Solution of pde $\frac{\partial z}{\partial x}\cdot \frac{\partial z}{\partial y}=1$ with $z(0,0)=0$

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!

• 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. – John Barber Jun 12 at 15:00
• You can immediately verify that $z(x,y)=\frac1{42}x+42y$ satisfies the pde and boundary condition, and gives you $z(0,1)=42$. – Hagen von Eitzen Jun 12 at 15:08
• Do you mean $z(0,1)=1/k \ne 0$? – Mathlover Jun 12 at 15:24
• @Hagen von Eitzen, I have made an edit. What is the value $z(0,1)$? – Mathlover Jun 12 at 15:39
• @John Barber, so can't we give the value of $z(0,1)$? – 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 $$2\sqrt{ab} + c = 0$$.

• Thanks @John Barber – 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.

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.

EXAMPLE :

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

$$x+\frac{y}{u^2}=a-\frac{b}{u^2}$$

$$u=\sqrt{\frac{-b-y}{x-a}}$$

$$z(x,y)=\int\sqrt{\frac{-b-y}{x-a}}dx=2\:\sqrt{(-b-y)(x-a)}+c$$

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.