Ordinary and Partial Differential Equations

All Products Maple MapleSim

Home : Support : Online Help : System : Information : Updates : Maple 2020 : Ordinary and Partial Differential Equations

Ordinary and Partial Differential Equations

Maple is the world leader in finding exact solutions to ordinary and partial differential equations. Maple 2020 extends that lead even further with new algorithms and techniques for solving more ODEs and PDEs, including general solutions, and solutions with initial conditions and/or boundary conditions.

For Maple 2020, there are significant improvements both in dsolve and in pdsolve for the exact solution of ODEs and PDEs, with and without initial or boundary conditions.

For ODEs, a new algorithm for computing hypergeometric solutions to $2^{nd}$ order linear ODEs is capable of solving new classes of problems that were previously out of reach.

For PDEs, improvements include a significantly extended ability to solve problems with boundary and initial conditions, also using integral Mellin and Hankel transforms, and two new methods for finding general PDE solutions, by mapping the PDE onto an ODE with auxiliary functions and by using decomposition into first integrals. As well, when the PDE depends on parameters, new symmetry techniques allow you to compute parameterized symmetries and to reduce the number of these parameters, something frequently convenient in different contexts.

Hypergeometric solutions for second-order linear ODEs

Exact solutions to PDEs with Boundary / Initial conditions

Mellin and Hankel transform solutions for PDE with Boundary Conditions

A general solution to a PDE calculated via rewriting the PDE as an ODE with arbitrary auxiliary functions

Solving a PDE by making use of first integrals

PDE parameterized symmetries

Parameter-continuous symmetry transformations

Hypergeometric solutions for second-order linear ODEs

Previous Maple releases already have algorithms for computing hypergeometric solutions for linear ODEs. The new algorithms implemented in Maple 2020, however, are more general. Suppose an ODE admits solutions of the form

$y (x) = r__0 (x) \cdot Y (x) + r__1 (x) \cdot Y' (x)$

where $r__0 (x)$ and $r__1 (x)$ are rational functions of x, and $Y (x)$ is the solution of some other linear ODE. Then the new algorithm can compute that other linear ODE and solve it, provided that it admits solutions of the form

$Y (x) = {&ExponentialE;}^{\int r (x) &DifferentialD; x} {}_{2}F_{1} (a, b &semi; c &semi; f (x))$

where $r (x)$ and $f (x)$ are rational functions of x, ${}_{2}F_{1}$ is the hypergeometric function and a, b, and c are arbitrary constants. The new algorithms use modular reduction, Hensel lifting, rational function reconstruction, and rational number reconstruction.

This algorithm is more general than previous algorithms in that it has no restrictions neither on the degree of the pullback function $f (x)$ nor the number of singularities of the input equation. The implementation follows the presentation by Imamoglu, E. and van Hoeij, M. "Computing Hypergeometric Solutions of Second Order Linear Differential Equations using Quotients of Formal Solutions and Integral Bases", Journal of Symbolic Computation, 83, (2017): 254-271.

Examples

A problem that admits solutions of the form $Y (x) = {&ExponentialE;}^{\int r (x) &DifferentialD; x} {}_{2}F_{1} (a, b &semi; c &semi; f (x))$

>	$restart &semi;$

>	$ode__1 ≔ \frac{{&DifferentialD;}^{2}}{&DifferentialD; x^{2}} y (x) + \frac{4 (24 x^{2} - 1) (\frac{&DifferentialD;}{&DifferentialD; x} y (x))}{x (16 x^{2} - 1)} + \frac{2 (192 x^{3} - 56 x^{2} - 6 x + 1) y (x)}{x^{2} (4 x - 1) (16 x^{2} - 1)} = 0 &colon;$

>	$dsolve (ode__1)$

$y (x) = \frac{_C1 {}_{2}F_{1} (- \frac{1}{2}, \frac{1}{2}; 1; \frac{{(4 x + 1)}^{2}}{{(4 x - 1)}^{2}})}{x^{2}} + \frac{_C2 {}_{2}F_{1} (\frac{1}{2}, \frac{3}{2}; 2; - \frac{16 x}{{(4 x - 1)}^{2}})}{x {(4 x - 1)}^{2}}$

(1)

An example where the solutions are of the form $y (x) = r__0 (x) \cdot Y (x) + r__1 (x) \cdot Y' (x)$

$ode__2 ≔ \frac{{&DifferentialD;}^{2}}{&DifferentialD; x^{2}} y (x) + \frac{2 (112 x^{3} - 168 x^{2} + 73 x - 9) (\frac{&DifferentialD;}{&DifferentialD; x} y (x))}{x (- 1 + 2 x) (16 x^{2} - 16 x + 3)} + \frac{4 (576 x^{5} - 1024 x^{4} + 488 x^{3} - 15 x^{2} - 27 x + 3) y (x)}{{(- 1 + 2 x)}^{2} x^{2} (1 + 4 x) (16 x^{2} - 16 x + 3)} = 0 &colon;$

>	$dsolve (ode__2)$

$y (x) =_C1 (- \frac{(4 x + 1) {}_{2}F_{1} (\frac{1}{2}, \frac{1}{2}; 1; 16 x^{2})}{(2 x - 1) x^{2}} + \frac{(- 16 x^{2} + 1) {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; 16 x^{2})}{2 x^{3} - x^{2}}) +_C2 (- \frac{(4 x + 1) {}_{2}F_{1} (\frac{1}{2}, \frac{1}{2}; 1; - 16 x^{2} + 1)}{(2 x - 1) x^{2}} + \frac{(16 x^{2} - 1) {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; - 16 x^{2} + 1)}{2 x^{3} - x^{2}})$

(2)

An example with a parameter a in the coefficients

$ode__3 ≔ (a x + 1) (\frac{{&DifferentialD;}^{2}}{&DifferentialD; x^{2}} y (x)) + \frac{(288 a x^{4} - 18 + (- 432 a + 224) x^{3} + (190 a - 336) x^{2} + (- 24 a + 146) x) (\frac{&DifferentialD;}{&DifferentialD; x} y (x))}{32 x^{4} - 48 x^{3} + 22 x^{2} - 3 x} + \frac{(4096 a x^{6} + 12 + (- 7232 a + 2304) x^{5} + (3568 a - 4096) x^{4} + (- 160 a + 1952) x^{3} + (- 218 a - 60) x^{2} + (30 a - 108) x) y (x)}{{(2 x - 1)}^{2} x^{2} (4 x + 1) (16 x^{2} - 16 x + 3)} = 0 &colon;$

>	$dsolve (ode__3)$

$y (x) = \frac{_C1 \sqrt{4 x - 3} (4 {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; 16 x^{2}) x - {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; 16 x^{2}) + {}_{2}F_{1} (\frac{1}{2}, \frac{1}{2}; 1; 16 x^{2})) (16 x^{2} - 1)}{\sqrt{4 x - 1} \sqrt{2 x - 1} (a x + 1) x^{2} \sqrt{32 x^{3} - 48 x^{2} + 22 x - 3}} + \frac{_C2 \sqrt{4 x - 3} (- 4 {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; - 16 x^{2} + 1) x + {}_{2}F_{1} (\frac{1}{2}, \frac{1}{2}; 1; - 16 x^{2} + 1) + {}_{2}F_{1} (\frac{3}{2}, \frac{3}{2}; 2; - 16 x^{2} + 1)) (16 x^{2} - 1)}{\sqrt{4 x - 1} \sqrt{2 x - 1} (a x + 1) x^{2} \sqrt{32 x^{3} - 48 x^{2} + 22 x - 3}}$

(3)

Verify this solution

>	$odetest (, ode__3)$

$0$

(4)

Some ODEs with 4 regular singular points that can be solved in terms of HeunG functions can also be solved using the new algorithm. This is possible whenever through a gauge transformation one of the singularities can be removed. The following ODE is of that kind, and the default solution returned by dsolve is in terms of HeunG

>	$ode__4 ≔ \frac{{&DifferentialD;}^{2}}{&DifferentialD; x^{2}} y (x) + \frac{(x^{4} - 44 x^{3} + 1206 x^{2} - 44 x + 1) y (x)}{4 {(x^{2} - 34 x + 1)}^{2} x^{2}} = 0 &colon;$

>	$dsolve (ode__4)$

$y (x) =_C1 \sqrt{x} {(- x^{2} + 34 x - 1)}^{\frac{3}{4}} HG (\frac{24 \sqrt{2}}{- 17 + 12 \sqrt{2}}, \frac{3 (36 \sqrt{2} - 25)}{4 {(- 17 + 12 \sqrt{2})}^{3} {(17 + 12 \sqrt{2})}^{2}}, \frac{3}{2}, \frac{3}{2}, \frac{3}{2}, 1, \frac{x - 17 + 12 \sqrt{2}}{- 17 + 12 \sqrt{2}}) +_C2 \sqrt{x} \sqrt{- x + 17 + 12 \sqrt{2}} {(- x^{2} + 34 x - 1)}^{\frac{1}{4}} HG (\frac{24 \sqrt{2}}{- 17 + 12 \sqrt{2}}, \frac{6 \sqrt{2} - 6}{{(- 17 + 12 \sqrt{2})}^{3} {(17 + 12 \sqrt{2})}^{2}}, 1, 1, \frac{1}{2}, 1, \frac{x - 17 + 12 \sqrt{2}}{- 17 + 12 \sqrt{2}})$

(5)

To get a solution in terms of hypergeometric functions you can indicate the method to be used, in this case the new method hypergeometricsols

>	$dsolve (ode__4, [hypergeometricsols])$

$y (x) =_C1 {&ExponentialE;}^{- \frac{(\int \frac{- x^{3} + 6 \sqrt{x^{2} - 34 x + 1} x + 15 x^{2} + 51 x - 1}{(x^{2} - 34 x + 1) x (x + 1)} &DifferentialD; x)}{2}} {}_{2}F_{1} (\frac{1}{3}, \frac{2}{3}; 1; \frac{x^{3} + 30 x^{2} - 24 x + 1 + (x^{2} - 7 x + 1) \sqrt{x^{2} - 34 x + 1}}{2 {(x + 1)}^{3}}) +_C2 {&ExponentialE;}^{- \frac{(\int \frac{- x^{3} + 6 \sqrt{x^{2} - 34 x + 1} x + 15 x^{2} + 51 x - 1}{(x^{2} - 34 x + 1) x (x + 1)} &DifferentialD; x)}{2}} {}_{2}F_{1} (\frac{1}{3}, \frac{2}{3}; 1; \frac{(- x^{2} + 7 x - 1) \sqrt{x^{2} - 34 x + 1} + x^{3} - 24 x^{2} + 30 x + 1}{2 {(x + 1)}^{3}})$

(6)

The integrals can actually be computed

>	$value ()$

$y (x) =_C1 {&ExponentialE;}^{\frac{arctanh (\frac{36 - 36 x}{12 \sqrt{{(x + 1)}^{2} - 36 x}})}{2} + \frac{\ln (x^{2} - 34 x + 1)}{4} - \frac{\ln (x + 1)}{2} + \frac{\ln (x)}{2}} {}_{2}F_{1} (\frac{1}{3}, \frac{2}{3}; 1; \frac{x^{3} + 30 x^{2} - 24 x + 1 + (x^{2} - 7 x + 1) \sqrt{x^{2} - 34 x + 1}}{2 {(x + 1)}^{3}}) +_C2 {&ExponentialE;}^{\frac{arctanh (\frac{36 - 36 x}{12 \sqrt{{(x + 1)}^{2} - 36 x}})}{2} + \frac{\ln (x^{2} - 34 x + 1)}{4} - \frac{\ln (x + 1)}{2} + \frac{\ln (x)}{2}} {}_{2}F_{1} (\frac{1}{3}, \frac{2}{3}; 1; \frac{(- x^{2} + 7 x - 1) \sqrt{x^{2} - 34 x + 1} + x^{3} - 24 x^{2} + 30 x + 1}{2 {(x + 1)}^{3}})$

(7)

Verify this solution

>	$odetest (, ode__4)$

$0$

(8)

Exact solutions to PDEs with Boundary / Initial conditions

Maple 2019 included a significant leap in the computation of exact solutions for PDE with Boundary / Initial conditions. For Maple 2020, another jump ahead in the solving capabilities happened. The examples below belong to different classes of problems out of reach of the Maple 2019 developments.

Examples

An example where the solution involves products of Bessel and Hankel functions

>	$pde__1 ≔ \frac{\partial}{\partial t} u (r, t) = \frac{\partial^{2}}{\partial r^{2}} u (r, t) + \frac{\frac{\partial}{\partial r} u (r, t)}{r} &colon;$

>	$iv__1 ≔ u (r, 0) = 1 - r, u (1, t) = 0 &colon;$

>	$pdsolve ([pde__1, iv__1])$

$u (r, t) = casesplit/ans (Sum (- \frac{BesselJ (0, λ_{n} r) π (BesselJ (1, λ_{n}) StruveH (0, λ_{n}) - BesselJ (0, λ_{n}) StruveH (1, λ_{n})) (\sinh (λ_{n}^{2} t) - \cosh (λ_{n}^{2} t))}{λ_{n}^{2} ({BesselJ (0, λ_{n})}^{2} + {BesselJ (1, λ_{n})}^{2})}, n = 1 .. \infty), \{And (λ_{n} = BesselJZeros (0, n), 0 < λ_{n})\})$

(9)

A similar problem but in three variables; the solution is a double infinite sum

>	$pde__2 ≔ \frac{\partial}{\partial t} u (r, z, t) = \frac{\partial^{2}}{\partial r^{2}} u (r, z, t) + \frac{\frac{\partial}{\partial r} u (r, z, t)}{r} + \frac{\partial^{2}}{\partial z^{2}} u (r, z, t) &colon;$

>	$iv__2 ≔ u (r, 0, t) = 0, u (r, 1, t) = 0, u (1, z, t) = 0, u (r, z, 0) = f (r, z) &colon;$

>	$pdsolve ([pde__2, iv__2])$

$u (r, z, t) = casesplit/ans (Sum (Sum (\frac{4 BesselJ (0, λ_{n1} r) \sin (n π z) \exp (- t (π^{2} n^{2} + λ_{n1}^{2})) (Int (BesselJ (0, λ_{n1} r) r (Int (\sin (n π z) f (r, z), z = 0 .. 1, AllSolutions)), r = 0 .. 1, AllSolutions))}{hypergeom ([\frac{1}{2}], [1, 2], - λ_{n1}^{2})}, n = 1 .. \infty), n1 = 1 .. \infty), \{And (λ_{n1} = BesselJZeros (0, n1), 0 \leq λ_{n1})\})$

(10)

A large number of new infinite series solutions are now computable for different classes of problems (the number of boundary or initial conditions, whether they are periodic or involve derivatives of the unknown, etc.)

>	$pde__3 ≔ \frac{\partial^{2}}{\partial t^{2}} u (x, t) = 4 (\frac{\partial^{2}}{\partial x^{2}} u (x, t)) &colon;$

>	$iv__3 ≔ u (x, 0) = 0, D_{2} (u) (x, 0) = \sin {(x)}^{2}, u (- π, t) = 0, u (π, t) = 0 &colon;$

>	$pdsolve ([pde__3, iv__3])$

$u (x, t) = \frac{315 (\sum_{n = 5}^{\infty} \frac{16 \sin (\frac{n (x + π)}{2}) \sin (n t) ({(−1)}^{n} - 1)}{π n^{2} (n^{2} - 16)}) π - 160 \cos (\frac{3 x}{2}) \sin (3 t) + 672 \cos (\frac{x}{2}) \sin (t)}{315 π}$

(11)

>	$pde__4 ≔ \frac{\partial^{2}}{\partial x^{2}} u (x, y) + \frac{\partial^{2}}{\partial y^{2}} u (x, y) = 0 &colon;$

>	$iv__4 ≔ u (0, y) = 0, u (x, 1) = A, u (x, 0) = 0 &colon;$

>	$pdsolve ([pde__4, iv__4], HINT = boundedseries (x = \infty)) assuming 0 < x, 0 < y$

$u (x, y) = A y + (\sum_{n = 1}^{\infty} \frac{2 {(−1)}^{n} A {&ExponentialE;}^{- n π x} \sin (n π y)}{n π})$

(12)

>	$pde__5 ≔ \frac{\partial}{\partial t} u (t, x) = \frac{\partial^{2}}{\partial x^{2}} u (t, x) &colon;$

>	$iv__5 ≔ u (0, x) = 1, u (t, - 1) = 0, u (t, 1) = 0 &colon;$

>	$pdsolve ([pde__5, iv__5])$

$u (t, x) = \sum_{n = 1}^{\infty} - \frac{2 {&ExponentialE;}^{- \frac{π^{2} n^{2} t}{4}} \sin (\frac{n π (x + 1)}{2}) ({(−1)}^{n} - 1)}{n π}$

(13)

An example with periodic conditions in both the unknown and its first derivative, where different methods combine resulting in a solution involving an arbitrary function of x

>	$pde__6 ≔ \frac{\partial}{\partial t} u (x, t) = \frac{\partial^{2}}{\partial x^{2}} u (x, t) &colon;$

>	$iv__6 ≔ u (- π, t) = u (π, t), D_{1} (u) (- π, t) = D_{1} (u) (π, t) &colon;$

>	$pdsolve ([pde__6, iv__6])$

$u (x, t) = \frac{2 (\sum_{n = 1}^{\infty} \frac{{&ExponentialE;}^{- n^{2} t} ((\int_{- π}^{π}_F1 (x) \sin (n x) &DifferentialD; x) \sin (n x) + (\int_{- π}^{π}_F1 (x) \cos (n x) &DifferentialD; x) \cos (n x))}{π}) π + \int_{- π}^{π}_F1 (x) &DifferentialD; x}{2 π}$

(14)

The detection of particular values of the summation index for which the solution branches with a different expression is now better:

>	$pde__7 ≔ \frac{\partial^{2}}{\partial t^{2}} u (x, t) + 2 (\frac{\partial}{\partial t} u (x, t)) = \frac{\partial^{2}}{\partial x^{2}} u (x, t) &colon;$

>	$iv__7 ≔ D_{2} (u) (x, 0) = 0, u (x, 0) = f (x), u (0, t) = 0, u (π, t) = 0 &colon;$

>	$pdsolve ([pde__7, iv__7], u (x, t)) assuming 0 < t$

$u (x, t) = \sum_{n = 1}^{\infty} \{\begin{array}{c} \frac{2 {&ExponentialE;}^{- t} (t + 1) (\int_{0}^{π} \sin (x) f (x) &DifferentialD; x) \sin (x)}{π} & n = 1 \\ \frac{((- 1 + \sqrt{- n^{2} + 1}) {&ExponentialE;}^{- (\sqrt{- n^{2} + 1} + 1) t} + {&ExponentialE;}^{(- 1 + \sqrt{- n^{2} + 1}) t} (\sqrt{- n^{2} + 1} + 1)) (\int_{0}^{π} \sin (n x) f (x) &DifferentialD; x) \sin (n x)}{\sqrt{- n^{2} + 1} π} & otherwise \end{array}$

(15)

A problem with two boundary and two initial conditions, all for the unknown $u (r, t)$ , where the solution method consists of splitting the problem into two PDE & BC problems and the solution is the sum of those two solutions

>	$pde__8 ≔ \frac{\partial^{2}}{\partial r^{2}} u (r, t) + \frac{\frac{\partial}{\partial r} u (r, t)}{r} + \frac{\frac{\partial^{2}}{\partial t^{2}} u (r, t)}{r^{2}} = 0 &colon;$

>	$iv__8 ≔ u (r, \frac{π}{6}) = \frac{π}{6}, u (r, \frac{π}{2}) = 0, u (1, t) = 0, u (2, t) = t &colon;$

>	$pdsolve ([pde__8, iv__8])$

$u (r, t) = \sum_{n1 = 1}^{\infty} (- \frac{(- r^{3 n1} + r^{- 3 n1}) 8^{n1} ({(−1)}^{n1} - \frac{1}{3}) \sin (\frac{n1 (- 6 t + π)}{2})}{(64^{n1} - 1) n1}) + (\sum_{n = 1}^{\infty} - \frac{({(−1)}^{n} - 1) \sin (\frac{n π \ln (r)}{\ln (2)}) ({&ExponentialE;}^{\frac{n π (7 π - 6 t)}{6 \ln (2)}} - {&ExponentialE;}^{\frac{n π (6 t + π)}{6 \ln (2)}})}{3 n ({&ExponentialE;}^{\frac{n π^{2}}{\ln (2)}} - {&ExponentialE;}^{\frac{n π^{2}}{3 \ln (2)}})})$

(16)

Similar example, but involving conditions for the two first derivatives of the unknown $u (r, t)$

>	$pde__9 ≔ \frac{\partial^{2}}{\partial r^{2}} u (r, t) + \frac{\frac{\partial}{\partial r} u (r, t)}{r} + \frac{\frac{\partial^{2}}{\partial t^{2}} u (r, t)}{r^{2}} = 0 &colon;$

>	$iv__9 ≔ u (r, \frac{π}{6}) = \frac{π}{6}, D_{2} (u) (r, \frac{π}{2}) = 0, u (1, t) = 0, D_{1} (u) (2, t) = t &colon;$

>	$pdsolve ([pde__9, iv__9])$

$u (r, t) = (\sum_{n1 = 0}^{\infty} \frac{16 \cos (\frac{(12 t - 2 π) n1}{4} + \frac{3 t}{2} + \frac{π}{4}) (- r^{\frac{3}{2} + 3 n1} + r^{- \frac{3}{2} - 3 n1}) ((π + 4 {(−1)}^{n1}) 2^{3 n1 + \frac{1}{2}} + 2^{\frac{3}{2} + 3 n1} π n1)}{9 (2^{6 n1 + 6} {n1}^{3} + (96 {n1}^{2} + 48 n1 + 8) 64^{n1} + 8 {(\frac{1}{2} + n1)}^{3}) π}) + (\sum_{n = 0}^{\infty} \frac{2 \sin (\frac{(1 + 2 n) π \ln (r)}{2 \ln (2)}) ({&ExponentialE;}^{\frac{(1 + 2 n) π (- 6 t + 5 π)}{12 \ln (2)}} + {&ExponentialE;}^{- \frac{(- 6 t + π) (1 + 2 n) π}{12 \ln (2)}})}{3 (1 + 2 n) ({&ExponentialE;}^{\frac{(1 + 2 n) π^{2}}{3 \ln (2)}} + 1)})$

(17)

A problem with initial conditions involving arbitrary functions evaluated at algebraic expressions, as $f (x, - \frac{a}{2} - \frac{b}{2})$ , and boundary conditions that are periodic as $u (a, t) = u (b, t)$

>	$pde__10 ≔ \frac{\partial}{\partial t} u (x, t) = k (\frac{\partial^{2}}{\partial x^{2}} u (x, t)) &colon;$

>	$iv__10 ≔ u (x, 0) = f (x, - \frac{a}{2} - \frac{b}{2}), u (a, t) = u (b, t), \frac{\partial}{\partial a} u (a, t) = \frac{\partial}{\partial b} u (b, t) &colon;$

>	$pdsolve ([pde__10, iv__10], u (x, t)) assuming a < b$

$u (x, t) = \frac{(- b + a) (\sum_{n = 1}^{\infty} - \frac{2 {&ExponentialE;}^{- \frac{4 k π^{2} n^{2} t}{{(- b + a)}^{2}}} ((\int_{a}^{b} f (x, - \frac{a}{2} - \frac{b}{2}) \sin (\frac{2 n π (- x + a)}{- b + a}) &DifferentialD; x) \sin (\frac{2 n π (- x + a)}{- b + a}) + (\int_{a}^{b} f (x, - \frac{a}{2} - \frac{b}{2}) \cos (\frac{2 n π (- x + a)}{- b + a}) &DifferentialD; x) \cos (\frac{2 n π (- x + a)}{- b + a}))}{- b + a}) - (\int_{a}^{b} f (x, - \frac{a}{2} - \frac{b}{2}) &DifferentialD; x)}{- b + a}$

(18)

Similar problem involving an arbitrary function in the initial conditions

>	$pde__11 ≔ \frac{\partial}{\partial t} υ (ξ, t) = k (\frac{\partial^{2}}{\partial ξ^{2}} υ (ξ, t)) &colon;$

>	$iv__11 ≔ D_{1} (υ) (a, t) = 0, υ (l + a, t) = 0, υ (ξ, 0) = f (ξ, a) &colon;$

>	$pdsolve ([pde__11, iv__11], υ (ξ, t)) assuming 0 < a, 0 < l$

$υ (ξ, t) = \sum_{n = 0}^{\infty} \frac{2 \cos (\frac{(1 + 2 n) π (- ξ + a)}{2 l}) {&ExponentialE;}^{- \frac{k π^{2} {(1 + 2 n)}^{2} t}{4 l^{2}}} (\int_{a}^{l + a} f (ξ, a) \cos (\frac{(1 + 2 n) π (- ξ + a)}{2 l}) &DifferentialD; ξ)}{l}$

(19)

Some simple, however tricky problems with nonlinear boundary conditions

>	$pde__12 ≔ - \frac{\partial^{2}}{\partial r^{2}} u (r, t) &colon;$

>	$iv__12 ≔ D_{1} (u) (0, t) + D_{1} (u) (0, t) u (0, t) = 0 &colon;$

>	$pdsolve ([pde__12, iv__12])$

$u (r, t) =_C2, u (r, t) =_C1 r - 1$

(20)

Problems with more than one evaluation point for the same variable, periodic, or involving piecewise initial conditions

>	$pde__13 ≔ \frac{\partial^{2}}{\partial t^{2}} υ (ξ, y, t) = \frac{\partial^{2}}{\partial ξ^{2}} υ (ξ, y, t) + \frac{\partial^{2}}{\partial y^{2}} υ (ξ, y, t) &colon;$

>	$iv__13 ≔ υ (\frac{π}{3}, y, t) = 0, υ (\frac{4 π}{3}, y, t) = 0, υ (ξ, 0, t) = 0, υ (ξ, π, t) = 0, D_{3} (υ) (ξ, y, 0) = 0, υ (ξ, y, 0) = (ξ - \frac{π}{3}) (\frac{4 π}{3} - ξ) y (π - y) &colon;$

>	$pdsolve ([pde__13, iv__13])$

$υ (ξ, y, t) = \sum_{n1 = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{16 (- {(−1)}^{n1 + n} + {(−1)}^{n1} + {(−1)}^{n} - 1) \sin (\frac{n (π - 3 ξ)}{3}) \sin (n1 y) \cos (\sqrt{n^{2} + {n1}^{2}} t)}{π^{2} n^{3} {n1}^{3}}$

(21)

>	$pde__14 ≔ \frac{\partial^{2}}{\partial t^{2}} u (x, t) + \frac{4 π (\frac{\partial}{\partial t} u (x, t))}{3} = 16 (\frac{\partial^{2}}{\partial x^{2}} u (x, t)) &colon;$

>	$iv__14 ≔ u (0, t) = 0, D_{1} (u) (3, t) = 0, D_{2} (u) (x, 0) = 0, u (x, 0) = \{\begin{array}{c} \frac{x}{10} & 0 \leq x and x \leq 1 \\ \frac{1}{10} & 1 < x and x \leq 3 \end{array} &colon;$

>	$pdsolve ([pde__14, iv__14])$

$u (x, t) = \sum_{n = 0}^{\infty} \{\begin{array}{c} \frac{4 \sin (\frac{π x}{6}) {&ExponentialE;}^{- \frac{2 π t}{3}} (π t + \frac{3}{2})}{5 π^{2}} & n = 0 \\ \frac{3 (\sqrt{3} \sin (\frac{π n}{3}) + \cos (\frac{π n}{3})) \sin (\frac{(1 + 2 n) π x}{6}) ((2 \sqrt{n} \sqrt{n + 1} + I) {&ExponentialE;}^{\frac{2 I}{3} π (- 2 \sqrt{n} \sqrt{n + 1} + I) t} - {&ExponentialE;}^{\frac{2 I}{3} π (2 \sqrt{n} \sqrt{n + 1} + I) t} (- 2 \sqrt{n} \sqrt{n + 1} + I))}{10 \sqrt{n + 1} \sqrt{n} π^{2} {(1 + 2 n)}^{2}} & otherwise \end{array}$

(22)

Mellin and Hankel transform solutions for PDE with Boundary Conditions

In previous Maple releases, the Fourier and Laplace transforms were used to compute exact solutions to PDE problems with boundary conditions. Now, Mellin and Hankel transforms are also used for that same purpose.

Examples:

>	$restart &semi;$

>	$pde__1 ≔ x^{2} (\frac{\partial^{2}}{\partial x^{2}} u (x, y)) + x (\frac{\partial}{\partial x} u (x, y)) + \frac{\partial^{2}}{\partial y^{2}} u (x, y) = 0 &colon;$ $iv__1 ≔ u (x, 0) = 0, u (x, 1) = \{\begin{array}{c} 1 & 0 \leq x and x < 1 \\ 0 & 1 < x \end{array} &colon;$

>	$sol ≔ pdsolve ([pde__1, iv__1])$

$u (x, y) = invmellin (\frac{\sin (s y)}{\sin (s) s}, s, x)$

(23)

As usual, you can let pdsolve choose the solving method, or indicate the method yourself:

>	$pde__2 ≔ \frac{\partial^{2}}{\partial r^{2}} u (r, t) + \frac{\frac{\partial}{\partial r} u (r, t)}{r} + \frac{\partial^{2}}{\partial t^{2}} u (r, t) = - Q__0 q (r) &colon;$ $iv__2 ≔ u (r, 0) = 0 &colon;$

>	$pdsolve ([pde__2, iv__2])$

$u (r, t) = Q__0 (- hankel (\frac{\exp (- s t) hankel (q (r), r, s, 0)}{s^{2}}, s, r, 0) + hankel (\frac{hankel (q (r), r, s, 0)}{s^{2}}, s, r, 0))$

(24)

It is sometimes preferable to see these solutions in terms of more familiar integrals. For that purpose, use

>	$convert (, Int, only = hankel)$

$u (r, t) = Q__0 (- (\int_{0}^{\infty} \frac{{&ExponentialE;}^{- s t} (\int_{0}^{\infty} q (r) r J_{0} (r s) &DifferentialD; r) J_{0} (r s)}{s} &DifferentialD; s) + \int_{0}^{\infty} \frac{(\int_{0}^{\infty} q (r) r J_{0} (r s) &DifferentialD; r) J_{0} (r s)}{s} &DifferentialD; s)$

(25)

An example where the hankel transform is computed:

>	$pde__3 ≔ c^{2} (\frac{\partial^{2}}{\partial r^{2}} u (r, t) + \frac{\frac{\partial}{\partial r} u (r, t)}{r}) = \frac{\partial^{2}}{\partial t^{2}} u (r, t) &colon;$ $iv__3 ≔ u (r, 0) = A a {(a^{2} + r^{2})}^{- \frac{1}{2}}, D_{2} (u) (r, 0) = 0 &colon;$

>	$pdsolve ([pde__3, iv__3], method = Hankel) assuming r > 0, t > 0, a > 0$

$u (r, t) = \frac{A a (\sqrt{- c^{2} t^{2} + 2 I a c t + a^{2} + r^{2}} + \sqrt{- c^{2} t^{2} - 2 I a c t + a^{2} + r^{2}})}{2 \sqrt{- c^{2} t^{2} - 2 I a c t + a^{2} + r^{2}} \sqrt{- c^{2} t^{2} + 2 I a c t + a^{2} + r^{2}}}$

(26)

A general solution to a PDE calculated via rewriting the PDE as an ODE with arbitrary auxiliary functions

Under certain conditions, it is possible to rewrite a PDE as an ODE for one unknown and some auxiliary arbitrary functions. When that is possible, the solving process involves as many steps as there are independent variables in the PDE. Each step consists of solving only a single intermediate ODE (and no PDEs). In this way, a general solution is obtained for the original PDE.

Below are four examples, all of them nonlinear. For the $4^{th}$ one the solving method is shown step by step.

Example 1:

>	$restart &semi;$

$pde__1 ≔ \frac{\partial^{2}}{\partial t \partial x} w (t, x) - \frac{a {(\frac{\partial}{\partial x} w (t, x))}^{2}}{w (t, x)} - (\frac{\frac{\partial}{\partial t} w (t, x)}{w (t, x)} + b + \frac{c}{w (t, x)}) (\frac{\partial}{\partial x} w (t, x)) - \frac{c (\frac{\partial}{\partial t} w (t, x))}{2 a w (t, x)} - k w (t, x) - \frac{b c}{2 a} - \frac{c^{2}}{4 a w (t, x)} &colon;$

>	$pdsolve (pde__1) &semi;$

$w (t, x) = (\int - \frac{c {&ExponentialE;}^{2 k (\int \frac{_F1 (x) \sqrt{- 4 k a + b^{2}} + {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t}}{4_F1 (x) a k -_F1 (x) b^{2} +_F1 (x) \sqrt{- 4 k a + b^{2}} b + {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t} b + \sqrt{- 4 k a + b^{2}} {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t}} &DifferentialD; x)}}{2 a} &DifferentialD; x +_F2 (t)) {&ExponentialE;}^{\int - \frac{2 k (_F1 (x) \sqrt{- 4 k a + b^{2}} + {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t})}{4_F1 (x) a k -_F1 (x) b^{2} +_F1 (x) \sqrt{- 4 k a + b^{2}} b + {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t} b + \sqrt{- 4 k a + b^{2}} {&ExponentialE;}^{- \sqrt{- 4 k a + b^{2}} t}} &DifferentialD; x}$

(27)

Example 2:

$pde__2 ≔ {w (t, x)}^{3} (\frac{\partial^{4}}{\partial t^{2} \partial x^{2}} w (t, x)) - 2 {w (t, x)}^{2} ((\frac{\partial}{\partial t} w (t, x)) (\frac{\partial^{3}}{\partial t \partial x^{2}} w (t, x)) + (\frac{\partial}{\partial x} w (t, x)) (\frac{\partial^{3}}{\partial t^{2} \partial x} w (t, x))) - 2 {(w (t, x) (\frac{\partial^{2}}{\partial t \partial x} w (t, x)) - 2 (\frac{\partial}{\partial t} w (t, x)) (\frac{\partial}{\partial x} w (t, x)))}^{2} + 2 (w (t, x) (\frac{\partial^{2}}{\partial x^{2}} w (t, x)) + {(\frac{\partial}{\partial x} w (t, x))}^{2}) {(\frac{\partial}{\partial t} w (t, x))}^{2} - w (t, x) (\frac{\partial^{2}}{\partial t^{2}} w (t, x)) (w (t, x) (\frac{\partial^{2}}{\partial x^{2}} w (t, x)) - 2 {(\frac{\partial}{\partial x} w (t, x))}^{2}) &colon;$

>	$pdsolve (pde__2) &semi;$

$w (t, x) =_F4 (t) {&ExponentialE;}^{_F1 (x) t -_F2 (x) -_F3 (t) x}$

(28)

Example 3:

$pde__3 ≔ 4 (\frac{\partial^{5}}{\partial t^{3} \partial x^{2}} y (x, t)) (\frac{\partial}{\partial t} y (x, t)) + 6 (\frac{\partial^{4}}{\partial t^{2} \partial x^{2}} y (x, t)) (\frac{\partial^{2}}{\partial t^{2}} y (x, t)) + 4 (\frac{\partial^{3}}{\partial t \partial x^{2}} y (x, t)) (\frac{\partial^{3}}{\partial t^{3}} y (x, t)) + (\frac{\partial^{2}}{\partial x^{2}} y (x, t)) (\frac{\partial^{4}}{\partial t^{4}} y (x, t)) + 6 {(\frac{\partial^{3}}{\partial t^{2} \partial x} y (x, t))}^{2} + 8 (\frac{\partial^{2}}{\partial t \partial x} y (x, t)) (\frac{\partial^{4}}{\partial t^{3} \partial x} y (x, t)) + 2 (\frac{\partial}{\partial x} y (x, t)) (\frac{\partial^{5}}{\partial t^{4} \partial x} y (x, t)) + y (x, t) (\frac{\partial^{6}}{\partial t^{4} \partial x^{2}} y (x, t)) &colon;$

>	$pdsolve (pde__3) &semi;$

$y (x, t) = RootOf (_F1 (x) t^{3} - 6_F2 (x) t^{2} - 6_F3 (x) t + 6_F5 (t) x + 3 {_Z}^{2} - 6_F4 (x) - 3_F6 (t))$

(29)

Example 4

$pde__4 ≔ \frac{\frac{\partial^{3}}{\partial t \partial x \partial z} E (t, x, z)}{E (t, x, z)} - \frac{(\frac{\partial^{2}}{\partial x \partial z} E (t, x, z)) (\frac{\partial}{\partial t} E (t, x, z))}{{E (t, x, z)}^{2}} - \frac{(\frac{\partial^{2}}{\partial t \partial x} E (t, x, z)) (\frac{\partial}{\partial z} E (t, x, z))}{{E (t, x, z)}^{2}} - \frac{(\frac{\partial}{\partial x} E (t, x, z)) (\frac{\partial^{2}}{\partial t \partial z} E (t, x, z))}{{E (t, x, z)}^{2}} + \frac{2 (\frac{\partial}{\partial x} E (t, x, z)) (\frac{\partial}{\partial z} E (t, x, z)) (\frac{\partial}{\partial t} E (t, x, z))}{{E (t, x, z)}^{3}} &colon;$

Step-by-step solving process for $pde__4$

Step 1. $pde__4$ can be written as an ODE with respect to t by taking the derivatives of $E (t, x, z)$ with respect to x and z as auxiliary arbitrary functions of t according to

>	$auxiliary_functions ≔ [A__x (t) = \frac{\partial}{\partial x} E (t, x, z), A__z (t) = \frac{\partial}{\partial z} E (t, x, z), A__xz (t) = \frac{\partial^{2}}{\partial x \partial z} E (t, x, z), E__0 (t) = E (t, x, z)] &colon;$

The resulting ODE is

$ODE__t ≔ \frac{\frac{&DifferentialD;}{&DifferentialD; t} A__xz (t)}{E__0 (t)} - \frac{A__xz (t) (\frac{&DifferentialD;}{&DifferentialD; t} E__0 (t))}{{E__0 (t)}^{2}} - \frac{(\frac{&DifferentialD;}{&DifferentialD; t} A__x (t)) A__z (t)}{{E__0 (t)}^{2}} + \frac{2 A__x (t) A__z (t) (\frac{&DifferentialD;}{&DifferentialD; t} E__0 (t))}{{E__0 (t)}^{3}} - \frac{A__x (t) (\frac{&DifferentialD;}{&DifferentialD; t} A__z (t))}{{E__0 (t)}^{2}} &colon;$

Despite the presence of the auxiliary arbitrary functions $A__x (t)$ , $A__z (t)$ and $A__xz (t)$ , the problem is in reach of dsolve's algorithms, so solve this $ODE__t$ for $E__0 (t)$

>	$dsolve (ODE__t, E__0 (t))$

$E__0 (t) = - \frac{A__xz (t) - \sqrt{{A__xz (t)}^{2} + 4_C1 A__x (t) A__z (t)}}{2_C1}, E__0 (t) = - \frac{A__xz (t) + \sqrt{{A__xz (t)}^{2} + 4_C1 A__x (t) A__z (t)}}{2_C1}$

(30)

The solution above reduced the order of the original problem $pde__4$ by one. Depending on the problem, more than one reduction of order may occur.

Step 2. We need to express the integration constant with respect to t in the context of a problem in t, x, and z, that is, _C1 is an arbitrary function of x and z. For illustration purposes take for instance the first solution

>	$_C1 = solve ([1],_C1) &semi;$

$_C1 = - \frac{A__xz (t) E__0 (t) - A__x (t) A__z (t)}{{E__0 (t)}^{2}}$

(31)

>	$subs (_C1 =_F1 (x, t),)$

$_F1 (x, t) = - \frac{A__xz (t) E__0 (t) - A__x (t) A__z (t)}{{E__0 (t)}^{2}}$

(32)

Restoring the original functions, we get

>	$subs (auxiliary_functions,)$

$_F1 (x, t) = - \frac{(\frac{\partial^{2}}{\partial x \partial z} E (t, x, z)) E (t, x, z) - (\frac{\partial}{\partial x} E (t, x, z)) (\frac{\partial}{\partial z} E (t, x, z))}{{E (t, x, z)}^{2}}$

(33)

This pde in t, x, and z can now be written in terms of two auxiliary functions just of x as

>	$auxiliary_functions ≔ [A__z (x) = \frac{\partial}{\partial z} E (t, x, z), E__0 (x) = E (t, x, z)] &colon;$

Resulting in

>	$ODE__x ≔_F1 (x, z) = - \frac{(\frac{&DifferentialD;}{&DifferentialD; x} A__z (x)) E__0 (x) - (\frac{&DifferentialD;}{&DifferentialD; x} E__0 (x)) A__z (x)}{{E__0 (x)}^{2}}$

$ODE__x ≔_F1 (x, z) = - \frac{(\frac{&DifferentialD;}{&DifferentialD; x} A__z (x)) E__0 (x) - (\frac{&DifferentialD;}{&DifferentialD; x} E__0 (x)) A__z (x)}{{E__0 (x)}^{2}}$

(34)

Solve this ODE

>	$dsolve (ODE__x, E__0 (x))$

$E__0 (x) = \frac{A__z (x)}{\int -_F1 (x, z) &DifferentialD; x +_C1}$

(35)

Step 3. Proceed as in the previous step, isolating _C1 , rewriting it as an arbitrary function of z and t and restoring the original functions

>	$_C1 = solve (,_C1) &semi;$

$_C1 = - \frac{E__0 (x) (\int -_F1 (x, z) &DifferentialD; x) - A__z (x)}{E__0 (x)}$

(36)

>	$subs (_C1 =_F2 (t, z),)$

$_F2 (t, z) = - \frac{E__0 (x) (\int -_F1 (x, z) &DifferentialD; x) - A__z (x)}{E__0 (x)}$

(37)

Although not necessary, a simplification is possible in that the integral of an arbitrary function is also an arbitrary function, so we can remove the integration by redefining $_F1 \to D [1] (_F1)$

>	$eval (, [Int = int,_F1 = D [1] (_F1)])$

$_F2 (t, z) = - \frac{- E__0 (x)_F1 (x, z) - A__z (x)}{E__0 (x)}$

(38)

Restore the original variables

>	$subs (auxiliary_functions,)$

$_F2 (t, z) = - \frac{- E (t, x, z)_F1 (x, z) - \frac{\partial}{\partial z} E (t, x, z)}{E (t, x, z)}$

(39)

This resulting pde is already an ODE in disguise, from where the general solution to the original problem $pde__4$ is

>	$pdsolve ()$

$E (t, x, z) =_F3 (t, x) {&ExponentialE;}^{\int (_F2 (t, z) -_F1 (x, z)) &DifferentialD; z}$

(40)

This solution can be simplified further by redefining $_F2 \to D [2] (_F2)$ and $_F1 \to D [2] (_F1)$

>	$eval (, [Int = int,_F2 = D [2] (_F2),_F1 = D [2] (_F1)])$

$E (t, x, z) =_F3 (t, x) {&ExponentialE;}^{_F2 (t, z) -_F1 (x, z)}$

(41)

Verify this solution with the original $pde__4$

>	$pdetest (, pde__4)$

$0$

(42)

Doing all these steps in one go

>	$pdsolve (pde__4)$

$E (t, x, z) =_F3 (t, x) {&ExponentialE;}^{_F2 (t, z) -_F1 (x, z)}$

(43)

Solving a PDE by making use of first integrals

Any PDE can be solved iterating the two steps below related to its first integrals, when they can be computed.

Step 1: Write the differential equation for a first integral - a function $Φ$ of the PDE's jet space - by equating to 0 its total derivative with respect to (any) one of the independent variables of the original problem. The resulting PDE is of first order and can be tackled using the characteristic strip method.

Step 2: When the PDE for $Φ$ can be solved, the solution is an arbitrary function of the differential invariants of the PDE for $Φ$ . Equate one such differential invariant to an arbitrary function of the remaining independent variables of the problem, resulting in a pde of reduced order with respect to the original PDE problem, a first integral. At this point one can either repeat the process to calculate more first integrals in order to attempt an algebraic elimination (multiple reduction of order) or move again to Step 1 with the reduced order PDE.

Examples

>	$restart &semi;$

>	$with (PDEtools) &colon;$

>	$pde__1 ≔ \frac{\partial}{\partial x} (u (x, y) - \frac{\partial}{\partial y} u (x, y))$

$pde__1 ≔ \frac{\partial}{\partial x} u (x, y) - \frac{\partial^{2}}{\partial x \partial y} u (x, y)$

(44)

>	$FirstIntegralSolver (pde__1, u (x, y)) &semi;$

$u (x, y) = (\int_F1 (y) {&ExponentialE;}^{- y} &DifferentialD; y +_F2 (x)) {&ExponentialE;}^{y}$

(45)

>	$pdetest (,)$

$0$

(46)

The solving process step-by-step

Construct an expression for the first integral $Φ (Q)$ , where $Q = (x, y, u, u_{x}, u_{y})$ , the jet space of $pde__1$

>	$Q ≔ x, y, u, u_{x}, u_{y}$

$Q ≔ x, y, u, u_{x}, u_{y}$

(47)

>	$declare (Φ (Q))$

$Phi (x, y, u, u_{x}, u_{y}) will now be displayed as Φ$

(48)

Take the total derivative of $Φ (Q)$ with respect to x

>	$D_Dx (Phi (Q), x, u (x, y)) = 0$

$diff (Φ (x, y, u, u_{x}, u_{y}), x) + (diff (Φ (x, y, u, u_{x}, u_{y}), u)) u_{x} + (diff (Φ (x, y, u, u_{x}, u_{y}), u_{x})) u_{x, x} + (diff (Φ (x, y, u, u_{x}, u_{y}), u_{y})) u_{x, y} = 0$

(49)

Take this PDE modulo $pde__1$ , and that is all the formulation of the problem

>	$isolate (ToJet (pde__1, u (x, y)), u [x, y])$

$u_{x, y} = u_{x}$

(50)

>	$PDE__Φ ≔ subs (,)$

$diff (Φ (x, y, u, u_{x}, u_{y}), x) + (diff (Φ (x, y, u, u_{x}, u_{y}), u)) u_{x} + (diff (Φ (x, y, u, u_{x}, u_{y}), u_{x})) u_{x, x} + (diff (Φ (x, y, u, u_{x}, u_{y}), u_{y})) u_{x} = 0$

(51)

Solve $PDE__Φ$ , and in doing so, the solution cannot depend on $u_{x, x}$ (it can only depend on $Q = (x, y, u, u_{x}, u_{y}))$ . You can achieve that using the ivars option of pdsolve

>	$pdsolve (PDE__Φ, ivars = u_{x, x})$

$\{Φ (x, y, u, u_{x}, u_{y}) =_F1 (y, - u + u_{y})\}$

(52)

Since $Φ$ is a first integral, the arguments of the arbitrary function entering this solution are also first integrals. Take the second one, which involves 1^st order derivatives, and equate it to an arbitrary function all the independent variables but x (the variable used when taking the total derivative of $Φ)$ . In this case, there is only one other variable, y.

>	$op ([1, 2, 2],) =_F1 (y)$

$- u + u_{y} =_F1 (y)$

(53)

Rewrite this result in function notation

>	$FromJet (, u (x, y))$

$- u (x, y) + diff (u (x, y), y) =_F1 (y)$

(54)

This result is already an ODE in disguise; solve it to arrive at the general solution of $pde__1$

>	$pdsolve ()$

$u (x, y) = (\int_F1 (y) {&ExponentialE;}^{- y} &DifferentialD; y +_F2 (x)) {&ExponentialE;}^{y}$

(55)

A more involved example:

$pde__2 ≔ A__1 (\frac{\partial^{2}}{\partial x__4 \partial x__1} w (x__1, x__2, x__3, x__4)) + A__2 (\frac{\partial^{2}}{\partial x__4 \partial x__2} w (x__1, x__2, x__3, x__4)) + A__3 (\frac{\partial^{2}}{\partial x__4 \partial x__3} w (x__1, x__2, x__3, x__4)) + A__6 + A__5 (\frac{\partial}{\partial x__4} w (x__1, x__2, x__3, x__4)) + A__7 (A__1 (\frac{\partial}{\partial x__1} w (x__1, x__2, x__3, x__4)) + A__2 (\frac{\partial}{\partial x__2} w (x__1, x__2, x__3, x__4)) + A__3 (\frac{\partial}{\partial x__3} w (x__1, x__2, x__3, x__4)) + A__5 w (x__1, x__2, x__3, x__4) + A__4) + A__8 {(A__1 (\frac{\partial}{\partial x__1} w (x__1, x__2, x__3, x__4)) + A__2 (\frac{\partial}{\partial x__2} w (x__1, x__2, x__3, x__4)) + A__3 (\frac{\partial}{\partial x__3} w (x__1, x__2, x__3, x__4)) + A__5 w (x__1, x__2, x__3, x__4) + A__4)}^{2} &colon;$

>	$PDEtools :- FirstIntegralSolver (pde__2, w (x__1, x__2, x__3, x__4)) &semi;$

$w (x__1, x__2, x__3, x__4) = - \frac{1}{2} (Intat (\frac{\exp (\frac{A__5_a}{A__1}) (2 A__3 A__4 A__8 + A__3 A__7 + \tan (\frac{1}{2} \frac{\sqrt{{A__3}^{2} (4 A__6 A__8 - {A__7}^{2})} (_F1 (_a, \frac{A__1 x__2 + A__2_a - A__2 x__1}{A__1}, \frac{A__1 x__3 + A__3_a - A__3 x__1}{A__1}) + x__4)}{A__3}) \sqrt{{A__3}^{2} (4 A__6 A__8 - {A__7}^{2})})}{A__1 A__3 A__8},_a = x__1) - 2_F2 (\frac{A__1 x__2 - A__2 x__1}{A__1}, \frac{A__1 x__3 - A__3 x__1}{A__1} &co))$

Maple

Add-ons Maple

MapleSim

Add-ons MapleSim

Ingénierie systèmes

Services-conseils

Produits d’enseignement en ligne

Enseignement

Industrie

Automobile et aéronautique

Robotique

Conception de machine & automatismes industriels

Autres

Domaines d’application

Prix et d'achats

Achats

Licences institutionnelles Etudiants

Vue d’ensemble du service EMP

Support

Formation produit

Aide en ligne Produit

Webinaires et évènements

Publications

Hubs de contenu

Exemples et applications

Communauté

A propos de Maplesoft

Ressources presse

Communauté

Contacts

Online Help

All Products Maple MapleSim

Maple

Logiciel de mathématiques puissant facile à utiliser

Add-ons Maple

MapleSim

Modélisation avancée au niveau système

Add-ons MapleSim

Ingénierie systèmes

Services-conseils

Produits d’enseignement en ligne

Enseignement

Industrie

Automobile et aéronautique

Robotique

Conception de machine & automatismes industriels

Autres

Domaines d’application

Prix et d'achats

Achats

Licences institutionnelles Etudiants

Vue d’ensemble du service EMP

Support

Formation produit

Aide en ligne Produit

Webinaires et évènements

Publications

Hubs de contenu

Exemples et applications

Communauté

A propos de Maplesoft

Ressources presse

Communauté

Contacts

Online Help

All Products Maple MapleSim