Mfcttrf

I am currently studying Calculus of Variations and have come up with this problem.

What functions $f:Bbb R^ntoBbb R$ satisfy $$sumlimits_ifracpartial fpartial x_i=0tag1$$ with $f$ of differentiability class $C^infty$?

It can be shown that if the total derivative is zero; that is, if $$fracdFdt=sum_ifracpartial fpartial x_icdotfracdx_idt=0$$ with $F(t)=f(x_1(t),cdots,x_n(t))$, then $f$ is constant. However, I cannot see how the same approach can be used for $(1)$.

The trivial function $f(x)=c$ satisfies $(1)$. For even $n>1$, a non-trivial solution is the function $$f(x_1,cdots,x_n)=expleft(sum_i(-1)^i+1x_iright)$$ since consecutive partial derivatives of $f$ cancel each other out. I suspect there are more solutions that invoke trigonometric expressions.

Is it possible to derive the entire family/families of functions $f$ that satisfy $(1)$?

Note that your equation is equivalent to
$$
(1,1,1,dots,1)cdotnabla f=0
$$
This means that $f$ is constant on lines parallel to $(1,1,1,dots,1)$; these lines are the characteristics of the partial differential equation.

Thus, we can define $f$ freely on $x_n=0$ and then
$$
f(x_1,x_2,x_3,dots,x_n)=f(x_1-x_n,x_2-x_n,x_3-x_n,dots,0)
$$

Define $$y_i:=x_i-x_n$$ for every $i=1,2,ldots,n-1$, and $$y_n:=x_n,.$$ Then, we have
$$x_i=y_i+y_n$$
for each $i=1,2,ldots,n-1$, and $$x_n=y_n,.$$ Observe that
$$fracpartialpartial y_n=sum_k=1^n,left(fracpartial x_kpartial y_nright),fracpartialpartial x_k=sum_k=1^n,fracpartialpartial x_k,.$$
Hence, the given partial differential equation is equivalent to $$fracpartial phipartial y_n(y_1,y_2,ldots,y_n-1,y_n)=0,,$$
where $$phi(y_1,y_2,ldots,y_n-1,y_n):=fleft(y_1+y_n,y_2+y_n,ldots,y_n-1+y_n,y_nright),.$$ Consequently, $phi(y_1,y_2,ldots,y_n-1,y_n)$ is a function of $y_1,y_2,ldots,y_n-1$. In other words, there exists a smooth function $Phi:mathbbR^n-1to mathbbR$ such that
$$fleft(x_1,x_2,ldots,x_n-1,x_nright)=Phileft(x_1-x_n,x_2-x_n,ldots,x_n-1-x_nright),.$$
We can check that such $f$ satisfies the condition. Your nontrivial example for even $n$ arises from taking
$$Phi(t_1,t_2,ldots,t_n-1):=expleft(sum_i=1^n-1,(-1)^i+1,t_iright),.$$

In general, let $alpha_1,alpha_2,ldots,alpha_ninmathbbR$ be arbitrary with $alpha_nneq 0$. Then, all differentiable functions $f:mathbbR^ntomathbbR$ that satisfy the partial differential equation
$$sum_i=1^n,alpha_i,fracpartial fpartial x_i(x_1,x_2,ldots,x_n)=0$$
for every $x_1,x_2,ldots,x_ninmathbbR$ take the form
$$f(x_1,x_2,ldots,x_n)=Phileft(alpha_n,x_1-alpha_1,x_n,alpha_n,x_2-alpha_2,x_n,ldots,alpha_n,x_n-1-alpha_n-1,x_nright),,$$
where $Phi:mathbbR^n-1tomathbbR$ is a differentiable function.