The problem is to solve $\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1$ with $z(0,0)=0$ and find $z(0,1)$.

I used Charpit's method to get $z(x,y)=kx + \frac{y}{k}+c$ as its solution. Now the initial condition gives $c=0$. This gives $z(x,y)=kx + \frac{y}{k}$ but the arbitrary constant is still present even after incorporating the initial condition and this is my point of confusion. Then $z(0,1)=\frac{1}{k}\ne 0$ if $k$ is non-zero. Added 1- I realized that the solution can also take the form $z=ky+\frac{x}{k}$ which gives me $z(0,1)=k$.

What is correct? $z(0,1)=\frac{1}{k}\ne 0$ or $z(0,1)=k$?

Am I correct? Any hints where I am wrong?

Edit- After few searches I found a way to determine the arbitrary constant 'k' which proceeds as follows:

Solve the $k$-quadratic $k^2x-kz+y=0$ for $k$ and substitute this $k$ in $z=k x+\frac{y}{k}$ which is now free from $k$.

I suspect this approach as $k$ thus obtained is no more a constant and the reason for determining $k$ in this way is not justified. I don't know it is correct or not. Hoping that someone answer my doubt!

  • 1
    $\begingroup$ The reason you still have the $k$ is because there are not enough boundary conditions to determine it. Your method (in the edit) for determining $k$ won't work because it is circular. $\endgroup$ – John Barber Jun 12 at 15:00
  • $\begingroup$ You can immediately verify that $z(x,y)=\frac1{42}x+42y$ satisfies the pde and boundary condition, and gives you $z(0,1)=42$. $\endgroup$ – Hagen von Eitzen Jun 12 at 15:08
  • $\begingroup$ Do you mean $z(0,1)=1/k \ne 0$? $\endgroup$ – Mathlover Jun 12 at 15:24
  • $\begingroup$ @Hagen von Eitzen, I have made an edit. What is the value $z(0,1)$? $\endgroup$ – Mathlover Jun 12 at 15:39
  • $\begingroup$ @John Barber, so can't we give the value of $z(0,1)$? $\endgroup$ – Mathlover Jun 12 at 15:40

There is another class of solutions to this PDE.

We try separation-of-variables and guess a solution of the form $z(x,y) = f(x) g(y) + c$: \begin{align*} \frac{\partial z}{\partial x} \cdot \frac{\partial z}{\partial y} &\;=\; f'(x) g(y)\cdot f(x) g'(y)\\ &\;=\; f'(x) f(x)\cdot g'(y) g(y) \;=\; 1 \end{align*} The last expression above can only be true for all $x$ and $y$ if the terms in $x$ are equal to some constant $k$ and the terms in $y$ are equal to $1/k$: \begin{align*} f'(x) f(x) &\;=\; k \;\;\;\quad\rightarrow\quad f(x) \;=\; \sqrt{2(k x \,+\, a)}\\[0.1in] g'(y) g(y) &\;=\; 1/k \quad\rightarrow\quad g(y) \;=\; \sqrt{2(y/k \,+\, b)} \end{align*} Here $a$ and $b$ are unknown constants. The solution to the PDE is thus: \begin{equation*} z(x,y) \;=\; f(x) g(y) + c \;=\; 2\sqrt{(k x \,+\, a)(y/k \,+\, b)} \,+\, c \end{equation*} The boundary condition $z(0,0) = 0$ implies that the unknown constants $a$, $b$, and $c$ must obey $2\sqrt{ab} + c = 0$.

  • $\begingroup$ Thanks @John Barber $\endgroup$ – Mathlover Jun 12 at 18:21

$$\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1\quad\text{ with } \quad z(0,0)=0 \tag 1$$ $$\frac{\partial^2 z}{\partial x^2}\cdot \frac{\partial z}{\partial y}+\frac{\partial z}{\partial x}\cdot \frac{\partial^2 z}{\partial x\partial y}=0$$ Change of function : $\quad u(x,y)=\frac{\partial z}{\partial x}$ $$\frac{\partial u}{\partial x}\left(-\frac{1}{u} \right)+u\frac{\partial u}{\partial y}=0$$ $$\frac{\partial u}{\partial x}-u^2\frac{\partial u}{\partial y}=0 \tag 2$$ system of characteristic ODEs : $\frac{dx}{1}=\frac{dy}{-u^2}=\frac{du}{0}$

First characteristic equation from $\frac{du}{0}=$finite function : $$u=c_1$$ Second characteristic equation from $\frac{dx}{1}=\frac{dy}{-c_1^2}\quad;\quad x+\frac{y}{c_1^2}=c_2$ $$x+\frac{y}{u^2}=c_2$$ General solution of Eq.$(2)$ on the form of implicite equation $c_2=F(c_1)$ : $$x+\frac{y}{u^2}=F(u)$$ $F$ is an arbitrary function to be determined by some boundary condition.


The only condition specified in the wording of the question is $z(0,0)=0$. Generally the boundary condition is given on a curve. On a point only is not sufficient to a well posed condition for Eq.$(2)$. Thus we expect an infinity of solutions.


Arbitrary choosing the function $F(X)=a+\frac{b}{X^2}$




This function is a solution of the PDE $(1)$ . We determine $c$ in order to satisfy the condition $z(0,0)=0$

$$z(x,y)=2\:\sqrt{(b+y)(a-x)}-2\:\sqrt{ab}$$ In particular $\quad z(0,1)=2\:\sqrt{b(a-1)}-2\:\sqrt{ab}$

Of course this is only an example. Other choice of the function $F(X)$ would lead to other results. Note that many would be not explicit and/or complicated depending on the kind of function $F(X)$.

In conclusion don't be surprized to not find a unique result for $z(0,1)$. Probably this is due to a typo in the wording of the problem. Especially a boundary condition such as $z(0,0)=0$ seems suspect.


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