+ j&^RIHt^RIHt!RR]4tRtR Rlt]]n]R8Xd^RIt]P!4R#R#) )bisect)xrangec]tRt^tRtR#) ODEMethodsN)__name__ __module__ __qualname____firstlineno____static_attributes__rU/home/luca/screen-checking/.venv/lib/python3.14/site-packages/mpmath/calculus/odes.pyrrsr rc  VP^V)4;rg\V4pV.p V.p Tp Tp VPp V ^V,,Vn\V4FrpV!W4p\ \V 44Uu.uFqV,WoV,,,NK! p pW, p V P V 4V P V 4Kt \V4Uu.uFp.NK pp\V^,4EFp^.V,pRV^,,p^p\V^,4Ffp\V4F,pVV;;,VW,V,,, uu&K. VVV, ^,,V),pV^, pKh VV),VP V4, p\V4F5pVV,V,VV&VV,P VV,4K7 EK WnVPpVFGpVR,'gK\VVPV\VR,4, V44pKI V^,pVVV,3#uupiuupi Yni;i)) ldexplenprecrangerappendfaconeminnthrootabs)ctxderivsx0y0tol_precnhtoldimxsysxyorigifxydserjsbkscaleradiustss&&&&&& r ode_taylorr5siiH9%%A b'C B B A A 88D1:qA,C(.s1v711aAhA7 FA IIaL IIaL !:&:ar:&qsACAQAA1Q3ZsAaDAaL(D$!A#a%[qb)Q  Gcggaj(E3Zte|!A ad# WWF b66SRV_a!@AF aKF 6 >98 's+AI;8%I1=I; I6%DI;1 I;;JNc aaaaaa a a aaaaaV'd%\SPV^4)4^ ,oMSP^ ,oS;'g+^\^SP,R, 4,oSP^(,o\ V4Ro\ SSSVSS4wrSV .oVSV 3.oV3Rlo VVVV VVVVV3 Rlo VV V VV3Rlp V # \ dSo T 3RloT.pRoLVi;i)a Returns a function `y(x) = [y_0(x), y_1(x), \ldots, y_n(x)]` that is a numerical solution of the `n+1`-dimensional first-order ordinary differential equation (ODE) system .. math :: y_0'(x) = F_0(x, [y_0(x), y_1(x), \ldots, y_n(x)]) y_1'(x) = F_1(x, [y_0(x), y_1(x), \ldots, y_n(x)]) \vdots y_n'(x) = F_n(x, [y_0(x), y_1(x), \ldots, y_n(x)]) The derivatives are specified by the vector-valued function *F* that evaluates `[y_0', \ldots, y_n'] = F(x, [y_0, \ldots, y_n])`. The initial point `x_0` is specified by the scalar argument *x0*, and the initial value `y(x_0) = [y_0(x_0), \ldots, y_n(x_0)]` is specified by the vector argument *y0*. For convenience, if the system is one-dimensional, you may optionally provide just a scalar value for *y0*. In this case, *F* should accept a scalar *y* argument and return a scalar. The solution function *y* will return scalar values instead of length-1 vectors. Evaluation of the solution function `y(x)` is permitted for any `x \ge x_0`. A high-order ODE can be solved by transforming it into first-order vector form. This transformation is described in standard texts on ODEs. Examples will also be given below. **Options, speed and accuracy** By default, :func:`~mpmath.odefun` uses a high-order Taylor series method. For reasonably well-behaved problems, the solution will be fully accurate to within the working precision. Note that *F* must be possible to evaluate to very high precision for the generation of Taylor series to work. To get a faster but less accurate solution, you can set a large value for *tol* (which defaults roughly to *eps*). If you just want to plot the solution or perform a basic simulation, *tol = 0.01* is likely sufficient. The *degree* argument controls the degree of the solver (with *method='taylor'*, this is the degree of the Taylor series expansion). A higher degree means that a longer step can be taken before a new local solution must be generated from *F*, meaning that fewer steps are required to get from `x_0` to a given `x_1`. On the other hand, a higher degree also means that each local solution becomes more expensive (i.e., more evaluations of *F* are required per step, and at higher precision). The optimal setting therefore involves a tradeoff. Generally, decreasing the *degree* for Taylor series is likely to give faster solution at low precision, while increasing is likely to be better at higher precision. The function object returned by :func:`~mpmath.odefun` caches the solutions at all step points and uses polynomial interpolation between step points. Therefore, once `y(x_1)` has been evaluated for some `x_1`, `y(x)` can be evaluated very quickly for any `x_0 \le x \le x_1`. and continuing the evaluation up to `x_2 > x_1` is also fast. **Examples of first-order ODEs** We will solve the standard test problem `y'(x) = y(x), y(0) = 1` which has explicit solution `y(x) = \exp(x)`:: >>> from mpmath import * >>> mp.dps = 15; mp.pretty = True >>> f = odefun(lambda x, y: y, 0, 1) >>> for x in [0, 1, 2.5]: ... print((f(x), exp(x))) ... (1.0, 1.0) (2.71828182845905, 2.71828182845905) (12.1824939607035, 12.1824939607035) The solution with high precision:: >>> mp.dps = 50 >>> f = odefun(lambda x, y: y, 0, 1) >>> f(1) 2.7182818284590452353602874713526624977572470937 >>> exp(1) 2.7182818284590452353602874713526624977572470937 Using the more general vectorized form, the test problem can be input as (note that *f* returns a 1-element vector):: >>> mp.dps = 15 >>> f = odefun(lambda x, y: [y[0]], 0, [1]) >>> f(1) [2.71828182845905] :func:`~mpmath.odefun` can solve nonlinear ODEs, which are generally impossible (and at best difficult) to solve analytically. As an example of a nonlinear ODE, we will solve `y'(x) = x \sin(y(x))` for `y(0) = \pi/2`. An exact solution happens to be known for this problem, and is given by `y(x) = 2 \tan^{-1}\left(\exp\left(x^2/2\right)\right)`:: >>> f = odefun(lambda x, y: x*sin(y), 0, pi/2) >>> for x in [2, 5, 10]: ... print((f(x), 2*atan(exp(mpf(x)**2/2)))) ... (2.87255666284091, 2.87255666284091) (3.14158520028345, 3.14158520028345) (3.14159265358979, 3.14159265358979) If `F` is independent of `y`, an ODE can be solved using direct integration. We can therefore obtain a reference solution with :func:`~mpmath.quad`:: >>> f = lambda x: (1+x**2)/(1+x**3) >>> g = odefun(lambda x, y: f(x), pi, 0) >>> g(2*pi) 0.72128263801696 >>> quad(f, [pi, 2*pi]) 0.72128263801696 **Examples of second-order ODEs** We will solve the harmonic oscillator equation `y''(x) + y(x) = 0`. To do this, we introduce the helper functions `y_0 = y, y_1 = y_0'` whereby the original equation can be written as `y_1' + y_0' = 0`. Put together, we get the first-order, two-dimensional vector ODE .. math :: \begin{cases} y_0' = y_1 \\ y_1' = -y_0 \end{cases} To get a well-defined IVP, we need two initial values. With `y(0) = y_0(0) = 1` and `-y'(0) = y_1(0) = 0`, the problem will of course be solved by `y(x) = y_0(x) = \cos(x)` and `-y'(x) = y_1(x) = \sin(x)`. We check this:: >>> f = odefun(lambda x, y: [-y[1], y[0]], 0, [1, 0]) >>> for x in [0, 1, 2.5, 10]: ... nprint(f(x), 15) ... nprint([cos(x), sin(x)], 15) ... print("---") ... [1.0, 0.0] [1.0, 0.0] --- [0.54030230586814, 0.841470984807897] [0.54030230586814, 0.841470984807897] --- [-0.801143615546934, 0.598472144103957] [-0.801143615546934, 0.598472144103957] --- [-0.839071529076452, -0.54402111088937] [-0.839071529076452, -0.54402111088937] --- Note that we get both the sine and the cosine solutions simultaneously. **TODO** * Better automatic choice of degree and step size * Make determination of Taylor series convergence radius more robust * Allow solution for `x < x_0` * Allow solution for complex `x` * Test for difficult (ill-conditioned) problems * Implement Runge-Kutta and other algorithms g@Tc$<S!W^,4.#rr)r'r(F_s&&rodefun..s"Q!+r Fc f<VUu.uFpSPVRRR1,V4NK! up#uupi)Nr)polyval)r-ar/rs&& rmpolyvalodefun..mpolyvals.145A AddGQ'555s%.ct< VS8d\h\S V4pV\S 48dS V^, ,#S R,wr#pS 'd\RW43,4S !W$V, 4pTp\ SSWES S4wr$S P V4S P W#V34W8:gKvS R,#)rz$Computing Taylor series for [%f, %f]r) ValueErrorrrprintr5r)r'r!r-xaxbr(Frdegreer?series_boundaries series_datar verbosers& r get_seriesodefun..get_seriess r6  $a ( s$% %qs# #%b/KCR<xGHe$AB a&AGC  $ $R (   } -w"2&r c<SPV4pSPpS SnS!V4wr#pS !W V, 4pVSnS 'dVUu.uFqf5NK up#V^,5# TSni;iuupir8)convertr) r'r)r-rDrEr(ykrrKr? return_vectorworkprecs & r interpolantodefun..interpolant sw KKNxx CH$Q-KCRd#ACH "#$!BC!$ $aD5L CH$s!A, A8, A5)intlogrdpsr TypeErrorr5)rrFrrr#rGmethodrJr-rErRr9rKr?rPrHrIr rQsfff&&f&f @@@@@@@@rodefunrY3sf Q'(+88B;  . .C#'' " --Fxx"}H B aR6:GCRR=/K6''$   W   &T s CC98C9__main__)NNtaylorF) r libmp.backendrobjectrr5rYrdoctesttestmodrr rr`sH"  *XgR  z OOr