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How to solve $f_1,f_2,f_3$ from the following simultaneous PDEs:

$\frac{\partial f_3(x,y)}{\partial y}-\frac{\partial f_2(z,x)}{\partial z}=y^2,$

$\frac{\partial f_1(y,z)}{\partial z}-\frac{\partial f_3(x,y)}{\partial x}=z^2,$

$\frac{\partial f_2(z,x)}{\partial x}-\frac{\partial f_1(y,z)}{\partial y}=x^2.$

Attempt:

I can intutively say that $f_1(y,z)=\frac{z^3}{3}$, $f_2(z,x)=\frac{x^3}{3}$ and $f_3(x,y)=\frac{y^3}{3}$ satisfy the above equations.

But how to say it in a generalized way?

user1942348
  • 4,256
  • You only found the particular solution, but not the homogeneous: $$f_{1}! \left(y,z\right) = \frac{z^{3}}{3}+\left(c_{1} y+c_{3}\right) z+c_{2} y+c_{4}, \ f_{2}! \left(x,z\right) = \frac{x^{3}}{3}+\left(c_{1} z+c_{2}\right) x+c_{5} z+c_{7}, \ f_{3}! \left(x,y\right) = \frac{y^{3}}{3}+\left(c_{1} x+c_{5}\right) y+c_{3} x+c_{6}$$ – gpmath Jun 05 '23 at 16:04
  • @gpmath Ok..How to get this particular solution, please elaborate – user1942348 Jun 05 '23 at 16:38
  • Please write in "Answer Your Question" style – user1942348 Jun 05 '23 at 17:03

2 Answers2

1

Note that you are looking for a field $F$ such that $$\nabla \times F=(y^2, z^2, x^2), $$ check this Anti-curl operator

Guillermo García Sáez
  • 8,159
1

Let's examine the first PDE, $$ \frac{\partial f_3(x,y)}{\partial y}-\frac{\partial f_2(z,x)}{\partial z}=y^2. \tag{1} $$ Since $f_2$ does not depend on $y$, if we differentiate both sides of $(1)$ with respect to $y$ we obtain $$ \frac{\partial^2f_3(x,y)}{\partial y^2}=2y. \tag{2} $$ Integrating the above equation twice, we obtain $$ f_3(x,y)=\frac{y^3}{3}+g_3(x)y+h_3(x), \tag{3} $$ where $g_3$ and $h_3$ are, for the moment, arbitrary functions of $x$. A similar reasoning with the other two PDEs shows that \begin{align} f_1(y,z)&=\frac{z^3}{3}+g_1(y)z+h_1(y), \tag{4} \\ f_2(z,x)&=\frac{x^3}{3}+g_2(z)x+h_2(z). \tag{5} \end{align} Now, let's return to the first PDE. Substituting $(3)$ and $(5)$ in $(1)$, we obtain $$ g_3(x)-g_2'(z)x-h_2'(z)=0. \tag{6} $$ Differentiating both sides of $(6)$ with respect to $x$ yields $g_3'(x)-g_2'(z)=0$. Since $g_3'$ is a function of $x$ and $g_2'$ is a function of $z$, this identity can only be true if $g_3'(x)=g_2'(z)=A$, where $A$ is a constant. Therefore, \begin{align} g_3(x)&=Ax+B, \tag{7} \\ g_2(z)&=Az+C. \tag{8} \end{align} Substituting $(7)$ and $(8)$ in $(6)$, we obtain $B-h_2'(z)=0$, from which follows $$ h_2(z)=Bz+D. \tag{9} $$ Repeating this procedure with the other two PDEs, one obtains $g_1(y)$, $h_1(y)$, and $h_3(x)$. This is left as an exercise.

Gonçalo
  • 15,869