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% -*- REDUCE -*- % The Postel/Zimmermann (11/4/96) ODE test examples. % Equation names from Postel/Zimmermann. % This version uses Maple-style functional notation wherever possible. % on trode; on div, intstr; off allfac; % to look prettier % 1 Single equations without initial conditions % ============================================== % 1.1 Linear equations % ==================== operator y; % (1) Linear Bernoulli 1 odesolve((x^4-x^3)*df(y(x),x) + 2*x^4*y(x) = x^3/3 + C, y(x), x); % (2) Linear Bernoulli 2 odesolve(-1/2*df(y(x),x) + y(x) = sin x, y(x), x); % (3) Linear change of variables (FJW: shifted Euler equation) odesolve(df(y(x),x,2)*(a*x+b)^2 + 4df(y(x),x)*(a*x+b)*a + 2y(x)*a^2 = 0, y(x), x); % (4) Adjoint odesolve((x^2-x)*df(y(x),x,2) + (2x^2+4x-3)*df(y(x),x) + 8x*y(x) = 1, y(x), x); % (5) Polynomial solutions % (FJW: Currently very slow, and fails anyway!) % odesolve((x^2-x)*df(y(x),x,2) + (1-2x^2)*df(y(x),x) + (4x-2)*y(x) = 0, % y(x), x); % (6) Dependent variable missing odesolve(df(y(x),x,2) + 2x*df(y(x),x) = 2x, y(x), x); % (7) Liouvillian solutions % (FJW: INTEGRATION IMPOSSIBLY SLOW WITHOUT EITHER ALGINT OR NOINT OPTION) begin scalar !*allfac; !*allfac := t; return odesolve((x^3/2-x^2)*df(y(x),x,2) + (2x^2-3x+1)*df(y(x),x) + (x-1)*y(x) = 0, y(x), x, noint); end; % WARNING: DO NOT RE-EVALUATE RESULT WITHOUT TURNING ON THE NOINT SWITCH % (8) Reduction of order % (FJW: Attempting to make explicit currently too slow.) odesolve(df(y(x),x,2) - 2x*df(y(x),x) + 2y(x) = 3, y(x), x); % (9) Integrating factors % (FJW: Currently very slow, and fails anyway!) % odesolve(sqrt(x)*df(y(x),x,2) + 2x*df(y(x),x) + 3y(x) = 0, y(x), x); % (10) Radical solution (FJW: omitted for now) % (11) Undetermined coefficients odesolve(df(y(x),x,2) - 2/x^2*y(x) = 7x^4 + 3*x^3, y(x), x); % (12) Variation of parameters odesolve(df(y(x),x,2) + y(x) = csc(x), y(x), x); % (13) Linear constant coefficients << factor exp(x); write odesolve(df(y(x),x,7) - 14df(y(x),x,6) + 80df(y(x),x,5) - 242df(y(x),x,4) + 419df(y(x),x,3) - 416df(y(x),x,2) + 220df(y(x),x) - 48y(x) = 0, y(x), x); remfac exp(x) >>; % (14) Euler odesolve(df(y(x),x,4) - 4/x^2*df(y(x),x,2) + 8/x^3*df(y(x),x) - 8/x^4*y(x) = 0, y(x), x); % (15) Exact n-th order odesolve((1+x+x^2)*df(y(x),x,3) + (3+6x)*df(y(x),x,2) + 6df(y(x),x) = 6x, y(x), x); % 1.2 Nonlinear equations % ======================= % (16) Integrating factors 1 odesolve(df(y(x),x) = y(x)/(y(x)*log y(x) + x), y(x), x); % (17) Integrating factors 2 odesolve(2y(x)*df(y(x),x)^2 - 2x*df(y(x),x) - y(x) = 0, y(x), x); % This parametric solution is correct, cf. Zwillinger (1989) p.168 (41.10) % (except that first edition is missing the constant C)! % (18) Bernoulli 1 odesolve(df(y(x),x) + y(x) = y(x)^3*sin x, y(x), x, explicit); expand_plus_or_minus ws; % (19) Bernoulli 2 operator P, Q; begin scalar soln, !*exp, !*allfac; % for a neat solution on allfac; soln := odesolve(df(y(x),x) + P(x)*y(x) = Q(x)*y(x)^n, y(x), x); off allfac; return soln end; odesolve(df(y(x),x) + P(x)*y(x) = Q(x)*y(x)^(2/3), y(x), x); % (20) Clairaut 1 odesolve((x^2-1)*df(y(x),x)^2 - 2x*y(x)*df(y(x),x) + y(x)^2 - 1 = 0, y(x), x, explicit); % (21) Clairaut 2 operator f, g; odesolve(f(x*df(y(x),x)-y(x)) = g(df(y(x),x)), y(x), x); % (22) Equations of the form y' = f(x,y) odesolve(df(y(x),x) = (3x^2-y(x)^2-7)/(exp(y(x))+2x*y(x)+1), y(x), x); % (23) Homogeneous odesolve(df(y(x),x) = (2x^3*y(x)-y(x)^4)/(x^4-2x*y(x)^3), y(x), x); % (24) Factoring the equation odesolve(df(y(x),x)*(df(y(x),x)+y(x)) = x*(x+y(x)), y(x), x); % (25) Interchange variables % (NB: Soln in Zwillinger (1989) wrong, as is last eqn in Table 68!) odesolve(df(y(x),x) = x/(x^2*y(x)^2+y(x)^5), y(x), x); % (26) Lagrange 1 odesolve(y(x) = 2x*df(y(x),x) - a*df(y(x),x)^3, y(x), x); odesolve(y(x) = 2x*df(y(x),x) - a*df(y(x),x)^3, y(x), x, implicit); % root_of quartic is VERY slow if explicit option used! % (27) Lagrange 2 odesolve(y(x) = 2x*df(y(x),x) - df(y(x),x)^2, y(x), x); odesolve(y(x) = 2x*df(y(x),x) - df(y(x),x)^2, y(x), x, implicit); % (28) Riccati 1 odesolve(df(y(x),x) = exp(x)*y(x)^2 - y(x) + exp(-x), y(x), x); % (29) Riccati 2 << factor x; write odesolve(df(y(x),x) = y(x)^2 - x*y(x) + 1, y(x), x); remfac x >>; % (30) Separable odesolve(df(y(x),x) = (9x^8+1)/(y(x)^2+1), y(x), x); % (31) Solvable for x odesolve(y(x) = 2x*df(y(x),x) + y(x)*df(y(x),x)^2, y(x), x); odesolve(y(x) = 2x*df(y(x),x) + y(x)*df(y(x),x)^2, y(x), x, implicit); % (32) Solvable for y begin scalar !*allfac; !*allfac := t; return odesolve(x = y(x)*df(y(x),x) - x*df(y(x),x)^2, y(x), x) end; % (33) Autonomous 1 odesolve(df(y(x),x,2)-df(y(x),x) = 2y(x)*df(y(x),x), y(x), x, explicit); % (34) Autonomous 2 (FJW: Slow without either algint or noint option.) odesolve(df(y(x),x,2)/y(x) - df(y(x),x)^2/y(x)^2 - 1 + 1/y(x)^3 = 0, y(x), x, noint); % (35) Differentiation method odesolve(2y(x)*df(y(x),x,2) - df(y(x),x)^2 = 1/3(df(y(x),x) - x*df(y(x),x,2))^2, y(x), x, explicit); % (36) Equidimensional in x odesolve(x*df(y(x),x,2) = 2y(x)*df(y(x),x), y(x), x, explicit); % (37) Equidimensional in y odesolve((1-x)*(y(x)*df(y(x),x,2)-df(y(x),x)^2) + x^2*y(x)^2 = 0, y(x), x); % (38) Exact second order odesolve(x*y(x)*df(y(x),x,2) + x*df(y(x),x)^2 + y(x)*df(y(x),x) = 0, y(x), x, explicit); % (39) Factoring differential operator odesolve(df(y(x),x,2)^2 - 2df(y(x),x)*df(y(x),x,2) + 2y(x)*df(y(x),x) - y(x)^2 = 0, y(x), x); % (40) Scale invariant (fails with algint option) odesolve(x^2*df(y(x),x,2) + 3x*df(y(x),x) = 1/(y(x)^3*x^4), y(x), x); % Revised scale-invariant example (hangs with algint option): ode := x^2*df(y(x),x,2) + 3x*df(y(x),x) + 2*y(x) = 1/(y(x)^3*x^4); % Choose full (explicit and expanded) solution: odesolve(ode, y(x), x, full); % or "explicit, expand" % Check it -- each solution should simplify to 0: foreach soln in ws collect trigsimp sub(soln, num(lhs ode - rhs ode)); % (41) Autonomous, 3rd order odesolve((df(y(x),x)^2+1)*df(y(x),x,3) - 3df(y(x),x)*df(y(x),x,2)^2 = 0, y(x), x); % odesolve((df(y(x),x)^2+1)*df(y(x),x,3) - 3df(y(x),x)*df(y(x),x,2)^2 = 0, % y(x), x, implicit); % Implicit form is currently too messy! % (42) Autonomous, 4th order odesolve(3*df(y(x),x,2)*df(y(x),x,4) - 5df(y(x),x,3)^2 = 0, y(x), x); % 1.3 Special equations % ===================== % (43) Delay odesolve(df(y(x),x) + a*y(x-1) = 0, y(x), x); % (44) Functions with several parameters odesolve(df(y(x,a),x) = a*y(x,a), y(x,a), x); % 2 Single equations with initial conditions % =========================================== % (45) Exact 4th order odesolve(df(y(x),x,4) = sin x, y(x), x, {x=0, y(x)=0, df(y(x),x)=0, df(y(x),x,2)=0, df(y(x),x,3)=0}); % (46) Linear polynomial coefficients -- Bessel J0 odesolve(x*df(y(x),x,2) + df(y(x),x) + 2x*y(x) = 0, y(x), x, {x=0, y(x)=1, df(y(x),x)=0}); % (47) Second-degree separable soln := odesolve(x*df(y(x),x)^2 - y(x)^2 + 1 = 0, y(x)=1, x=0, explicit); % Alternatively ... soln where e^~x => cosh x + sinh x; % but this works ONLY with `on div, intstr; off allfac;' % A better alternative is ... trigsimp(soln, hyp, combine); expand_plus_or_minus ws; % (48) Autonomous odesolve(df(y(x),x,2) + y(x)*df(y(x),x)^3 = 0, y(x), x, {x=0, y(x)=0, df(y(x),x)=2}); %% Only one explicit solution satisfies the conditions: begin scalar !*trode, !*fullroots; !*fullroots := t; return odesolve(df(y(x),x,2) + y(x)*df(y(x),x)^3 = 0, y(x), x, {x=0, y(x)=0, df(y(x),x)=2}, explicit); end; % 3 Systems of equations % ======================= % (49) Integrable combinations operator x, z; odesolve({df(x(t),t) = -3y(t)*z(t), df(y(t),t) = 3x(t)*z(t), df(z(t),t) = -x(t)*y(t)}, {x(t),y(t),z(t)}, t); % (50) Matrix Riccati operator a, b; odesolve({df(x(t),t) = a(t)*(y(t)^2-x(t)^2) + 2b(t)*x(t)*y(t) + 2c*x(t), df(y(t),t) = b(t)*(y(t)^2-x(t)^2) - 2a(t)*x(t)*y(t) + 2c*y(t)}, {x(t),y(t)}, t); % (51) Triangular odesolve({df(x(t),t) = x(t)*(1 + cos(t)/(2+sin(t))), df(y(t),t) = x(t) - y(t)}, {x(t),y(t)}, t); % (52) Vector odesolve({df(x(t),t) = 9x(t) + 2y(t), df(y(t),t) = x(t) + 8y(t)}, {x(t),y(t)}, t); % (53) Higher order odesolve({df(x(t),t) - x(t) + 2y(t) = 0, df(x(t),t,2) - 2df(y(t),t) = 2t - cos(2t)}, {x(t),y(t)}, t); % (54) Inhomogeneous system equ := {df(x(t),t) = -1/(t*(t^2+1))*x(t) + 1/(t^2*(t^2+1))*y(t) + 1/t, df(y(t),t) = -t^2/(t^2+1)*x(t) + (2t^2+1)/(t*(t^2+1))*y(t) + 1}; odesolve(equ, {x(t),y(t)}, t); end;