Mathematica Asked on April 8, 2021
I’m trying to solve a simple system of second order PDEs with Mathematica 11.1.
Here is the system:
DSolve[{
D[f[x, y, z], x, y] == 0,
D[f[x, y, z], x, z] == 0,
D[f[x, y, z], y, z] == 0
}, f[x, y, z], {x, y, z}]
Mathematica returns it unevaluated…
I expect the solution to be
f[x,y,z] -> C[1][x] + C[2][y] + C[3][z]
Any idea what I can try?
Thanks!
Update:
We can differentiate a third time:
DSolve[{D[f[x, y, z], x, y, z] == 0}, f[x, y, z], {x, y, z}]
and then Mathematica gives the following answer:
f[x, y, z] -> C[1][y, z] + C[2][x, z] + C[3][x, y]
Maybe this can be somehow combined with the answer by bbgodfrey to give the correct answer for the system above?
The comment by user64494 suggested to me the following
sxy = Flatten@DSolve[{D[f[x, y, z], x, y] == 0}, f[x, y, z], {x, y, z}] /.
C[n_][z][v_] -> d[n][v, z];
sxz = Flatten@DSolve[{D[d[1][x, z], x, z] == 0}, d[1][x, z], {x, z}] /. C -> c;
syz = Flatten@DSolve[{D[d[2][y, z], y, z] == 0}, d[2][y, z], {y, z}] /. C -> b;
sxy /. sxz /. syz
(* {f[x, y, z] -> b[1][y] + b[2][z] + c[1][x] + c[2][z]} *)
Not at all satisfying but perhaps useful in some circumstances.
Correct answer by bbgodfrey on April 8, 2021
First we solve the system:
s = Flatten @@
DSolve[{D[f[x, y, z], x, y, z] == 0}, f[x, y, z], {x, y, z}] /.
C[n_][a_, b_] -> d[n][a, b]
getting:
{f[x, y, z] -> d[1][y, z] + d[2][x, z] + d[3][x, y]}
Then we solve the original system equation for equation:
sxyz = Join[
Flatten @@
DSolve[D[d[1][y, z] + d[2][x, z] + d[3][x, y], x, y] == 0,
d[3][x, y], {x, y}] /. C[n_][a_] -> f1[n][a],
Flatten @@
DSolve[D[d[1][y, z] + d[2][x, z] + d[3][x, y], x, z] == 0,
d[2][x, z], {x, z}] /. C[n_][a_] -> f2[n][a],
Flatten @@
DSolve[D[d[1][y, z] + d[2][x, z] + d[3][x, y], y, z] == 0,
d[1][y, z], {y, z}] /. C[n_][a_] -> f3[n][a]
]
getting:
{d[3][x, y] -> f1[1][x] + f1[2][y],
d[2][x, z] -> f2[1][x] + f2[2][z],
d[1][y, z] -> f3[1][y] + f3[2][z]}
The final solution comes then from:
(s /. sxyz) //. f_[n_][a_] + g_[m_][a_] -> F[n][m][a]
as:
{f[x, y, z] -> F[1][1][x] + F[2][1][y] + F[2][2][z]}
Answered by C. B. on April 8, 2021
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