Consider the Runge-Kutta method 1 Yi+1 = Y¡ + hf (z¡ + = h, y¡ + hß21 ƒ (zi, Yi)) with a parameter B21ER. For which of the following choices of B21 is the local error of this method 0(h³)?
Consider the Runge-Kutta method 1 Yi+1 = Y¡ + hf (z¡ + = h, y¡ + hß21 ƒ (zi, Yi)) with a parameter B21ER. For which of the following choices of B21 is the local error of this method 0(h³)?
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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Transcribed Image Text:Consider the Runge-Kutta method
with a parameter B21ER. For which of the following choices of B21 is the local error of this method 0(h³)?
Hint: Use Remark 8.12 and Taylor expansion.
r
a. B21=0
b. 21=1/6
C. B21=1/4
d. B21=1/3
e. B21=1/2
f. 21-2/3
1
Yi+1 = y¡ + hf(z; + = h, y¡ + hß21 ƒ (zi, yi))
f
2
g. 21=3/4
h. B21=5/6
i. B21=1
![Remark 8.12: Taylor expansion of the solution
Let [a, b] C R. be an interval with a ≤ b, and let ƒ € C²([a, b] × Rd, Rª).
Taking into account that by the chain rule, the solution y of the initial value
problem (8.3) satisfies
y'(x) = f(x, y(x)),
d
af
3^(2) = f(x, y(x)) = f (2,3(2)) + f(x,v(x))/(x)
ду
y"(x)
af
of
-(x, y(x)) + -(x, y(x))ƒ(x, y(x)),
ду
?х
Taylor's theorem yields
=
h²
y(x + h) = y(x) + hy'(x) + 2y"(x) + 0(h³)
h² of
= y(x) + hf(x, y(x)) +- -(x, y(x)) + −(x, y(x))ƒ (x, y(x)) + 0(h³).
2 მყ
h² Of
2 дх
In principle, the solution can be expanded as far as the degree of smoothness
of y allows, but the expressions become complicated very soon.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3cb672f7-47ed-4ee3-be4e-71db737c6150%2Fc20c35b4-6ed2-4a9d-955b-1b637eba3ac2%2Fxu49jexp_processed.png&w=3840&q=75)
Transcribed Image Text:Remark 8.12: Taylor expansion of the solution
Let [a, b] C R. be an interval with a ≤ b, and let ƒ € C²([a, b] × Rd, Rª).
Taking into account that by the chain rule, the solution y of the initial value
problem (8.3) satisfies
y'(x) = f(x, y(x)),
d
af
3^(2) = f(x, y(x)) = f (2,3(2)) + f(x,v(x))/(x)
ду
y"(x)
af
of
-(x, y(x)) + -(x, y(x))ƒ(x, y(x)),
ду
?х
Taylor's theorem yields
=
h²
y(x + h) = y(x) + hy'(x) + 2y"(x) + 0(h³)
h² of
= y(x) + hf(x, y(x)) +- -(x, y(x)) + −(x, y(x))ƒ (x, y(x)) + 0(h³).
2 მყ
h² Of
2 дх
In principle, the solution can be expanded as far as the degree of smoothness
of y allows, but the expressions become complicated very soon.
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