Answer to Question #252328 in Differential Equations for pde

Question

Given that u=u(x,y,z)=c₁,v=v(x,y,z)=c₂ are solutions of dx/P=dy/Q=dz/R. Show that F(u,v)=0 is a general solution of the Lagrange's equation

Pp+Qq=R

Answer 1

Given Lagrange’s partial differential equation is

$$𝑃𝑝 + 𝑄𝑞 = 𝑅$$

Let

$$F(u,v)=0$$

be the solution of the given Lagrange’s equation .

Differentiating partially with respect to $x$ , we have

$$(\dfrac{\partial F}{\partial u}\dfrac{\partial u}{\partial x}+\dfrac{\partial F}{\partial u}\dfrac{\partial u}{\partial z}\dfrac{\partial z}{\partial x})$$

$$+(\dfrac{\partial F}{\partial v}\dfrac{\partial v}{\partial x}+\dfrac{\partial F}{\partial v}\dfrac{\partial v}{\partial z}\dfrac{\partial z}{\partial x})=0$$

Or

$$\dfrac{\partial F}{\partial u}(\dfrac{\partial u}{\partial x}+p\dfrac{\partial u}{\partial z})+\dfrac{\partial F}{\partial v}(\dfrac{\partial v}{\partial x}+p\dfrac{\partial v}{\partial z})=0$$

Similary, differentiating partially with respect to $x$ , we have

$$\dfrac{\partial F}{\partial u}(\dfrac{\partial u}{\partial y}+q\dfrac{\partial u}{\partial z})+\dfrac{\partial F}{\partial v}(\dfrac{\partial v}{\partial y}+q\dfrac{\partial v}{\partial z})=0$$

Eliminating $\dfrac{\partial F}{\partial u}$ and $\dfrac{\partial F}{\partial v},$ we get

$$(\dfrac{\partial u}{\partial x}+p\dfrac{\partial u}{\partial z})(\dfrac{\partial v}{\partial y}+q\dfrac{\partial v}{\partial z})$$

$$-(\dfrac{\partial u}{\partial y}+q\dfrac{\partial u}{\partial z})(\dfrac{\partial v}{\partial x}+p\dfrac{\partial v}{\partial z})=0$$

Then

$$(\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial z}-\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial y})p+(\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial z})q$$

$$=\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial y}-\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$

which can also be put in the form

$$\dfrac{\partial (u,v)}{\partial (y,z)}p+\dfrac{\partial (u,v)}{\partial (z,x)}q=\dfrac{\partial (u,v)}{\partial (x,y)}$$

or

$$Pp+Qq=R$$

where

$$P=\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial z}-\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial y}=\dfrac{\partial (u,v)}{\partial (y,z)}$$

$$Q=\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial z}=\dfrac{\partial (u,v)}{\partial (z,x)}$$

$$R=\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial y}-\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x}=\dfrac{\partial (u,v)}{\partial (x,y)}$$

Thus, the equation $𝑃𝑝 + 𝑄𝑞 = 𝑅$ is a partial differential equation of order one and degree one for which $F(u,v)$ is a solution.

Now taking the differentials of two independent solutions $𝑢(𝑥, 𝑦, 𝑧) = 𝑐_1$ and $𝑣(𝑥, 𝑦, 𝑧) = 𝑐_2 ,$ we get

$$\dfrac{\partial u}{\partial x}dx+\dfrac{\partial u}{\partial y}dy+\dfrac{\partial u}{\partial z}dz=0$$

$$\dfrac{\partial v}{\partial x}du+\dfrac{\partial v}{\partial y}dy+\dfrac{\partial v}{\partial z}dz=0$$

Since $u$ and $v$ are independent functions, therefore, solving equations for the ratios $𝑑𝑥, 𝑑𝑦,𝑑𝑧,$ we get

$$\dfrac{dx}{\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial z}-\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial y}}=\dfrac{dy}{\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial z}}$$

$$=\dfrac{dz}{\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial y}-\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x}}$$

Given that $u=u(x,y,z)=c_1,v=v(x,y,z)=c_2$ are solutions of

$$dx/P=dy/Q=dz/R$$

Then

$$\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial z}-\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial y}=kP$$

$$\dfrac{\partial u}{\partial z}\dfrac{\partial v}{\partial x}-\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial z}=kQ$$

$$\dfrac{\partial u}{\partial x}\dfrac{\partial v}{\partial y}-\dfrac{\partial u}{\partial y}\dfrac{\partial v}{\partial x}=kR$$

Substitute in$(*)$

$$kPp+kQq=kR$$

We get

$$Pp+Qq=R$$

which is the given partial differential equation.

Therefore, if $𝑢(𝑥, 𝑦, 𝑧) = 𝑐_1$ and $𝑣(𝑥, 𝑦, 𝑧) = 𝑐_2$ are two independent solutions of the system of differential equations $dx/P=dy/Q=dz/R,$ then $F(u,v)=0$ is a solution of $𝑃𝑝 + 𝑄𝑞 = 𝑅,$ where $F$ is an arbitrary function.