Runge Kutta Order 1 and Stage 2
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I am looking for examples of Runge-Kutta method with order 1 and stage 2, Then if it is possible, can you exemplify order 1 and stage 3 RK?
I am really confused with the stage and order in runge kutta. It would be great to see examples only on Butcher tableau so that it makes it easier for me to understand. Thanks in advance.
runge-kutta-methods
asked Nov 14 '18 at 20:26
enesenes
245
$endgroup$
add a comment |
$begingroup$
I am looking for examples of Runge-Kutta method with order 1 and stage 2, Then if it is possible, can you exemplify order 1 and stage 3 RK?
I am really confused with the stage and order in runge kutta. It would be great to see examples only on Butcher tableau so that it makes it easier for me to understand. Thanks in advance.
runge-kutta-methods
asked Nov 14 '18 at 20:26
enesenes
245
$endgroup$
$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
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@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21
add a comment |
$begingroup$
I am looking for examples of Runge-Kutta method with order 1 and stage 2, Then if it is possible, can you exemplify order 1 and stage 3 RK?
I am really confused with the stage and order in runge kutta. It would be great to see examples only on Butcher tableau so that it makes it easier for me to understand. Thanks in advance.
runge-kutta-methods
asked Nov 14 '18 at 20:26
enesenes
245
$endgroup$
I am looking for examples of Runge-Kutta method with order 1 and stage 2, Then if it is possible, can you exemplify order 1 and stage 3 RK?
I am really confused with the stage and order in runge kutta. It would be great to see examples only on Butcher tableau so that it makes it easier for me to understand. Thanks in advance.
runge-kutta-methods
runge-kutta-methods
asked Nov 14 '18 at 20:26
enesenes
245
asked Nov 14 '18 at 20:26
enesenes
245
asked Nov 14 '18 at 20:26
enesenes
245
asked Nov 14 '18 at 20:26
enesenes
245
asked Nov 14 '18 at 20:26
enesenes
245
245
$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
$begingroup$
@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21
add a comment |
$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
$begingroup$
@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21
$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
$begingroup$
@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21
add a comment |
1 Answer
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As an example, any explicit two-stage Runge-Kutta (RK) method with Butcher table
begin{equation}
begin{array}{c|cc}
0 & 0 & 0\
a & a & 0\
hline
& b & 1-b
end{array}
quad a, b in mathbb{R},
end{equation}
is first-order consistent, and it is second-order consistent if and only if $b = 1 - frac{1}{2a}$. This follows from the order conditions.
With $a = frac{1}{2}$ and $b=1 - frac{1}{2a} = 0$ we obtain the (explicit) midpoint method, which is second-order consistent, and there are of course other two-stage RK methods which are second-order consistent. However for, say, $a=1$ and $b=0neq 1-frac{1}{2a}$ you obtain a two-stage RK method which is only first-order consistent.
In general all order conditions for a given order $p$ must be satisfied by the entries of the Butcher table, and since the number of conditions grows with $p$, the number of entries in the Butcher table must also increase in order to satisfy all of the conditions. The only way to achieve this is to use additional stages. Therefore, higher-order methods require more stages.
On the other hand, it is rather easy to give examples of RK methods whose order is lower than the maximum, by choosing entries in the Butcher table which do not satisfy some of the order conditions.
The order conditions are tough to derive for $p geq 5$, and their number grows rather quickly, but it has been done of course.
answered Jan 1 at 11:34
ChristophChristoph
4366
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$begingroup$
As an example, any explicit two-stage Runge-Kutta (RK) method with Butcher table
begin{equation}
begin{array}{c|cc}
0 & 0 & 0\
a & a & 0\
hline
& b & 1-b
end{array}
quad a, b in mathbb{R},
end{equation}
is first-order consistent, and it is second-order consistent if and only if $b = 1 - frac{1}{2a}$. This follows from the order conditions.
With $a = frac{1}{2}$ and $b=1 - frac{1}{2a} = 0$ we obtain the (explicit) midpoint method, which is second-order consistent, and there are of course other two-stage RK methods which are second-order consistent. However for, say, $a=1$ and $b=0neq 1-frac{1}{2a}$ you obtain a two-stage RK method which is only first-order consistent.
In general all order conditions for a given order $p$ must be satisfied by the entries of the Butcher table, and since the number of conditions grows with $p$, the number of entries in the Butcher table must also increase in order to satisfy all of the conditions. The only way to achieve this is to use additional stages. Therefore, higher-order methods require more stages.
On the other hand, it is rather easy to give examples of RK methods whose order is lower than the maximum, by choosing entries in the Butcher table which do not satisfy some of the order conditions.
The order conditions are tough to derive for $p geq 5$, and their number grows rather quickly, but it has been done of course.
answered Jan 1 at 11:34
ChristophChristoph
4366
$endgroup$
add a comment |
$begingroup$
As an example, any explicit two-stage Runge-Kutta (RK) method with Butcher table
begin{equation}
begin{array}{c|cc}
0 & 0 & 0\
a & a & 0\
hline
& b & 1-b
end{array}
quad a, b in mathbb{R},
end{equation}
is first-order consistent, and it is second-order consistent if and only if $b = 1 - frac{1}{2a}$. This follows from the order conditions.
With $a = frac{1}{2}$ and $b=1 - frac{1}{2a} = 0$ we obtain the (explicit) midpoint method, which is second-order consistent, and there are of course other two-stage RK methods which are second-order consistent. However for, say, $a=1$ and $b=0neq 1-frac{1}{2a}$ you obtain a two-stage RK method which is only first-order consistent.
In general all order conditions for a given order $p$ must be satisfied by the entries of the Butcher table, and since the number of conditions grows with $p$, the number of entries in the Butcher table must also increase in order to satisfy all of the conditions. The only way to achieve this is to use additional stages. Therefore, higher-order methods require more stages.
On the other hand, it is rather easy to give examples of RK methods whose order is lower than the maximum, by choosing entries in the Butcher table which do not satisfy some of the order conditions.
The order conditions are tough to derive for $p geq 5$, and their number grows rather quickly, but it has been done of course.
answered Jan 1 at 11:34
ChristophChristoph
4366
$endgroup$
add a comment |
$begingroup$
As an example, any explicit two-stage Runge-Kutta (RK) method with Butcher table
begin{equation}
begin{array}{c|cc}
0 & 0 & 0\
a & a & 0\
hline
& b & 1-b
end{array}
quad a, b in mathbb{R},
end{equation}
is first-order consistent, and it is second-order consistent if and only if $b = 1 - frac{1}{2a}$. This follows from the order conditions.
With $a = frac{1}{2}$ and $b=1 - frac{1}{2a} = 0$ we obtain the (explicit) midpoint method, which is second-order consistent, and there are of course other two-stage RK methods which are second-order consistent. However for, say, $a=1$ and $b=0neq 1-frac{1}{2a}$ you obtain a two-stage RK method which is only first-order consistent.
In general all order conditions for a given order $p$ must be satisfied by the entries of the Butcher table, and since the number of conditions grows with $p$, the number of entries in the Butcher table must also increase in order to satisfy all of the conditions. The only way to achieve this is to use additional stages. Therefore, higher-order methods require more stages.
On the other hand, it is rather easy to give examples of RK methods whose order is lower than the maximum, by choosing entries in the Butcher table which do not satisfy some of the order conditions.
The order conditions are tough to derive for $p geq 5$, and their number grows rather quickly, but it has been done of course.
answered Jan 1 at 11:34
ChristophChristoph
4366
$endgroup$
As an example, any explicit two-stage Runge-Kutta (RK) method with Butcher table
begin{equation}
begin{array}{c|cc}
0 & 0 & 0\
a & a & 0\
hline
& b & 1-b
end{array}
quad a, b in mathbb{R},
end{equation}
is first-order consistent, and it is second-order consistent if and only if $b = 1 - frac{1}{2a}$. This follows from the order conditions.
With $a = frac{1}{2}$ and $b=1 - frac{1}{2a} = 0$ we obtain the (explicit) midpoint method, which is second-order consistent, and there are of course other two-stage RK methods which are second-order consistent. However for, say, $a=1$ and $b=0neq 1-frac{1}{2a}$ you obtain a two-stage RK method which is only first-order consistent.
In general all order conditions for a given order $p$ must be satisfied by the entries of the Butcher table, and since the number of conditions grows with $p$, the number of entries in the Butcher table must also increase in order to satisfy all of the conditions. The only way to achieve this is to use additional stages. Therefore, higher-order methods require more stages.
On the other hand, it is rather easy to give examples of RK methods whose order is lower than the maximum, by choosing entries in the Butcher table which do not satisfy some of the order conditions.
The order conditions are tough to derive for $p geq 5$, and their number grows rather quickly, but it has been done of course.
answered Jan 1 at 11:34
ChristophChristoph
4366
answered Jan 1 at 11:34
ChristophChristoph
4366
answered Jan 1 at 11:34
ChristophChristoph
4366
answered Jan 1 at 11:34
ChristophChristoph
4366
4366
add a comment |
add a comment |
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$begingroup$
Up to order 4 you can achieve stages = order, the typical examples start with order 5 where the usual methods have 6 or 7 stages (but usually only 6 function evaluations per time step). See math.stackexchange.com/q/1213614/115115 for some numerical experiments with (nominally) 2-stage methods that include implicit and explicit Euler as extreme cases.
$endgroup$
– LutzL
Nov 14 '18 at 20:53
$begingroup$
@LutzL thanks for the link. Also is there any other example for RK 1 order 2 stage as embedded explicit Rk?
$endgroup$
– enes
Nov 14 '18 at 21:17
$begingroup$
In principle in any explicit method, the first stage provides an explicit Euler step.
$endgroup$
– LutzL
Nov 14 '18 at 21:21