Sec:5.4 RUNGE-KUTTA METHODS.

Sec:5.4 RUNGE-KUTTA METHODS

Sec:5.4 RUNGE-KUTTA METHODS
1) Euler Method 2) Modified Euler Method 3) Midpoint Method 4) Heun's Method 5) RK4 Method Computational Examples MATLAB Derivations

Initial Value Problem Euler’s Method 𝑦 ′ =𝑓 𝑡,𝑦 𝑦(𝑎) =𝛼 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑤 =𝛼 𝑎≤ 𝑡 ≤𝑏, Modified Euler Method Euler’s Method 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 2 𝑲 𝟏 + 𝑲 𝟐 𝑤 =𝛼 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑲 𝟏 𝑤 =𝛼 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 +ℎ, 𝑤 𝑖 + 𝑲 𝟏 ℎ Modified Euler Method 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 2 𝒇 𝒕 𝒊 , 𝒘 𝒊 +𝑓 𝑡 𝑖 +ℎ, 𝑤 𝑖 +ℎ𝒇 𝒕 𝒊 , 𝒘 𝒊 𝑤 =𝛼

Modified Euler Method Example 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 2 𝑲 𝟏 + 𝑲 𝟐 𝑤 =𝛼 Use the Modified Euler method h = 0.5, to approximate 𝑦(1) 𝑦 ′ =𝑦− 𝑡 2 +1 0≤ 𝑡 ≤4, 𝑦(0) = 0.5 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 +ℎ, 𝑤 𝑖 + 𝑲 𝟏 ℎ 𝑓(𝑡,𝑦)=𝑦− 𝑡 2 +1 𝑲 𝟏 =𝑓 𝑡 1 , 𝑤 1 𝑲 𝟏 =𝑓 𝑡 0 , 𝑤 0 =𝑓 0.5,1.375 =2.125 𝑲 𝟐 =𝑓 𝑡 1 +ℎ, 𝑤 1 + 𝑲 𝟏 ℎ =𝑓 0,0.5 =1.5 =𝑓 1,1.375+𝟐.𝟏𝟐𝟓∗0.5 𝑲 𝟐 =𝑓 𝑡 0 +ℎ, 𝑤 0 + 𝑲 𝟏 ℎ =2.4375 =𝑓 0.5,0.5+𝟏.𝟓∗0.5 =2 𝒘 𝟐 =𝟏.𝟑𝟕𝟓+ 𝟏 𝟐 ∗𝟐.𝟏𝟐𝟓 + 𝟏 𝟐 ∗𝟐.𝟒𝟑𝟕𝟓 𝟎.𝟓 𝑤 1 = ∗𝟏.𝟓+ 1 2 ∗𝟐 0.5 𝑤 2 = 𝑤 1 =1.375 →𝒚(𝟏)≈𝟐.𝟓𝟏𝟓𝟔𝟐𝟓 →𝒚(𝟎.𝟓)≈𝟏.𝟑𝟕𝟓

1) Euler Method 2) Modified Euler Method 3) Midpoint Method 4) Heun's Method 5) RK4 Method

Example Midpoint method Use the Midpoint method h = 0.5, to approximate 𝑦(0.5) 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ𝑲 𝟐 𝑤 =𝛼 𝑦 ′ =𝑦− 𝑡 2 +1 0≤ 𝑡 ≤4, 𝑦(0) = 0.5 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 + ℎ 2 , 𝑤 𝑖 + ℎ 2 𝑲 𝟏 Midpoint method 𝑤 𝑖+1 = 𝑤 𝑖 +𝒉𝑓 𝑡 𝑖 + ℎ 2 , 𝑤 𝑖 + ℎ 2 𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑤 =𝛼 𝑲 𝟐 =𝑓 𝑡 ∗ℎ, 𝑤 ∗𝑲 𝟏 ℎ 𝑓(𝑡,𝑦)=𝑦− 𝑡 2 +1 𝑲 𝟏 =𝑓 𝑡 0 , 𝑤 0 =𝑓 0.5, ∗𝟏.𝟓∗0.5 =1.8125 =𝑓 0,0.5 =1.5 𝑤 1 =0.5+ 𝟏.𝟖𝟏𝟐𝟓 0.5 = →𝒚(𝟎.𝟓)≈𝟏.𝟒𝟎𝟔𝟐𝟓

1) Euler Method 2) Modified Euler Method 3) Midpoint Method 4) Heun's Method 5) RK4 Method

Heun's Method Example 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 4 𝑲 𝟏 +3 𝑲 𝟑 𝑤 =𝛼 Use the Heun's method with h = 0.2, to approximate 𝑦(0.2) 𝑦 ′ =𝑦− 𝑡 2 +1 0≤ 𝑡 ≤2, 𝑦(0) = 0.5 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 + ℎ 3 , 𝑤 𝑖 + ℎ 3 𝑲 𝟏 𝑲 𝟑 =𝑓 𝑡 𝑖 + 2ℎ 3 , 𝑤 𝑖 + 2ℎ 3 𝑲 𝟐 𝑓(𝑡,𝑦)=𝑦− 𝑡 2 +1 𝑲 𝟑 =𝑓 𝑡 0 +2/15, 𝑤 0 +(2/15)(1.5956) 𝑲 𝟏 =𝑓 𝑡 0 , 𝑤 0 = =𝑓 0,0.5 =1.5 𝑤 1 = 𝟏.𝟓+3∗𝟏 𝑲 𝟐 =𝑓 𝑡 0 +1/15, 𝑤 /15 =1.5956 𝑤 1 = →𝒚(𝟎.𝟐)≈𝟎.𝟖𝟐𝟗𝟐𝟒

𝑶(𝒉) 1) Euler Method 𝑶( 𝒉 𝟐 ) 2) Modified Euler Method 𝑶( 𝒉 𝟐 ) 3) Midpoint Method 𝑶( 𝒉 𝟑 ) 4) Heun's Method 𝑶( 𝒉 𝟒 ) 5) RK4 Method

Modified Euler Method Euler’s Method 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑲 𝟏 𝑤 =𝛼 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 2 𝑲 𝟏 + 𝑲 𝟐 𝑤 =𝛼 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 Heun's Method 𝑲 𝟐 =𝑓 𝑡 𝑖 +ℎ, 𝑤 𝑖 + 𝑲 𝟏 ℎ 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 4 𝑲 𝟏 +3 𝑲 𝟑 𝑤 =𝛼 Midpoint method 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ𝑲 𝟐 𝑤 =𝛼 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 + ℎ 3 , 𝑤 𝑖 + ℎ 3 𝑲 𝟏 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 + ℎ 2 , 𝑤 𝑖 + ℎ 2 𝑲 𝟏 𝑲 𝟑 =𝑓 𝑡 𝑖 + 2ℎ 3 , 𝑤 𝑖 + 2ℎ 3 𝑲 𝟐

clear; clc; y-t.^2+1; h=0.2; alpha = 0.5; tspan = 0:h:4 [w] = midpoint(f,tspan,alpha); [tspan;w]' plot(tspan,w, '-ro'); grid on function [w] = midpoint(f,tspan,alpha) w(1)=alpha; t(1)=tspan(1); n1=length(tspan)-1; h=tspan(2)-tspan(1); for i=1:n1 K1 = f(t(i),w(i)); K2 = f(t(i)+0.5*h,w(i)+0.5*h*K1) w(i+1)=w(i)+(K2)*h t(i+1)=t(i)+h; end; end

1) Euler Method 2) Modified Euler Method 3) Midpoint Method 4) Heun's Method 5) RK4 Method Computational Examples MATLAB Derivations

Fourth-Order Runge-Kutta Methods
Sec:25.3 RUNGE-KUTTA METHODS 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏 The most popular method is the fourth order Runge-Kutta method, or RK4 method Fourth-Order Runge-Kutta Methods 𝑊 𝑖+1 = 𝑊 𝑖 + 𝟏 𝟔 𝑲 𝟏 + 𝟐𝑲 𝟐 + 𝟐𝑲 𝟑 + 𝑲 𝟒 𝑊 =𝛼 𝑲 𝟏 =ℎ 𝐹 𝑡 𝑖 , 𝑊 𝑖 𝑲 𝟐 =ℎ 𝐹 𝑡 𝑖 + ℎ 2 , 𝑊 𝑖 𝑲 𝟏 𝑲 𝟑 =ℎ 𝐹 𝑡 𝑖 + ℎ 2 , 𝑊 𝑖 𝑲 𝟐 𝑲 𝟒 =ℎ 𝐹 𝑡 𝑖+1 , 𝑊 𝑖 + 𝑲 𝟑

Fourth-Order Runge-Kutta Methods
Sec:5.4 RUNGE-KUTTA METHODS Fourth-Order Runge-Kutta Methods Example Use the RK4 method with h = 0.2, to approximate 𝑦(0.2) 𝑊 𝑖+1 = 𝑊 𝑖 + 𝟏 𝟔 𝑲 𝟏 + 𝟐𝑲 𝟐 + 𝟐𝑲 𝟑 + 𝑲 𝟒 𝑊 =𝛼 𝑲 𝟏 =ℎ 𝐹 𝑡 𝑖 , 𝑊 𝑖 𝑦 ′ =𝑦− 𝑡 2 +1 0≤ 𝑡 ≤2, 𝑦(0) = 0.5 𝑲 𝟐 =ℎ 𝐹 𝑡 𝑖 + ℎ 2 , 𝑊 𝑖 𝑲 𝟏 𝑲 𝟑 =ℎ 𝐹 𝑡 𝑖 + ℎ 2 , 𝑊 𝑖 𝑲 𝟐 𝑲 𝟒 =ℎ 𝐹 𝑡 𝑖+1 , 𝑊 𝑖 + 𝑲 𝟑

Implicit Euler’s Method Explicit Euler’s Method
Sec:5.4 RUNGE-KUTTA METHODS Explicit and Implicit Euler Implicit Euler’s Method called implicit because the unknown appears on both sides of the equation. 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝒊+𝟏 , 𝑤 𝒊+𝟏 𝑤 =𝛼 Example Use Implicit Euler’s method with h = 1, to approximate 𝑦(2) Explicit Euler’s Method 𝑦 ′ = 𝑡 1+ 𝑦 ≤ 𝑡 ≤4, 𝑦(1) =1 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝒊 , 𝑤 𝒊 𝑤 =𝛼 Implicit Euler 𝑤 𝑖+1 = 𝑤 𝑖 + ℎ 𝑡 𝑖 𝑤 𝑖+1 2 𝑤 1 = 𝑤 1 2 you need to solve nonlinear equation

Explicit and Implicit Euler Explicit Euler’s Method 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝒊 , 𝑤 𝒊 𝑤 =𝛼 Implicit Euler’s Method Implicit 2ed order Method Implicit 2ed order Method 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝟏 𝟐 𝑓 𝑡 𝒊 , 𝑤 𝒊 + 𝟏 𝟐 𝑓 𝑡 𝒊+𝟏 , 𝑤 𝒊+𝟏 𝑤 =𝛼 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝒊+𝟏 , 𝑤 𝒊+𝟏 𝑤 =𝛼 predictor-corrector method we will use a predicted value, 𝒘 𝒊+𝟏 ∗ obtained with Explicit Euler to evaluate an approximation to the value 𝒘 𝒊+𝟏 at the new point. Predictor: 𝒘 𝒊+𝟏 ∗ = 𝑤 𝑖 +ℎ𝑓 𝑡 𝒊 , 𝑤 𝒊 𝒘 𝒊+𝟏 = 𝑤 𝑖 +ℎ 𝟏 𝟐 𝑓 𝑡 𝒊 , 𝑤 𝒊 + 𝟏 𝟐 𝑓 𝑡 𝒊+𝟏 , 𝒘 𝒊+𝟏 ∗ Corrector:

ODE Solver in MATLAB Ode45, ode23, ode113, ode15s, ode23s, ode23t, ode23tb, ode15i, Example Example 𝑦 ′ =𝒚− 𝒕 𝟐 +𝟏 𝟎≤ 𝒕 ≤𝟒, 𝑦(0) = 𝟎.𝟓 [t,y] = ode23(odefun,tspan,y0) y-t^2+1,[0 4],0.5); plot(t,y);grid on

ODE Solver in MATLAB Ode45, ode23, ode113, ode15s, ode23s, ode23t, ode23tb, ode15i, Example Example 𝒚 𝟏 ′ = 𝒚 𝟐 [t,y] = ode45(odefun,tspan,y0) 𝒚 𝟐 ′ = 𝟏− 𝒚 𝟏 𝟐 𝒚 𝟐 − 𝒚 𝟐 𝟎≤ 𝒕 ≤𝟐𝟎, 𝒚 𝟏 (𝟎) 𝒚 𝟐 (𝟎) = 𝟐 𝟎 F [y(2); (1-y(1)^2)*y(2)-y(1)]; [t,y] = ode45(F,[0 20],[2; 0]); plot(t,y(:,1),'-o',t,y(:,2),'-o')

Solver Problem Type Accuracy When to Use ode45 Nonstiff Medium Most of the time. ode45 should be the first solver you try. ode23 Low ode23 can be more efficient than ode45 at problems with crude tolerances, or in the presence of moderate stiffness. Low to High ode113 can be more efficient than ode45 at problems with stringent error tolerances, or when the ODE function is expensive to evaluate. ode113 ode15s Stiff Low to Medium Try ode15s when ode45 fails or is inefficient and you suspect that the problem is stiff. Also use ode15s when solving differential algebraic equations (DAEs). ode23s can be more efficient than ode15s at problems with crude error tolerances. It can solve some stiff problems for which ode15s is not effective. ode23s computes the Jacobian in each step, so it is beneficial to provide the Jacobian via odesetto maximize efficiency and accuracy. If there is a mass matrix, it must be constant. ode23s ode23t Use ode23t if the problem is only moderately stiff and you need a solution without numerical damping. ode23t can solve differential algebraic equations (DAEs). ode23tb Like ode23s, the ode23tb solver might be more efficient than ode15s at problems with crude error tolerances. ode15i Fully implicit Use ode15i for fully implicit problems f(t,y,y’) = 0 and for differential algebraic equations (DAEs) of index 1. Sec:25.3 RUNGE-KUTTA METHODS

The motion of a swinging pendulum under certain simplifying assumptions is described by the second-order differential equation 𝒅 𝟐 𝜽 𝒅 𝒕 𝟐 + 𝒈 𝑳 𝐬𝐢𝐧 𝜽=𝟎 where L=2 is the length of the pendulum, g ≈ ft/s2 is the gravitational constant of the earth, and θ is the angle the pendulum makes with the vertical. If, in addition, we specify the position of the pendulum when the motion begins, 𝜃( 𝑡 0 ) = 𝜋 6 , and its velocity at that point, 𝜃′( 𝑡 0 ) =0, we have what is called an initial-value problem. Use midpoint method with h = 0.1 s, compute the angle θ obtained for this initial-value problems at t = 0, 1, and 2 s.

Convert into system of 1st order ode
Sec:5.4 RUNGE-KUTTA METHODS 𝒅 𝟐 𝜽 𝒅 𝒕 𝟐 + 𝒈 𝑳 𝐬𝐢𝐧 𝜽=𝟎 Convert into system of 1st order ode 𝑦 1 =𝜃 𝑦 2 = 𝑑𝜃 𝑑𝑡 𝑑 𝑦 1 𝑑𝑡 = 𝑑𝜃 𝑑𝑡 𝑑 𝑦 2 𝑑𝑡 = 𝑑 2 𝜃 𝑑 𝑡 2 𝑑 𝑦 1 𝑑𝑡 = 𝑦 2 𝑑 𝑦 2 𝑑𝑡 =− 𝑔 𝐿 sin 𝑦 1 𝑌′ 𝑡 = 𝑑 𝑦 1 𝑑𝑡 𝑑 𝑦 2 𝑑𝑡 𝑌 𝑡 = 𝑦 1 (𝑡) 𝑦 2 (𝑡) 𝑌′ 𝑡 =𝐹(𝑡,𝑌) 𝜃(0) = 𝜋 6 𝐹 𝑡,𝑌 = 𝑦 2 − 𝑔 𝐿 sin 𝑦 1 𝜃′(0) =0 𝑌 0 = 𝜋/6 0

𝒅 𝟐 𝜽 𝒅 𝒕 𝟐 + 𝒈 𝑳 𝐬𝐢𝐧 𝜽=𝟎 𝑌′ 𝑡 =𝐹(𝑡,𝑌) 𝐹 𝑡,𝑌 = 𝑦 2 − 𝑔 𝐿 sin 𝑦 1 𝜽(𝟎) = 𝝅 𝟔 𝜽′(𝟎) =𝟎 𝑌 0 = 𝜋/6 0 Midpoint method 𝑊 𝑖+1 = 𝑊 𝑖 + 𝑲 𝟐 ℎ 𝑊 =𝛼 𝑊 𝑖 is a vector of size 2×1 𝑲 𝟏 , 𝑲 𝟐 and 𝜶 are vector of size 2×1 𝑲 𝟏 =𝐹 𝑡 𝑖 , 𝑊 𝑖 ℎ and 𝑡 𝑖 are scalars 𝑲 𝟐 =𝐹 𝑡 𝑖 ℎ, 𝑊 𝑖 𝑲 𝟏 ℎ 𝐹: 𝑡, 𝑦 1 𝑦 →

Remark: use this modified for system
Sec:5.4 RUNGE-KUTTA METHODS Midpoint method clear; clc; L = 2; g=32.17; [ Y(2) ; ... -(g/L)*sin(Y(1))]; t0=0; tf=2; n=1000; h=(tf-t0)/n; alpha = [ pi/6 ; 0 ]; tspan = t0:h:tf; [w] = midpoint(F,tspan,alpha); [tspan',w']; plot(tspan,w(1,:));grid on 𝑊 𝑖+1 = 𝑊 𝑖 + 𝑲 𝟐 ℎ 𝑊 =𝛼 𝑲 𝟏 =𝐹 𝑡 𝑖 , 𝑊 𝑖 𝑲 𝟐 =𝐹 𝑡 𝑖 ℎ, 𝑊 𝑖 𝑲 𝟏 ℎ function [w] = midpoint(f,tspan,alpha) w(:,1)=alpha; t(1)=tspan(1); n1=length(tspan)-1; h=tspan(2)-tspan(1); for i=1:n1 K1 = f(t(i),w(:,i)); K2 = f(t(i)+0.5*h,w(:,i)+0.5*h*K1); w(:,i+1)=w(:,i)+(K2)*h; t(i+1)=t(i)+h; end; end 𝑌′ 𝑡 =𝐹(𝑡,𝑌) 𝐹 𝑡,𝑌 = 𝑦 2 − 𝑔 𝐿 sin 𝑦 1 Remark: use this modified for system 𝑌 0 = 𝜋/6 0

𝒅 𝟐 𝜽 𝒅 𝒕 𝟐 + 𝒈 𝑳 𝐬𝐢𝐧 𝜽=𝟎 t(:) w(1,:) w(2,:) 𝜽(𝟎) = 𝝅 𝟔 𝜽′(𝟎) =𝟎 𝜽 𝜽′

clear; clc; L = 2; g=32.17; [ Y(2) ; ... -(g/L)*sin(Y(1))]; t0=0; tf=10; n=1000; h=(tf-t0)/n; alpha = [ pi/4 ; 0 ]; tspan = t0:h:tf; [w] = midpoint(F,tspan,alpha); [tspan',w']; plot(tspan,w(1,:));grid on function [w] = midpoint(f,tspan,alpha) w(:,1)=alpha; t(1)=tspan(1); n1=length(tspan)-1; h=tspan(2)-tspan(1); for i=1:n1 K1 = f(t(i),w(:,i)); K2 = f(t(i)+0.5*h,w(:,i)+0.5*h*K1); w(:,i+1)=w(:,i)+(K2)*h; t(i+1)=t(i)+h; end; end 𝑌′ 𝑡 =𝐹(𝑡,𝑌) 𝐹 𝑡,𝑌 = 𝑦 2 − 𝑔 𝐿 sin 𝑦 1 𝑌 0 = 𝝅/𝟒 0 𝒕 𝒇 =𝟏𝟎

Theorem 513 (nth Taylor polynomial in two variables)
Sec:5.4 RUNGE-KUTTA METHODS nth Taylor polynomial in two variables 𝑓 𝑥 =𝑓 𝑥 0 +(𝑥− 𝑥 0 )𝑓′ 𝑥 (𝑥− 𝑥 0 ) 2 𝑓′′ 𝑥 (𝑥− 𝑥 0 ) 3 𝑓′′′ 𝑥 0 +⋯ Theorem 513 (nth Taylor polynomial in two variables) 1st derivatives 2ed nth derivatives

𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏 Taylor’s Series Methods Initial Value Problem Euler’s Method 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑤 =𝛼 𝑦 ′ =𝑓 𝑡,𝑦 𝑦(𝑎) =𝛼 (*) 𝑎≤ 𝑡 ≤𝑏, We will use Taylor’s Theorem to derive Euler’s method. Suppose that 𝒚(𝒕), the unique solution to (*) Taylor’s Series Methods 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 𝒕 𝒊+𝟏 −𝒕 𝒊 𝒚 ′ 𝒕 𝒊 𝑡 𝑖+1 −𝑡 𝑖 𝟐 𝑦 ′′ 𝒕 𝒊 +⋯ 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 𝒕 𝒊+𝟏 −𝒕 𝒊 𝒚 ′ 𝒕 𝒊 + 𝑡 𝑖+1 −𝑡 𝑖 𝟐 𝑦 ′′ 𝜉 𝑖 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 +𝒉𝒚′ 𝒕 𝒊 𝒉 2 𝒚′′ 𝒕 𝒊 +⋯ 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 +𝒉𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 𝒉 𝟐 𝒇′ 𝒕 𝒊 ,𝑦 𝑡 𝑖 +⋯ 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 +𝒉𝒚′ 𝒕 𝒊 + 𝒉 2 𝑦 ′′ 𝜉 𝑖 𝒇 ′ 𝒕 𝒊 ,𝒚 𝒕 𝒊 = 𝝏 𝝏𝒕 𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 = 𝒇 𝒕 + 𝒇 𝒚 𝒚′ 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 +𝒉𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 + 𝒉 𝟐 𝑦 ′′ 𝜉 𝑖 𝑦 𝑡 𝑖+1 ≈𝑦 𝑡 𝑖 +𝒉𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 𝑦 𝑡 𝑖+1 =𝑦 𝑡 𝑖 +𝒉𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 𝒉 𝟐 𝒇 𝒕 + 𝒇 𝒚 𝒇 +⋯ 𝑤 𝑖+1 = 𝑤 𝑖 + 𝒉𝒇 𝒕 𝒊 , 𝒘 𝒊 𝑦 𝑡 𝑖+1 ≈𝑦 𝑡 𝑖 +𝒉𝒇 𝒕 𝒊 ,𝑦 𝑡 𝑖 𝒉 𝟐 𝒇 𝒕 + 𝒇 𝒚 𝒇 One way to reduce the error of Euler’s method would be to include higher-order terms of the Taylor series expansion in the solution. 𝑤 𝑖+1 = 𝑤 𝑖 +𝒉𝒇 𝒕 𝒊 , 𝑤 𝑖 𝒉 𝟐 𝒇 𝒕 + 𝒇 𝒚 𝒇 𝒕 𝒊 , 𝑤 𝑖 Higher-order derivatives become increasingly more complicated

Taylor’s Series Methods 𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏 Taylor’s Series Methods 𝒚 𝒕 𝒊+𝟏 =𝒚 𝒕 𝒊 +𝒉 𝒚 ′ 𝒕 𝒊 + 𝟏 𝟐 𝒉 𝟐 𝒚 ′′ 𝒕 𝒊 + 𝟏 𝟔 𝒉 𝟑 𝒚 ′′′ 𝒕 𝒊 +⋯ 𝒚 𝒕 𝒊+𝟏 =𝒚 𝒕 𝒊 +𝒉𝒇+ 𝟏 𝟐 𝒉 𝟐 𝒇 ′ + 𝟏 𝟔 𝒉 𝟑 𝒇 ′′ +⋯ 𝒚 𝒕 𝒊+𝟏 =𝒚 𝒕 𝒊 +𝒉 𝒇+ 𝟏 𝟐 𝒉 𝒇 ′ + 𝟏 𝟔 𝒉 𝟐 𝒇 ′′ +⋯ Taylor’s Series Method of order one 𝒚 𝒕 𝒊+𝟏 =𝒚 𝒕 𝒊 +𝒉 𝒇+ 𝟏 𝟐 𝒉 𝒇 ′ + 𝟏 𝟔 𝒉 𝟐 𝒇 ′′ +⋯ Euler’s Method Taylor’s Series Method of order two 𝒇 ′ = 𝒇 𝒕 + 𝒇 𝒚 𝒚′ = 𝒇 𝒕 + 𝒇 𝒚 𝒇 𝒕 𝒊 ,𝒚 𝒕 𝒊 𝒚 𝒕 𝒊+𝟏 =𝒚 𝒕 𝒊 +𝒉 𝒇+ 𝟏 𝟐 𝒉 𝒇 ′ + 𝟏 𝟔 𝒉 𝟐 𝒇 ′′ +⋯ 𝒘 𝒊+𝟏 = 𝒘 𝒊 +𝒉 𝒇+ 𝟏 𝟐 𝒉 𝒇 𝒕 + 𝟏 𝟐 𝒉𝒇 𝒇 𝒚 𝒕 𝒊 , 𝒘 𝒊

Taylor series in 2-dimension
Sec:5.4 RUNGE-KUTTA METHODS Midpoint method Mod EulerMethod 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑲 𝟏 𝑲 𝟐 𝑤 𝑖+1 = 𝑤 𝑖 +ℎ 𝑲 𝟐 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟏 =𝑓 𝑡 𝑖 , 𝑤 𝑖 𝑲 𝟐 =𝑓 𝑡 𝑖 ℎ, 𝑤 𝑖 𝑲 𝟏 ℎ 𝑲 𝟐 =𝑓 𝑡 𝑖 +ℎ, 𝑤 𝑖 + 𝑲 𝟏 ℎ 𝑲 𝟏 =𝒇 𝒕 𝒊 , 𝒘 𝒊 Taylor’s Series Method of order two 𝑲 𝟐 =𝒇 𝒕 𝒊 + 𝒑 𝟏 𝒉, 𝒘 𝒊 + 𝒒 𝟏𝟏 𝑲 𝟏 𝒉 𝒘 𝒊+𝟏 = 𝒘 𝒊 +𝒉 𝒇+ 𝟏 𝟐 𝒉 𝒇 𝒕 + 𝟏 𝟐 𝒉𝒇 𝒇 𝒚 𝒕 𝒊 , 𝒘 𝒊 𝒘 𝒊+𝟏 = 𝒘 𝒊 +𝒉 𝒂 𝟏 𝑲 𝟏 + 𝒂 𝟐 𝑲 𝟐 Taylor series in 2-dimension + <2ed deriv>+⋯ =𝒇 𝒕 𝒊 , 𝒘 𝒊 + 𝒑 𝟏 𝒉 𝒇 𝒕 𝒕 𝒊 , 𝒘 𝒊 + 𝒒 𝟏𝟏 𝑲 𝟏 𝒉 𝒇 𝒚 𝒕 𝒊 , 𝒘 𝒊 𝑲 𝟐 𝒂 𝟏 𝑲 𝟏 + 𝒂 𝟐 𝑲 𝟐 = 𝒂 𝟏 𝒇+ 𝒂 𝟐 𝒇+ 𝒂 𝟐 𝒑 𝟏 𝒉 𝒇 𝒕 + 𝒂 𝟐 𝒒 𝟏𝟏 𝑲 𝟏 𝒉 𝒇 𝒚 +⋯ Matching the coefficients of f and its derivatives 𝒂 𝟐 𝒑 𝟏 = 𝟏 𝟐 𝒂 𝟐 𝒒 𝟏𝟏 = 𝟏 𝟐 𝒂 𝟏 + 𝒂 𝟐 =𝟏

𝑫𝒆𝒓𝒊𝒗𝒂𝒕𝒊𝒐𝒏 𝑲 𝟏 =𝒇 𝒕 𝒊 , 𝒘 𝒊 Matching the coefficients of f and its derivatives 𝑲 𝟐 =𝒇 𝒕 𝒊 + 𝒑 𝟏 𝒉, 𝒘 𝒊 + 𝒒 𝟏𝟏 𝑲 𝟏 𝒉 𝒂 𝟏 + 𝒂 𝟐 =𝟏 𝒘 𝒊+𝟏 = 𝒘 𝒊 +𝒉 𝒂 𝟏 𝑲 𝟏 + 𝒂 𝟐 𝑲 𝟐 𝒂 𝟐 𝒑 𝟏 = 𝟏 𝟐 three equations four unknown We have infinite number of solutions , there are an infinite number of second-order RK methods. 𝒂 𝟐 𝒒 𝟏𝟏 = 𝟏 𝟐 We present two of the most commonly used and preferred versions: 𝒂 𝟐 = 𝟏 𝟐 𝒂 𝟐 =𝟏 Modefied Euler Method Midpoint method

Sec:5.4 RUNGE-KUTTA METHODS.

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Sec:5.4 RUNGE-KUTTA METHODS.

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Presentation on theme: "Sec:5.4 RUNGE-KUTTA METHODS."— Presentation transcript:

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