that is precise enough.



Just to summarize. You turn the boundary value problem into an initial value problem with two parameters. You tune the two parameters such that you get closer to the solution of the boundary value problem.






share|cite|improve this answer












You can apply the shooting method. So you will start with the initial conditions $w(0)=w'(0)=0$ and $w''(0)=a_0$ and $w'''(0)=b_0$. Then simulate the system and check the values $w''(1)$ and $w'''(1)$. I would create a grid of $a_0,b_0$ values and let multiple simulations run. Then look what is closest to your boundary conditions, then refine the grid points of $a_0,b_0$ and simulate the system again. Proceed until you have found a solution that is precise enough.



Just to summarize. You turn the boundary value problem into an initial value problem with two parameters. You tune the two parameters such that you get closer to the solution of the boundary value problem.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 10 '17 at 9:48









MrYouMath

13.8k31236




13.8k31236












  • How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
    – Nima
    Nov 22 '17 at 9:13




















  • How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
    – Nima
    Nov 22 '17 at 9:13


















How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
– Nima
Nov 22 '17 at 9:13






How to apply on this $frac{Ad^6w}{zeta^6}+ frac{Bd^5w}{zeta^5}+frac{Cd^4w}{zeta^4}+frac{Dd^3w}{zeta^3}+{Cw}+frac{E}{(1-w)^2}+frac{F}{(1-w)^3}=0$ $Boundaryspace Conditions:$ $zeta=0 rightarrow begin{cases} w(0)=0 \ frac{dw(0)}{dzeta}=0 & \ frac{d^3w(0)}{dzeta^3}=0 & text{} end{cases} $ $zeta=1 rightarrow begin{cases} w(1)=0 \ frac{dw(1)}{dzeta}=0 &\ frac{d^3w(1)}{dzeta^3}=0 & text{} end{cases} $ is it possible to consider w(1)=0 as$frac{d^2w(0)}{zeta^2}=a0, w'(1) as frac{d^4w(0)}{zeta^4}=b0, w'''(0)as frac{d^5w(0)}{zeta^5}=c0$
– Nima
Nov 22 '17 at 9:13




















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