How to check a solution of a partial differential equations?

Question

Here is a system which Maple can solve, but Mathematica cannot.

FF = f[a, b, c, d]

eq1 = D[FF, d] + (1 - a)*D[FF, a]/d == 0
eq2 = D[FF, c] + (b - (a - 1)*d)*D[FF, a]/(c*d) == 0
eq3 = D[FF, b] + D[FF, a]/d == 0
DSolve[{eq1, eq2, eq3}, {FF}, {a, b, c, d}]

DSolve does solve them individually, but can not solve them as a system, which is something I really need.

DSolve[eq1, f, {a, b, c, d}]
(* {{f -> Function[{a, b, c, d}, C[1][b, c][-(-1 + a) d]]}} *)
DSolve[eq2, f, {a, b, c, d}]
(* {{f -> Function[{a, b, c, d}, C[1][b, d][c (b + d - a d)]]}} *)
DSolve[eq3, f, {a, b, c, d}]
(* {{f -> Function[{a, b, c, d}, C[1][c, d][b - a d]]}} *)

Here is how it works in Maple:

So the solution is a function that takes parameters ${a,b,c,d}$, and the function is constructed somehow in this way $c(-b+(a-1)d)$.

How do I use _Mathematica to check, that it is a solution?

And similarly, how to do use MMA to construct a function from

(* {{f -> Function[{a, b, c, d}, C[1][b, c][-(-1 + a) d]]}} *)

That is, how do I construct a function with parameters ${a,b,c,d}$, and the body is separate in $b$ and $c$, but $a$ and $d$ are involved like $-(-1 + a) d$?

Thanks

Answer 1

You can check your solution by substituting it back into the equations:

{eq1, eq2, eq3} /. {
    f -> Function[{a, b, c, d}, g[c (d (a - 1) - b)]]}
// Simplify

(* {True, True, True} *)

Note that we substituted f with a Function of four variables, equal to some function g of the quantity c(d(a - 1) - b).

This answers your second question: you can construct such a function like so:

Function[{a, b, c, d}, h[b, c, (a-1)d]]

Where h represents some unknown functional dependence. In fact, DSolve already returned such a function:

Function[{a, b, c, d}, C[1][b, c][-(-1 + a) d]]

Instead of h, it uses a function C[1] that takes two parameters (b and c), and returns another function that takes one parameter (-(-1 + a) d). You can turn this into a more standard form by applying the following rule:

{C[n_][x1__][x2__] :> C[n][x1, x2]}

This turns, for example, the first solution into:

{f -> Function[{a, b, c, d}, C[1][b, c, -(-1 + a) d]]}

However, note that each of the three solutions uses the same name (C[1]) for its unknown function, but each of those functions will be different.