Solve the system of partial differential equations

How to solve the system of partial differential equations $$u_{1x_1}=u_{2x_2}=u_{3x_3=0}$$$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0$$ for $i=1,2,3$

I tried so hard and it becomes so complicated. I first conclude $u_1=u_1(x_2,x_3), u_2=u_2(x_1,x_3),u_3=u_3(x_1,x_2)$. Then by $\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}=0$, I get $u_1(x_2,x_3)=c_1(x_3)x_2+b_1(x_3)$, and $u_2(x_1,x_3)=-c_1(x_3)x_1+b_2(x_3)$. I did these for three of the system of equations and in the end it becomes very complicated. I cannot solve for the coefficient $c_1,b_1,c_2,b_2$.etc. at all.

Could someone kindely provide some help ? Thanks!

• What about $j$ ? . I suppose that $j\neq i$ in the second equation. – JJacquelin Sep 16 '15 at 21:50

In THIS ANSWER, I showed that each component $u_i$ is a linear function of $x_j$ and $x_k$, with $i$, $j$ and $k$ distinct. In that development, we first used that since the diagonal terms add to zero, we obtain

$$\frac{\partial u_i}{\partial x_i}=0 \tag 1$$

which implies that the $i$th component of $u$ is independent of the $i$th coordinate variable. We can write, therefore

\begin{align} u_1=u_1(x_2,x_3)\\\\ u_2=u_2(x_1,x_3)\\\\ u_3=u_3(x_1,x_2)\\\\ \end{align}

In this problem, we are given that

$$\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0 \tag 2$$

Observe $(2)$ for the case for which $i=1$ and $j=2$. Then, we have

$$\frac{\partial u_1(x_2,x_3)}{\partial x_2}=-\frac{\partial u_2(x_1,x_3)}{\partial x_1} \tag 3$$

The left-hand side of $(3)$ is independent of $x_1$ while the right-hand side of $(3)$ is independent of $x_2$. But this implies that $u_1(x_2,x_3)$ must be a linear function of $x_2$; seen by straightforward integration, $u_1(x_2,x_3)=A_1(x_3)x_2+B_1(x_3)$. And similarly, $u_2$ must be a linear function of $x_1$ with $u_2(x_1,x_3)=A_2(x_3)x_1+B_2(x_3)$. Continuing to use $(2)$ with various combinations of $i$ and $j$ gives us the result

\begin{align} u_1&=A_{12}x_2+A_{13}x_3+Bx_2x_3+C_1\\\\ u_2&=A_{12}x_1+A_{23}x_3+Bx_1x_3+C_2\\\\ u_3&=A_{13}x_1+A_{23}x_2+Bx_1x_2+C_3 \end{align}

Without additional information on $\vec u$, the constants are arbitrary.

• In equation (3). I think the coefficient can depend on $x_3$. So it may not be linear? – Sherry Sep 16 '15 at 22:09
• In $\frac{\partial u_1(x_2,x_3)}{\partial x_2}=-\frac{\partial u_2(x_1,x_3)}{\partial x_1}$. I don't understand why we can conclude LHS is a linear function of $x_2$, since the coefficient can depend on $x_3$? Could you pls help explain that? Thanks so much! – Sherry Sep 16 '15 at 22:32
• @Sherry Good catch. I edited to correct. – Mark Viola Sep 16 '15 at 23:51
• You're welcome. My pleasure. – Mark Viola Sep 17 '15 at 13:53

I prefer to change the notations in order to make it clearer : NOTE :

in the wording of the question, it is written : $\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}=0$ for $i=1,2,3$. But nothing is specified about $j$. In my answer above I supposed $j=1,2,3$ with $j\neq i$ because if the equations with $i=j$ are allowed, the problem becomes trivial (too simple).

• We are not given that the second partial derivatives exist. And the i th component cannot depend on the i th coordinate variable. – Mark Viola Sep 17 '15 at 4:19
• We are not given that the second partial derivatives exist. But we don't care in so far as the functions finally obtained fullfill the PDEs. It's better than to remain blocked. In any case, you can suppose that the second partial derivatives exists, you solve the equations, you test the result and see that they do exist. So, you obtained a set of functions $u_1,u_2,u_3$ wich are derivable since they are made of arbitrary derivable functions $F_1,F_2,F_3,F_4$. – JJacquelin Sep 17 '15 at 6:14
• Unfortunately, the answer posted is not the correct solution. $u_i$ cannot be a function of $x_i$ for $i=1,2,3$ since $\frac{\partial u_i}{\partial x_i}=0$. And if the second partials exist, all of them are consequently $0$. – Mark Viola Sep 17 '15 at 14:42