Ordinary Differential Equations Scientific Computing Numerical Solution Of Ordinary Differential Equations - Further Analysis

Solving First Order ODE’s A general form for a first order ODE is Or alternatively We desire a solution x(t) which satisfies this ODE and one specified boundary condition. x(a) = c

Numerical Solution The value of the true solution is approximated at a set of n values of t. We denote the approximation at these pts by so that

Numerical Solution Let x(t) be the actual solution for dx/dt at the step values. Then, assuming no roundoff error, the difference in the calculated and true value is the truncation error,

Taylor Series Last time we looked at expressing the solution x(t) about some starting point t0 using a Taylor expansion. The ODE is given by

Error Analysis for Taylor Approximation Then, if we use the first n terms of the series as an approximation for the solution, we get:

Error Analysis for Taylor Approximation We saw that the local truncation error is bounded as follows: where

Truncation Error vs Roundoff Error

Stability of solutions Solution of ODE is Stable if solutions resulting from perturbations of initial value remain close to original solution Asymptotically stable if solutions resulting from perturbations converge back to original solution Unstable if solutions resulting from perturbations diverge away from original solution without bound

Example: Stable Solutions

Example: Unstable solution ODE x’ = x is unstable (solution is x(t)=cet ) we show solutions with Euler’s method

Example: Unstable solution

Example: asymptotically stable solutions ODE x’ = -x is stable (solution is x(t)= ce-t ) (solution -> 0) if h too large, numerical solution is unstable we show solutions with Euler’s method in red

x’=-x, stable but slow solution

x’=-x, stable, rather inaccurate solution for h large

x’=-x, stable but poor solution

Stability Analysis The solution to satisfying some initial condition x(a) = s will be: unstable if fx ≥ δ for some positive δ and stable if fx ≤ -δ for some positive δ. Proof: Let x(t,s) be the solution to the IVP above. Then Since t and s are independent variables, then dt/ds=0, so ->

Stability Analysis Proof(continued): So, if we assume x is C1 Let The solution to this diff eq is u(t) = c eQ(t) , where If q(r) = fx(r,x(r,s)) ≥ δ >0 then Q-> ∞ so, u-> ∞ and the solution is unstable. If q(r) = fx(r,x(r,s)) ≤ -δ <0 then Q-> -∞ so, u-> 0 and the solution is stable. QED

Euler’s Method Review: The simple Euler method is a linear approximation, and is a perfect solution only if the function is linear (or at least linear in the interval). This is inherently inaccurate. Because of this inaccuracy small step sizes are required when using the algorithm.

Better Methods If we took more terms in the Taylor’s Series expansion of f(t,x), we would get a more accurate solution algorithm. Problem: It is difficult to compute higher-order derivatives numerically. Idea: Approximate the value of higher derivatives of f(t,x) by evaluating f several times between iterates: ti and ti+1 .

Better Methods The ODE +IV x(a)= x0 =s can be alternatively solved as an integral equation. For example, on the first interval from x0 to x1 :

Better Methods In general: The problem is that we don’t know x(t), so evaluating the integral is not trivial. We assume that if h is small enough, f(t,x) will not change that much over the interval [ti, ti+1]. Using “left-rectangular integration,” This is the Euler Method. A family of improved versions are called Runge-Kutta methods

Euler Method Euler Solution x xi true solution h ti ti+1 = ti + h Tuncation error: true solution

Euler vs Actual Solution Euler Solution RK methods differ in how they estimate fi For Euler Method (a type of RK method), fi ≈ f(ti, xi) x xi fi true solution h ti ti+1 = ti + h

Runge-Kutta Order 2 Method x xi fi h Weighted average of two slopes x ti ti+αh ti+1