General Solutions’ Laws of Nonlinear Partial Differential Equations

Honglai Zhu

School of Physics and Electronic Information, Huaibei Normal University, Huaibei, China.

DOI: 10.4236/jamp.2025.131008                

Abstract

In previous papers, we proposed the important $Z$ transformations and obtained general solutions to a large number of linear and quasi-linear partial differential equations for the first time. In this paper, we will use the $Z_1$ transformation to get the general solutions of some nonlinear partial differential equations for the first time, and use the general solutions to obtain the exact solutions of some typical definite solution problems.

Keywords

Transformation, Nonlinear Partial Differential Equations, Analytical Solution, General Solution, Definite Solution Problems

1. Introduction

The general solution of nonlinear ordinary differential equations (ODEs) is a field which has been studied in depth [1], and many research results have been acquired, such as Abel equation [2]-[5], Riccati equation [6] and so on.

Since the birth of the discipline of partial differential equations (PDEs), there are very few cases that general solutions of linear PDEs can be obtained [7] [8], and the general solution of nonlinear PDEs is one of the most mysterious areas of mathematics in which few mathematicians have studied [8] [9]. Current research directions for nonlinear PDEs are mainly:

1) Use diversified analysis methods to get exact solutions [10]-[14], such as tanh-coth expansion method [15], exp-function method [16]-[18], tanh-expansion method [19], homogeneous balance method [20]-[22] and so on. Among them, the study of solitary wave solutions is one of the most concerned focuses.

2) Use various numerical methods to study the definite solution problems [23]-[27].

3) Using qualitative theory to analyze the problem of definite solutions, such as the existence [28]-[30], uniqueness [31] [32], asymptotic behavior of solutions [33] [34] and so on.

4) Exact solutions and qualitative theory for some fractional nonlinear PDEs [35].

In our previous paper [36]-[41], general solutions of many differential equation were obtained using the newly proposed $Z$ transformations and $Z_A$ method. In this paper, we will use $Z_1$ transformation to solve some typical non-linear PDEs, and analyze some definite solution problems.

2. General Solutions of Some Nonlinear PDEs and Exact Solutions of Some Definite Solution Problems

In the previous paper [36],

$a_1{u}_{x_1}+a_2{u}_{x_2}+a_3{u}_{x_3}=\mathrm{0,}$ (1)

we used to get the solution of Equation (1) in ${ℝ}^3$ is

$u=f\left(\frac{-a_2{c}_2-a_3{c}_3}{a_1}{x}_1+c_2{x}_2+c_3{x}_3\mathrm{,}\frac{-a_2{c}_5-a_3{c}_6}{a_1}{x}_1+c_5{x}_2+c_6{x}_3\right)\mathrm{.}$ (2)

In the following, we will use the $Z_1$ transformation to obtain analytical solutions similar to Equation (2) for nonlinear partial differential equations. Theorem 1 is presented first.

Theorem 1. In ${ℝ}^3$ , if

$\Theta \left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)+\Lambda \left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)=\mathrm{0,}$ (3)

where $a_i$ are any known constants ($1\le i\le 9$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $\Lambda =\Lambda \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , then the analytical solution of Equation (3) is

$u=f\left(v\mathrm{,}w\right)\mathrm{,}$ (4)

$v=k_{1}t+k_{2}x+k_{3}y+k_4\mathrm{,}$ (5)

$w=k_{5}t+k_{6}x+k_{7}y+k_8\mathrm{,}$ (6)

where $f$ is an arbitrary smooth function, $v$ and $w$ are independent of each other, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_5\mathrm{,}{k}_6\mathrm{,}{k}_7$ need satisfy

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$ (7)

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7=\mathrm{0,}$ (8)

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)=\mathrm{0,}\end{array}$ (9)

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$ (10)

$a_7{k}_5+a_8{k}_6+a_9{k}_7=0.$ (11)

Proof. According to $Z_1$ transformation, set $u=f\left(v\mathrm{,}w\right)$ , $v=k_{1}t+k_{2}x+k_{3}y+k_4$ , $w=k_{5}t+k_{6}x+k_{7}y+k_8$ . $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_8$ are undetermined constants, $v$ and $w$ are independent of each other, so

$\begin{array}{l}\Theta \left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)+\Lambda \left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)\\ =\Theta a_1{\left(k_1{f}_v+k_5{f}_w\right)}^2+\Theta a_2{\left(k_2{f}_v+k_6{f}_w\right)}^2+\Theta a_3{\left(k_3{f}_v+k_7{f}_w\right)}^2\\ +\Theta a_4\left(k_1{f}_v+k_5{f}_w\right)\left(k_2{f}_v+k_6{f}_w\right)+\Theta a_5\left(k_2{f}_v+k_6{f}_w\right)\left(k_3{f}_v+k_7{f}_w\right)\\ +\Theta a_6\left(k_3{f}_v+k_7{f}_w\right)\left(k_1{f}_v+k_5{f}_w\right)+\Lambda a_7\left(k_1{f}_v+k_5{f}_w\right)\\ +\Lambda a_8\left(k_2{f}_v+k_6{f}_w\right)+\Lambda a_9\left(k_3{f}_v+k_7{f}_w\right)\\ =0.\end{array}$

Namely

$\begin{array}{l}\Theta f_v^2\left(a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3\right)\\ +\Theta f_w^2\left(a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7\right)\\ +\Theta f_v{f}_w(2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right))\\ +\Lambda f_v\left(a_7{k}_1+a_8{k}_2+a_9{k}_3\right)+\Lambda f_w\left(a_7{k}_5+a_8{k}_6+a_9{k}_7\right)=0.\end{array}$

Set

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)=\mathrm{0,}\end{array}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

$a_7{k}_5+a_8{k}_6+a_9{k}_7=0.$

Therefore, the analytical solution of Equation (3) is

$u=f\left(v\mathrm{,}w\right)$

The theorem is proven. $\square$

In Theorem 1, if all partial derivatives in $\Theta$ and $\Lambda$ are of first order, then (4) is a general solution of (3). Next, we use Theorem 1 to analyze a definite solution problem.

Example 1. In ${ℝ}^3$ , use Theorem 1 to obtain the analytical solution of

$\begin{array}{l}\left(u_t^5+u_x^3\right)\left(3u_t^2+3u_x^2+3u_y^2-10u_t{u}_x+6u_x{u}_y-10u_y{u}_t\right)\\ -\left(u_t^2+u_y^2\right)\left(9u_t+3u_x+3u_y\right)=\mathrm{0,}\end{array}$ (12)

in the condition of $u\left(\mathrm{0,}x\mathrm{,}y\right)=g\left(x\mathrm{,}y\right)$ , $g$ is an arbitrary known first differentiable function.

Solution. According to Theorem 1, the general solution of (12) is

$u\left(t\mathrm{,}x\mathrm{,}y\right)=f\left(t+k_{2}x+\left(3-k_2\right)y\mathrm{,}t+k_{6}x+\left(3-k_6\right)y\right)\mathrm{.}$ (13)

So

$u\left(\mathrm{0,}x\mathrm{,}y\right)=f\left(k_{2}x+\left(3-k_2\right)y\mathrm{,}{k}_{6}x+\left(3-k_6\right)y\right)=g\left(x\mathrm{,}y\right)\mathrm{.}$

Set

$k_{2}x+\left(3-k_2\right)y=\beta \mathrm{,}{k}_{6}x+\left(3-k_6\right)y=\gamma \mathrm{.}$ (14)

We obtain

$y=\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}\mathrm{,}$

$x=\frac{\beta }{k_2}+\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}-\frac{k_6\beta -k_2\gamma }{k_2{k}_6-k_2^2}\mathrm{.}$

Namely

$\begin{array}{c}u\left(0,x,y\right)=f\left(k_{2}x+\left(3-k_2\right)y,k_{6}x+\left(3-k_6\right)y\right)\\ =g\left(\frac{\beta }{k_2}+\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}-\frac{k_6\beta -k_2\gamma }{k_2{k}_6-k_2^2},\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}\right).\end{array}$ (15)

Set

$t+k_{2}x+\left(3-k_2\right)y=\beta ,t+k_{6}x+\left(3-k_6\right)y=\gamma .$

We get

$\frac{\beta }{k_2}+\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}-\frac{k_6\beta -k_2\gamma }{k_2{k}_6-k_2^2}=\frac{t+3x}{3}\mathrm{,}$

$\frac{k_6\beta -k_2\gamma }{3k_6-3k_2}=\frac{t+3y}{3}\mathrm{.}$

Then the analytical solution of the definite solution problem is

$u\left(t\mathrm{,}x\mathrm{,}y\right)=g\left(\frac{t+3x}{3}\mathrm{,}\frac{t+3y}{3}\right)\mathrm{.}$ (16)

According to Example 1, we can directly obtain the analytical solution of Equation (12) in various initial value conditions. If the initial value condition is $u\left(\mathrm{0,}x\mathrm{,}y\right)=\sin \left(2x+y\right)+{\text{e}}^{4x-y}$ , the corresponding analytical solution is $u=\sin \left(t+2x+y\right)+{\text{e}}^{t+4x-y}$ .

According to Theorem 1, set $k_5=k_6=k_7=k_8=0$ , we can get Theorem 2.

Theorem 2. In ${ℝ}^3$ , if

$\Theta \left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)+\Lambda \left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)=\mathrm{0,}$

where $a_i$ are any known constants ($1\le i\le 9$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $\Lambda =\Lambda \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , then the analytical solution of Equation (3) is

$u=f\left(v\right)\mathrm{,}$ (17)

where $f$ is an arbitrary smooth function, $v=k_{1}t+k_{2}x+k_{3}y+k_4$ , and the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3$ need satisfy

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=0.$

The reason why we propose Theorem 2 is that there is no an analytical solution in the form of $u=f\left(v\mathrm{,}w\right)$ of Equation (3), such as examples 2 and 3.

Example 2. Prove that $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t=0$ does not have an analytical solution in the form of $u=f\left(v\mathrm{,}w\right)$ , $a_i$ are any known constants ($1\le i\le 7$ ), $a_7\ne 0$ .

Proof. According to Theorem 1, if $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t=0$ has an analytical solution in the form of $u=f\left(v\mathrm{,}w\right)$ , and

$v=k_{1}t+k_{2}x+k_{3}y+k_4\mathrm{,}w=k_{5}t+k_{6}x+k_{7}y+k_8\mathrm{,}$

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)=\mathrm{0,}\end{array}$

$a_7{k}_1=\mathrm{0,}$

$a_7{k}_5=0.$

For $a_7\ne 0$ , we get

$k_1=k_5=0.$

So

$a_2{k}_2^2+a_3{k}_3^2+a_5{k}_2{k}_3=\mathrm{0,}$

$a_2{k}_6^2+a_3{k}_7^2+a_5{k}_6{k}_7=\mathrm{0,}$

$2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_5\left(k_2{k}_7+k_3{k}_6\right)=0.$

Since $f$ is an arbitrary first differentiable function, we may set

$k_2=k_6=1.$

Then

$a_2+a_3{k}_3^2+a_5{k}_3=\mathrm{0,}$

$a_2+a_3{k}_7^2+a_5{k}_7=\mathrm{0,}$

$2a_2+2a_3{k}_3{k}_7+a_5\left(k_7+k_3\right)=0.$

Set

$k_3=\frac{-a_5+\sqrt{a_5^2-4a_2{a}_3}}{2a_3}\mathrm{,}$

$k_7=\frac{-a_5-\sqrt{a_5^2-4a_2{a}_3}}{2a_3}\mathrm{.}$

So

$2a_2+2a_3{k}_3{k}_7+a_5\left(k_7+k_3\right)=2a_2+2a_2-\frac{a_5^2}{a_3}=0.$

Namely

$a_5^2=4a_2{a}_3\mathrm{.}$

We obtain $k_3=k_7$ , $v$ and $w$ are not independent of each other. That is, $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t=0$ does not have an analytical solution similar to the $u=f\left(v\mathrm{,}w\right)$ form of Theorem 1.

Example 3. Prove that $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t-a_7{u}_x=0$ does not have an analytical solution similar to Theorem 1, $a_i$ are any known constants ($1\le i\le 7$ ), $a_7\ne 0$ .

Proof. According to Theorem 1, if $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t-a_7{u}_x=0$ has an analytical solution in the form of $u=f\left(v\mathrm{,}w\right)$ , and

$v=k_{1}t+k_{2}x+k_{3}y+k_4\mathrm{,}w=k_{5}t+k_{6}x+k_{7}y+k_8\mathrm{,}$

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)=\mathrm{0,}\end{array}$

$k_1=k_2\mathrm{,}$

$k_5=k_6\mathrm{.}$

Since $f$ is an arbitrary first differentiable function, we may set

$k_1=k_2=k_5=k_6=1.$

So

$\begin{array}{l}{a}_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3\\ =a_3{k}_3^2+\left(a_5+a_6\right)k_3+a_1+a_2+a_4=\mathrm{0,}\end{array}$

$\begin{array}{l}{a}_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7\\ =a_3{k}_7^2+\left(a_5+a_6\right)k_7+a_1+a_2+a_4=0.\end{array}$

Set

$k_3=\frac{-a_5-a_6+\sqrt{{\left(a_5+a_6\right)}^2-4a_3\left(a_1+a_2+a_4\right)}}{2a_3}\mathrm{,}$

$k_7=\frac{-a_5-a_6-\sqrt{{\left(a_5+a_6\right)}^2-4a_3\left(a_1+a_2+a_4\right)}}{2a_3}\mathrm{.}$

Then

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)\\ =2a_1+2a_2+2a_3{k}_3{k}_7+2a_4+a_5\left(k_7+k_3\right)+a_6\left(k_3+k_7\right)\\ =4\left(a_1+a_2+a_4\right)-\frac{{\left(a_5+a_6\right)}^2}{a_3}=0.\end{array}$

Namely

$4a_3\left(a_1+a_2+a_4\right)={\left(a_5+a_6\right)}^2\mathrm{.}$

We obtain $k_3=k_7$ , $v$ and $w$ are not independent of each other. That is, $a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t+a_7{u}_t-a_7{u}_x=0$ does not have an analytical solution similar to the $u=f\left(v\mathrm{,}w\right)$ form of Theorem 1.

Next, we use Theorem 2 to analyze a definite solution problem.

Example 4. In ${ℝ}^3$ , use Theorem 2 to obtain the analytical solution of

$\left(u_t^2+u_x{u}_y\right)\left(u_t^2+u_x^2-u_y^2-u_t{u}_x+u_x{u}_y-u_y{u}_t\right)+\left(u_x^2-u_t{u}_x\right){\left(u_t-u_x\right)}^2=\mathrm{0,}$ (18)

in the condition of $u\left(\mathrm{0,}y\mathrm{,}z\right)={\displaystyle \sum _i{\phi }_i\left({\kappa }_{i}x-{\kappa }_{i}y+{\lambda }_i\right)}$ , ${\phi }_i$ are arbitrary known first differentiable functions, ${\kappa }_i$ and ${\lambda }_i$ are any known constants.

Solution. According to Theorem 2, the general solution of (18) is

$u=f\left(k_{1}t+k_{2}x+k_{3}y+k_4\right)\mathrm{,}$

and

$k_1^2+k_2^2-k_3^2-k_1{k}_2+k_2{k}_3-k_1{k}_3=\mathrm{0,}$

$k_1=k_2\mathrm{.}$

Then

$k_1^2+k_2^2-k_3^2-k_1{k}_2+k_2{k}_3-k_1{k}_3=k_1^2-k_3^2=\mathrm{0,}$

$k_3=\pm k_1\mathrm{.}$

That is, the solution of (18) is

$u=f\left(k_{1}t+k_{1}x+k_{1}y+k_4\right)={\displaystyle \sum _i{f}_i}\left(k_{1_i}t+k_{1_i}x+k_{1_i}y+k_{4_i}\right).$ (19)

Or

$u=f\left(k_{1}t+k_{1}x-k_{1}y+k_4\right)={\displaystyle \sum _i{f}_i}\left(k_{1_i}t+k_{1_i}x-k_{1_i}y+k_{4_i}\right).$ (20)

Since the initial value condition is

$u\left(\mathrm{0,}x\mathrm{,}y\right)={\displaystyle \sum _i{\phi }_i}\left({\kappa }_{i}x-{\kappa }_{i}y+{\lambda }_i\right)\mathrm{.}$

Then the corresponding general solution of (18) is (20), set

$f_i={\phi }_i,k_{1_i}={\kappa }_i,k_{4_i}={\lambda }_i.$

So the analytical solution of the definite solution problem is

$u\left(t,x,y\right)={\displaystyle \sum _i{\phi }_i}\left({\kappa }_{i}t+{\kappa }_{i}x-{\kappa }_{i}y+{\lambda }_i\right).$ (21)

According to Theorem 2, we can obtain Theorem 3.

Theorem 3. In ${ℝ}^2$ , if

$\Theta \left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_t{u}_x\right)+\Lambda \left(a_4{u}_t+a_5{u}_x\right)=\mathrm{0,}$ (22)

where $a_i$ are known constants ($1\le i\le 5$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{tx}\mathrm{,}\cdots \right)$ , $\Lambda =\Lambda \left(t\mathrm{,}x\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{tx}\mathrm{,}\cdots \right)$ , then the analytical solution of Equation (22) is

$u=f\left(\frac{-a_5{k}_2}{a_4}t+k_{2}x+k_3\right)\mathrm{,}$ (23)

where $f$ is an arbitrary smooth function, $k_2$ and $k_3$ are arbitrary constants, and $a_i$ need satisfy

$a_1{a}_5^2+a_2{a}_4^2-a_3{a}_4{a}_5=0.$ (24)

Proof. According to Theorem 2, the analytical solution of (22) is

$u=f\left(k_{1}t+k_{2}x+k_3\right)\mathrm{,}$

$k_1\mathrm{,}{k}_2$ and $k_3$ satisfy

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_1{k}_2=\mathrm{0,}$

$a_4{k}_1+a_5{k}_2=0.$

So

$k_1=\frac{-a_5{k}_2}{a_4}\mathrm{,}$

$\begin{array}{l}{a}_1{k}_1^2+a_2{k}_2^2+a_3{k}_1{k}_2=a_1\frac{a_5^2}{a_4^2}{k}_2^2+a_2{k}_2^2-\frac{a_3{a}_5{k}_2^2}{a_4}=0\\ \Rightarrow a_1{a}_5^2+a_2{a}_4^2-a_3{a}_4{a}_5=0.\end{array}$

Therefore, the analytical solution of Equation (22) is

$u=f\left(\frac{-a_5{k}_2}{a_4}t+k_{2}x+k_3\right)\mathrm{,}$

and $a_i$ need satisfy

$a_1{a}_5^2+a_2{a}_4^2-a_3{a}_4{a}_5=0.$

The theorem is proven. $\square$

If the initial value condition of (22) is

$u\left(\mathrm{0,}x\right)=g\left(x\right)\mathrm{.}$

Set $k_3=0$ , so

$u\left(t\mathrm{,}x\right)=f\left(\frac{-a_5{k}_2}{a_4}t+k_{2}x\right)=g\left(\frac{-a_5}{a_4}t+x\right),$

that is, the exact solution of the definite solution problem is $u\left(t\mathrm{,}x\right)=g\left(\frac{-a_5}{a_4}t+x\right)$ .

Next we propose Theorem 4.

Theorem 4. In ${ℝ}^3$ , if

$\Theta {\left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)}^n+\Lambda {\left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)}^m=\mathrm{0,}$ (25)

where $a_i$ are any known constants ($1\le i\le 9$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $\Lambda =\Lambda \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $n\ge 1$ , $m\ge 1$ , then the analytical solution of Equation (25) is

$u=f\left(v\mathrm{,}w\right)\mathrm{,}$

$v=k_{1}t+k_{2}x+k_{3}y+k_4,w=k_{5}t+k_{6}x+k_{7}y+k_8,$

where $f$ is an arbitrary smooth function, $v$ and $w$ are independent of each other, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_5\mathrm{,}{k}_6\mathrm{,}{k}_7$ need satisfy

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_5{k}_6+a_5{k}_6{k}_7+a_6{k}_5{k}_7=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+a_4\left(k_1{k}_6+k_2{k}_5\right)\\ +a_5\left(k_2{k}_7+k_3{k}_6\right)+a_6\left(k_3{k}_5+k_1{k}_7\right)=\mathrm{0,}\end{array}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

$a_7{k}_5+a_8{k}_6+a_9{k}_7=0.$

Proof. Set

${\Theta }^{\prime }=\Theta {\left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)}^{n-1}\mathrm{,}$

${\Lambda }^{\prime }=\Lambda {\left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)}^{m-1}\mathrm{.}$

Then

$\begin{array}{l}\Theta {\left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)}^n+\Lambda {\left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)}^m\\ ={\Theta }^{\prime }\left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)+{\Lambda }^{\prime }\left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)=0.\end{array}$

Obviously ${\Theta }^{\prime }={\Theta }^{\prime }\left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , ${\Lambda }^{\prime }={\Lambda }^{\prime }\left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , according to Theorem 1, the analytical solution of the above equation is (4), so the theorem is proved. $\square$

Theorem 4 explains that the analytical solution of (25) is independent of $\Theta$ , $\Lambda$ , $n$ and $m$ , that is, analytical solutions of these infinitely many nonlinear PDEs are the same. Theorems 2 and 3 also have similar laws, which we will not elaborate here.

Next we propose Theorem 5.

Theorem 5. In ${ℝ}^3$ , if

$\Theta \left(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_t{u}_x+a_5{u}_x{u}_y+a_6{u}_y{u}_t\right)+A\left(a_7{u}_t+a_8{u}_x+a_9{u}_y\right)=B\mathrm{,}$ (26)

where $a_i$ are any known constants ($1\le i\le 9$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $A=A\left(t\mathrm{,}x\mathrm{,}y\right)$ , $B=B\left(t\mathrm{,}x\mathrm{,}y\right)$ , then the analytical solution of Equation (26) is

$u=f\left(p\mathrm{,}q\right)+\frac{\underset{}{{\displaystyle {\displaystyle \int \frac{B\left(p\mathrm{,}q\mathrm{,}r\right)}{A\left(p\mathrm{,}q\mathrm{,}r\right)}\text{d}r}}}}{a_7{k}_7+a_8{k}_8+a_9{k}_9}\mathrm{,}$ (27)

$p=k_{1}t+k_{2}x+k_{3}y,q=k_{4}t+k_{5}x+k_{6}y,r=k_{7}t+k_{8}x+k_{9}y,$ (28)

where $f$ is an arbitrary smooth function, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_9$ need satisfy

$-k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8-k_1{k}_6{k}_8-k_2{k}_4{k}_9+k_1{k}_5{k}_9\ne \mathrm{0,}$ (29)

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$ (30)

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$ (31)

$a_1{k}_7^2+a_2{k}_8^2+a_3{k}_9^2+a_4{k}_7{k}_8+a_5{k}_8{k}_9+a_6{k}_7{k}_9=\mathrm{0,}$ (32)

$\begin{array}{l}2a_1{k}_1{k}_4+2a_2{k}_2{k}_5+2a_3{k}_3{k}_6+a_4\left(k_1{k}_5+k_2{k}_4\right)\\ +a_5\left(k_2{k}_6+k_3{k}_5\right)+a_6\left(k_3{k}_4+k_1{k}_6\right)=\mathrm{0,}\end{array}$ (33)

$\begin{array}{l}2a_1{k}_4{k}_7+2a_2{k}_5{k}_8+2a_3{k}_6{k}_9+a_4\left(k_4{k}_8+k_5{k}_7\right)\\ +a_5\left(k_5{k}_9+k_6{k}_8\right)+a_6\left(k_6{k}_7+k_4{k}_9\right)=\mathrm{0,}\end{array}$ (34)

$\begin{array}{l}2a_1{k}_1{k}_7+2a_2{k}_2{k}_8+2a_3{k}_3{k}_9+a_4\left(k_1{k}_8+k_2{k}_7\right)\\ +a_5\left(k_2{k}_9+k_3{k}_8\right)+a_6\left(k_3{k}_7+k_1{k}_9\right)=\mathrm{0,}\end{array}$ (35)

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$ (36)

$a_7{k}_4+a_8{k}_5+a_9{k}_6=0.$ (37)

Proof. According to $Z_1$ transformation, set $u=u\left(p\mathrm{,}q\mathrm{,}r\right)$ , $p=k_{1}t+k_{2}x+k_{3}y$ , $q=k_{4}t+k_{5}x+k_{6}y$ , $r=k_{7}t+k_{8}x+k_{9}y$ . $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_9$ are undetermined constants, $p\mathrm{,}q$ and $r$ are independent of each other, so

$-k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8-k_1{k}_6{k}_8-k_2{k}_4{k}_9+k_1{k}_5{k}_9\ne \mathrm{0,}$

and

Set

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

$a_7{k}_4+a_8{k}_5+a_9{k}_6=0.$

We get

(38)

The analytical solution of (38) is

so the theorem is proved.

Similar to the proof method of Theorem 5, we can obtain Theorem 6.

Theorem 6. In , if

(39)

where are any known constants (), , , , then the analytical solution of Equation (39) is

(40)

where is an arbitrary smooth function, and the constants need satisfy

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

Proof. According to transformation, set , , , . are undetermined constants, and are independent of each other, so

and

Set

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

We get

(41)

The general solution of (41) is

where is an arbitrary smooth function, so the theorem is proved.

Next, we use Theorem 5 to analyze a definite solution problem.

Example 5. In , use Theorem 5 to obtain the analytical solution of

(42)

in the condition of , is an arbitrary known first differentiable function.

Solution. According to Theorem 5, the general solution of (42) is

So

Namely

Set

We obtain

That is

Set

Then

So

According to Example 5, we can directly obtain the analytical solution of Equation (42) in various initial value conditions. If the initial value condition is , the analytical solution is

.

For

(43)

(44)

where are any known constants (), and readers can try to obtain their analytical solutions by the method similar to Theorem 5.

Next we propose Theorem 7.

Theorem 7. In , if

(45)

where are any known constants (), , , , then the analytical solution of Equation (45) is

(46)

where is an arbitrary smooth function, and the constants need satisfy

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

(47)

(48)

(49)

Proof. According to transformation, set , , , . are undetermined constants, and are independent of each other, so

and

Set

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

We get

(50)

The analytical solution of Equation (50) is

so the theorem is proved.

Similar to the proof method of Theorem 7, we can obtain Theorem 8.

Theorem 8. In , if

(51)

where are any known constants ($1\le i\le 13$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , $\Lambda \mathrm{<mo>=</mo>}\Lambda \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txy}\mathrm{,}\cdots \right)$ , then the analytical solution of Equation (51) is

$u={\text{e}}^{\frac{-a_{13}r}{a_{10}{k}_7+a_{11}{k}_8+a_{12}{k}_9}}f\left(p\mathrm{,}q\right)\mathrm{,}$ (52)

$p=k_{1}t+k_{2}x+k_{3}y,q=k_{4}t+k_{5}x+k_{6}y,r=k_{7}t+k_{8}x+k_{9}y,$

where $f$ is an arbitrary smooth function, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_9$ need satisfy

$-k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8-k_1{k}_6{k}_8-k_2{k}_4{k}_9+k_1{k}_5{k}_9\ne \mathrm{0,}$

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_1{k}_2+a_5{k}_2{k}_3+a_6{k}_1{k}_3=\mathrm{0,}$

$a_1{k}_4^2+a_2{k}_5^2+a_3{k}_6^2+a_4{k}_4{k}_5+a_5{k}_5{k}_6+a_6{k}_4{k}_6=\mathrm{0,}$

$a_1{k}_7^2+a_2{k}_8^2+a_3{k}_9^2+a_4{k}_7{k}_8+a_5{k}_8{k}_9+a_6{k}_7{k}_9=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1{k}_4+2a_2{k}_2{k}_5+2a_3{k}_3{k}_6+a_4\left(k_1{k}_5+k_2{k}_4\right)\\ +a_5\left(k_2{k}_6+k_3{k}_5\right)+a_6\left(k_3{k}_4+k_1{k}_6\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_4{k}_7+2a_2{k}_5{k}_8+2a_3{k}_6{k}_9+a_4\left(k_4{k}_8+k_5{k}_7\right)\\ +a_5\left(k_5{k}_9+k_6{k}_8\right)+a_6\left(k_6{k}_7+k_4{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_1{k}_7+2a_2{k}_2{k}_8+2a_3{k}_3{k}_9+a_4\left(k_1{k}_8+k_2{k}_7\right)\\ +a_5\left(k_2{k}_9+k_3{k}_8\right)+a_6\left(k_3{k}_7+k_1{k}_9\right)=\mathrm{0,}\end{array}$

$a_7{k}_1+a_8{k}_2+a_9{k}_3=\mathrm{0,}$

$a_7{k}_4+a_8{k}_5+a_9{k}_6=\mathrm{0,}$

$a_7{k}_7+a_8{k}_8+a_9{k}_9=\mathrm{0,}$

$a_{10}{k}_1+a_{11}{k}_2+a_{12}{k}_3=\mathrm{0,}$

$a_{10}{k}_4+a_{11}{k}_5+a_{12}{k}_6=0.$

Next, we use Theorem 8 to analyze a definite solution problem.

Example 6. In ${ℝ}^3$ , use Theorem 8 to obtain the analytical solution of

$2u_t^2+u_x^2+u_y^2+3u_t{u}_x+3u_t{u}_y+2u_x{u}_y+u\left(u_t+u_x+u_y\right)+2u_t+u_x+u_y+u=\mathrm{0,}$ (53)

in the condition of $u\left(\mathrm{0,}x\mathrm{,}y\right)=g\left(x\mathrm{,}y\right)$ , $g$ is an arbitrary known first differentiable function.

Solution. According to Theorem 8, the general solution of (53) is

$u={\text{e}}^{\frac{-t}{2}}f\left(\frac{\left(-c_1-c_2\right)t}{2}+c_{1}x+c_{2}y\mathrm{,}\frac{\left(-c_3-c_4\right)t}{2}+c_{3}x+c_{4}y\right)\mathrm{,}$ (54)

or

$u={\text{e}}^{\frac{-x}{2}}f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)\mathrm{,}$ (55)

where $c_1\mathrm{,}{c}_2\mathrm{,}{c}_3$ and $c_4$ are arbitrary constants. If the solution is (54), so

$u\left(\mathrm{0,}x\mathrm{,}y\mathrm{,}z\right)=u\left(\mathrm{0,}x\mathrm{,}y\right)=f\left(c_{1}x+c_{2}y\mathrm{,}{c}_{3}x+c_{4}y\right)=g\left(x\mathrm{,}y\right)\mathrm{.}$

Set

$c_{1}x+c_{2}y=\beta ,c_{3}x+c_{4}y=\gamma .$

We obtain

$x=\frac{c_4\beta -c_2\gamma }{c_1{c}_4-c_2{c}_3},y=\frac{c_1\gamma -c_3\beta }{c_1{c}_4-c_2{c}_3}.$

Namely

$f\left(c_{1}x+c_{2}y\mathrm{,}{c}_{3}x+c_{4}y\right)=g\left(\frac{c_4\beta -c_2\gamma }{c_1{c}_4-c_2{c}_3}\mathrm{,}\frac{c_1\gamma -c_3\beta }{c_1{c}_4-c_2{c}_3}\right)\mathrm{.}$

Set

$\frac{\left(-c_1-c_2\right)t}{2}+c_{1}x+c_{2}y=\beta ,\frac{\left(-c_3-c_4\right)t}{2}+c_{3}x+c_{4}y=\gamma .$

Then

$\frac{c_4\beta -c_2\gamma }{c_1{c}_4-c_2{c}_3}=-\frac{t}{2}+x\mathrm{,}$

$\frac{c_1\gamma -c_3\beta }{c_1{c}_4-c_2{c}_3}=-\frac{t}{2}+y\mathrm{.}$

So the analytical solution of the definite solution problem is

$u\left(t\mathrm{,}x\mathrm{,}y\right)=g\left(-\frac{t}{2}+x\mathrm{,}-\frac{t}{2}+y\right)\mathrm{.}$ (56)

If the solution is (55), in a similar way, we can get

$u\left(t\mathrm{,}x\mathrm{,}y\right)=g\left(-t+x\mathrm{,}-t+y\right)\mathrm{.}$ (57)

That is, if the general solution of a PDE is not unique, the analytical solution of its definite solution problem may not be unique either. Such as $u\left(\mathrm{0,}x\mathrm{,}y\right)=g\left(x\mathrm{,}y\right)={\text{e}}^{x+2y}$ , then

$u\left(t\mathrm{,}x\mathrm{,}y\right)=g\left(-\frac{t}{2}+x\mathrm{,}-\frac{t}{2}+y\right)={\text{e}}^{-\frac{3t}{2}+x+2y}\mathrm{,}$

$u\left(t\mathrm{,}x\mathrm{,}y\right)=g\left(-t+x\mathrm{,}-t+y\right)={\text{e}}^{-3t+x+2y}\mathrm{,}$

and

$u\left(t\mathrm{,0,}y\right)={\text{e}}^{-\frac{3t}{2}+2y}\mathrm{,}$

$u\left(t\mathrm{,}x\mathrm{,0}\right)={\text{e}}^{-3t+x}\mathrm{.}$

So two definite solution conditions are needed to make the analytical solutions of the definite solution problem unique.

Next we propose Theorem 9.

Theorem 9. In ${ℝ}^4$ , if

$\begin{array}{l}\Theta (a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_z^2+a_5{u}_t{u}_x+a_6{u}_t{u}_y+a_7{u}_t{u}_z+a_8{u}_x{u}_y\\ +a_9{u}_x{u}_z+a_{10}{u}_y{u}_z)+\Lambda \left(a_{11}{u}_t+a_{12}{u}_x+a_{13}{u}_y+a_{14}{u}_z\right)=\mathrm{0,}\end{array}$ (58)

where $a_i$ are any known constants ($1\le i\le 14$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txyz}\mathrm{,}\cdots \right)$ , $\Lambda =\Lambda \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txyz}\mathrm{,}\cdots \right)$ , then the analytical solution of Equation (58) is

$u=f\left(p\mathrm{,}q\mathrm{,}r\right)\mathrm{,}$ (59)

$p=k_{1}t+k_{2}x+k_{3}y+k_{4}z\mathrm{,}$ (60)

$q=k_{5}t+k_{6}x+k_{7}y+k_{8}z\mathrm{,}$ (61)

$r=k_{9}t+k_{10}x+k_{11}y+k_{12}z\mathrm{,}$ (62)

where $f$ is an arbitrary smooth function; $p\mathrm{,}q$ and $r$ are independent of each other, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_{12}$ need satisfy

$\begin{array}{l}{a}_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_4^2+a_5{k}_1{k}_2+a_6{k}_1{k}_3+a_7{k}_1{k}_4\\ +a_8{k}_2{k}_3+a_9{k}_2{k}_4+a_{10}{k}_3{k}_4=\mathrm{0,}\end{array}$ (63)

$\begin{array}{l}{a}_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_8^2+a_5{k}_5{k}_6+a_6{k}_5{k}_7+a_7{k}_5{k}_8\\ +a_8{k}_6{k}_7+a_9{k}_6{k}_8+a_{10}{k}_7{k}_8=\mathrm{0,}\end{array}$ (64)

$\begin{array}{l}{a}_1{k}_9^2+a_2{k}_{10}^2+a_3{k}_{11}^2+a_4{k}_{12}^2+a_5{k}_9{k}_{10}+a_6{k}_9{k}_{11}+a_7{k}_9{k}_{12}\\ +a_8{k}_{10}{k}_{11}+a_9{k}_{10}{k}_{12}+a_{10}{k}_{11}{k}_{12}=\mathrm{0,}\end{array}$ (65)

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+2a_4{k}_4{k}_8+a_5\left(k_1{k}_6+k_2{k}_5\right)+a_6\left(k_1{k}_7+k_3{k}_5\right)\\ +a_7\left(k_1{k}_8+k_4{k}_5\right)+a_8\left(k_2{k}_7+k_3{k}_6\right)+a_9\left(k_2{k}_8+k_4{k}_6\right)+a_{10}\left(k_3{k}_8+k_4{k}_7\right)=\mathrm{0,}\end{array}$ (66)

$\begin{array}{l}2a_1{k}_1{k}_9+2a_2{k}_2{k}_{10}+2a_3{k}_3{k}_{11}+2a_4{k}_4{k}_{12}+a_5\left(k_1{k}_{10}+k_2{k}_9\right)+a_6\left(k_1{k}_{11}+k_3{k}_9\right)\\ +a_7\left(k_1{k}_{12}+k_4{k}_9\right)+a_8\left(k_2{k}_{11}+k_3{k}_{10}\right)+a_9\left(k_2{k}_{12}+k_4{k}_{10}\right)+a_{10}\left(k_3{k}_{12}+k_4{k}_{11}\right)=\mathrm{0,}\end{array}$ (67)

$\begin{array}{l}2a_1{k}_5{k}_9+2a_2{k}_6{k}_{10}+2a_3{k}_7{k}_{11}+2a_4{k}_8{k}_{12}+a_5\left(k_5{k}_{10}+k_6{k}_9\right)+a_6\left(k_5{k}_{11}+k_7{k}_9\right)\\ +a_7\left(k_5{k}_{12}+k_8{k}_9\right)+a_8\left(k_6{k}_{11}+k_7{k}_{10}\right)+a_9\left(k_6{k}_{12}+k_8{k}_{10}\right)+a_{10}\left(k_7{k}_{12}+k_8{k}_{11}\right)=\mathrm{0,}\end{array}$ (68)

$a_{11}{k}_1+a_{12}{k}_2+a_{13}{k}_3+a_{14}{k}_4=\mathrm{0,}$ (69)

$a_{11}{k}_5+a_{12}{k}_6+a_{13}{k}_7+a_{14}{k}_8=\mathrm{0,}$ (70)

$a_{11}{k}_9+a_{12}{k}_{10}+a_{13}{k}_{11}+a_{14}{k}_{12}=0.$ (71)

Proof. According to $Z_1$ transformation, set $u=f\left(p\mathrm{,}q\mathrm{,}r\right)$ , $p=k_{1}t+k_{2}x+k_{3}y+k_{4}z$ , $q=k_{5}t+k_{6}x+k_{7}y+k_{8}z$ , $r=k_{9}t+k_{10}x+k_{11}y+k_{12}z$ ; $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_{12}$ are undetermined constants, $p\mathrm{,}q$ and $r$ are independent of each other, so

(72)

Set

$a_1{k}_1^2+a_2{k}_2^2+a_3{k}_3^2+a_4{k}_4^2+a_5{k}_1{k}_2+a_6{k}_1{k}_3+a_7{k}_1{k}_4+a_8{k}_2{k}_3+a_9{k}_2{k}_4+a_{10}{k}_3{k}_4=\mathrm{0,}$

$a_1{k}_5^2+a_2{k}_6^2+a_3{k}_7^2+a_4{k}_8^2+a_5{k}_5{k}_6+a_6{k}_5{k}_7+a_7{k}_5{k}_8+a_8{k}_6{k}_7+a_9{k}_6{k}_8+a_{10}{k}_7{k}_8=\mathrm{0,}$

$\begin{array}{l}{a}_1{k}_9^2+a_2{k}_{10}^2+a_3{k}_{11}^2+a_4{k}_{12}^2+a_5{k}_9{k}_{10}+a_6{k}_9{k}_{11}+a_7{k}_9{k}_{12}\\ +a_8{k}_{10}{k}_{11}+a_9{k}_{10}{k}_{12}+a_{10}{k}_{11}{k}_{12}=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_1{k}_5+2a_2{k}_2{k}_6+2a_3{k}_3{k}_7+2a_4{k}_4{k}_8+a_5\left(k_1{k}_6+k_2{k}_5\right)+a_6\left(k_1{k}_7+k_3{k}_5\right)\\ +a_7\left(k_1{k}_8+k_4{k}_5\right)+a_8\left(k_2{k}_7+k_3{k}_6\right)+a_9\left(k_2{k}_8+k_4{k}_6\right)+a_{10}\left(k_3{k}_8+k_4{k}_7\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_1{k}_9+2a_2{k}_2{k}_{10}+2a_3{k}_3{k}_{11}+2a_4{k}_4{k}_{12}+a_5\left(k_1{k}_{10}+k_2{k}_9\right)+a_6\left(k_1{k}_{11}+k_3{k}_9\right)\\ +a_7\left(k_1{k}_{12}+k_4{k}_9\right)+a_8\left(k_2{k}_{11}+k_3{k}_{10}\right)+a_9\left(k_2{k}_{12}+k_4{k}_{10}\right)+a_{10}\left(k_3{k}_{12}+k_4{k}_{11}\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_5{k}_9+2a_2{k}_6{k}_{10}+2a_3{k}_7{k}_{11}+2a_4{k}_8{k}_{12}+a_5\left(k_5{k}_{10}+k_6{k}_9\right)+a_6\left(k_5{k}_{11}+k_7{k}_9\right)\\ +a_7\left(k_5{k}_{12}+k_8{k}_9\right)+a_8\left(k_6{k}_{11}+k_7{k}_{10}\right)+a_9\left(k_6{k}_{12}+k_8{k}_{10}\right)+a_{10}\left(k_7{k}_{12}+k_8{k}_{11}\right)=\mathrm{0,}\end{array}$

$a_{11}{k}_1+a_{12}{k}_2+a_{13}{k}_3+a_{14}{k}_4=\mathrm{0,}$

$a_{11}{k}_5+a_{12}{k}_6+a_{13}{k}_7+a_{14}{k}_8=\mathrm{0,}$

$a_{11}{k}_9+a_{12}{k}_{10}+a_{13}{k}_{11}+a_{14}{k}_{12}=0.$

Therefore, the analytical solution of Equation (58) is

$u=f\left(p\mathrm{,}q\mathrm{,}r\right)$

The theorem is proven. $\square$

In ${ℝ}^4$ , if

$\begin{array}{l}\Theta (a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_z^2+a_5{u}_t{u}_x+a_6{u}_t{u}_y+a_7{u}_t{u}_z+a_8{u}_x{u}_y\\ +a_9{u}_x{u}_z+a_{10}{u}_y{u}_z)+A\left(a_{11}{u}_t+a_{12}{u}_x+a_{13}{u}_y+a_{14}{u}_z+u\right)=B\mathrm{,}\end{array}$ (73)

$\begin{array}{l}A(a_1{u}_t^2+a_2{u}_x^2+a_3{u}_y^2+a_4{u}_z^2+a_5{u}_t{u}_x+a_6{u}_t{u}_y+a_7{u}_t{u}_z+a_8{u}_x{u}_y\\ +a_9{u}_x{u}_z+a_{10}{u}_y{u}_z)+\Theta \left(a_{11}{u}_t+a_{12}{u}_x+a_{13}{u}_y+a_{14}{u}_z+u\right)=B\mathrm{,}\end{array}$ (74)

where $a_i$ are any known constants ($1\le i\le 14$ ), $\Theta =\Theta \left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\mathrm{,}u\mathrm{,}{u}_t\mathrm{,}\cdots ,u_{txyz}\mathrm{,}\cdots \right)$ , $A=A\left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\right)$ , $B=B\left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\right)$ , set

$T=k_{1}t+k_{2}x+k_{3}y+k_{4}z\mathrm{,}$ (75)

$X=k_{5}t+k_{6}x+k_{7}y+k_{8}z\mathrm{,}$ (76)

$Y=k_{9}t+k_{10}x+k_{11}y+k_{12}z\mathrm{,}$ (77)

$Z=k_{13}t+k_{14}x+k_{15}y+k_{16}z\mathrm{,}$ (78)

$\frac{\partial \left(X\mathrm{,}Y\mathrm{,}Z\mathrm{,}T\right)}{\partial \left(x\mathrm{,}y\mathrm{,}z\mathrm{,}t\right)}\ne 0.$ (79)

Similar to the calculation of Theorem 5, the analytical solutions of (73), (74) can be obtained, and readers can try it by themselves.

Next, we use Theorem 9 to analyze a definite solution problem.

Example 7. In ${ℝ}^4$ , use Theorem 9 to obtain the analytical solution of

${\left(u_t^2+u_x^2-2u_t{u}_x+u_t{u}_y+u_t{u}_z-u_x{u}_y-u_x{u}_z\right)}^2+{\left(u_t-u_x\right)}^3=\mathrm{0,}$ (80)

in the condition of $u\left(\mathrm{0,}x\mathrm{,}y\mathrm{,}z\right)=g\left(x\mathrm{,}y\mathrm{,}z\right)$ , $g$ is an arbitrary known first differentiable function.

Solution. According to Theorem 9, the general solution of (80) is

$u\left(x\mathrm{,}y\mathrm{,}z\right)=f\left(t+x+k_{3}y+k_{4}z\mathrm{,}t+x+k_{7}y+k_{8}z\mathrm{,}t+x+k_{11}y+k_{12}z\right)\mathrm{,}$ (81)

where $k_3\mathrm{,}{k}_4\mathrm{,}{k}_7\mathrm{,}{k}_8\mathrm{,}{k}_{11}$ and $k_{12}$ are arbitrary constants, so

$u\left(\mathrm{0,}x\mathrm{,}y\mathrm{,}z\right)=f\left(x+k_{3}y+k_{4}z\mathrm{,}x+k_{7}y+k_{8}z\mathrm{,}x+k_{11}y+k_{12}z\right)=g\left(x\mathrm{,}y\mathrm{,}z\right)\mathrm{.}$

Set

$x+k_{3}y+k_{4}z=\alpha ,x+k_{7}y+k_{8}z=\beta ,x+k_{11}y+k_{12}z=\gamma .$

We obtain

$x=-\frac{-\gamma k_4{k}_7+\gamma k_3{k}_8+\beta k_4{k}_{11}-\alpha k_8{k}_{11}-\beta k_3{k}_{12}+\alpha k_7{k}_{12}}{k_4{k}_7-k_3{k}_8-k_4{k}_{11}+k_8{k}_{11}+k_3{k}_{12}-k_7{k}_{12}}\mathrm{,}$ (82)

$y=-\frac{-\beta k_4+\gamma k_4+\alpha k_8-\gamma k_8-\alpha k_{12}+\beta k_{12}}{k_4{k}_7-k_3{k}_8-k_4{k}_{11}+k_8{k}_{11}+k_3{k}_{12}-k_7{k}_{12}}\mathrm{,}$ (83)

$z=-\frac{-\beta k_3+\gamma k_3+\alpha k_7-\gamma k_7-\alpha k_{11}+\beta k_{11}}{-k_4{k}_7+k_3{k}_8+k_4{k}_{11}-k_8{k}_{11}-k_3{k}_{12}+k_7{k}_{12}}\mathrm{.}$ (84)

Namely

$\begin{array}{l}u\left(\mathrm{0,}x\mathrm{,}y\mathrm{,}z\right)=f\left(x+k_{3}y+k_{4}z\mathrm{,}x+k_{7}y+k_{8}z\mathrm{,}x+k_{11}y+k_{12}z\right)\\ =g\left(-\frac{-\gamma k_4{k}_7+\cdots +\alpha k_7{k}_{12}}{k_4{k}_7-\cdots -k_7{k}_{12}}\mathrm{,}-\frac{-\beta k_4+\cdots +\beta k_{12}}{k_4{k}_7-\cdots -k_7{k}_{12}}\mathrm{,}-\frac{-\beta k_3+\cdots +\beta k_{11}}{-k_4{k}_7+\cdots +k_7{k}_{12}}\right)\mathrm{.}\end{array}$

Set

$t+x+k_{3}y+k_{4}z=\alpha ,t+x+k_{7}y+k_{8}z=\beta ,t+x+k_{11}y+k_{12}z=\gamma .$

Then

$-\frac{-\gamma k_4{k}_7+\gamma k_3{k}_8+\beta k_4{k}_{11}-\alpha k_8{k}_{11}-\beta k_3{k}_{12}+\alpha k_7{k}_{12}}{k_4{k}_7-k_3{k}_8-k_4{k}_{11}+k_8{k}_{11}+k_3{k}_{12}-k_7{k}_{12}}=t+x\mathrm{,}$

$-\frac{-\beta k_4+\gamma k_4+\alpha k_8-\gamma k_8-\alpha k_{12}+\beta k_{12}}{k_4{k}_7-k_3{k}_8-k_4{k}_{11}+k_8{k}_{11}+k_3{k}_{12}-k_7{k}_{12}}=y\mathrm{,}$

$-\frac{-\beta k_3+\gamma k_3+\alpha k_7-\gamma k_7-\alpha k_{11}+\beta k_{11}}{-k_4{k}_7+k_3{k}_8+k_4{k}_{11}-k_8{k}_{11}-k_3{k}_{12}+k_7{k}_{12}}=z\mathrm{.}$

So the analytical solution of the definite solution problem is

$u\left(t\mathrm{,}x\mathrm{,}y\mathrm{,}z\right)=g\left(t+x\mathrm{,}y\mathrm{,}z\right)$

According to Example 7, if the initial value condition is $u\left(\mathrm{0,}x\mathrm{,}y\mathrm{,}z\right)={\left(x+2y+z\right)}^3+\cos \left(x+y+2z\right)+\text{tan}\left(2x-y-z\right)$ , the analytical solution is $u={\left(t+x+2y+z\right)}^3+\cos \left(t+x+y+2z\right)+\text{tan}\left(2t+2x-y-z\right)$ .

Theorem 10 is presented below.

Theorem 10. In ${ℝ}^3$ , if

$\begin{array}{l}{a}_1{u}_t^3+a_2{u}_x^3+a_3{u}_y^3+a_4{u}_t^2{u}_x+a_5{u}_t^2{u}_y+a_6{u}_x^2{u}_t+a_7{u}_x^2{u}_y\\ +a_8{u}_y^2{u}_t+a_9{u}_y^2{u}_x+a_{10}{u}_t{u}_x{u}_y=\mathrm{0,}\end{array}$ (85)

where $a_i$ are any known constants ($1\le i\le 9$ ), then the general solution of Equation (85) is

$u=f\left(v\mathrm{,}w\right)$

$v=k_{1}t+k_{2}x+k_{3}y+k_4,w=k_{5}t+k_{6}x+k_{7}y+k_8,$

where $f$ is an arbitrary first differentiable function, $v$ and $w$ are independent of each other, and the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_5\mathrm{,}{k}_6\mathrm{,}{k}_7$ need satisfy

$\begin{array}{l}{a}_1{k}_1^3+a_2{k}_2^3+a_3{k}_3^3+a_4{k}_1^2{k}_2+a_5{k}_1^2{k}_3+a_6{k}_1{k}_2^2+a_7{k}_2^2{k}_3\\ +a_8{k}_1{k}_3^2+a_9{k}_2{k}_3^2+a_{10}{k}_1{k}_2{k}_3=\mathrm{0,}\end{array}$ (86)

$\begin{array}{l}{a}_1{k}_5^3+a_2{k}_6^3+a_3{k}_7^3+a_4{k}_5^2{k}_6+a_5{k}_5^2{k}_7+a_6{k}_5{k}_6^2+a_7{k}_6^2{k}_7\\ +a_8{k}_5{k}_7^2+a_9{k}_6{k}_7^2+a_{10}{k}_5{k}_6{k}_7=\mathrm{0,}\end{array}$ (87)

$\begin{array}{l}3a_1{k}_1^2{k}_5+3a_2{k}_2^2{k}_6+3a_3{k}_3^2{k}_7+a_4\left(k_1^2{k}_6+2k_1{k}_2{k}_5\right)+a_5\left(k_1^2{k}_7+2k_1{k}_3{k}_5\right)\\ +a_6\left(k_2^2{k}_5+2k_1{k}_2{k}_6\right)+a_7\left(k_2^2{k}_7+2k_2{k}_3{k}_6\right)+a_8\left(k_3^2{k}_5+2k_1{k}_3{k}_7\right)\\ +a_9\left(k_2{k}_7^2+2k_2{k}_3{k}_7\right)+a_{10}\left(k_2{k}_3{k}_5+k_1{k}_3{k}_6+k_1{k}_2{k}_7\right)=\mathrm{0,}\end{array}$ (88)

$\begin{array}{l}3a_1{k}_1{k}_5^2+3a_2{k}_2{k}_6^2+3a_3{k}_3{k}_7^2+a_4\left(k_2{k}_5^2+2k_1{k}_5{k}_6\right)+a_5\left(k_3{k}_5^2+2k_1{k}_5{k}_7\right)\\ +a_6\left(k_1{k}_6^2+2k_2{k}_5{k}_6\right)+a_7\left(k_3{k}_6^2+2k_2{k}_6{k}_7\right)+a_8\left(k_1{k}_7^2+2k_3{k}_5{k}_7\right)\\ +a_9\left(k_3^2{k}_6+2k_3{k}_6{k}_7\right)+a_{10}\left(k_3{k}_5{k}_6+k_2{k}_5{k}_7+k_1{k}_6{k}_7\right)=0.\end{array}$ (89)

Proof. According to $Z_1$ transformation, set $u=f\left(v\mathrm{,}w\right)$ , $v=k_{1}t+k_{2}x+k_{3}y+k_4$ , $w=k_{5}t+k_{6}x+k_{7}y+k_8$ . $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_8$ are undetermined constants, $v$ and $w$ are independent of each other, so

Set

$\begin{array}{l}{a}_1{k}_1^3+a_2{k}_2^3+a_3{k}_3^3+a_4{k}_1^2{k}_2+a_5{k}_1^2{k}_3+a_6{k}_1{k}_2^2+a_7{k}_2^2{k}_3\\ +a_8{k}_1{k}_3^2+a_9{k}_2{k}_3^2+a_{10}{k}_1{k}_2{k}_3=\mathrm{0,}\end{array}$

$\begin{array}{l}{a}_1{k}_5^3+a_2{k}_6^3+a_3{k}_7^3+a_4{k}_5^2{k}_6+a_5{k}_5^2{k}_7+a_6{k}_5{k}_6^2+a_7{k}_6^2{k}_7\\ +a_8{k}_5{k}_7^2+a_9{k}_6{k}_7^2+a_{10}{k}_5{k}_6{k}_7=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_1^2{k}_5+3a_2{k}_2^2{k}_6+3a_3{k}_3^2{k}_7+a_4\left(k_1^2{k}_6+2k_1{k}_2{k}_5\right)+a_5\left(k_1^2{k}_7+2k_1{k}_3{k}_5\right)\\ +a_6\left(k_2^2{k}_5+2k_1{k}_2{k}_6\right)+a_7\left(k_2^2{k}_7+2k_2{k}_3{k}_6\right)+a_8\left(k_3^2{k}_5+2k_1{k}_3{k}_7\right)\\ +a_9\left(k_2{k}_7^2+2k_2{k}_3{k}_7\right)+a_{10}\left(k_2{k}_3{k}_5+k_1{k}_3{k}_6+k_1{k}_2{k}_7\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_1{k}_5^2+3a_2{k}_2{k}_6^2+3a_3{k}_3{k}_7^2+a_4\left(k_2{k}_5^2+2k_1{k}_5{k}_6\right)+a_5\left(k_3{k}_5^2+2k_1{k}_5{k}_7\right)\\ +a_6\left(k_1{k}_6^2+2k_2{k}_5{k}_6\right)+a_7\left(k_3{k}_6^2+2k_2{k}_6{k}_7\right)+a_8\left(k_1{k}_7^2+2k_3{k}_5{k}_7\right)\\ +a_9\left(k_3^2{k}_6+2k_3{k}_6{k}_7\right)+a_{10}\left(k_3{k}_5{k}_6+k_2{k}_5{k}_7+k_1{k}_6{k}_7\right)=0.\end{array}$

Therefore, the analytical solution of Equation (85) is

$u=f\left(v\mathrm{,}w\right)\mathrm{.}$

The theorem is proven. $\square$

Next, we use Theorem 10 to analyze two definite solution problems.

Example 8. In ${ℝ}^3$ , use Theorem 10 to obtain the analytical solution of

$u_x^3+u_y^3+u_t^2{u}_x+u_t^2{u}_y-u_x^2{u}_t+u_x^2{u}_y+u_y^2{u}_t+u_y^2{u}_x=\mathrm{0,}$ (90)

in the condition of $u\left(t\mathrm{,0,}y\right)=g\left(t\mathrm{,}y\right)$ , $g$ is an arbitrary known first differentiable function.

Solution. According to Theorem 10, the general solution of (90) is

$u=f\left(t+x-y\mathrm{,}t-x+y\right)\mathrm{.}$ (91)

Then

$u\left(t\mathrm{,0,}y\right)=f\left(t-y\mathrm{,}t+y\right)=g\left(t\mathrm{,}y\right)\mathrm{.}$

Set

$t-y=\beta ,t+y=\gamma .$

We obtain

$t=\frac{\beta +\gamma }{2},y=\frac{-\beta +\gamma }{2}.$

Namely

$u\left(t,0,y\right)=f\left(t-y,t+y\right)=g\left(\frac{\beta +\gamma }{2},\frac{-\beta +\gamma }{2}\right).$

Set

$t+x-y=\beta ,t-x+y=\gamma .$

We get

$\frac{\beta +\gamma }{2}=\frac{t+x-y+t-x+y}{2}=t,$

$\frac{-\beta +\gamma }{2}=\frac{-t-x+y+t-x+y}{2}=-x+y.$

So the analytical solution of the definite solution problem is

$u\left(t,x,y\right)=f\left(t+x-y,t-x+y\right)=g\left(t,-x+y\right).$ (92)

Example 9. In ${ℝ}^3$ , prove that the exact solution of

$u_t^3+2u_x^3+2u_y^3+4u_t^2{u}_x+4u_t^2{u}_y+5u_t{u}_x^2+7u_x^2{u}_y+5u_t{u}_y^2+7u_x{u}_y^2+11u_t{u}_x{u}_y=\mathrm{0,}$ (93)

in the condition of $u\left(\mathrm{0,}x\mathrm{,}y\right)=g\left(x\mathrm{,}y\right)$ is

$u=g\left(-t+x\mathrm{,}-t+y\right)\mathrm{,}$ (94)

or

$u=g\left(-t+x\mathrm{,}-2t+y\right)\mathrm{,}$ (95)

or

$u=g\left(-2t+x\mathrm{,}-t+y\right)\mathrm{,}$ (96)

where $g$ is an arbitrary known first differentiable function

Proof. According to Theorem 10, the general solution of (93) is

$u=f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)\mathrm{,}$ (97)

or

$u=f\left(\left(-c_1-2c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-2c_4\right)t+c_{3}x+c_{4}y\right)\mathrm{,}$ (98)

or

$u=f\left(\left(-2c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-2c_3-c_4\right)t+c_{3}x+c_{4}y\right)\mathrm{.}$ (99)

If $u=f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)$ , then

$u\left(\mathrm{0,}x\mathrm{,}y\right)=f\left(c_{1}x+c_{2}y\mathrm{,}{c}_{3}x+c_{4}y\right)=g\left(x\mathrm{,}y\right)\mathrm{.}$

Set

$c_{1}x+c_{2}y=\beta ,c_{3}x+c_{4}y=\gamma .$

We obtain

$x=\frac{\beta c_4-\gamma c_2}{c_1{c}_4-c_2{c}_3},y=\frac{\gamma c_1-\beta c_3}{c_1{c}_4-c_2{c}_3}.$

That is

$u\left(0,x,y\right)=f\left(c_{1}x+c_{2}y,c_{3}x+c_{4}y\right)=g\left(\frac{\beta c_4-\gamma c_2}{c_1{c}_4-c_2{c}_3},\frac{\gamma c_1-\beta c_3}{c_1{c}_4-c_2{c}_3}\right).$

Set

$\left(-c_1-c_2\right)t+c_{1}x+c_{2}y=\beta ,\left(-c_3-c_4\right)t+c_{3}x+c_{4}y=\gamma .$

Then

$\frac{\beta c_4-\gamma c_2}{c_1{c}_4-c_2{c}_3}=\frac{\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\right)c_4-\left(\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)c_2}{c_1{c}_4-c_2{c}_3}=-t+x,$

$\frac{\gamma c_1-\beta c_3}{c_1{c}_4-c_2{c}_3}=\frac{\left(\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)c_1-\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\right)c_3}{c_1{c}_4-c_2{c}_3}=-t+y.$

So the analytical solution of the definite solution problem is

$u\left(t,x,y\right)=f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y,\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)=g\left(-t+x,-t+y\right).$

If $u=f\left(\left(-c_1-2c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-2c_4\right)t+c_{3}x+c_{4}y\right)$ , or $u=f\left(\left(-2c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-2c_3-c_4\right)t+c_{3}x+c_{4}y\right)$ , similar to the above method, we can obtain

$\begin{array}{c}u\left(t\mathrm{,}x\mathrm{,}y\right)=f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y\mathrm{,}\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)\\ =g\left(-t+x\mathrm{,}-2t+y\right)\mathrm{,}\end{array}$

$\begin{array}{c}u\left(t,x,y\right)=f\left(\left(-c_1-c_2\right)t+c_{1}x+c_{2}y,\left(-c_3-c_4\right)t+c_{3}x+c_{4}y\right)\\ =g\left(-2t+x,-t+y\right).\end{array}$

That is, different exact solutions of the same definite solution problem may correspond to different general solutions! Such as

$u\left(0,x,y\right)=g\left(x,y\right)=\sin xy.$

Then

$g\left(-t+x\mathrm{,}-t+y\right)=\text{sin}\left(-t+x\right)\left(-t+y\right)\mathrm{,}$

$g\left(-t+x\mathrm{,}-2t+y\right)=\text{sin}\left(-t+x\right)\left(-2t+y\right)\mathrm{,}$

$g\left(-2t+x\mathrm{,}-t+y\right)=\text{sin}\left(-2t+x\right)\left(-t+y\right)\mathrm{,}$

are the analytical solutions of (93) under $u\left(0,x,y\right)=\sin xy$ . For

$u\left(t,0,y\right)=\sin \left(-t\right)\left(-t+y\right)=\sin \left(-t\right)\left(-2t+y\right)=\sin \left(-2t\right)\left(-t+y\right),$

$u\left(t,x,0\right)=\sin \left(-t+x\right)\left(-t\right)=\sin \left(-t+x\right)\left(-2t\right)=\sin \left(-2t+x\right)\left(-t\right).$

Therefore, for this case and other more complex cases of first-order equations, two definite solution conditions are generally required to specify the unique analytical solution.

Theorem 11. In ${ℝ}^3$ , if

$\begin{array}{l}{a}_1{u}_t^3+a_2{u}_x^3+a_3{u}_y^3+a_4{u}_t^2{u}_x+a_5{u}_t^2{u}_y+a_6{u}_x^2{u}_t+a_7{u}_x^2{u}_y\\ +a_8{u}_y^2{u}_t+a_9{u}_y^2{u}_x+a_{10}{u}_t{u}_x{u}_y=A\left(t\mathrm{,}x\mathrm{,}y\right)\mathrm{,}\end{array}$ (100)

where $a_i$ are any known constants ($1\le i\le 10$ ), then the general solution of Equation (100) is

$u=f\left(q\mathrm{,}r\right)+{\displaystyle \int \sqrt[3]{\frac{A\left(t\mathrm{,}x\mathrm{,}y\right)}{B_1}}\text{d}p}\mathrm{,}$ (101)

$\begin{array}{c}{B}_1=a_1{k}_1^3+a_2{k}_2^3+a_3{k}_3^3+a_4{k}_1^2{k}_2+a_5{k}_1^2{k}_3+a_6{k}_1{k}_2^2+a_7{k}_2^2{k}_3\\ +a_8{k}_1{k}_3^2+a_9{k}_2{k}_3^2+a_{10}{k}_1{k}_2{k}_3\mathrm{,}\end{array}$ (102)

where $f$ is an arbitrary first differentiable function, and

$p=k_{1}t+k_{2}x+k_{3}y,q=k_{4}t+k_{5}x+k_{6}y,r=k_{7}t+k_{8}x+k_{9}y,$

the constants $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_9$ need satisfy

$-k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8-k_1{k}_6{k}_8-k_2{k}_4{k}_9+k_1{k}_5{k}_9\ne \mathrm{0,}$

$\begin{array}{l}3a_1{k}_1^2{k}_4+3a_2{k}_2^2{k}_5+3a_3{k}_3^2{k}_6+a_4\left(2k_1{k}_2{k}_4+k_1^2{k}_5\right)+a_5\left(2k_1{k}_3{k}_4+k_1^2{k}_6\right)\\ +a_6\left(k_2^2{k}_4+2k_1{k}_2{k}_5\right)+a_7\left(2k_2{k}_3{k}_5+k_2^2{k}_6\right)+a_8\left(k_3^2{k}_4+2k_1{k}_3{k}_6\right)\\ +a_9\left(k_3^2{k}_5+2k_2{k}_3{k}_6\right)+a_{10}\left(k_2{k}_3{k}_4+k_1{k}_3{k}_5+k_1{k}_2{k}_6\right)=\mathrm{0,}\end{array}$ (103)

$\begin{array}{l}3a_1{k}_1{k}_4^2+3a_2{k}_2{k}_5^2+3a_3{k}_3{k}_6^2+a_4\left(k_2{k}_4^2+2k_1{k}_4{k}_5\right)+a_5\left(k_3{k}_4^2+2k_1{k}_4{k}_6\right)\\ +a_6\left(2k_2{k}_4{k}_5+k_1{k}_5^2\right)+a_7\left(k_3{k}_5^2+2k_2{k}_5{k}_6\right)+a_8\left(2k_3{k}_4{k}_6+k_1{k}_6^2\right)\\ +a_9\left(2k_3{k}_5{k}_6+k_2{k}_6^2\right)+a_{10}\left(k_3{k}_4{k}_5+k_2{k}_4{k}_6+k_1{k}_5{k}_6\right)=\mathrm{0,}\end{array}$ (104)

$\begin{array}{l}{a}_1{k}_4^3+a_2{k}_5^3+a_3{k}_6^3+a_4{k}_4^2{k}_5+a_5{k}_4^2{k}_6+a_6{k}_4{k}_5^2+a_7{k}_5^2{k}_6\\ +a_8{k}_4{k}_6^2+a_9{k}_5{k}_6^2+a_{10}{k}_4{k}_5{k}_6=\mathrm{0,}\end{array}$ (105)

$\begin{array}{l}3a_1{k}_1^2{k}_7+3a_2{k}_2^2{k}_8+3a_3{k}_3^2{k}_9+a_4\left(2k_1{k}_2{k}_7+k_1^2{k}_8\right)+a_5\left(2k_1{k}_3{k}_7+k_1^2{k}_9\right)\\ +a_6\left(k_2^2{k}_7+2k_1{k}_2{k}_8\right)+a_7\left(2k_2{k}_3{k}_8+k_2^2{k}_9\right)+a_8\left(k_3^2{k}_7+2k_1{k}_3{k}_9\right)\\ +a_9\left(k_3^2{k}_8+2k_2{k}_3{k}_9\right)+a_{10}\left(k_2{k}_3{k}_7+k_1{k}_3{k}_8+k_1{k}_2{k}_9\right)=\mathrm{0,}\end{array}$ (106)

$\begin{array}{l}6a_1{k}_1{k}_4{k}_7+6a_2{k}_2{k}_5{k}_8+6a_3{k}_3{k}_6{k}_9+2a_4\left(k_2{k}_4{k}_7+k_1{k}_5{k}_7+k_1{k}_4{k}_8\right)\\ +2a_5\left(k_3{k}_4{k}_7+k_1{k}_6{k}_7+k_1{k}_4{k}_9\right)+2a_6\left(k_2{k}_5{k}_7+k_2{k}_4{k}_8+k_1{k}_5{k}_8\right)\\ +2a_7\left(k_3{k}_5{k}_8+k_2{k}_6{k}_8+k_2{k}_5{k}_9\right)+2a_8\left(k_3{k}_6{k}_7+k_3{k}_4{k}_9+k_1{k}_6{k}_9\right)\\ +2a_9\left(k_3{k}_6{k}_8+k_3{k}_5{k}_9+k_2{k}_6{k}_9\right)\\ +a_{10}\left(k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8+k_1{k}_6{k}_8+k_2{k}_4{k}_9+k_1{k}_5{k}_9\right)=\mathrm{0,}\end{array}$ (107)

$\begin{array}{l}3a_1{k}_4^2{k}_7+3a_2{k}_5^2{k}_8+3a_3{k}_6^2{k}_9+a_4\left(2k_4{k}_5{k}_7+k_4^2{k}_8\right)+a_5\left(2k_4{k}_6{k}_7+k_4^2{k}_9\right)\\ +a_6\left(k_5^2{k}_7+2k_4{k}_5{k}_8\right)+a_7\left(2k_5{k}_6{k}_8+k_5^2{k}_9\right)+a_8\left(k_6^2{k}_7+2k_4{k}_6{k}_9\right)\\ +a_9\left(k_6^2{k}_8+2k_5{k}_6{k}_9\right)+a_{10}\left(k_5{k}_6{k}_7+k_4{k}_6{k}_8+k_4{k}_5{k}_9\right)=\mathrm{0,}\end{array}$ (108)

$\begin{array}{l}3a_1{k}_1{k}_7^2+3a_2{k}_2{k}_8^2+3a_3{k}_3{k}_9^2+a_4\left(k_2{k}_7^2+2k_1{k}_7{k}_8\right)+a_5\left(k_3{k}_7^2+2k_1{k}_7{k}_9\right)\\ +a_6\left(2k_2{k}_7{k}_8+k_1{k}_8^2\right)+a_7\left(k_3{k}_8^2+2k_2{k}_8{k}_9\right)+a_8\left(2k_3{k}_7{k}_9+k_1{k}_9^2\right)\\ +a_9\left(2k_3{k}_8{k}_9+k_2{k}_9^2\right)+a_{10}\left(k_3{k}_7{k}_8+k_2{k}_7{k}_9+k_1{k}_8{k}_9\right)=\mathrm{0,}\end{array}$ (109)

$\begin{array}{l}3a_1{k}_4{k}_7^2+3a_2{k}_5{k}_8^2+3a_3{k}_6{k}_9^2+a_4\left(k_5{k}_7^2+2k_4{k}_7{k}_8\right)+a_5\left(k_6{k}_7^2+2k_4{k}_7{k}_9\right)\\ +a_6\left(2k_5{k}_7{k}_8+k_4{k}_8^2\right)+a_7\left(k_6{k}_8^2+2k_5{k}_8{k}_9\right)+a_8\left(2k_6{k}_7{k}_9+k_4{k}_9^2\right)\\ +a_9\left(2k_6{k}_8{k}_9+k_5{k}_9^2\right)+a_{10}\left(k_6{k}_7{k}_8+k_5{k}_7{k}_9+k_4{k}_8{k}_9\right)=\mathrm{0,}\end{array}$ (110)

$\begin{array}{l}{a}_1{k}_7^3+a_2{k}_8^3+a_3{k}_9^3+a_4{k}_7^2{k}_8+a_5{k}_7^2{k}_9+a_6{k}_7{k}_8^2+a_7{k}_8^2{k}_9\\ +a_8{k}_7{k}_9^2+a_9{k}_8{k}_9^2+a_{10}{k}_7{k}_8{k}_9=0.\end{array}$ (111)

Proof. According to $Z_1$ transformation, set $u=u\left(p\mathrm{,}q\mathrm{,}r\right)$ , $p=k_{1}t+k_{2}x+k_{3}y$ , $q=k_{4}t+k_{5}x+k_{6}y$ ,$r=k_{7}t+k_{8}x+k_{9}y$ . $k_1\mathrm{,}{k}_2\mathrm{,}\cdots \mathrm{,}{k}_9$ are undetermined constants, $p\mathrm{,}q$ and $r$ are independent of each other, so

$-k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8-k_1{k}_6{k}_8-k_2{k}_4{k}_9+k_1{k}_5{k}_9\ne \mathrm{0,}$

and

$\begin{array}{l}+a_8\left(k_6^2{k}_7+2k_4{k}_6{k}_9\right)u_q^2{u}_r+a_9\left(k_6^2{k}_8+2k_5{k}_6{k}_9\right)u_q^2{u}_r+a_{10}\left(k_5{k}_6{k}_7+k_4{k}_6{k}_8+k_4{k}_5{k}_9\right)u_q^2{u}_r\\ +\left(3a_1{k}_1{k}_7^2+3a_2{k}_2{k}_8^2+3a_3{k}_3{k}_9^2\right)u_p{u}_r^2+a_4\left(k_2{k}_7^2+2k_1{k}_7{k}_8\right)u_p{u}_r^2+a_5\left(k_3{k}_7^2+2k_1{k}_7{k}_9\right)u_p{u}_r^2\\ +a_6\left(2k_2{k}_7{k}_8+k_1{k}_8^2\right)u_p{u}_r^2+a_7\left(k_3{k}_8^2+2k_2{k}_8{k}_9\right)u_p{u}_r^2+a_8\left(2k_3{k}_7{k}_9+k_1{k}_9^2\right)u_p{u}_r^2\\ +a_9\left(2k_3{k}_8{k}_9+k_2{k}_9^2\right)u_p{u}_r^2+a_{10}\left(k_3{k}_7{k}_8+k_2{k}_7{k}_9+k_1{k}_8{k}_9\right)u_p{u}_r^2\\ +\left(3a_1{k}_4{k}_7^2+3a_2{k}_5{k}_8^2+3a_3{k}_6{k}_9^2\right)u_q{u}_r^2+a_4\left(k_5{k}_7^2+2k_4{k}_7{k}_8\right)u_q{u}_r^2\\ +a_5\left(k_6{k}_7^2+2k_4{k}_7{k}_9\right)u_q{u}_r^2+a_6\left(2k_5{k}_7{k}_8+k_4{k}_8^2\right)u_q{u}_r^2+a_7\left(k_6{k}_8^2+2k_5{k}_8{k}_9\right)u_q{u}_r^2\\ +a_8\left(2k_6{k}_7{k}_9+k_4{k}_9^2\right)u_q{u}_r^2+a_9\left(2k_6{k}_8{k}_9+k_5{k}_9^2\right)u_q{u}_r^2+a_{10}\left(k_6{k}_7{k}_8+k_5{k}_7{k}_9+k_4{k}_8{k}_9\right)u_q{u}_r^2\\ +\left(a_1{k}_7^3+a_2{k}_8^3+a_3{k}_9^3+a_4{k}_7^2{k}_8+a_5{k}_7^2{k}_9+a_6{k}_7{k}_8^2+a_7{k}_8^2{k}_9+a_8{k}_7{k}_9^2+a_9{k}_8{k}_9^2+a_{10}{k}_7{k}_8{k}_9\right)u_r^3\\ =A\left(t,x,y\right).\end{array}$ (112)

Set

$\begin{array}{l}3a_1{k}_1^2{k}_4+3a_2{k}_2^2{k}_5+3a_3{k}_3^2{k}_6+a_4\left(2k_1{k}_2{k}_4+k_1^2{k}_5\right)+a_5\left(2k_1{k}_3{k}_4+k_1^2{k}_6\right)\\ +a_6\left(k_2^2{k}_4+2k_1{k}_2{k}_5\right)+a_7\left(2k_2{k}_3{k}_5+k_2^2{k}_6\right)+a_8\left(k_3^2{k}_4+2k_1{k}_3{k}_6\right)\\ +a_9\left(k_3^2{k}_5+2k_2{k}_3{k}_6\right)+a_{10}\left(k_2{k}_3{k}_4+k_1{k}_3{k}_5+k_1{k}_2{k}_6\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_1{k}_4^2+3a_2{k}_2{k}_5^2+3a_3{k}_3{k}_6^2+a_4\left(k_2{k}_4^2+2k_1{k}_4{k}_5\right)+a_5\left(k_3{k}_4^2+2k_1{k}_4{k}_6\right)\\ +a_6\left(2k_2{k}_4{k}_5+k_1{k}_5^2\right)+a_7\left(k_3{k}_5^2+2k_2{k}_5{k}_6\right)+a_8\left(2k_3{k}_4{k}_6+k_1{k}_6^2\right)\\ +a_9\left(2k_3{k}_5{k}_6+k_2{k}_6^2\right)+a_{10}\left(k_3{k}_4{k}_5+k_2{k}_4{k}_6+k_1{k}_5{k}_6\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}{a}_1{k}_4^3+a_2{k}_5^3+a_3{k}_6^3+a_4{k}_4^2{k}_5+a_5{k}_4^2{k}_6+a_6{k}_4{k}_5^2+a_7{k}_5^2{k}_6\\ +a_8{k}_4{k}_6^2+a_9{k}_5{k}_6^2+a_{10}{k}_4{k}_5{k}_6=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_1^2{k}_7+3a_2{k}_2^2{k}_8+3a_3{k}_3^2{k}_9+a_4\left(2k_1{k}_2{k}_7+k_1^2{k}_8\right)+a_5\left(2k_1{k}_3{k}_7+k_1^2{k}_9\right)\\ +a_6\left(k_2^2{k}_7+2k_1{k}_2{k}_8\right)+a_7\left(2k_2{k}_3{k}_8+k_2^2{k}_9\right)+a_8\left(k_3^2{k}_7+2k_1{k}_3{k}_9\right)\\ +a_9\left(k_3^2{k}_8+2k_2{k}_3{k}_9\right)+a_{10}\left(k_2{k}_3{k}_7+k_1{k}_3{k}_8+k_1{k}_2{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}6a_1{k}_1{k}_4{k}_7+6a_2{k}_2{k}_5{k}_8+6a_3{k}_3{k}_6{k}_9+2a_4\left(k_2{k}_4{k}_7+k_1{k}_5{k}_7+k_1{k}_4{k}_8\right)\\ +2a_5\left(k_3{k}_4{k}_7+k_1{k}_6{k}_7+k_1{k}_4{k}_9\right)+2a_6\left(k_2{k}_5{k}_7+k_2{k}_4{k}_8+k_1{k}_5{k}_8\right)\\ +2a_7\left(k_3{k}_5{k}_8+k_2{k}_6{k}_8+k_2{k}_5{k}_9\right)+2a_8\left(k_3{k}_6{k}_7+k_3{k}_4{k}_9+k_1{k}_6{k}_9\right)\\ +2a_9\left(k_3{k}_6{k}_8+k_3{k}_5{k}_9+k_2{k}_6{k}_9\right)\\ +a_{10}\left(k_3{k}_5{k}_7+k_2{k}_6{k}_7+k_3{k}_4{k}_8+k_1{k}_6{k}_8+k_2{k}_4{k}_9+k_1{k}_5{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_4^2{k}_7+3a_2{k}_5^2{k}_8+3a_3{k}_6^2{k}_9+a_4\left(2k_4{k}_5{k}_7+k_4^2{k}_8\right)+a_5\left(2k_4{k}_6{k}_7+k_4^2{k}_9\right)\\ +a_6\left(k_5^2{k}_7+2k_4{k}_5{k}_8\right)+a_7\left(2k_5{k}_6{k}_8+k_5^2{k}_9\right)+a_8\left(k_6^2{k}_7+2k_4{k}_6{k}_9\right)\\ +a_9\left(k_6^2{k}_8+2k_5{k}_6{k}_9\right)+a_{10}\left(k_5{k}_6{k}_7+k_4{k}_6{k}_8+k_4{k}_5{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_1{k}_7^2+3a_2{k}_2{k}_8^2+3a_3{k}_3{k}_9^2+a_4\left(k_2{k}_7^2+2k_1{k}_7{k}_8\right)+a_5\left(k_3{k}_7^2+2k_1{k}_7{k}_9\right)\\ +a_6\left(2k_2{k}_7{k}_8+k_1{k}_8^2\right)+a_7\left(k_3{k}_8^2+2k_2{k}_8{k}_9\right)+a_8\left(2k_3{k}_7{k}_9+k_1{k}_9^2\right)\\ +a_9\left(2k_3{k}_8{k}_9+k_2{k}_9^2\right)+a_{10}\left(k_3{k}_7{k}_8+k_2{k}_7{k}_9+k_1{k}_8{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}3a_1{k}_4{k}_7^2+3a_2{k}_5{k}_8^2+3a_3{k}_6{k}_9^2+a_4\left(k_5{k}_7^2+2k_4{k}_7{k}_8\right)+a_5\left(k_6{k}_7^2+2k_4{k}_7{k}_9\right)\\ +a_6\left(2k_5{k}_7{k}_8+k_4{k}_8^2\right)+a_7\left(k_6{k}_8^2+2k_5{k}_8{k}_9\right)+a_8\left(2k_6{k}_7{k}_9+k_4{k}_9^2\right)\\ +a_9\left(2k_6{k}_8{k}_9+k_5{k}_9^2\right)+a_{10}\left(k_6{k}_7{k}_8+k_5{k}_7{k}_9+k_4{k}_8{k}_9\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}{a}_1{k}_7^3+a_2{k}_8^3+a_3{k}_9^3+a_4{k}_7^2{k}_8+a_5{k}_7^2{k}_9+a_6{k}_7{k}_8^2+a_7{k}_8^2{k}_9\\ +a_8{k}_7{k}_9^2+a_9{k}_8{k}_9^2+a_{10}{k}_7{k}_8{k}_9=0.\end{array}$

We get

$\begin{array}{l}{a}_1{u}_t^3+a_2{u}_x^3+a_3{u}_y^3+a_4{u}_t^2{u}_x+a_5{u}_t^2{u}_y+a_6{u}_x^2{u}_t\\ +a_7{u}_x^2{u}_y+a_8{u}_y^2{u}_t+a_9{u}_y^2{u}_x+a_{10}{u}_t{u}_x{u}_y\\ =(a_1{k}_1^3+a_2{k}_2^3+a_3{k}_3^3+a_4{k}_1^2{k}_2+a_5{k}_1^2{k}_3+a_6{k}_1{k}_2^2\\ +a_7{k}_2^2{k}_3+a_8{k}_1{k}_3^2+a_9{k}_2{k}_3^2+a_{10}{k}_1{k}_2{k}_3)u_p^3\\ =A\left(t\mathrm{,}x\mathrm{,}y\right)\mathrm{.}\end{array}$

Namely

$u_p=\sqrt[3]{\frac{A\left(t\mathrm{,}x\mathrm{,}y\right)}{B_1}}\mathrm{,}$

$\begin{array}{c}{B}_1=a_1{k}_1^3+a_2{k}_2^3+a_3{k}_3^3+a_4{k}_1^2{k}_2+a_5{k}_1^2{k}_3+a_6{k}_1{k}_2^2{u}_p^3\\ +a_7{k}_2^2{k}_3+a_8{k}_1{k}_3^2{u}_p^3+a_9{k}_2{k}_3^2+a_{10}{k}_1{k}_2{k}_3\mathrm{.}\end{array}$

So the general solution of (100) is

$u=f\left(q\mathrm{,}r\right)+{\displaystyle \int \sqrt[3]{\frac{A\left(t\mathrm{,}x\mathrm{,}y\right)}{B_1}}\text{d}p}\mathrm{.}$

The theorem is proved. $\square$

Theorem 12. In ${ℝ}^2$ , if

$a_1{u}_t^4+a_2{u}_x^4+a_3{u}_t^3{u}_x+a_4{u}_t^2{u}_x^2+a_5{u}_t{u}_x^3=A\left(t\mathrm{,}x\right)\mathrm{,}$ (113)

where $a_i$ are any known constants ($1\le i\le 5$ ), then the general solution of Equation (113) is

$u=f\left(q\right)+\underset{}{{\displaystyle {\displaystyle \int }}}\sqrt[4]{\frac{A\left(p\mathrm{,}q\right)}{a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^3{k}_2+a_4{k}_1^2{k}_2^2+a_5{k}_1{k}_2^3}}\text{d}p\mathrm{,}$ (114)

where $f$ is an arbitrary first differentiable function, and

$p=k_{1}t+k_{2}x,q=k_{3}t+k_{4}x,$

the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_4$ need satisfy

$k_1{k}_4-k_2{k}_3\ne \mathrm{0,}$

$\begin{array}{l}4a_1{k}_1^3{k}_3+4a_2{k}_2^3{k}_4+a_3\left(3k_1^2{k}_2{k}_3+k_1^3{k}_4\right)+a_4\left(2k_1{k}_2^2{k}_3+2k_1^2{k}_2{k}_4\right)\\ +a_5\left(k_2^3{k}_3+3k_1{k}_2^2{k}_4\right)=\mathrm{0,}\end{array}$ (115)

$\begin{array}{l}6a_1{k}_1^2{k}_3^2+6a_2{k}_2^2{k}_4^2+a_3\left(3k_1{k}_2{k}_3^2+3k_1^2{k}_3{k}_4\right)+a_4\left(k_2^2{k}_3^2+4k_1{k}_2{k}_3{k}_4+k_1^2{k}_4^2\right)\\ +a_5\left(3k_2^2{k}_3{k}_4+3k_1{k}_2{k}_4^2\right)=\mathrm{0,}\end{array}$ (116)

$\begin{array}{l}4a_1{k}_1{k}_3^3+4a_2{k}_2{k}_4^3+a_3\left(k_2{k}_3^3+3k_1{k}_3^2{k}_4\right)+a_4\left(2k_2{k}_3^2{k}_4+2k_1{k}_3{k}_4^2\right)\\ +a_5\left(3k_2{k}_3{k}_4^2+k_1{k}_4^3\right)=\mathrm{0,}\end{array}$ (117)

$a_1{k}_3^4+a_2{k}_4^4+a_3{k}_3^3{k}_4+a_4{k}_3^2{k}_4^2+a_5{k}_3{k}_4^3=0.$ (118)

Proof. By $Z_1$ transformation, set

$u\left(t\mathrm{,}x\right)=u\left(p\mathrm{,}q\right)\mathrm{,}$

$p=k_{1}t+k_{2}x,q=k_{3}t+k_{4}x,$

and

$t=\frac{p{k}_4-q{k}_2}{k_1{k}_4-k_2{k}_3},x=\frac{q{k}_1-p{k}_3}{k_1{k}_4-k_2{k}_3},$

$k_1{k}_4-k_2{k}_3\ne 0.$

Then

(119)

Set

$\begin{array}{l}4a_1{k}_1^3{k}_3+4a_2{k}_2^3{k}_4+a_3\left(3k_1^2{k}_2{k}_3+k_1^3{k}_4\right)+a_4\left(2k_1{k}_2^2{k}_3+2k_1^2{k}_2{k}_4\right)\\ +a_5\left(k_2^3{k}_3+3k_1{k}_2^2{k}_4\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}6a_1{k}_1^2{k}_3^2+6a_2{k}_2^2{k}_4^2+a_3\left(3k_1{k}_2{k}_3^2+3k_1^2{k}_3{k}_4\right)+a_4\left(k_2^2{k}_3^2+4k_1{k}_2{k}_3{k}_4+k_1^2{k}_4^2\right)\\ +a_5\left(3k_2^2{k}_3{k}_4+3k_1{k}_2{k}_4^2\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}4a_1{k}_1{k}_3^3+4a_2{k}_2{k}_4^3+a_3\left(k_2{k}_3^3+3k_1{k}_3^2{k}_4\right)+a_4\left(2k_2{k}_3^2{k}_4+2k_1{k}_3{k}_4^2\right)\\ +a_5\left(3k_2{k}_3{k}_4^2+k_1{k}_4^3\right)=\mathrm{0,}\end{array}$

$a_1{k}_3^4+a_2{k}_4^4+a_3{k}_3^3{k}_4+a_4{k}_3^2{k}_4^2+a_5{k}_3{k}_4^3=0.$

So

$\begin{array}{l}{a}_1{u}_t^4+a_2{u}_x^4+a_3{u}_t^3{u}_x+a_4{u}_t^2{u}_x^2+a_5{u}_t{u}_x^3\\ =\left(a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^3{k}_2+a_4{k}_1^2{k}_2^2+a_5{k}_1{k}_2^3\right)u_p^4=A\left(p\mathrm{,}q\right)\mathrm{.}\end{array}$ (120)

So the general solution of (113) is

$u=f\left(q\right)+{\displaystyle \int \sqrt[4]{\frac{A\left(p\mathrm{,}q\right)}{a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^3{k}_2+a_4{k}_1^2{k}_2^2+a_5{k}_1{k}_2^3}}\text{d}p}\mathrm{.}$

The theorem is proved. $\square$

In the following, we use $Z_1$ transformation to study the general solutions of some second-order nonlinear partial differential equations.

Theorem 13. In ${ℝ}^2$ , if

$a_1{u}_{tt}^2+a_2{u}_{xx}^2+a_3{u}_{tx}^2+a_4{u}_{tt}{u}_{xx}+a_5{u}_{tt}{u}_{tx}+a_6{u}_{xx}{u}_{tx}=A\left(t\mathrm{,}x\right)\mathrm{,}$ (121)

where $a_i$ are any known constants ($1\le i\le 6$ ), then the general solution of Equation (121) is

$u=f\left(q\right)+pg\left(q\right)+{\displaystyle \iint \sqrt{B_1^{-1}A\left(p\mathrm{,}q\right)}\text{d}p\text{d}p}\mathrm{,}$ (122)

where $f$ and $g$ are arbitrary second differentiable functions, and

$p=k_{1}t+k_{2}x,q=k_{3}t+k_{4}x,$

$k_1{k}_4-k_2{k}_3\ne \mathrm{0,}$

$B_1=a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^2{k}_2^2+a_4{k}_1^2{k}_2^2+a_5{k}_1^3{k}_2+a_6{k}_1{k}_2^3\mathrm{,}$ (123)

the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_4$ need satisfy

$a_1{k}_3^4+a_2{k}_4^4+a_3{k}_3^2{k}_4^2+a_4{k}_3^2{k}_4^2+a_5{k}_3^3{k}_4+a_6{k}_3{k}_4^3=\mathrm{0,}$ (124)

$\begin{array}{l}4a_1{k}_1^2{k}_3^2+4a_2{k}_2^2{k}_4^2+a_3\left(k_1^2{k}_4^2+2k_1{k}_2{k}_3{k}_4+k_2^2{k}_3^2\right)+4a_4{k}_1{k}_2{k}_3{k}_4\\ +2a_5{k}_1{k}_3\left(k_1{k}_4+k_2{k}_3\right)+2a_6{k}_2{k}_4\left(k_1{k}_4+k_2{k}_3\right)=\mathrm{0,}\end{array}$ (125)

$\begin{array}{l}2a_1{k}_1^2{k}_3^2+2a_2{k}_2^2{k}_4^2+2a_3{k}_1{k}_2{k}_3{k}_4+a_4\left(k_1^2{k}_4^2+k_2^2{k}_3^2\right)\\ +a_5\left(k_1{k}_2{k}_3^2+k_1^2{k}_3{k}_4\right)+a_6\left(k_1{k}_2{k}_4^2+k_2^2{k}_3{k}_4\right)=\mathrm{0,}\end{array}$ (126)

$\begin{array}{l}4a_1{k}_1^3{k}_3+4a_2{k}_2^3{k}_4+2a_3{k}_1{k}_2\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_1^2{k}_2{k}_4+k_2^2{k}_1{k}_3\right)\\ +a_5\left(2k_1^2{k}_2{k}_3+k_1^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_1{k}_2^2{k}_4+k_2^2\left(k_1{k}_4+k_2{k}_3\right)\right)=\mathrm{0,}\end{array}$ (127)

$\begin{array}{l}4a_1{k}_1{k}_3^3+4a_2{k}_2{k}_4^3+2a_3{k}_3{k}_4\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_3^2{k}_2{k}_4+k_4^2{k}_1{k}_3\right)\\ +a_5\left(2k_1{k}_3^2{k}_4+k_3^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_2{k}_3{k}_4^2+k_4^2\left(k_1{k}_4+k_2{k}_3\right)\right)=0.\end{array}$ (128)

Proof. By $Z_1$ transformation, set $u=f\left(p\mathrm{,}q\right)$ , $p=k_{1}t+k_{2}x$ , $q=k_{3}t+k_{4}x$ , and $k_1{k}_4-k_2{k}_3\ne 0$ , so

$\begin{array}{l}+2a_3{k}_3{k}_4\left(k_1{k}_4+k_2{k}_3\right)f_{qq}{f}_{pq}+2a_4\left(k_3^2{k}_2{k}_4+k_4^2{k}_1{k}_3\right)f_{qq}{f}_{pq}\\ +a_5\left(2k_1{k}_3^2{k}_4+k_3^2\left(k_1{k}_4+k_2{k}_3\right)\right)f_{qq}{f}_{pq}+a_6\left(2k_2{k}_3{k}_4^2+k_4^2\left(k_1{k}_4+k_2{k}_3\right)\right)f_{qq}{f}_{pq}\\ =A\left(p\mathrm{,}q\right)\mathrm{.}\end{array}$ (129)

Set

$a_1{k}_3^4+a_2{k}_4^4+a_3{k}_3^2{k}_4^2+a_4{k}_3^2{k}_4^2+a_5{k}_3^3{k}_4+a_6{k}_3{k}_4^3=\mathrm{0,}$

$\begin{array}{l}4a_1{k}_1^2{k}_3^2+4a_2{k}_2^2{k}_4^2+a_3\left(k_1^2{k}_4^2+2k_1{k}_2{k}_3{k}_4+k_2^2{k}_3^2\right)+4a_4{k}_1{k}_2{k}_3{k}_4\\ +2a_5{k}_1{k}_3\left(k_1{k}_4+k_2{k}_3\right)+2a_6{k}_2{k}_4\left(k_1{k}_4+k_2{k}_3\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}2a_1{k}_1^2{k}_3^2+2a_2{k}_2^2{k}_4^2+2a_3{k}_1{k}_2{k}_3{k}_4+a_4\left(k_1^2{k}_4^2+k_2^2{k}_3^2\right)\\ +a_5\left(k_1{k}_2{k}_3^2+k_1^2{k}_3{k}_4\right)+a_6\left(k_1{k}_2{k}_4^2+k_2^2{k}_3{k}_4\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}4a_1{k}_1^3{k}_3+4a_2{k}_2^3{k}_4+2a_3{k}_1{k}_2\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_1^2{k}_2{k}_4+k_2^2{k}_1{k}_3\right)\\ +a_5\left(2k_1^2{k}_2{k}_3+k_1^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_1{k}_2^2{k}_4+k_2^2\left(k_1{k}_4+k_2{k}_3\right)\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}4a_1{k}_1{k}_3^3+4a_2{k}_2{k}_4^3+2a_3{k}_3{k}_4\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_3^2{k}_2{k}_4+k_4^2{k}_1{k}_3\right)\\ +a_5\left(2k_1{k}_3^2{k}_4+k_3^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_2{k}_3{k}_4^2+k_4^2\left(k_1{k}_4+k_2{k}_3\right)\right)=0.\end{array}$

Then

$\begin{array}{l}{a}_1{u}_{tt}^2+a_2{u}_{xx}^2+a_3{u}_{tx}^2+a_4{u}_{tt}{u}_{xx}+a_5{u}_{tt}{u}_{tx}+a_6{u}_{xx}{u}_{tx}\\ =\left(a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^2{k}_2^2+a_4{k}_1^2{k}_2^2+a_5{k}_1^3{k}_2+a_6{k}_1{k}_2^3\right)u_{pp}^2=A\left(p\mathrm{,}q\right)\mathrm{.}\end{array}$ (130)

Namely

$u_{pp}=\sqrt{B_1^{-1}A\left(p\mathrm{,}q\right)}\mathrm{,}$

$B_1=a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^2{k}_2^2+a_4{k}_1^2{k}_2^2+a_5{k}_1^3{k}_2+a_6{k}_1{k}_2^3\mathrm{.}$

So the general solution of (121) is

$u=f\left(q\right)+pg\left(q\right)+{\displaystyle \iint \sqrt{B_1^{-1}A\left(p\mathrm{,}q\right)}\text{d}p\text{d}p}\mathrm{.}$

The theorem is proved. $\square$

A similar proof of Theorem 13 leads us to Theorem 14.

Theorem 14. In ${ℝ}^2$ , if

$a_1{u}_{tt}^2+a_2{u}_{xx}^2+a_3{u}_{tx}^2+a_4{u}_{tt}{u}_{xx}+a_5{u}_{tt}{u}_{tx}+a_6{u}_{xx}{u}_{tx}=A\left(t\mathrm{,}x\right)\mathrm{,}$

the general solution of Equation (131) is

$u=f\left(p\right)+g\left(q\right)+{\displaystyle \iint \sqrt{B_2^{-1}A\left(p\mathrm{,}q\right)}\text{d}p\text{d}q}\mathrm{,}$ (131)

where $f$ and $g$ are arbitrary second differentiable functions, and

$p=k_{1}t+k_{2}x\mathrm{,}q=k_{3}t+k_{4}x\mathrm{,}$

$k_1{k}_4-k_2{k}_3\ne \mathrm{0,}$

$\begin{array}{c}{B}_2=4a_1{k}_1^2{k}_3^2+4a_2{k}_2^2{k}_4^2+a_3\left(k_1^2{k}_4^2+2k_1{k}_2{k}_3{k}_4+k_2^2{k}_3^2\right)+4a_4{k}_1{k}_2{k}_3{k}_4\\ +2a_5{k}_1{k}_3\left(k_1{k}_4+k_2{k}_3\right)+2a_6{k}_2{k}_4\left(k_1{k}_4+k_2{k}_3\right)\mathrm{,}\end{array}$

the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_4$ need satisfy

$a_1{k}_1^4+a_2{k}_2^4+a_3{k}_1^2{k}_2^2+a_4{k}_1^2{k}_2^2+a_5{k}_1^3{k}_2+a_6{k}_1{k}_2^3=\mathrm{0,}$

$a_1{k}_3^4+a_2{k}_4^4+a_3{k}_3^2{k}_4^2+a_4{k}_3^2{k}_4^2+a_5{k}_3^3{k}_4+a_6{k}_3{k}_4^3=\mathrm{0,}$

$\begin{array}{l}2a_1{k}_1^2{k}_3^2+2a_2{k}_2^2{k}_4^2+2a_3{k}_1{k}_2{k}_3{k}_4+a_4\left(k_1^2{k}_4^2+k_2^2{k}_3^2\right)\\ +a_5\left(k_1{k}_2{k}_3^2+k_1^2{k}_3{k}_4\right)+a_6\left(k_1{k}_2{k}_4^2+k_2^2{k}_3{k}_4\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}4a_1{k}_1^3{k}_3+4a_2{k}_2^3{k}_4+2a_3{k}_1{k}_2\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_1^2{k}_2{k}_4+k_2^2{k}_1{k}_3\right)\\ +a_5\left(2k_1^2{k}_2{k}_3+k_1^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_1{k}_2^2{k}_4+k_2^2\left(k_1{k}_4+k_2{k}_3\right)\right)=\mathrm{0,}\end{array}$

$\begin{array}{l}4a_1{k}_1{k}_3^3+4a_2{k}_2{k}_4^3+2a_3{k}_3{k}_4\left(k_1{k}_4+k_2{k}_3\right)+2a_4\left(k_3^2{k}_2{k}_4+k_4^2{k}_1{k}_3\right)\\ +a_5\left(2k_1{k}_3^2{k}_4+k_3^2\left(k_1{k}_4+k_2{k}_3\right)\right)+a_6\left(2k_2{k}_3{k}_4^2+k_4^2\left(k_1{k}_4+k_2{k}_3\right)\right)=0.\end{array}$

Next, we use Theorem 14 to analyze a definite solution problem.

Example 10. In ${ℝ}^2$ , use Theorem 10 to obtain the analytical solution of

$u_{tt}^2+6u_{xx}^2-7u_{tt}{u}_{xx}+u_{tt}{u}_{tx}-u_{xx}{u}_{tx}=\mathrm{0,}$ (132)

in the conditions of $u\left(\mathrm{0,}x\right)=\Phi \left(x\right)$ and $u_t\left(\mathrm{0,}x\right)=\Psi \left(x\right)$ . $\Phi$ is an arbitrary known second differentiable function, $\Psi$ is an arbitrary known first differentiable function.

Solution. According to Theorem 14, the general solution of (132) is

$u=f\left(t+x\right)+g\left(t-x\right)\mathrm{,}$ (133)

or

$u=f\left(2t+x\right)+g\left(3t-x\right)\mathrm{.}$ (134)

If $u=f\left(t+x\right)+g\left(t-x\right)$ , then

$u\left(\mathrm{0,}x\right)=f\left(x\right)+g\left(-x\right)=\Phi \left(x\right)\mathrm{,}$

$u_t\left(\mathrm{0,}x\right)=f_p\left(x\right)+g_q\left(-x\right)=\Psi \left(x\right)\mathrm{.}$

When $t=0,p=x,q=-x$ , so

$f_p\left(x\right)-g_q\left(-x\right)={\Phi }^{\prime }\left(x\right)\mathrm{.}$

Therefore

$2f_p\left(x\right)=\Psi \left(x\right)+{\Phi }^{\prime }\left(x\right)\mathrm{,}$

$f\left(x\right)=\frac{1}{2}\left(\Phi \left(x\right)+{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}$

$\begin{array}{c}g\left(-x\right)=\Phi \left(x\right)-f\left(x\right)=\Phi \left(x\right)-\frac{1}{2}\left(\Phi \left(x\right)+{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\right)\\ =\frac{1}{2}\left(\Phi \left(x\right)-{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}\end{array}$

$g\left(x\right)=\frac{1}{2}\left(\Phi \left(-x\right)-{\displaystyle {\int }_{x_0}^{-x}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{.}$

Whereupon

$f\left(t+x\right)=\frac{1}{2}\left(\Phi \left(t+x\right)+{\displaystyle {\int }_{x_0}^{t+x}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}$

$g\left(t-x\right)=\frac{1}{2}\left(\Phi \left(-\left(t-x\right)\right)-{\displaystyle {\int }_{x_0}^{-\left(t-x\right)}\psi \left(\xi \right)\text{d}\xi }\right)=\frac{1}{2}\left(\Phi \left(x-t\right)-{\displaystyle {\int }_{x_0}^{x-t}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}$

$\begin{array}{c}u=f\left(t+x\right)+g\left(t-x\right)\\ =\frac{1}{2}\left(\Phi \left(t+x\right)+{\displaystyle {\int }_{x_0}^{t+x}\psi \left(\xi \right)\text{d}\xi }\right)+\frac{1}{2}\left(\Phi \left(x-t\right)-{\displaystyle {\int }_{x_0}^{x-t}\psi \left(\xi \right)\text{d}\xi }\right)\\ =\frac{1}{2}\left(\Phi \left(t+x\right)+\Phi \left(x-t\right)+{\displaystyle {\int }_{x-t}^{t+x}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{.}\end{array}$

So the analytical solution of the definite solution problem is

$u=\frac{1}{2}\left(\Phi \left(t+x\right)+\Phi \left(x-t\right)+{\displaystyle {\int }_{x-t}^{t+x}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{.}$ (135)

If $u=f\left(2t+x\right)+g\left(3t-x\right)$ , then

$u\left(\mathrm{0,}x\right)=f\left(x\right)+g\left(-x\right)=\Phi \left(x\right)\mathrm{,}$

$u_t\left(\mathrm{0,}x\right)=2f_p\left(x\right)+3g_q\left(-x\right)=\Psi \left(x\right)\mathrm{.}$

When $t=\mathrm{0,}p=x\mathrm{,}q=-x$ , then

$f_p\left(x\right)-g_q\left(-x\right)={\Phi }^{\prime }\left(x\right)\mathrm{.}$

Thereupon

$5f_p\left(x\right)=\Psi \left(x\right)+3{\Phi }^{\prime }\left(x\right)\mathrm{,}$

$f\left(x\right)=\frac{1}{5}\left(3\Phi \left(x\right)+{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}$

$\begin{array}{c}g\left(-x\right)=\Phi \left(x\right)-f\left(x\right)=\Phi \left(x\right)-\frac{1}{5}\left(3\Phi \left(x\right)+{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\right)\\ =\frac{2}{5}\Phi \left(x\right)-\frac{1}{5}{\displaystyle {\int }_{x_0}^x\psi \left(\xi \right)\text{d}\xi }\mathrm{,}\end{array}$

$g\left(x\right)=\frac{2}{5}\Phi \left(-x\right)-\frac{1}{5}{\displaystyle {\int }_{x_0}^{-x}\psi \left(\xi \right)\text{d}\xi }\mathrm{.}$

So

$f\left(2t+x\right)=\frac{1}{5}\left(3\Phi \left(2t+x\right)+{\displaystyle {\int }_{x_0}^{2t+x}\psi \left(\xi \right)\text{d}\xi }\right)\mathrm{,}$

$\begin{array}{c}g\left(3t-x\right)=\frac{2}{5}\Phi \left(-\left(3t-x\right)\right)-\frac{1}{5}{\displaystyle {\int }_{x_0}^{-\left(3t-x\right)}\psi \left(\xi \right)\text{d}\xi }\\ =\frac{2}{5}\Phi \left(x-3t\right)-\frac{1}{5}{\displaystyle {\int }_{x_0}^{x-3t}\psi \left(\xi \right)\text{d}\xi }\mathrm{,}\end{array}$

$\begin{array}{c}u=f\left(2t+x\right)+g\left(3t-x\right)\\ =\frac{3}{5}\Phi \left(2t+x\right)+\frac{1}{5}{\displaystyle {\int }_{x_0}^{2t+x}\psi \left(\xi \right)\text{d}\xi }+\frac{2}{5}\Phi \left(x-3t\right)-\frac{1}{5}{\displaystyle {\int }_{x_0}^{x-3t}\psi \left(\xi \right)\text{d}\xi }\\ =\frac{3}{5}\Phi \left(2t+x\right)+\frac{2}{5}\Phi \left(x-3t\right)+\frac{1}{5}{\displaystyle {\int }_{x-3t}^{2t+x}\psi \left(\xi \right)\text{d}\xi }\mathrm{.}\end{array}$

Then the analytical solution of the definite solution problem is

$u=\frac{3}{5}\Phi \left(2t+x\right)+\frac{2}{5}\Phi \left(x-3t\right)+\frac{1}{5}{\displaystyle {\int }_{x-3t}^{2t+x}\psi \left(\xi \right)\text{d}\xi }\mathrm{.}$ (136)

Example 10 illustrates that the analytical solution to this definite solution problem is not unique. If one needs to determine exactly which analytical solution is the case, more conditions are needed. For example, if

$u\left(t\mathrm{,0}\right)=\frac{1}{2}\left(\Phi \left(t\right)+\Phi \left(-t\right)+{\displaystyle {\int }_{-t}^t\psi \left(\xi \right)\text{d}\xi }\right)$ is also known, then the analytical solution of this definite solution problem can be determined as $u=\frac{1}{2}\left(\Phi \left(t+x\right)+\Phi \left(x-t\right)+{\displaystyle {\int }_{x-t}^{t+x}\psi \left(\xi \right)\text{d}\xi }\right)$ .

Theorem 15. In ${ℝ}^2$ , if

$\begin{array}{l}{a}_1{u}_{ttt}^2+a_2{u}_{xxx}^2+a_3{u}_{ttx}^2+a_4{u}_{txx}^2+a_5{u}_{ttt}{u}_{xxx}+a_6{u}_{ttt}{u}_{ttx}+a_7{u}_{ttt}{u}_{txx}\\ +a_8{u}_{xxx}{u}_{ttx}+a_9{u}_{xxx}{u}_{txx}+a_{10}{u}_{ttx}{u}_{txx}=A\left(t\mathrm{,}x\right)\mathrm{,}\end{array}$ (137)

where $a_i$ are any known constants ($1\le i\le 10$ ), then the general solution of Equation (137) is

$u=f\left(q\right)+pg\left(q\right)+p^{2}h\left(q\right)+{\displaystyle \iiint \sqrt{B_1^{-1}A\left(p\mathrm{,}q\right)}\text{d}{p}^3}\mathrm{,}$ (138)

where $f\mathrm{,}g\mathrm{,}h$ are arbitrary third differentiable functions, and

$p=k_{1}t+k_{2}x,q=k_{3}t+k_{4}x,$

$k_1{k}_4-k_2{k}_3\ne \mathrm{0,}$

$\begin{array}{c}{B}_1=a_1{k}_1^6+a_2{k}_2^6+a_3{k}_1^4{k}_2^2+a_4{k}_1^2{k}_2^4+a_5{k}_1^3{k}_2^3+a_6{k}_1^5{k}_2\\ +a_7{k}_1^4{k}_2^2+a_8{k}_1^2{k}_2^4+a_9{k}_1{k}_2^5+a_{10}{k}_1^3{k}_2^3\mathrm{,}\end{array}$ (139)

the constants $k_1\mathrm{,}{k}_2\mathrm{,}{k}_3\mathrm{,}{k}_4$ need satisfy

$\begin{array}{l}{a}_1{k}_3^6+a_2{k}_4^6+a_3{k}_3^4{k}_4^2+a_4{k}_3^2{k}_4^4+a_5{k}_3^3{k}_4^3+a_6{k}_3^5{k}_4\\ +a_7{k}_3^4{k}_4^2+a_8{k}_3^2{k}_4^4+a_9{k}_3{k}_4^5+a_{10}{k}_3^3{k}_4^3=\mathrm{0,}\end{array}$ (140)

$\begin{array}{l}6a_1{k}_1^5{k}_3+6a_2{k}_2^5{k}_4+4a_3{k}_1^3{k}_2^2{k}_3+2a_3{k}_1^4{k}_2{k}_4+2a_4{k}_1{k}_2^4{k}_3+4a_4{k}_1^2{k}_2^3+3a_5{k}_1^2{k}_2^3{k}_3\\ +3a_5{k}_1^3{k}_2^2{k}_4+5a_6{k}_1^4{k}_2{k}_3+a_6{k}_1^5{k}_4+4a_7{k}_1^3{k}_2^2{k}_3+2a_7{k}_1^4{k}_2{k}_4+2a_8{k}_1{k}_2^4{k}_3\\ +4a_8{k}_1^2{k}_2^3{k}_4+a_9{k}_2^5{k}_3{u}_{p^3}{u}_{p^{2}q}+5a_9{k}_1{k}_2^4{k}_4+3a_{10}{k}_1^2{k}_2^3{k}_3+3a_{10}{k}_1^3{k}_2^2{k}_4=0,\end{array}$ (141)

$\begin{array}{l}9a_1{k}_1^4{k}_3^2+9a_2{k}_2^4{k}_4^2+4a_3{k}_1^2{k}_2^2{k}_3^2+4a_3{k}_1^3{k}_2{k}_3{k}_4+a_3{k}_1^4{k}_4^2+a_4{k}_2^4{k}_3^2\\ +4a_4{k}_1{k}_2^3{k}_3{k}_4+4a_4{k}_1^2{k}_2^2{k}_4^2+9a_5{k}_1^2{k}_2^2{k}_3{k}_4+6a_6{k}_1^3{k}_2{k}_3^2+3a_6{k}_1^4{k}_3{k}_4\\ +3a_7{k}_1^2{k}_2^2{k}_3^2+6a_7{k}_1^3{k}_2{k}_3{k}_4+6a_8{k}_1{k}_2^3{k}_3{k}_4+3a_8{k}_1^2{k}_2^2{k}_4^2+3a_9{k}_2^4{k}_3{k}_4\\ +6a_9{k}_1{k}_2^3{k}_4^2+2a_{10}{k}_1{k}_2^3{k}_3^2+5a_{10}{k}_1^2{k}_2^2{k}_3{k}_4+2a_{10}{k}_1^3{k}_2{k}_4^2=0,\end{array}$ (142)

$\begin{array}{l}6a_1{k}_1^4{k}_3^2+6a_2{k}_2^4{k}_4^2+2a_3{k}_1^2{k}_2^2{k}_3^2+4a_3{k}_1^3{k}_2{k}_3{k}_4+4a_4{k}_1{k}_2^3{k}_3{k}_4+2a_4{k}_1^2{k}_2^2{k}_4^2\\ +3a_5{k}_1{k}_2^3{k}_3^2+3a_5{k}_1^3{k}_2{k}_4^2+4a_6{k}_1^3{k}_2{k}_3^2+2a_6{k}_1^4{k}_3{k}_4+3a_7{k}_1^2{k}_2^2{k}_3^2+2a_7{k}_1^3{k}_2{k}_3{k}_4\\ +a_7{k}_1^4{k}_4^2+a_8{k}_2^4{k}_3^2+2a_8{k}_1{k}_2^3{k}_3{k}_4+3a_8{k}_1^2{k}_2^2{k}_4^2+2a_9{k}_2^4{k}_3{k}_4{u}_{p^3}{u}_{p{q}^2}+4a_9{k}_1{k}_2^3{k}_4^2\\ +a_{10}{k}_1{k}_2^3{k}_3^2+4a_{10}{k}_1^2{k}_2^2{k}_3{k}_4+a_{10}{k}_1^3{k}_2{k}_4^2=0,\end{array}$ (143)

$\begin{array}{l}18a_1{k}_1^3{k}_3^3+18a_2{k}_2^3{k}_4^3+4a_3{k}_1{k}_2^2{k}_3^3+10a_3{k}_1^2{k}_2{k}_3^2{k}_4+4a_3{k}_1^3{k}_3{k}_4^2+4a_4{k}_2^3{k}_3^2{k}_4\\ +10a_4{k}_1{k}_2^2{k}_3{k}_4^2+4a_4{k}_1^2{k}_2{k}_4^3+9a_5{k}_1{k}_2^2{k}_3^2{k}_4+9a_5{k}_1^2{k}_2{k}_3{k}_4^2+9a_6{k}_1^2{k}_2{k}_3^3\\ +9a_6{k}_1^3{k}_3^2{k}_4+3a_7{k}_1{k}_2^2{k}_3^3+12a_7{k}_1^2{k}_2{k}_3^2{k}_4+3a_7{k}_1^3{k}_3{k}_4^2+3a_8{k}_2^3{k}_3^2{k}_4\\ +12a_8{k}_1{k}_2^2{k}_3{k}_4^2+3a_8{k}_1^2{k}_2{k}_4^3+9a_9{k}_2^3{k}_3{k}_4^2+9a_9{k}_1{k}_2^2{k}_4^3+a_{10}{k}_2^3{k}_3^3\\ +8a_{10}{k}_1{k}_2^2{k}_3^2{k}_4+8a_{10}{k}_1^2{k}_2{k}_3{k}_4^2+a_{10}{k}_1^3{k}_4^3=0,\end{array}$ (144)

$\begin{array}{l}9a_1{k}_1^2{k}_3^4+9a_2{k}_2^2{k}_4^4+a_3{k}_2^2{k}_3^4+4a_3{k}_1{k}_2{k}_3^3{k}_4+4a_3{k}_1^2{k}_3^2{k}_4^2+4a_4{k}_2^2{k}_3^2{k}_4^2\\ +4a_4{k}_1{k}_2{k}_3{k}_4^3+a_4{k}_1^2{k}_4^4+9a_5{k}_1{k}_2{k}_3^2{k}_4^2+3a_6{k}_1{k}_2{k}_3^4+6a_6{k}_1^2{k}_3^3{k}_4+6a_7{k}_1{k}_2{k}_3^3{k}_4\\ +3a_7{k}_1^2{k}_3^2{k}_4^2+3a_8{k}_2^2{k}_3^2{k}_4^2+6a_8{k}_1{k}_2{k}_3{k}_4^3+6a_9{k}_2^2{k}_3{k}_4^3+3a_9{k}_1{k}_2{k}_4^4+2a_{10}{k}_2^2{k}_3^3{k}_4\\ +5a_{10}{k}_1{k}_2{k}_3^2{k}_4^2+2a_{10}{k}_1^2{k}_3{k}_4^3=0,\end{array}$ (145)

$\begin{array}{l}2a_1{k}_1^3{k}_3^3+2a_2{k}_2^3{k}_4^3+2a_3{k}_1^2{k}_2{k}_3^2{k}_4+2a_4{k}_1{k}_2^2{k}_3{k}_4^2+a_5{k}_2^3{k}_3^3+a_5{k}_1^3{k}_4^3+a_6{k}_1^2{k}_2{k}_3^3\\ +a_6{k}_1^3{k}_3^2{k}_4+a_7{k}_1{k}_2^2{k}_3^3+a_7{k}_1^3{k}_3{k}_4^2+a_8{k}_2^3{k}_3^2{k}_4+a_8{k}_1^2{k}_2{k}_4^3+a_9{k}_2^3{k}_3{k}_4^2+a_9{k}_1{k}_2^2{k}_4^3\\ +a_{10}{k}_1{k}_2^2{k}_3^2{k}_4+a_{10}{k}_1^2{k}_2{k}_3{k}_4^2=\mathrm{0,}\end{array}$ (146)

$\begin{array}{l}6a_1{k}_1^2{k}_3^4+6a_2{k}_2^2{k}_4^4+4a_3{k}_1{k}_2{k}_3^3{k}_4+2a_3{k}_1^2{k}_3^2{k}_4^2+2a_4{k}_2^2{k}_3^2{k}_4^2+4a_4{k}_1{k}_2{k}_3{k}_4^3\\ +3a_5{k}_2^2{k}_3^3{k}_4+3a_5{k}_1^2{k}_3{k}_4^3+2a_6{k}_1{k}_2{k}_3^4+4a_6{k}_1^2{k}_3^3{k}_4+a_7{k}_2^2{k}_3^4+2a_7{k}_1{k}_2{k}_3^3{k}_4\\ +3a_7{k}_1^2{k}_3^2{k}_4^2+3a_8{k}_2^2{k}_3^2{k}_4^2+2a_8{k}_1{k}_2{k}_3{k}_4^3+a_8{k}_1^2{k}_4^4+4a_9{k}_2^2{k}_3{k}_4^3+2a_9{k}_1{k}_2{k}_4^4\\ +a_{10}{k}_2^2{k}_3^3{k}_4+4a_{10}{k}_1{k}_2{k}_3^2{k}_4^2+a_{10}{k}_1^2{k}_3{k}_4^3=\mathrm{0,}\end{array}$ (147)

$\begin{array}{l}6a_1{k}_1{k}_3^5+6a_2{k}_2{k}_4^5+2a_3{k}_2{k}_3^4{k}_4+4a_3{k}_1{k}_3^3{k}_4^2+4a_4{k}_2{k}_3^2{k}_4^3+2a_4{k}_1{k}_3{k}_4^4\\ +3a_5{k}_2{k}_3^3{k}_4^2+3a_5{k}_1{k}_3^2{k}_4^3+a_6{k}_2{k}_3^5+5a_6{k}_1{k}_3^4{k}_4+2a_7{k}_2{k}_3^4{k}_4+4a_7{k}_1{k}_3^3{k}_4^2\\ +4a_8{k}_2{k}_3^2{k}_4^3+2a_8{k}_1{k}_3{k}_4^4+5a_9{k}_2{k}_3{k}_4^4+a_9{k}_1{k}_4^5+3a_{10}{k}_2{k}_3^3{k}_4^2+3a_{10}{k}_1{k}_3^2{k}_4^3=0.\end{array}$ (148)

Proof. By $Z_1$ transformation, set $u=u\left(p\mathrm{,}q\right)$ , and $p=k_{1}t+k_{2}x$ , $q=k_{3}t+k_{4}x$ , then

(149)

Set

Thereupon

(150)

Namely

So the general solution of (137) is

The theorem is proved.

A similar proof of Theorem 15 leads to other general solutions of (137) as

(151)

(152)

(153)

(154)

(155)

(156)

3. Discussion and Summary

In this paper, we demonstrate through specific cases that transformation is an important method for obtaining analytical solutions and general solutions of nonlinear partial differential equations, and that using such solutions to gain analytical solutions of definite solution problems is also a very effective method. In practical cases, we find that the analytical solutions of some definite solution problems of first- and second-order nonlinear partial differential equations may not be unique, and more definite solution conditions are needed to make the analytical solutions of these definite solution problems unique.

The transformation is a completely new method that we have proposed to obtain general solutions to linear partial differential equations. Having successfully used it to obtain general solutions to a wide variety of linear partial differential equations, this paper is the first to use the transformation to study the current hot topic of nonlinear partial differential equations. A large number of new laws have been discovered and a large number of new cases have been solved, demonstrating the novelty and importance of this approach.

For methods capable of studying both linear and non-linear partial differential equations, analytical solutions are not available for qualitative analysis, and numerical methods are often only capable of studying local behaviour, and it is virtually impossible to obtain solutions involving arbitrary functions. The analytical solutions obtained by various existing analytical methods generally do not contain arbitrary functions, which makes it difficult to study definite solution problems and does not explain the infinitely variable physical phenomena in nature. The ability of the transformation to obtain analytical solutions containing arbitrary functions for many kinds of linear and nonlinear partial differential equations demonstrates its unique importance.

Now we all know that the universe is moving and changing. Physicists have discovered that the physical laws behind many natural phenomena are differential equations, especially infinitely variable physical phenomena such as light waves, sound waves, water waves and so on. In mathematics, perhaps only arbitrary functions can accurately describe physical phenomena with infinite variations, so proposing correct differential equations and obtaining analytical solutions of these equations containing arbitrary functions are of great theoretical and practical significance for human beings to understand, use and transform nature. The unveiling of the transformation has opened this door, and we can expect even greater progress and success to follow!

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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