Problem 40
Hard First-Order PDEs

Find the general solution of (y+z) (∂ u)/(∂ x) +(z+x) (∂…

Find the general solution of \[ (y+z)\,\frac{\partial u}{\partial x} +(z+x)\,\frac{\partial u}{\partial y} +(x+y)\,\frac{\partial u}{\partial z} = u, \] where $u = u(x,y,z)$.

Explanation of Method

A Lagrange PDE $Pu_x+Qu_y+Ru_z=S$ with $P=y+z$, $Q=z+x$, $R=x+y$, $S=u$. Pairwise differences of the first three ratios give two independent first integrals in $x,y,z$ alone; their sum, combined with the fourth ratio, gives a third integral involving $u$.

Step 1 · Characteristic equations

\[ \frac{dx}{y+z} = \frac{dy}{z+x} = \frac{dz}{x+y} = \frac{du}{u}. \tag{1} \]

Step 2 · First integral

Subtracting the 2nd ratio from the 1st, and the 3rd from the 2nd, both equal $dx/(y+z)$: \[ -\frac{d(x-y)}{x-y} = -\frac{d(y-z)}{y-z} \;\Longrightarrow\; \frac{d(x-y)}{x-y}=\frac{d(y-z)}{y-z}. \] Integrating: $\dfrac{x-y}{y-z}=c_1$. \tag{I}

Step 3 · Second integral

Similarly, the 1st-minus-3rd combination equals the 1st-minus-2nd: \[ \frac{d(x-y)}{x-y} = \frac{d(x-z)}{x-z}. \] Integrating: $\dfrac{x-y}{x-z}=c_2$. \tag{II} (Independent of $c_1$.)

Step 4 · Third integral

Summing the first three ratios with multipliers $(1,1,1)$: \[ \frac{dx+dy+dz}{2(x+y+z)} = \frac{d(x+y+z)}{2(x+y+z)} = \frac{du}{u}. \] Integrating: $\ln u=\tfrac12\ln(x+y+z)+\text{const}$, so \[ \frac{u}{\sqrt{x+y+z}} = c_3. \tag{III} \]

Step 5 · General solution

The general solution $c_3=\Psi(c_1,c_2)$ is: \[ u = \sqrt{x+y+z}\;\Psi\!\left(\frac{x-y}{y-z},\;\frac{x-y}{x-z}\right), \] where $\Psi$ is an arbitrary differentiable function of two arguments.


Verification

The condition for $F(x,y,z,u)$ to be a first integral of the system $dx/P=dy/Q=dz/R=du/S$ is $PF_x+QF_y+RF_z+SF_u=0$ (the $+$ sign, since $u$ is the dependent variable and $dF=F_x\,dx+\cdots+F_u\,du=0$ along a characteristic where $dx=P\,d\tau,\dots,du=S\,d\tau$).
Check $c_1=(x-y)/(y-z)$: \[ P(c_1)_x+Q(c_1)_y+R(c_1)_z = \frac{(y+z)(y-z)-(z+x)(x-z)+(x+y)(x-y)}{(y-z)^2} = \frac{0}{(y-z)^2}=0. \checkmark \] Check $c_2=(x-y)/(x-z)$: similarly simplifies to $0$. $\checkmark$
Check $c_3=u/\sqrt{x+y+z}$: with $(c_3)_x=(c_3)_y=(c_3)_z=-u/[2(x+y+z)^{3/2}]$ and $(c_3)_u=1/\sqrt{x+y+z}$: \[ P(c_3)_x+Q(c_3)_y+R(c_3)_z+S(c_3)_u = -\frac{u\cdot2(x+y+z)}{2(x+y+z)^{3/2}} + \frac{u}{\sqrt{x+y+z}} = -\frac{u}{\sqrt{x+y+z}}+\frac{u}{\sqrt{x+y+z}} = 0. \checkmark \]

Result

The general solution is:

\[ \boxed{u = \sqrt{x+y+z}\;\Psi\!\left(\frac{x-y}{y-z},\;\frac{x-y}{x-z}\right)} \] where $\Psi$ is an arbitrary differentiable function of two arguments.