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My students found an interesting PDE today,

$$ u_{xxx}+u_{yyy}+u_{zzz}-3u_{xyz} = 0. $$

Does anyone happen to know if this PDE has a name?

Incidentally, one curious solution I just found is $u(x,y,z) = e^{x+y-2z}$.

Connection with other question: in my other question "Is this a wave equation" I asked about solving a system of first order ODEs in dependent variables $u,v,w$ and dependent variables $x,y,z$. I believe that solutions of that system are likewise solutions of this system. However, the curious solution (and many others Will has elucidated) are not solutions to the first order system. The third order PDE considered in this question has additional, and from my viewpoint, extraneous solutions. Moreover, the connection with circulant matrices is not at accidental! This PDE was constructed from the left-regular representation of the group algebra of the cyclic group of order $3$. In view of this origin, I find the comments here delightful.

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James S. Cook
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    More generally it has special solutions of the form $e^{ax+by+cz}$ where $a^3+b^3+c^3=3abc$. – mjqxxxx Sep 19 '12 at 02:16
  • One of the lesser known items is that $a^3 + b^3 + c^3 - 3 a bc$ is divisible by $a+b+c.$ I'm just saying. – Will Jagy Sep 19 '12 at 02:19
  • thanks for the edit. Interesting, I wonder are there other solutions? – James S. Cook Sep 19 '12 at 02:22
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    Yes, any polynomial in $x,y,z$ with no term of degree over 2, so $xy$ is a legal term but not $xyz.$ You might also have them find all solutions that are homogeneous polynomials of degree 3. $$ a x^3 + b y^3 + c z^3 + d xyz + e y^2 z + f z^2 y + g z^2 x + h x^2 z + i x^2 y + j y^2 x. $$ – Will Jagy Sep 19 '12 at 02:40
  • Factorize $x^3+y^3+z^3-3xyz$ in $\mathbb{Z}$ will get $(x+y+z)(x^2+y^2+z^2-xy-xz-yz)$ . But how to factorize $x^2+y^2+z^2-xy-xz-yz$ in $\mathbb{C}$ ? – doraemonpaul Sep 19 '12 at 03:10
  • @doraemonpaul $$ ; \left( \frac{1}{4} \right) ; \left( x + y - 2 z + i \sqrt 3 (x-y) \right) ; \left( x + y - 2 z - i \sqrt 3 (x-y) \right)$$ – Will Jagy Sep 19 '12 at 03:23
  • @Will Jagy, how to derive to this? – doraemonpaul Sep 19 '12 at 03:42
  • @doraemonpaul, it's long. A homogeneous quadratic form with real coefficients in three variables can only be reducible over $\mathbb C$ if, when you write the Hessian matrix $H$ of second partial derivatives, you have $\det H = 0.$ I'm not sure that always suffices. Maybe. It does in the example above. I will fiddle with it some more tomorrow. – Will Jagy Sep 19 '12 at 04:45
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    @doraemonpaul: the expression $x^3 + y^3 + z^3 - 3xyz$ has a symmetry $(x, y, z) \mapsto (x, \omega y, \omega^2 z)$ where $\omega$ is a primitive third root of unity. If $x + y + z$ is a factor it follows that $(x + \omega y + \omega^2 z)$ and $(x + \omega^2 y + \omega z)$ must also be factors. In fact this polynomial is the determinant of a circulant matrix of order $3$ with entries $x, y, z$ and it is straightforward to write down the eigenvectors and eigenvalues of such a thing. This is a special case of the discrete Fourier transform. – Qiaochu Yuan Sep 19 '12 at 04:48
  • @doraemonpaul, it is necessary and sufficient, if the coefficient of $x^2$ is positive, that $H$ be positive semi-definite but with $\det H = 0,$ so not definite. Easier than I expected. With that, demanding the coefficient of $x^2$ be $1,$ and nonzero complex numbers $\alpha, \beta$ you get $$ x^2 + \mbox{other} = (x + y \alpha + z \beta)(x + y \bar{\alpha} + z \bar{\beta}) $$ – Will Jagy Sep 19 '12 at 04:55
  • James, after your edit: we are here to give you delight. It is our mission. – Will Jagy Sep 19 '12 at 05:11

2 Answers2

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I don't know if it has a name but here are a bunch of solutions. Write $D_x$ for the operator which partially differentiates by $x$, etc. Then the equation can be written

$$(D_x^3 + D_y^3 + D_z^3 - 3D_x D_y D_z) u = 0.$$

This operator happens to be a product of linear factors as was observed in the comments. Letting $\omega$ be a primitive third root of unity we can write it as

$$(D_x + D_y + D_z)(D_x + \omega D_y + \omega^2 D_z)(D_x + \omega^2 D_y + \omega D_z) u = 0.$$

Consequently a solution to any of the equations

$$u_x + u_y + u_z = 0$$ $$u_x + \omega u_y + \omega^2 u_z = 0$$ $$u_x + \omega^2 u_y + \omega u_z = 0$$

gives a solution to the original equation, and one might hope that linear combinations of such solutions give all solutions to the original equation (I don't know if this is true). I believe the solutions to the first equation are all linear combinations of solutions of the form

$$u(x, y, z) = f(ax + by + cz)$$

whenever $a + b + c = 0$. If we want $u$ to be real-valued, then the second two equations are equivalent but impose two conditions on $u$ by taking real and imaginary parts, namely

$$u_x - \frac{u_y + u_z}{2} = 0$$ $$u_y - u_z = 0.$$

This is equivalent to requiring $u_x = u_y = u_z$, which I believe is equivalent to

$$u(x, y, z) = g(x + y + z).$$

Qiaochu Yuan
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Here is a general solution,

$$ u \left( x,y,z \right) ={\it F_1} \left( y-x,z-x \right) +{\it F_2} \left( y- \left( -\frac{1}{2}+\frac{i\sqrt {3}}{2} \right) x,z- \left( -\frac{1}{2}-\frac{i \sqrt {3}}{2} \right) x \right) + {\it F_3} \left( y- \left( -\frac{1}{2}-\frac{i \sqrt {3}}{2} \right) x, z- \left( -\frac{1}{2}+\frac{i\sqrt {3}}{2} \right) x \right) \,.$$

Note that your solution is a special case of this solution.

Mhenni Benghorbal
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    How about your steps? Why the arbitrary functions of the general solution of the 3-independent-variables PDE are 2-parameters functions? – doraemonpaul Sep 21 '12 at 02:18