Approximate integral using the trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ))] , t = t + h. n+1 n 2 n n n+1 n+1 n+1 n
Use Euler’s method to approximate Y (tn+1) ≈ Y (tn) + hf (tn, Y (tn)) in trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ) + hf (t , Y (t )))] . n+1 n 2 n n n+1 n n n Hence the modified Euler’s scheme
K1 = hf (tn, yn) h y = y + [f (t , y ) + f (t , y + hf (t , y ))] ⇔ K2 = hf (tn+1, yn + K1) n+1 n 2 n n n+1 n n n K1 + K2 y = y + n+1 n 2
5.3.1 Modified Euler Method
Numerical solution of Initial Value Problem: dY Z tn+1 = f (t, Y ) ⇔ Y (tn+1) = Y (tn) + f (t, Y (t)) dt. dt tn Use Euler’s method to approximate Y (tn+1) ≈ Y (tn) + hf (tn, Y (tn)) in trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ) + hf (t , Y (t )))] . n+1 n 2 n n n+1 n n n Hence the modified Euler’s scheme
K1 = hf (tn, yn) h y = y + [f (t , y ) + f (t , y + hf (t , y ))] ⇔ K2 = hf (tn+1, yn + K1) n+1 n 2 n n n+1 n n n K1 + K2 y = y + n+1 n 2
5.3.1 Modified Euler Method
Numerical solution of Initial Value Problem: dY Z tn+1 = f (t, Y ) ⇔ Y (tn+1) = Y (tn) + f (t, Y (t)) dt. dt tn Approximate integral using the trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ))] , t = t + h. n+1 n 2 n n n+1 n+1 n+1 n Hence the modified Euler’s scheme
K1 = hf (tn, yn) h y = y + [f (t , y ) + f (t , y + hf (t , y ))] ⇔ K2 = hf (tn+1, yn + K1) n+1 n 2 n n n+1 n n n K1 + K2 y = y + n+1 n 2
5.3.1 Modified Euler Method
Numerical solution of Initial Value Problem: dY Z tn+1 = f (t, Y ) ⇔ Y (tn+1) = Y (tn) + f (t, Y (t)) dt. dt tn Approximate integral using the trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ))] , t = t + h. n+1 n 2 n n n+1 n+1 n+1 n
Use Euler’s method to approximate Y (tn+1) ≈ Y (tn) + hf (tn, Y (tn)) in trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ) + hf (t , Y (t )))] . n+1 n 2 n n n+1 n n n 5.3.1 Modified Euler Method
Numerical solution of Initial Value Problem: dY Z tn+1 = f (t, Y ) ⇔ Y (tn+1) = Y (tn) + f (t, Y (t)) dt. dt tn Approximate integral using the trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ))] , t = t + h. n+1 n 2 n n n+1 n+1 n+1 n
Use Euler’s method to approximate Y (tn+1) ≈ Y (tn) + hf (tn, Y (tn)) in trapezium rule: h Y (t ) ≈ Y (t ) + [f (t , Y (t )) + f (t , Y (t ) + hf (t , Y (t )))] . n+1 n 2 n n n+1 n n n Hence the modified Euler’s scheme
K1 = hf (tn, yn) h y = y + [f (t , y ) + f (t , y + hf (t , y ))] ⇔ K2 = hf (tn+1, yn + K1) n+1 n 2 n n n+1 n n n K1 + K2 y = y + n+1 n 2 Taylor Series of f (tn + h, Y (tn) + K1) in two variables:
∂ ∂ K = h f (t , Y (t )) + h f (t , Y (t )) + K f (t , Y (t )) + O h2, K 2 . 2 n n ∂t n n 1 ∂Y n n 1
Since K1 = hf (tn, Y (tn)) = O(h),
1 (K + K ) = hf (t , Y (t )) 2 1 2 n n h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 , 2 ∂t n n n n ∂Y n n
Expression to be compared with Taylor expansion of Y (tn+1)
5.3.1 Modified Euler Method — Local truncation error (1/3)
Local truncation error due to the approximation: 1 Y (t ) ≈ Y (t ) + (K + K ) n+1 n 2 1 2
where K1 = hf (tn, Y (tn)) and K2 = hf (tn + h, Y (tn) + K1). Since K1 = hf (tn, Y (tn)) = O(h),
1 (K + K ) = hf (t , Y (t )) 2 1 2 n n h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 , 2 ∂t n n n n ∂Y n n
Expression to be compared with Taylor expansion of Y (tn+1)
5.3.1 Modified Euler Method — Local truncation error (1/3)
Local truncation error due to the approximation: 1 Y (t ) ≈ Y (t ) + (K + K ) n+1 n 2 1 2
where K1 = hf (tn, Y (tn)) and K2 = hf (tn + h, Y (tn) + K1).
Taylor Series of f (tn + h, Y (tn) + K1) in two variables:
∂ ∂ K = h f (t , Y (t )) + h f (t , Y (t )) + K f (t , Y (t )) + O h2, K 2 . 2 n n ∂t n n 1 ∂Y n n 1 5.3.1 Modified Euler Method — Local truncation error (1/3)
Local truncation error due to the approximation: 1 Y (t ) ≈ Y (t ) + (K + K ) n+1 n 2 1 2
where K1 = hf (tn, Y (tn)) and K2 = hf (tn + h, Y (tn) + K1).
Taylor Series of f (tn + h, Y (tn) + K1) in two variables:
∂ ∂ K = h f (t , Y (t )) + h f (t , Y (t )) + K f (t , Y (t )) + O h2, K 2 . 2 n n ∂t n n 1 ∂Y n n 1
Since K1 = hf (tn, Y (tn)) = O(h),
1 (K + K ) = hf (t , Y (t )) 2 1 2 n n h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 , 2 ∂t n n n n ∂Y n n
Expression to be compared with Taylor expansion of Y (tn+1) 0 Substitute Y (tn) = f (tn, Y (tn)) and
00 d ∂ d ∂ Y (tn)= f (t, Y (t)) = f (tn, Y (tn)) + Y (tn) f (tn, Y (tn)), dt ∂t dt ∂Y tn ∂ ∂ = f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )), ∂t n n n n ∂Y n n to get
Y (tn + h) = Y (tn) + hf (tn, Y (tn)) h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 . (5.10) 2 ∂t n n n n ∂Y n n
5.3.1 Modified Euler Method — Local truncation error (2/3)
Taylor Series of Y (tn+1) = Y (tn + h):
h2 Y (t + h) = Y (t ) + hY 0(t )+ Y 00(t )+ O h3 . n n n 2 n and
00 d ∂ d ∂ Y (tn)= f (t, Y (t)) = f (tn, Y (tn)) + Y (tn) f (tn, Y (tn)), dt ∂t dt ∂Y tn ∂ ∂ = f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )), ∂t n n n n ∂Y n n to get
Y (tn + h) = Y (tn) + hf (tn, Y (tn)) h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 . (5.10) 2 ∂t n n n n ∂Y n n
5.3.1 Modified Euler Method — Local truncation error (2/3)
Taylor Series of Y (tn+1) = Y (tn + h):
h2 Y (t + h) = Y (t ) + hY 0(t )+ Y 00(t )+ O h3 . n n n 2 n 0 Substitute Y (tn) = f (tn, Y (tn)) to get
Y (tn + h) = Y (tn) + hf (tn, Y (tn)) h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 . (5.10) 2 ∂t n n n n ∂Y n n
5.3.1 Modified Euler Method — Local truncation error (2/3)
Taylor Series of Y (tn+1) = Y (tn + h):
h2 Y (t + h) = Y (t ) + hY 0(t )+ Y 00(t )+ O h3 . n n n 2 n 0 Substitute Y (tn) = f (tn, Y (tn)) and
00 d ∂ d ∂ Y (tn)= f (t, Y (t)) = f (tn, Y (tn)) + Y (tn) f (tn, Y (tn)), dt ∂t dt ∂Y tn ∂ ∂ = f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )), ∂t n n n n ∂Y n n 5.3.1 Modified Euler Method — Local truncation error (2/3)
Taylor Series of Y (tn+1) = Y (tn + h):
h2 Y (t + h) = Y (t ) + hY 0(t )+ Y 00(t )+ O h3 . n n n 2 n 0 Substitute Y (tn) = f (tn, Y (tn)) and
00 d ∂ d ∂ Y (tn)= f (t, Y (t)) = f (tn, Y (tn)) + Y (tn) f (tn, Y (tn)), dt ∂t dt ∂Y tn ∂ ∂ = f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )), ∂t n n n n ∂Y n n to get
Y (tn + h) = Y (tn) + hf (tn, Y (tn)) h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 . (5.10) 2 ∂t n n n n ∂Y n n and 1 (K + K ) = hf (t , Y (t )) 2 1 2 n n h2 ∂ ∂ + f (t , Y (t )) + f (t , Y (t )) f (t , Y (t )) + O h3 2 ∂t n n n n ∂Y n n 1 imply that Y (t ) = Y (t ) + (K + K ) + O h3 . n+1 n 2 1 2