How to derive PDE for a given surface?

Question

The title may be ambiguous, therefore I will illustrate it using an example. Let's say we have a curve $y=\sin \left( x\right)$. We can say that corresponding ODE is $\dfrac{dy}{dx}=\cos \left( x\right)$. i.e. $y=\sin \left( x\right)$ is solution of $\dfrac{dy}{dx}=\cos \left( x\right)$. Now, lets say I have a surface in 3d i.e $z=\sin \left( x\right) +\cos \left( y\right)$. Is it possible to come up with PDE whose solution is $z=\sin \left( x\right) +\cos \left( y\right)$.

Answer 1

Say we are given the function \begin{align} y=f(c_1,c_2,x), \end{align} where $c_1$ and $c_2$ are parameters. Taking a derivative we find that \begin{align} y'=f'(c_1,c_2,x). \end{align} We can treat these two equations as a system for the parameters $c_1$ and $c_2$. Assuming that it could be solved, we'll denote the solutions as \begin{align} c_1=\varphi_1(x,y,y') \quad \text{and}\quad c_2=\varphi_2(x,y,y'). \end{align} Now consider the equation of the form \begin{align} F(\varphi_1,\varphi_2)=0, \end{align} where $F$ is an arbitrary equation of $x$, $y$, and $y'$. The equation $F=0$ is solved by the function $y=f(c_1,c_2,x)$, with the constraint on the constants $F(c_1,c_2)=0$.

$F$ can be any function, It will have the solution $y=f$! You can use this method with an implicitly defined function like $f(c_1,c_2,x,y)=0$, and you can add constants to arrive at higher order ODEs. The text I am pulling this from does not claim this is the most general equation with solution $y=f$, nor am I. (Though it doesn't seem like the craziest notion).

Let's alter your example to have two constants so we can use this method: \begin{align} y=c_1\sin(x)+c_2,\quad \rightarrow \quad y'=c_1\cos(x). \end{align} Solving for our parameters we get that \begin{align} c_1=\sec(x)y'\equiv\varphi_1,\quad c_2=y-\tan(x)y'\equiv \varphi_2. \end{align} So any equation of the form \begin{align} F(\sec(x)y',y-\tan(x)y')=0 \end{align} has the solution \begin{align} y=c_1\sin(x)+c_2,\quad\text{w/ constraint}\quad F(c_1,c_2)=0 \end{align}

Can we extend this to multivariable functions/surfaces? I don't see why not. Given the function \begin{align} z=f(c_1,c_2,c_3,x,y) \end{align} we'll take two derivatives, \begin{align} z_x=f_x(c_1,c_2,c_3,x,y),\quad z_y=f_y(c_1,c_2,c_3,x,y). \end{align} Again assuming we can solve these equations for the constants we would have the (3) equations \begin{align} c_k=\varphi_k(x,y,z,z_x,z_y). \end{align} Considering the function \begin{align} F(\varphi_1,\varphi_2,\varphi_3)=0, \end{align} we again see that $z=f(c_1,c_2,c_3,x,y)$ solves it, with the constraint that $F(c_1,c_2,c_3)=0$.

Let us alter another one of your examples: \begin{align} z=c_1\sin(x)+c_2\cos(y)+c_3. \end{align} Utilizing the method described above we find that \begin{align} c_1=\sec(x)z_x\equiv \varphi_1,\quad c_2=-\csc(y)z_y\equiv \varphi_2,\quad c_3=z-\tan(x)z_x+\cot(y)z_y\equiv\varphi_3. \end{align}

So then, any equation of the form \begin{align} F\big(\sec(x)z_x, -\csc(y)z_y, z-\tan(x)z_x+\cot(y)z_y\big)=0 \end{align} has the solution \begin{align} z=c_1\sin(x)+c_2\cos(x)+c_3,\quad\text{w/ constraint}\quad F(c_1,c_2,c_3)=0. \end{align}

I pulled this method (for the ODEs) from the text Handbook of exact solutions for ordinary differential equations / Andrei D. Polyanin, Valentin F. Zaitsev.--2nd ed.