A simple Partial Differential Equation? I thought…

I was looking at the Partial Differential Equation involving function:

$$z(x,y)$$

$$\frac{\partial z}{\partial x} + c \frac{\partial z}{\partial y} = 0$$

Which fairly intuitively has a solution:

$$z = f\left( x - \frac{1}{c} y \right)$$

Now I was considering the generalization of this equation to:

$$\frac{\partial z}{\partial x} + (c_1x + c_2) \frac{\partial z}{\partial y} = 0$$

But instead of starting too large I would focus first on:

$$\frac{\partial z}{\partial x} + c_1 x\frac{\partial z}{\partial y} = 0$$

Which itself i focused on:

$$\frac{\partial z}{\partial x} + x\frac{\partial z}{\partial y} = 0$$

I did some algebraic debauchery and have concluded that:

$$g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right]$$

Is a solution to this. But I do not know its closed form.

Work So Far:

One notes that the solution $z = f\left( x - \frac{1}{c} y \right)$ to the first problem can be derived by noting that:

$$z = 1, x - \frac{1}{c}y, \left(x - \frac{1}{c}y\right)^2 ...$$

Are all solutions and the transform is linear so any function of the form:

$$F = a_0(1) + a_1 \left(x - \frac{1}{c}y \right) + a_2 \left(x - \frac{1}{c}y\right)^2 ...$$ Is a solution, which is the general power series of all functions of the form

$$f \left( x - \frac{1}{c}y\right)$$

Using that same intuition note:

$$g_0 = a_0$$

Is a solution... then we can consider $$g_1 = a_0 + a_1 x$$ and ask what term should be added to $g_1$ to make $g_2$ such that $$\frac{\partial g_1}{\partial x} + x \frac{\partial g_2}{\partial y} = 0$$. Then recursively generate $g_3, g_4 ...$

By doing so we conclude that one solution is

$$g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right]$$

What is the closed form for this guy? From here it becomes clear that if we consider $g^r$

$$\frac{\partial g^r}{\partial x} + x \frac{\partial g^r}{\partial y} =$$ $$r g^{r-1} \left( \frac{\partial g}{\partial x} + x \frac{\partial g}{\partial y} \right) = r g^{r-1}(0) = 0$$

And since our transform is linear:

$$z = e_0 + e_1g + e_2 g^2 + e_3 g^3 ...$$

Is a solution to the equation meaning that the general solution is

$$z = f(g)$$

• Not that it directly answers your question about the closed form of that guy, but if you change variables to $u=x$, $v=y/2x^2$, you'll find that $z=f(v)=f(y/2x^2)$ is the general solution to $z_x+xz_y=0$ (in the region $x>0$, say). – Hans Lundmark Jan 5 '15 at 12:37
• (By the way, in your approach, I think you need $a_1=0$, otherwise $g_1$ isn't a solution...) – Hans Lundmark Jan 5 '15 at 12:38
• $g_1, g_2 ....$ all aren't solutions, $g_{i+1}$ is simply a correction to $g_{i}$ but in the limit to infinity they form the infinite sum – frogeyedpeas Jan 5 '15 at 14:04

For $\dfrac{\partial z}{\partial x}+(c_1x+c_2)\dfrac{\partial z}{\partial y}=0$ ,

$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$

$\dfrac{dy}{dt}=c_1x+c_2=c_1t+c_2$ , letting $y(0)=y_0$ , we have $y=\dfrac{c_1t^2}{2}+c_2t+y_0=\dfrac{c_1x^2}{2}+c_2x+y_0$

$\dfrac{dz}{dt}=0$ , letting $z(0)=f(y_0)$ , we have $z(x,y)=f(y_0)=f\left(y-\dfrac{c_1x^2}{2}-c_2x\right)$

After playing around with wolfram alpha I found that

$$\sum_{i=1}^{\infty} \left[ \frac{1}{i} \begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} u^i \right] = \frac{1}{2} (1 - \sqrt{1 - 4u} )$$

Therefore

$$g = a_0 + a_1 x - 2a_1 x \frac{1}{2} \left(1 - \sqrt{1 - 4\frac{y}{2x^2}} \right) =$$ $$a_0 + a_1x - a_1x \left(1 - \sqrt{1 - 2 \frac{y}{x^2}} \right) =$$ $$a_0 + a_1x - a_1x \left(1 - \frac{\sqrt{x^2 - 2 y}}{x} \right) =$$ $$a_0 + a_1x - a_1 \left(x - \sqrt{x^2 - 2 y} \right) =$$ $$a_0 + a_1 \sqrt{x^2 - 2y}$$

We verify:

$$\frac{\partial g}{\partial x} = a_1 \frac{x}{\sqrt{x^2 - 2y}}$$ $$x\frac{\partial g}{\partial y} = -a_1 \frac{x}{\sqrt{x^2 - 2y}}$$

$$\frac{\partial g}{\partial x} + x\frac{\partial g}{\partial y} = 0$$

• This is only defined however if $\left|\frac{y}{2x^2} \right|< \frac{1}{4}$ – frogeyedpeas Jan 5 '15 at 14:17
• Except, even if that property is violated, if one allows complex roots then the equation is still well defined – frogeyedpeas Jan 5 '15 at 14:20