[h&SSKJr SSKJr "SS\5rSrS Sjr\\l\S :XaSSKr\R"5 gg) )bisect)xrangec\rSrSrSrg) ODEMethodsN)__name__ __module__ __qualname____firstlineno____static_attributes__r L/var/www/auris/envauris/lib/python3.13/site-packages/mpmath/calculus/odes.pyrrsrrc URSU*5=pg[U5nU/n U/n Un Un URn U SU--Ul[U5H_nU"X5n[ [U 55Vs/sHoUXoU--PM n nX- n U R U 5 U R U 5 Ma [U5Vs/sHn/PM nn[US-5HnS/U-nSUS--nSn[US-5H>n[U5HnUU==UXU-- ss'M UUU- S--U*-nUS- nM@ UU*-UR U5- n[U5H%nUUU-UU'UUR UU5 M' M XlURnUH:nUS(dM[UURU[US5- U55nM< US-nUUU-4$s snfs snf!Xlf=f)Nrr) ldexplenprecrangerappendfaconeminnthrootabs)ctxderivsx0y0tol_precnhtoldimxsysxyorigifxydserjsbkscaleradiustss r ode_taylorr7siiH9%%A b'C B B A A 88D1:qA,C(.s1v711aAhA7 FA IIaL IIaL !:&:ar:&qsACAQAA1Q3ZsAaDAaL(D$!A#a%[qb)Q  Gcggaj(E3Zte|!A ad# WWF b66SRV_a!@AF aKF 6 >98 's*9G60G,9G6? G1 CG6, G66G>Nc ^^^^^^ ^ ^ ^^^^^U(a [TRUS5*5S-mOTRS-mT=(d S[STR-S- 5-mTRS-m[ U5 Sm[ TTTUTT5upTU /mUTU 4/mU4S jm UUUU UUUUU4 S jm UU U UU4S jn U $![ a Tm U 4SjmU/nSmNUf=f) 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 r g@(Tc>T"XS5/$Nrr )r)r*F_s rodefun..s"Q!+rFc `>UVs/sHnTRUSSS2U5PM sn$s snf)Nr)polyval)r/ar1rs rmpolyvalodefun..mpolyvals.145A AddGQ'555s"+c<> UT:a[e[T U5nU[T 5:aT US- $T Sup#nT (a[SX44-5 T "X$U- 5nUn[ TTXET T5up$T R U5 T R X#U45 X::aT S$Ml)Nrrz$Computing Taylor series for [%f, %f]) ValueErrorrrprintr7r)r)r#r/xaxbr*FrdegreerDseries_boundaries series_datar"verboser s r get_seriesodefun..get_seriess r6  $a ( s$% %qs# #%b/KCR<xGHe$AB a&AGC  $ $R (   } -w"2&rc>TRU5nTRnT TlT"U5up#nT "X U- 5nUTlT (aUVs/sHof7PM sn$US7$!UTlf=fs snfr=)convertr) r)r+r/rIrJr*ykrrPrD return_vectorworkprecs r interpolantodefun..interpolant sw KKNxx CH$Q-KCRd#ACH "#$!BC!$ $aD5L CH$sA$ A0$ A-)intlogrdpsr TypeErrorr7)rrKr r!r%rLmethodrOr/rJrWr>rPrDrUrMrNr"rVs``` ` ` @@@@@@@@rodefunr^3sf Q'(+88B;  .C#'' " --Fxx"}H B aR6:GCRR=/K6''$   W   &T s; CCC__main__)NNtaylorF) r libmp.backendrobjectrr7r^r doctesttestmodr rrresH"  *XgR  z OOr