# Lagrange Method for Linear PDE with 3 independent variables

I have 2 questions

1. Suppose i have this pde

$$x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}+t\dfrac{\partial z}{\partial t}=xyt$$

And the auxiliary equation is

$$\dfrac{\partial x}{x}=\dfrac{\partial y}{y}=\dfrac{\partial t}{t}=\dfrac{\partial z}{xyt}$$

I know that the last solution will be has this form :

$$\Phi(a,b,c)=0$$

So, i have to find out what $$a$$, $$b$$, and $$c$$ are.

From $$\dfrac{\partial x}{x}=\dfrac{\partial y}{y}$$, i got $$a=\dfrac{x}{y}$$

From $$\dfrac{\partial y}{y}=\dfrac{\partial t}{t}$$, i got $$b=\dfrac{y}{t}$$

1. And i want to write $$\Phi(a,b,c)=0$$ as explicit in $$z$$ so, i have to solve for $$dz$$

Suppose i want to solve for

$$\dfrac{\partial t}{t}=\dfrac{\partial z}{xyt}$$

And i'm stuck here. I was confused with the denominator $$xyt$$ that is not has $$z$$ term. How do i solve this? And we know, if i have a solution $$F(a,b)=0$$, it can be written in explicit term as $$F(a)=b$$.

But how about 3 variables that is in my case i have a solution $$\Phi(a,b,c)=0$$?

• 1. What does it matter if the denominator of the fraction containing $dz$ has no $z$ term? If you want to solve $$\frac{dz}{xyt} = \frac{dt}{t}$$ then if you know what $x, y$ are as functions of $t$, then you can just integrate straight away. 2. \begin{align} \Phi(a,b,c) &= 0 \\ \implies a &= \phi(b,c) \end{align} Just so you know, the solution is given by $$z(x,y,t) = \frac{xyt}{3} + f \left(\frac{x}{t}, \frac{y}{t} \right)$$ Commented Jun 8, 2019 at 13:54
• Actually i can't explain correctly. Besides maybe i don't understand with this concept. Ok, and i know that $z$ isn't has effect on $dz$, and consider from $\frac{\partial t}{t}=\frac{\partial z}{xyt}$ i got $c=\frac{z}{xy}-t$. Then the complete solution is $\Phi(x/y, y/t, z/xy - t)$ but when i rewrite this as $z=xyt- \frac{xy^2}{t}+xyf\left(\frac{x}{y}\right)\,$.it doesn't satisfied the equation. Commented Jun 8, 2019 at 14:01
• It is wrong because, as I wrote in my previous comment, when you integrate $$\frac{dz}{xyt} = \frac{dt}{t}$$ you need to know what $x, y$ are in terms of $t$ i.e $x = x(t), y = y(t)$. You can find both of these by integrating $$\frac{dx}{x} = \frac{dt}{t}$$ and $$\frac{dy}{y} = \frac{dt}{t}$$ Commented Jun 8, 2019 at 14:11

$$x\dfrac{\partial z}{\partial x}+y\dfrac{\partial z}{\partial y}+t\dfrac{\partial z}{\partial t}=xyt$$

$$\dfrac{d x}{x}=\dfrac{d y}{y}=\dfrac{d t}{t}=\dfrac{d z}{xyt}$$

A first characteristic equation comes from $$\dfrac{d x}{x}=\dfrac{d y}{y}$$ $$\frac{x}{y}=c_1 \tag 1$$ A second characteristic equation comes from $$\dfrac{d y}{y}=\dfrac{d t}{t}$$ $$\frac{y}{t}=c_2 \tag 2$$

A third characteristic equation comes from

$$\dfrac{d x}{x}=\dfrac{d y}{y}=\dfrac{d t}{t}= \frac{(yt)dx+(xt)dy+(xy)dt}{(yt)x+(xt)y+(xy)t}=\dfrac{d (xyt)}{3(xyt)}=\dfrac{d z}{xyt}\qquad\text{See note below.}$$ $$z-\frac13 xyt=c_3 \tag 3$$

General solution of the PDE on the form of implicit equation $$\Phi(c_1, c_2,c_3)=0$$

Or equivalently on the form : $$c_3=F(c_1, c_2)=z-\frac13 xyt=F\left(\frac{x}{y}\:,\:\frac{y}{t} \right)$$

$$\boxed{z(x,y,t)=\frac13 xyt+F\left(\frac{x}{y}\:,\:\frac{y}{t} \right)}$$ This consistent with the result from Mathos $$z(x,y,t)=\frac13 xyt+f\left(\frac{x}{t}\:,\:\frac{y}{t} \right)$$ since $$F$$ and $$f$$ are both arbitrary functions : $$F(X,Y)=f(XY,Y)$$.

NOTE : The relationship between equal fractions is a well known property in elementary fraction calculus. If $$\frac{A}{B}=\frac{C}{D}$$ then $$\frac{A}{B}=\frac{C}{D}=\frac{k_1A+k_2C}{k_1B+k_2D}$$ with any $$k_1, k_2$$ not both nul.