Mixing
Stiffness and stability - proof of order stages
$begingroup$
I've got this theorem in my book which say:
"The order of a Runge–Kutta method with $s$ stages cannot exceed
$2s$"
Proof: Assume that $p ≥ 2s + 1$. The corresponding quadrature formula
begin{equation}int_{t_m}^{t_{m+1}} f(t) dt approx hsum_{i=1}^{s} b_if(t_m+c_i h) end{equation}
is then exact for all polynomials of degree ≤ 2s: From the “bushy” trees we obtain the order conditions
begin{equation}sum_{i=1}^s b_ic_i^{l-1}=frac{1}{l} end{equation}
for $l=1,...,2s+1$
This implying
begin{equation}int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt=frac{1}{l}h^l=hsum_{i=1}^s b_i(c_ih)^{l-1}=frac{1}{l}h^l end{equation}
for $l=1,...,2s+1$
Choose
begin{equation}f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
Then $int_{t_m}^{t_{m+1}}f(t) dt >0$, but $hsum_{i=1}^s b_if(t_m+c_ih)=0$ which is a contradiciton!
--
I can't figure it out, so can anyone here clearify it for me? Thanks in advance.
Update: Why do we choose
begin{equation} f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
and why the integral
begin{equation} int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt end{equation}
ordinary-differential-equations
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
$endgroup$
add a comment |
$begingroup$
I've got this theorem in my book which say:
"The order of a Runge–Kutta method with $s$ stages cannot exceed
$2s$"
Proof: Assume that $p ≥ 2s + 1$. The corresponding quadrature formula
begin{equation}int_{t_m}^{t_{m+1}} f(t) dt approx hsum_{i=1}^{s} b_if(t_m+c_i h) end{equation}
is then exact for all polynomials of degree ≤ 2s: From the “bushy” trees we obtain the order conditions
begin{equation}sum_{i=1}^s b_ic_i^{l-1}=frac{1}{l} end{equation}
for $l=1,...,2s+1$
This implying
begin{equation}int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt=frac{1}{l}h^l=hsum_{i=1}^s b_i(c_ih)^{l-1}=frac{1}{l}h^l end{equation}
for $l=1,...,2s+1$
Choose
begin{equation}f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
Then $int_{t_m}^{t_{m+1}}f(t) dt >0$, but $hsum_{i=1}^s b_if(t_m+c_ih)=0$ which is a contradiciton!
--
I can't figure it out, so can anyone here clearify it for me? Thanks in advance.
Update: Why do we choose
begin{equation} f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
and why the integral
begin{equation} int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt end{equation}
ordinary-differential-equations
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
$endgroup$
add a comment |
$begingroup$
I've got this theorem in my book which say:
"The order of a Runge–Kutta method with $s$ stages cannot exceed
$2s$"
Proof: Assume that $p ≥ 2s + 1$. The corresponding quadrature formula
begin{equation}int_{t_m}^{t_{m+1}} f(t) dt approx hsum_{i=1}^{s} b_if(t_m+c_i h) end{equation}
is then exact for all polynomials of degree ≤ 2s: From the “bushy” trees we obtain the order conditions
begin{equation}sum_{i=1}^s b_ic_i^{l-1}=frac{1}{l} end{equation}
for $l=1,...,2s+1$
This implying
begin{equation}int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt=frac{1}{l}h^l=hsum_{i=1}^s b_i(c_ih)^{l-1}=frac{1}{l}h^l end{equation}
for $l=1,...,2s+1$
Choose
begin{equation}f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
Then $int_{t_m}^{t_{m+1}}f(t) dt >0$, but $hsum_{i=1}^s b_if(t_m+c_ih)=0$ which is a contradiciton!
--
I can't figure it out, so can anyone here clearify it for me? Thanks in advance.
Update: Why do we choose
begin{equation} f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
and why the integral
begin{equation} int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt end{equation}
ordinary-differential-equations
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
$endgroup$
I've got this theorem in my book which say:
"The order of a Runge–Kutta method with $s$ stages cannot exceed
$2s$"
Proof: Assume that $p ≥ 2s + 1$. The corresponding quadrature formula
begin{equation}int_{t_m}^{t_{m+1}} f(t) dt approx hsum_{i=1}^{s} b_if(t_m+c_i h) end{equation}
is then exact for all polynomials of degree ≤ 2s: From the “bushy” trees we obtain the order conditions
begin{equation}sum_{i=1}^s b_ic_i^{l-1}=frac{1}{l} end{equation}
for $l=1,...,2s+1$
This implying
begin{equation}int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt=frac{1}{l}h^l=hsum_{i=1}^s b_i(c_ih)^{l-1}=frac{1}{l}h^l end{equation}
for $l=1,...,2s+1$
Choose
begin{equation}f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
Then $int_{t_m}^{t_{m+1}}f(t) dt >0$, but $hsum_{i=1}^s b_if(t_m+c_ih)=0$ which is a contradiciton!
--
I can't figure it out, so can anyone here clearify it for me? Thanks in advance.
Update: Why do we choose
begin{equation} f(t)=prod_{i=1}^s(t-t_m-c_ih)^2 end{equation}
and why the integral
begin{equation} int_{t_m}^{t_{m+1}}(t-t_m)^{l-1}dt end{equation}
ordinary-differential-equations
ordinary-differential-equations
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
asked Jan 16 at 0:45
Joey AdamsJoey Adams
457
457
add a comment |
add a comment |
1 Answer
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$begingroup$
This is a proof by contradiction, so they aim to construct a polynomial of degree $leq 2s$ for which the quadrature formula is not exact.
The function $f$ chosen here is clearly a polynomial of degree $2s$ (the highest power of $t$ is $2s$), and it has double zeros at each $t = t_m + c_i h$, $i=1,2,dots,s$. Therefore, the quadrature formula yields $0$ for this polynomial, and it remains to show that the integral of $f$ over $(t_m,t_{m+1})$ is not zero.
This is obvious since $f$, as a product of squares, is a non-negative function, and it's not the zero function. Therefore, the integral of $f$ over $(t_m,t_{m+1})$ must be positive.
I don't really see why we need the integrals of $(t-t_m)^{ell-1}$, but I guess the author's idea could be that by expanding $f$ we can obtain a sum of powers of $t-t_m$, and then the integral of $f$ can be written as a sum of integrals of $(t-t_m)^{ell-1}$, each of which is positive.
answered Jan 16 at 4:47
ChristophChristoph
58116
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add a comment |
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$begingroup$
This is a proof by contradiction, so they aim to construct a polynomial of degree $leq 2s$ for which the quadrature formula is not exact.
The function $f$ chosen here is clearly a polynomial of degree $2s$ (the highest power of $t$ is $2s$), and it has double zeros at each $t = t_m + c_i h$, $i=1,2,dots,s$. Therefore, the quadrature formula yields $0$ for this polynomial, and it remains to show that the integral of $f$ over $(t_m,t_{m+1})$ is not zero.
This is obvious since $f$, as a product of squares, is a non-negative function, and it's not the zero function. Therefore, the integral of $f$ over $(t_m,t_{m+1})$ must be positive.
I don't really see why we need the integrals of $(t-t_m)^{ell-1}$, but I guess the author's idea could be that by expanding $f$ we can obtain a sum of powers of $t-t_m$, and then the integral of $f$ can be written as a sum of integrals of $(t-t_m)^{ell-1}$, each of which is positive.
answered Jan 16 at 4:47
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
This is a proof by contradiction, so they aim to construct a polynomial of degree $leq 2s$ for which the quadrature formula is not exact.
The function $f$ chosen here is clearly a polynomial of degree $2s$ (the highest power of $t$ is $2s$), and it has double zeros at each $t = t_m + c_i h$, $i=1,2,dots,s$. Therefore, the quadrature formula yields $0$ for this polynomial, and it remains to show that the integral of $f$ over $(t_m,t_{m+1})$ is not zero.
This is obvious since $f$, as a product of squares, is a non-negative function, and it's not the zero function. Therefore, the integral of $f$ over $(t_m,t_{m+1})$ must be positive.
I don't really see why we need the integrals of $(t-t_m)^{ell-1}$, but I guess the author's idea could be that by expanding $f$ we can obtain a sum of powers of $t-t_m$, and then the integral of $f$ can be written as a sum of integrals of $(t-t_m)^{ell-1}$, each of which is positive.
answered Jan 16 at 4:47
ChristophChristoph
58116
$endgroup$
add a comment |
$begingroup$
This is a proof by contradiction, so they aim to construct a polynomial of degree $leq 2s$ for which the quadrature formula is not exact.
The function $f$ chosen here is clearly a polynomial of degree $2s$ (the highest power of $t$ is $2s$), and it has double zeros at each $t = t_m + c_i h$, $i=1,2,dots,s$. Therefore, the quadrature formula yields $0$ for this polynomial, and it remains to show that the integral of $f$ over $(t_m,t_{m+1})$ is not zero.
This is obvious since $f$, as a product of squares, is a non-negative function, and it's not the zero function. Therefore, the integral of $f$ over $(t_m,t_{m+1})$ must be positive.
I don't really see why we need the integrals of $(t-t_m)^{ell-1}$, but I guess the author's idea could be that by expanding $f$ we can obtain a sum of powers of $t-t_m$, and then the integral of $f$ can be written as a sum of integrals of $(t-t_m)^{ell-1}$, each of which is positive.
answered Jan 16 at 4:47
ChristophChristoph
58116
$endgroup$
This is a proof by contradiction, so they aim to construct a polynomial of degree $leq 2s$ for which the quadrature formula is not exact.
The function $f$ chosen here is clearly a polynomial of degree $2s$ (the highest power of $t$ is $2s$), and it has double zeros at each $t = t_m + c_i h$, $i=1,2,dots,s$. Therefore, the quadrature formula yields $0$ for this polynomial, and it remains to show that the integral of $f$ over $(t_m,t_{m+1})$ is not zero.
This is obvious since $f$, as a product of squares, is a non-negative function, and it's not the zero function. Therefore, the integral of $f$ over $(t_m,t_{m+1})$ must be positive.
I don't really see why we need the integrals of $(t-t_m)^{ell-1}$, but I guess the author's idea could be that by expanding $f$ we can obtain a sum of powers of $t-t_m$, and then the integral of $f$ can be written as a sum of integrals of $(t-t_m)^{ell-1}$, each of which is positive.
answered Jan 16 at 4:47
ChristophChristoph
58116
edited Jan 16 at 7:14
answered Jan 16 at 4:47
ChristophChristoph
58116
answered Jan 16 at 4:47
ChristophChristoph
58116
answered Jan 16 at 4:47
ChristophChristoph
58116
58116
add a comment |
add a comment |
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