General solution of quasi-linear equations

Theorem 1   The general solution of a quasi-linear equation

$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u),$$ (1.7)

can be written in form

$$F(\psi_1,\psi_2)=0,$$

where $F$ is an arbitrary function and $\psi_1(x,y,u)=c_1$ and $\psi_2(x,y,u)=c_2$ form a solution of the equation

$$\frac{dx}{a(x,y,u)}=\frac{dy}{b(x,y,u)}=\frac{du}{c(x,y,u)}.$$ (1.8)

(PDEs of the type () are also known as Lagrange's equations. Eq. () is known as the auxiliary equation of given PDE ())

Proof On the integral surface the total differential $du$ is given by

$$du=\frac{\partial u}{\partial x}dx+\frac{\partial u}{\partial y}dy.$$ (1.9)

Rewriting Eq.() and Eq.() in matrix form

$$\begin{bmatrix} a & b \\ dx & dy \end{bmatrix}\begin{bmatrix} u_x \\ u_y \end{bmatrix} = \begin{bmatrix} c \\ du \end{bmatrix}.$$

On expanding the determinants, one get

$$\frac{dx}{a(x,y,u)}=\frac{dy}{b(x,y,u)}=\frac{du}{c(x,y,z)}.$$

To complete the proof, one shows that any surface generated by the integral curves of Eq.() has equation of the form $F_1(\psi_1,\psi_2)=0.$

Let $\psi_1(x,y,u)=c_1$ and $\psi_2(x,y,u)=c_2$ be two independent integrals curves of Eq.(). Therefore,

$$d\psi_1=\frac{\partial \psi_1}{\partial x}dx+\frac{\partial \psi_1}{\partial y}dy+\frac{\partial \psi_1}{\partial u}du=0,$$

$$d\psi_2=\frac{\partial \psi_2}{\partial x}dx+\frac{\partial \psi_2}{\partial y}dy+\frac{\partial \psi_2}{\partial u}du=0.$$

Solving this gives

$$\frac{dx}{ \begin{vmatrix} \frac{\partial \psi_1}{\partial y} & ... ...rtial \psi_2}{\partial x} & \frac{\partial \psi_2}{\partial y} \end{vmatrix}},$$

$$\frac{dx}{\frac{\partial (\psi_1,\psi_2)}{\partial (y,u)}}= \frac... ...}{\partial (u,x)}}= \frac{du}{\frac{\partial (\psi_1,\psi_2)}{\partial (x,y)}}.$$ (1.10)

Recall: $F(\psi_1,\psi_2)=0$, where $F$ is an arbitrary function of $\psi_1$ and $\psi_2$ gives PDE

$$\frac{\partial (\psi_1,\psi_2)}{\partial (y,u)}u_x+\frac{\partial... ..._1,\psi_2)}{\partial (u,x)}u_y=\frac{\partial (\psi_1,\psi_2)}{\partial (x,y)}.$$

Therefore, one can write Eq.() as

$$\frac{dx}{a}=\frac{dy}{b}=\frac{du}{c}.$$

The solutions of these equations are $\psi_1(x,y,u)=c_1$ and $\psi_2(x,y,u)=c_2$. Hence $F(\psi_1,\psi_2)$ is the required solution of Eq.() if $\psi_1$ and $\psi_2$ are two independent integral curves of Eq.().

Example 1.3.1   Find the general solution of

$$(y+ux)u_x-(x+yu)u_y=x^2-y^2.$$

Answer The given equation is first order quasi-linear PDE . Comparing it with the standard form

$$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u),$$

one has

$$a=y+ux,~b=-(x+uy)~$$and$$~c=x^2-y^2.$$

The integral surface of given PDE is generated by integral curves of auxiliary equation

$$\frac{dx}{y+ux}=\frac{dy}{-(x+uy)}=\frac{du}{x^2-y^2}.$$

To get the first integral curve:

$$\frac{xdx+ydy}{xy+ux^2-xy-uy^2}=\frac{du}{x^2-y^2},$$

$$\implies~xdx+ydy=udu,$$

Therefore,

$$x^2+y^2-u^2=c_1.$$

To get the second integral curve, consider the combination

$$\begin{split} \frac{ydx+xdy}{y^2+uxy-x^2-uxy} &=\frac{du}{x^2... ...lies~ ydx+xdy&=-du,\nonumber\\ \implies~xy+u&=c_2. \end{split}$$

Thus required integral surface if given by the arbitrary function defined as

$$F(x^2+y^2-u^2,xy+u)=0.$$

Example 1.3.2   Find the integral surface of

$$x(y^2+u)u_x-y(x^2+u)u_y=(x^2-y^2)u.$$

Answer The required integral surface is generated by the integral curves of the auxiliary equation

$$\frac{dx}{x(y^2+u)}=\frac{dy}{-y(x^2+u)}=\frac{du}{(x^2-y^2)u}.$$

To get the first integral curve

$$\frac{xdx+ydy}{x^2y^2+ux^2-x^2y^2-uy^2}=\frac{du}{(x^2-y^2)u},$$

$$\implies~xdx+ydy=du,$$

Therefore,

$$x^2+y^2-2u=c_1.$$

To get the second integral curve, consider the combination

$$\begin{split} \implies~\frac{yudx+xudy}{y^2-x^2}&=\frac{xydu}... ...s~ yudx+xudy&=-xydu,\nonumber\\ \implies~xyu&=c_2. \end{split}$$

Thus required integral surface if given by the arbitrary function defined as

$$F(x^2+y^2-2u,xyu)=0.$$