# Geometric interpretation and solution of Lagrange-Charpit equations

What is the geometric meaning of the Lagrange-Charpit equation, $$P\dfrac{\partial z}{\partial x} + Q \dfrac {\partial z}{\partial y}= R \\Pp+ Qq= R$$ where $$P$$,$$Q$$,and $$R$$ are functions of $$x\,$$,$$\,y\,$$ and $$z$$. How do we arrive at the solution of the same? I've got a feeling that it has something to do with direction cosines and stuff. But I just cant relate them with the equation. We are being taught just to write the auxiliary equation, $$\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}$$ group, and integrate or find two set of "multipliers" and finally write $$\phi (u,v)=0$$. I've got no clue, what's happening behind the curtains.

P.S: I'm relatively new to the subject

• You may have a look at the method of characteristics Wikipedia article. Commented Jul 29, 2019 at 8:45

First Interpretation:

Consider $$P(x,y,z)p+Q(x,y,z)q=R(x,y,z)\tag1$$ and$$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\tag2$$ Let $$z=\phi(x,y)\tag3$$ represents the solution of equation $$(1)$$. Then $$(3)$$ represents a surface whose normal at any point $$~(x,~y,~z)~$$ has direction ratios $$~\frac{\partial z}{\partial x},~\frac{\partial z}{\partial y},~-1~$$ i.e., $$~p,~q,-1~$$.

Also we know that the simultaneous equation $$(2)$$ represents a family of curves such that the tangent at any point has direction ratios $$~P,~Q,~R~$$.

Rewriting $$(1)$$ , we have $$P(x,y,z)p+Q((x,y,z)q+R(x,y,z)~(-1)=0\tag4$$ showing that the normal to surface $$(3)$$ at any point is perpendicular to the number of family of curves $$(2)$$ through that point. Hence the member must touch the surface at that point. Since this holds for each point on $$(3)$$, we conclude that the curves $$(2)$$ lie completely on the surface $$(3)$$ whose differential equation is $$(1)$$.

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Second Interpretation:

We know that the curve whose equations are solutions of $$\frac{dx}{P}=\frac{dy}{Q}=\frac{dz}{R}\tag{a}$$ are orthogonal to the system of the surfaces whose equation satisfies $$P(x,y,z)~dx+Q(x,y,z)~dy+R(x,y,z)~dz=0~.\tag{b}$$

From the previous discussion we conclude that the curve of $$(a)$$ lie completely on the surface represented by $$P(x,y,z)p+Q((x,y,z)q=R(x,y,z)~.\tag{c}$$

Hence surfaces represented by $$(b)$$ and $$(c)$$ are orthogonal.

References:

$$1.~~$$ "Introductory Course in Differential Equations" by Daniel A. Murray

$$2.~~$$ "Ordinary and Partial Differential Equations" by M.D. Raisinghania