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$x_X(X,Y)=1+x(X,Y)+y(X,Y)$ $y_Y(X,Y)=1+x(X,Y)+y(X,Y)$

where subscripts are partial derivatives. To look for non-constant x, y solutions, is it possible to get an analytic solution of this PDE in mathematica?

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  • $\begingroup$ What did you try so far? Without boundary-conditions your pde-system has solution X[x,y]==const,Y[x,y]==-(1+const) $\endgroup$ Commented Mar 14, 2023 at 10:25
  • $\begingroup$ I just tried dsolve. How did you get this solution in inverse functions? $\endgroup$
    – feynman
    Commented Mar 14, 2023 at 11:35
  • $\begingroup$ Systematic guessing! It's not an inverse function, I took pde {Derivative[1, 0][X][x, y] == 1 + X[x, y] + Y[x, y], Derivative[0, 1 ][Y][x, y] == 1 + X[x, y] + Y[x, y]} $\endgroup$ Commented Mar 14, 2023 at 11:39
  • $\begingroup$ sorry don't get what you mean $\endgroup$
    – feynman
    Commented Mar 14, 2023 at 12:10
  • $\begingroup$ In your notation possible solutions are x[X,Y]==const,y[X,Y]==-(1+const) $\endgroup$ Commented Mar 14, 2023 at 12:17

2 Answers 2

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According to your equation, X[x,y] and Y[x,y] have the same derivative. Therefore, they can only differ by a constant. This then changes your equations to:

eq= Derivative[1, 0][X][x, y] == X[x, y] + const

And we have the solution:

DSolve[Derivative[1, 0][X][x, y] == X[x, y] + const, X, {x, y}]

{{X -> Function[{x, y}, -const + E^x C[1][y]]}}
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  • $\begingroup$ x and y are dependent variable, X and Y are independent ones. $\endgroup$
    – feynman
    Commented Mar 15, 2023 at 8:31
  • $\begingroup$ Does it matter? $\endgroup$ Commented Mar 15, 2023 at 9:45
  • $\begingroup$ sorry don't get what you mean by 'X[x,y] and Y[x,y] have the same derivative'. They have a common partial derivative but not necessarily all partial derivatives. How did you get your 2 equations from my original ones? $\endgroup$
    – feynman
    Commented Mar 16, 2023 at 8:11
  • $\begingroup$ The general solution of both functions are the same. $\endgroup$ Commented Mar 16, 2023 at 8:19
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The equation does not qualify as a well defined, solvable PDE problem.

Sum and difference give

d_x Y + d_y X == 2(1+ X + Y)
d_x Y - d_y X == 0

By changing names it looks like the condition of rot V = 0. So, if a solution exists loacally, there exists a potential function V such that

{X,Y} = Grad V

and

 d_x,y V + d_y,x V = 2 d_xy  == 2 (1 + (d_x + d_y) V

As a system of first order, its approximation on a lattice of points { m dx, n dy } reads as an evolution displacement

f[x+dx,y] :> f[x,y] + dx  (1+ f[x,y] +g[x,y] )
g[x,y+dy] :> g[x,y] + dy  (1+ f[x,y] +g[x,y] )

there is no rule to get the missing points

f[x,y+dy] and g[x+dx,y]?

As it stands, it is a differential condition, the has not solutions, but may select from an infinity of expressions parametrized by free functions instead of free constants as in algebraic linear systems of equations.

First Order PDE

Such a differential conditions may arise e.g. for the flow of a vector field between to parallel planes and may serve a rude replacement of surface bundary conditions.

For the more deeply interested only:

It is always a good idea in first order PDE theory to have two models in mind:

1: The 2d-Cauchy-Riemann equations for holomorphic functions with singularities and

2: The 4d-Maxwell Lorentz system of first order equations for differential 2-forms of force fields and their duals, the charge and current induction fields

The process of solving the Cauchy-Riemann equations, effectively, yields a function (and all its derivatives and integrals) in a domain bounded by an integrable curve by an integral over the data on the boundary. The so called Greens function, the kernel if this map is 1/z, the famous Cauchy integral theorem.

The Maxwell equations say nearly nothing besides the existence of a 4-vector potential and give receipes for selecting a meaningful sets of initial and boundary conditions in arbitrary geometries.

For the inhomgenous equations they yield conserved currents of the charges sitting in the discontinuities and boundaries.

Again the Greens function is 1/Abs[{x,y,z}] exhibitin the common features of both theories.

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  • $\begingroup$ Can you ground your claim "By changing names it looks like the condition of rot V = 0"? $\endgroup$
    – user64494
    Commented Apr 14, 2023 at 6:41
  • $\begingroup$ The difference of both equations yields zero. The zero sum of the two partial derivatves of two components of a vector may be interpretes always as the commutativity of the second order mixed deriviative of a function, always $\endgroup$
    – Roland F
    Commented Apr 14, 2023 at 7:08

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