# Finding unkown function

I want to find a function t3

t3[x1, x2, x3]


that fulfills these equations:

2 D[t3[x1, x2, x3], x3] + 6 D[t3[x1, x2, x3], x2] - 9 x2^2 D[t3[x1, x2, x3], x1] !=0


for any real x

D[t3[x1, x2, x3], x3] + 3 D[t3[x1, x2, x3], x2] +
(-1 - 9/2 x1^2) D[t3[x1, x2, x3], x1] == 0


How can I do this (with Mathematica)?

A simpler example

• The != means that the expression should not be identically equal to zero, or it should for no values x1, x2, x3 take the value zero? – corey979 Jan 13 '17 at 22:45
• it means that the expression should never be 0 for any real x1,x2,x3 – OhmSweetOhm Jan 14 '17 at 15:24

Let's first solve the equation:

sol[x1_, x2_, x3_] =
t3[x1, x2, x3] /.
DSolve[D[t3[x1, x2, x3], x3] +
3 D[t3[x1, x2, x3], x2] + (-1 - 9/2 x1^2) D[t3[x1, x2, x3], x1] == 0,
t3[x1, x2, x3], {x1, x2, x3}][[1]]


C[1] here is an arbitrary function of two variables.

Let's take the second condition:

sol2[x1_, x2_, x3_] =
FullSimplify[
D[sol[x1, x2, x3], x3] + 6 D[sol[x1, x2, x3], x2] - 9 x2^2 D[sol[x1, x2, x3], x1]
]


Now, I'm not sure what is != supposed to mean: for the expression to not be identically equal to zero, or to never attain the value zero? If the former, then with

f[a_, b_] := a^2 + b^2


for example

sol2[1, 1, 1] /. C[1] -> f // N


3.71851

so the expression is not identically zero. Thence, the solution is

sol[x1, x2, x3] /. C[1] -> f


Here

FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> f, {x1, x2, x3}, Reals]


{{x1 -> 0, x2 -> 0, x3 -> 0}}

On the other hand, if C[1] == ArcTan, then

sol2[1, 1, 1] /. C[1] -> ArcTan // N


-0.365403

so the solution is not identically equal to zero. Moreover,

FindInstance[sol2[x1, x2, x3] == 0 /. C[1] -> ArcTan, {x1, x2, x3}, Reals]


didn't give any instance for several minutes, so it's likely there is none and the function is non-zero everywhere.