{-# LANGUAGE NoImplicitPrelude #-} {- | Lazy evaluation allows for the solution of differential equations in terms of power series. Whenever you can express the highest derivative of the solution as explicit expression of the lower derivatives where each coefficient of the solution series depends only on lower coefficients, the recursive algorithm will work. -} module MathObj.PowerSeries.DifferentialEquation where import qualified MathObj.PowerSeries as PS import qualified MathObj.PowerSeries.Example as PSE import qualified Algebra.Field as Field import qualified Algebra.ZeroTestable as ZeroTestable import NumericPrelude import PreludeBase {- | Example for a linear equation: Setup a differential equation for @y@ with > y t = (exp (-t)) * (sin t) > y' t = -(exp (-t)) * (sin t) + (exp (-t)) * (cos t) > y'' t = -2 * (exp (-t)) * (cos t) Thus the differential equation > y'' = -2 * (y' + y) holds. The following function generates a power series for @exp (-t) * sin t@ by solving the differential equation. -} solveDiffEq0 :: (Field.C a) => [a] solveDiffEq0 = let -- the initial conditions are passed to "PS.integrate" y = PS.integrate 0 y' y' = PS.integrate 1 y'' y'' = PS.scale (-2) (PS.add y' y) in y verifyDiffEq0 :: (Field.C a) => [a] verifyDiffEq0 = PS.mul (zipWith (*) (iterate negate 1) PSE.exp) PSE.sin propDiffEq0 :: Bool propDiffEq0 = solveDiffEq0 == (verifyDiffEq0 :: [Rational]) {- | We are not restricted to linear equations! Let the solution be y with y t = (1-t)^-1 y' t = (1-t)^-2 y'' t = 2*(1-t)^-3 then it holds y'' = 2 * y' * y -} solveDiffEq1 :: (ZeroTestable.C a, Field.C a) => [a] solveDiffEq1 = let -- the initial conditions are passed to "PS.integrate" y = PS.integrate 1 y' y' = PS.integrate 1 y'' y'' = PS.scale 2 (PS.mul y' y) in y verifyDiffEq1 :: (ZeroTestable.C a, Field.C a) => [a] verifyDiffEq1 = PS.divide [1] [1, -1] propDiffEq1 :: Bool propDiffEq1 = solveDiffEq1 == (verifyDiffEq1 :: [Rational])