Example 3. Determine the type of the equation

$$
u_{x x}+2 u_{x y}+2 u_{y y}+6 u_{x}+6 u_{y}-3 u=x+y^{2}
$$

and reduce it to the canonical form.
Solution.

1. We determine the coefficients of the equation $a_{11}, a_{12}$, and $a_{22}$. We have

$$
a_{11}=1, \quad a_{12}=1, \quad a_{22}=2
$$

2. We compute the expression

$$
a_{12}^{2}-a_{11} a_{12}=1-2=-1<0
$$

3. Since $a_{12}^{2}-a_{11} a_{12}<0$ for all $x, y$, equation (8) is of elliptic type in the entire $X O Y$ plane.
4. We find the general integrals of the characteristic equation:

$$
(d y)^{2}-2 d y d x+2(d x)^{2}=0
$$

Solving this equation for $d y$, we get

$$
d y=(1+i) d x \quad \text { and } \quad d y=(1-i) d x
$$

Thus, the characteristic equation (9) has two complex integrals:

$$
f_{1}(x, y)=y-x-i x \quad \text { and } \quad f_{2}(x, y)=y-x+i x
$$

Let $h_{1}(x, y)=\operatorname{Re} f_{2}(x, y)=y-x$ and $h_{2}(x, y)=\operatorname{Im} f_{2}(x, y)=x$. (We could have taken $\operatorname{Re} f_{1}(x, y)$ and $\operatorname{Im} f_{1}(x, y)$ as $h_{1}$ and $h_{2}$.)

5. In equation (8), we make the change of variables:

$$
x, y \longmapsto \xi=h_{1}(x, y)=y-x, \quad \eta=x .
$$

By the chain rule for differentiation,

$$
\begin{aligned}
& u_{x} \longmapsto u_{\xi}+u_{\eta}, \quad u_{y} \longmapsto u_{\xi} \\
& u_{x x} \longmapsto u_{\xi \xi}-2 u_{\xi \eta}+u_{\eta \eta}, \quad u_{x y} \longmapsto u_{\xi \xi}+u_{\xi \eta}, \quad u_{y y} \longmapsto u_{\xi \xi}
\end{aligned}
$$

As a result of the change of variables, equation (8) takes the form

$u_{\xi \xi}-2 u_{\xi \eta}+u_{\eta \eta}+2 u_{\xi \xi}+2 u_{\xi \eta}+2 u_{\xi \xi}+6\left(-u_{\xi}+u_{\eta}\right)+6 u_{\xi}-3 u=\eta+(\xi+\eta)^{2}$, i.e., equation (8) is reduced to the canonical form

$$
u_{\xi \xi}+u_{\eta \eta}=-2 u_{\eta}+u+\eta+(\xi+\eta)^{2}
$$

Answer. Equation (8) is of elliptic type in the entire $X O Y$ plane and has the canonical form

$$
u_{\xi \xi}+u_{\eta \eta}=-2 u_{\eta}+u+\eta+(\xi+\eta)^{2}
$$

where $\xi=y-x$ and $\eta=x$.

Conditions of the Problems. Determine the type of the equation and bring it to the canonical form.

1. $u_{x x}+2 u_{x y}-3 u_{y y}+2 u_{x}+7 u_{y}-3 u=0$.
2. $4 u_{x x}-8 u_{x y}+u_{y y}-2 u_{x}+2 u_{y}-3 u=0$.
3. $8 u_{x x}-6 u_{x y}-u_{y y}-u_{x}-3 u_{y}-u=0$.
4. $u_{x x}-2 u_{x y}+u_{y y}-3 u_{x}+2 u_{y}-5 u=0$.
5. $2 u_{x x}+8 u_{x y}+8 u_{y y}-u_{x}-2 u_{y}-5 u=0$.
6. $u_{x x}-6 u_{x y}+9 u_{y y}+4 u_{x}-3 u_{y}-7 u=0$.
7. $u_{x x}-4 u_{x y}+4 u_{y y}+4 u_{x}-9 u_{y}-3 u=0$.
8. $2 u_{x x}+6 u_{x y}+8 u_{y y}+u_{x}+5 u_{y}-2 u=0$.
9. $3 u_{x x}-8 u_{x y}+7 u_{y y}+3 u_{x}-u_{y}+2 u=0$.
10. $3 u_{x x}+8 u_{x y}+6 u_{y y}+3 u_{x}+u_{y}-2 u=0$.

Answers.

1. The equation is of hyperbolic type in the entire $X O Y$ plane, $u_{\xi \eta}=\left(3 u-u_{\xi}-9 u_{\eta}\right) / 16, \quad \xi=y-3 x, \quad \eta=y+x$.
2. The equation is of hyperbolic type in the entire $X O Y$ plane, $u_{\xi \eta}=\left(-3 u+\sqrt{3} u_{\xi}-\sqrt{3} u_{\eta}\right) / 12$ $\xi=y+(1-\sqrt{3} / 2) x, \quad \eta=y+(1+\sqrt{3} / 2) x$.
3. The equation is of hyperbolic type in the entire $X O Y$ plane, $u_{\xi \eta}=\left[-8 u-(27-\sqrt{17}) u_{\xi}-(27+\sqrt{17}) u_{\eta}\right] / 68$, $\xi=y+(3 / 8-\sqrt{17} / 8) x, \quad \eta=y+(3 / 8+\sqrt{17} / 8) x$.
4. The equation is of parabolic type in the entire $X O Y$ plane, $u_{\eta \eta}=5 u+u_{\xi}-2 u_{\eta}, \quad \xi=y+x, \quad \eta=y$.
5. The equation is of parabolic type in the entire $X O Y$ plane, $u_{\eta \eta}=\left(5 u+2 u_{\eta}\right) / 8, \quad \xi=y-2 x, \quad \eta=y$.
6. The equation is of parabolic type in the entire $X O Y$ plane, $u_{\eta \eta}=\left(7 u-9 u_{\xi}+3 u_{\eta}\right) / 9, \quad \xi=y+3 x, \quad \eta=y$.
7. The equation is of parabolic type in the entire $X O Y$ plane, $u_{\eta \eta}=\left(3 u+u_{\xi}+9 u_{\eta}\right) / 4, \quad \xi=y+2 x, \quad \eta=y$.
8. The equation is of elliptic type in the entire $X O Y$ plane, $u_{\xi \xi}+u_{\eta \eta}=\left(4 u-7 u_{\xi}-\sqrt{7} u_{\eta}\right) / 7, \quad \xi=y-3 x / 2, \quad \eta=\sqrt{7} x / 2$.
9. The equation is of elliptic type in the entire $X O Y$ plane, $u_{\xi \xi}+u_{\eta \eta}=\left(-6 u-9 u_{\xi}-3 \sqrt{5} u_{\eta}\right) / 5, \quad \xi=y+4 x / 3, \quad \eta=\sqrt{5} x / 3$.
10. The equation is of elliptic type in the entire $X O Y$ plane, $u_{\xi \xi}+u_{\eta \eta}=\left(6 u+9 u_{\xi}-3 \sqrt{2} u_{\eta}\right) / 2, \quad \xi=y-4 x / 3, \quad \eta=\sqrt{2} x / 3$.

## 5.2. General Solution of a Hyperbolic Equation

Problem Statement. Find the general solution of the hyperbolic equation

$$
u_{x x}+2 b u_{x y}+\left(b^{2}-a^{2}\right) u_{y y}=0
$$

## PLAN OF SOLUTION

1. We reduce equation (1) to the canonical form. We obtain

$$
u_{\xi \eta}=0
$$

where $\xi=y-(b+a) x, \eta=y-(b-a) x$ (Problem 5.1).

2. Integrating equation (2) with respect to $\eta$, we get

$$
u_{\xi}=C_{0}(\xi)
$$

Integrating equation (3) with respect to $\xi$, we get

$$
u(\xi, \eta)=\int C_{0}(\xi) d \xi+C_{2}(\eta) \equiv C_{1}(\xi)+C_{2}(\eta)
$$

Thus, the general solution of equation (2) is

$$
u(\xi, \eta)=C_{1}(\xi)+C_{2}(\eta)
$$

where $C_{1}(\xi)$ and $C_{2}(\eta)$ are arbitrary twice differentiable functions.

3. Substituting $\xi=y-(b+a) x$ and $\eta=y-(b-a) x$ into (4), we get the general solution of equation (1):

$$
u(x, y)=C_{1}(y-(b+a) x)+C_{2}(y-(b-a) x) .
$$

We perform the check and write the answer.