First Order Pratial Differential Equation
We will prove that we can use First Integral Method of ODEs to solve Quasi-linear First order PDE (一阶拟线性偏微分方程).
First Integral Method
Definition of First Intergal
Consider ODEs
\[
\begin{equation}
\frac{d y_j}{dx}=f_{j}(x,y_1,\cdots,y_n),\quad j=1,2\cdots, n\label{ODEs}
\end{equation}
\]
where \(f_j(x,y_1,\cdots,y_n) (j=1,2,\cdots,n)\in C(D)\) , with \(D\subset \mathbb{R}^{n+1}\) , and are continuously differentiable with respect to \(y_1,y_2,\cdots, y_n\) .
Assume \(V=V(x,y_1,\cdots,y_n)\in C^1(G)\) , where \(G\subset D\) . If \(V\) is not a constant function, but constant when following an arbitrary integral curve of the above ODEs \(\ref{ODEs}\) , i.e.
\[
V(x,y_1(x),\cdots,y_n(x))\equiv c_0,\quad x\in J
\]
then we call \(V(x,y_1,\cdots,y_2)=c\) is a First Integral of ODEs \(\ref{ODEs}\) .
Property of First Integral
Property of First Integral (From curve to Points)
Assume non-constant function \(\varPhi(x,y_1,\cdots,y_n)\in C^1(G)\) , then
\[
\varPhi(x,y_1,\cdots,y_n) = c
\]
is a first integral of ODEs \(\ref{ODEs}\) , iff
\[
\begin{equation}
\frac{\partial \varPhi}{\partial x} + \frac{\partial \varPhi}{\partial y_1}f_1+\cdots + \frac{\partial \varPhi}{\partial y_n}f_n =0, \quad \forall (x,y_1,\cdots, y_n)\in G\label{Property}
\end{equation}
\]
Proof
"\(\Rightarrow\) ". Assume \(\varPhi(x,y_1,\cdots,y_n)\) is a first integral of ODEs \(\ref{ODEs}\) , then \(\forall (x_0,y_{10},\cdots,y_{n0})\in G\) , which an integral curve passes is denoted as
\[
y_1(x),y_2(x),\cdots, y_n(x),\forall x\in J
\]
we have
\[
\varPhi(x,y_1(x),\cdots,y_n(x))=c
\]
take partial derivatives from both sides and let \(x=x_0\) , we have
\[
\frac{\partial \varPhi}{\partial x} + \frac{\partial \varPhi}{\partial y_1}f_1+\cdots + \frac{\partial \varPhi}{\partial y_n}f_n =0
\]
which holds at point \((x_0,y_{10},\cdots,y_{n0})\) . But due to \((x_0,y_{10},\cdots,y_{n0})\) is arbitrary, so the above equation holds forall points in \(G\) .
"\(\Leftarrow\) ". Assume equation \(\ref{Property}\) holds, then it holds for arbitrary integral curve, i.e. denote \(\Gamma\in G\)
\[
\Gamma: y_1=\phi_1(x),y_2=\phi_2(x),\cdots,y_n=\phi_n(x),\quad x\in J
\]
and equation \(\ref{Property}\) holds for \(\Gamma\) . That is,
\[
\frac{\partial \varPhi}{\partial x} + \frac{\partial \varPhi}{\partial y_1}f_1+\cdots + \frac{\partial \varPhi}{\partial y_n}f_n =0, \quad \forall (x,\phi_1,\cdots, \phi_n)\in G
\]
which means
\[
\frac{d}{dx}\varPhi(x,\phi_1(x),\cdots, \phi_n(x))=0
\]
So integrate we have
\[
\varPhi(x,\phi_1(x),\cdots,\phi_n(x))=c_0, \quad x\in J.
\]
If we have a first integral for an ODEs, we can reduce the order for that ODEs. It is obvious if we can solve one variable \(y_i\) out and substitute it into ODEs \(\ref{ODEs}\) , then total number of variables are reduced to \(n-1\) .
Naively, if we have enough number of first integrals, we can solve total ODEs out. But we have to be careful, because some first integral are of the same type, or to be more specific, if a first integral could be expressed by another one, than we call them relevant.
Independence of First Integrals
We call \(n\) first Integrals in region \(G\)
\[
V_j(x,y_1,\cdots,y_n)=c_j,\quad j=1,1\cdots,n.
\]
are independent, if
\[
\text{det}\frac{\partial (V_1,V_2,\cdots,V_n)}{\partial(y_1,y_2,\cdots,y_n)}\neq 0,\quad \forall(x,y_1,\cdots,y_n)\in G.
\]
Then we have the following theorem.
Solution by First Integral Method
If
\[
V_j(x,y_1,\cdots,y_n)=c_j,\quad j=1,2\cdots,n
\]
are \(n\) independent first integrals in region \(G\) for ODEs \(\ref{ODEs}\) . Then the solution to ODEs \(\ref{ODEs}\) is a function
\[
y_j=\phi_j(x,c_1,c_2,\cdots,c_n),\quad j=1,2,\cdots,n
\]
specified by these \(n\) first integrals. And it expresses all the solution of ODEs \(\ref{ODEs}\) on region \(G\) .
Proof
(i) Cause \(V_1,V_2,\cdots,V_n\) are independent, so
\[
\text{det}\frac{\partial (V_1,V_2,\cdots,V_n)}{\partial(y_1,y_2,\cdots,y_n)}\neq 0,\quad \forall(x,y_1,\cdots,y_n)\in G.
\]
By theorem of implicit function , from
\[
V_j(x,y_1,\cdots,y_n)=c_j,\quad j=1,2\cdots,n
\]
solve
\[
y_j=\phi_j(x,c_1,\cdots,c_n),\quad j=1,2\cdots,n
\]
(ii) Then we have to prove \(\phi_j (j=1,2,\cdots,n)\) are solution to original ODEs \(\ref{ODEs}\) .
Notice that
\[
\begin{equation}
V_j(x,\phi_1(x,c_1,\cdots,c_n),\cdots,\phi_n(x,c_1,\cdots,c_n))\equiv c_j,\quad j=1,2\cdots,n\label{eq3}
\end{equation}
\]
take derivative with respect to \(x\) on both sides, we have
\[
\begin{equation}
\frac{\partial V_j}{\partial x}+\frac{\partial V_j}{\partial y_1}\phi_1'+\cdots+\frac{\partial V_j}{\partial y_n}\phi_n'=0,\quad j=1,2\cdots,n\label{eq1}
\end{equation}
\]
by Property of First Integral , we also have
\[
\begin{equation}
\frac{\partial V_j}{\partial x} + \frac{\partial V_j}{\partial y_1}f_1+\cdots + \frac{\partial V_j}{\partial y_n}f_n =0, \quad \forall (x,y_1,\cdots, y_n)\in G \label{eq2}
\end{equation}
\]
So subtract equation \(\ref{eq1}\) to equation \(\ref{eq2}\) and get
\[
\frac{\partial V_j}{\partial y_1}(\phi_1'-f_1)+\cdots+\frac{\partial V_j}{\partial y_n}(\phi_n'-f_n)=0,\quad j=1,2\cdots,n
\]
If we define
\[
A=\frac{\partial (V_1,V_2,\cdots,V_n)}{\partial(y_1,y_2,\cdots,y_n)}
\]
\[
\pmb{y}=[(\phi_1'-f_1),(\phi_2'-f_2),\cdots,(\phi_n'-f_n)]^T
\]
then we have
\[
A\pmb{y}=\pmb{0}
\]
Because \(\text{det}A\neq 0\) , so it follows
\[
\pmb{y}=\pmb{0}
\]
which means
\[
\phi_j=f_j,\quad j=1,2,\cdots,n.
\]
The above equation is exactly original ODEs. So \(\phi_j (j=1,2,\cdots,n)\) are solution to ODEs \(\ref{ODEs}\) .
(iii) Next, we have to prove one part of the last statement: \(\phi_j (j=1,2,\cdots,n)\) are general solution of ODEs \(\ref{ODEs}\) . Initially, these \(n\) solutions are independent, i.e.
\[
\text{det}\frac{\partial (\phi_1,\phi_2,\cdots,\phi_n)}{\partial(c_1,c_2,\cdots,c_n)}\neq 0
\]
Actually, from equation \(\ref{eq3}\) , we take partial derivatives with respect to \(c_k\) , we have
\[
\frac{\partial V_j}{\partial y_1}\frac{\partial \phi_1}{\partial c_k}+\cdots+\frac{\partial V_j}{\partial y_n}\frac{\partial \phi_n}{\partial c_k}=\delta_{jk}=\begin{cases}0,\quad j\neq k\\ 1,\quad j=k.\end{cases},\quad j,k=1,2,\cdots,n
\]
which means one matrices times another equals to \(E\) , i.e.
\[
\text{det}\frac{\partial (\phi_1,\phi_2,\cdots,\phi_n)}{\partial(c_1,c_2,\cdots,c_n)}=\left[\text{det}\frac{\partial (V_1,V_2,\cdots,V_n)}{\partial(y_1,y_2,\cdots,y_n)}\right]^{-1}\neq 0
\]
(iv) Lastly these solutions could express all other solutions. That is, \(\forall y_1=z_1(x),y_2=z_2(x),\cdots,y_n=z_n(x)\) are solution to ODEs \(\ref{ODEs}\) , could we express them with \(\phi_1,\phi_2,\cdots,\phi_n\) ?
If we denote \(y_1^0=z_1(x_0),\cdots,y_n^0=z_n(x_0)\) , and from this point we could determine
\[
c_1^0=V_1(x_0,y_1^0,\cdots,y_n^0), \cdots,c_n^0=V_n(x_0,y_1^0,\cdots,y_n^0)
\]
Solve
\[
V_j(x,y_1,\cdots,y_n)=c_j^0,\quad j=1,2,\cdots,n
\]
out with \(n\) answers
\[
\phi^0_j(x,c_1^0,\cdots,c_n^0),\quad j=1,2\cdots,n
\]
which follows the whole above proof property, i.e. it is solution to ODEs \(\ref{ODEs}\) and satisfies initial condition
\[
y_j^0=\phi^0_j(x_0,c_1^0,\cdots,c_n^0),\quad j=1,2\cdots,n
\]
Due to Uniqueness of Solution, we have
\[
z_j(x)\equiv \phi_j(x),\quad j=1,2\cdots,n
\]
which means \(z_j\) could be expressed by \(\phi_j\) .
Existence of First Integral
Practically speaking, it is really hard to find a first integral of ODEs \(\ref{ODEs}\) . Here we prove that in a sufficiently wide condition, first integral exists (locally).
Existence of First Integral
Assume \(P_0(x_0,y_1^0,y_2^0,\cdots, y_n^0)\in G\) , then there exists a neighborhood of \(P_0\) , denoted as \(G_1\subset G\) , such that ODEs \(\ref{ODEs}\) has \(n\) independent first integrals in \(G_1\) .
Proof
Choose a small neighborhood \(G_1=O_\delta(P_0)\subset G\) , with a point \((x_0,c_1,\cdots,c_n)\) , consider with initial condition
\[
y_1(x_0)=c_1,\cdots,y_n(x_0)=c_n.
\]
ODEs has unique solution
\[
y_1=\phi_1(x,c_1,\cdots,c_n),\cdots,y_n=\phi_n(x,c_1,\cdots,c_n).
\]
where \(\phi_j(x,c_1,\cdots,c_n) (j=1,2,\cdots,n)\) are continuously differentiable with respect to \((x,c_1,\cdots,c_n)\) . Using Continuous Differentiable dependence of solution on intial condition , we have
\[
\text{det}\frac{\partial (\phi_1,\cdots,\phi_n)}{\partial(c_1,\cdots,c_n)}\Bigg|_{x=x_0}=1
\]
So there exists function group \(V_1=V_1(x,y_1,\cdots,y_n),\cdots,V_n(x,y_1,\cdots,y_n)\) such that
\[
V_1(x,y_1,\cdots,y_n)=c_1,\cdots,V_n(x,y_1,\cdots,y_n)=c_n
\]
and
\[
\text{det}\frac{\partial(V_1,\cdots,V_n)}{\partial(y_1,\cdots,y_n)}\Bigg|_{x=x_0}=1
\]
So there exists a neighborhood \(G_1\subset G\) in which
\[
\text{det}\frac{\partial(V_1,\cdots,V_n)}{\partial(y_1,\cdots,y_n)}\neq 0
\]
meaning \(V_1,\cdots,V_n\) are \(n\) independent first integrals.
The following theorem is similar to linear space.
Maximum number of independent integrals
ODEs \(\ref{ODEs}\) has at most \(n\) independent first integrals.
Expression of First Integral
The following theorem states the expresstion of integrals, which is like integral factors .
Expresstion of First Integral
Assume \(V_1=V_1(x,y_1,\cdots,y_n)\) , \(\cdots\) , \(V_n=V_n(x,y_1,\cdots,y_n)\) are \(n\) independent first integrals of ODEs \(\ref{ODEs}\) . Then for an arbitrary first integral \(V=V(x,y_1,\cdots,y_n)\subset G\) , there exists \(C^1\) function \(h\) such that
\[
V(x,y_1,\cdots,y_n)=h(V_1,\cdots,V_n).
\]
Proof
Cause \(V_1(x,y_1,\cdots,y_n),\cdots, V_n(x,y_1,\cdots,y_n)\) are independent first integrals, we have
\[
J=\text{det}\frac{\partial (V_1,\cdots,V_n)}{\partial (y_1,\cdots,y_n)}\neq 0
\]
By theorem of implicite function, from ODEs \(\ref{ODEs}\) , we could solve
\[
y_j=y_j(x,V_1,\cdots,V_n),\quad j=1,2\cdots,n.
\]
For an arbitrary first integral \(V(x,V_1,\cdots,V_n)\) , let
\[
h(x,V_1,\cdots,V_n)=V(x,y_1(x,V_1,\cdots,V_n),\cdots,y_n(x,V_1,\cdots, V_n))
\]
Now we prove that \(h\) is irrelevant with \(x\) , i.e.
\[
\frac{\partial h}{\partial x}=0.
\]
Actually,
\[
\begin{align}
\frac{\partial h}{\partial x}=\frac{\partial V}{\partial x}+\sum_{j=1}^n\frac{\partial V}{\partial y_j}\frac{\partial y_j}{\partial x}. \label{h-x}
\end{align}
\]
By definition of first integral, we have
\[
\frac{\partial V_j}{\partial x}+\sum_{k=1}^n \frac{\partial V_j}{\partial y_k}\frac{\partial y_k}{\partial x}=0, \quad j=1,2,\cdots,n
\]
that is,
\[
\left[\begin{array}{cccc}
\displaystyle \frac{\partial V_1}{\partial y_1} & \displaystyle \frac{\partial V_1}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_1}{\partial y_n} \\
\displaystyle \frac{\partial V_2}{\partial y_1} & \displaystyle \frac{\partial V_2}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_2}{\partial y_n}\\
\displaystyle \vdots& \vdots& \ddots &\vdots\\
\displaystyle \frac{\partial V_n}{\partial y_1} & \displaystyle \frac{\partial V_n}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_n}{\partial y_n}
\end{array}\right] \cdot \left[\begin{array}{c}
\displaystyle \frac{\partial y_1}{\partial x}\\
\displaystyle \frac{\partial y_2}{\partial x}\\ \displaystyle \vdots\\
\displaystyle \frac{\partial y_n}{\partial x}
\end{array}\right]=\left[\begin{array}{c}
\displaystyle \frac{\partial V_1}{\partial x}\\
\displaystyle \frac{\partial V_2}{\partial x}\\
\displaystyle \vdots \\
\displaystyle \frac{\partial y_n}{\partial x}\end{array}\right]
\]
By Cramer principle, we have
\[
\begin{align}
\frac{\partial y_j}{\partial x}=\frac{1}{J}\text{det}\frac{\partial(V_1,\cdots,V_n)}{\partial (y_1,\cdots,y_{j-1},x,y_{j+1},\cdots,y_n)}\label{Cramer}
\end{align}
\]
Substitute equation \(\ref{Cramer}\) back into equation \(\ref{h-x}\) and we have
\[
\begin{align*}
\frac{\partial h}{\partial x}&=\frac{\partial V}{\partial x}+\sum_{j=1}^n\frac{\partial V}{\partial y_j}\frac{1}{J}\text{det}\frac{\partial(V_1,\cdots,V_n)}{\partial (y_1,\cdots,y_{k-1},x,y_{k+1},\cdots,y_n)}\\
&=\frac{1}{J}\left[ J \frac{\partial V}{\partial x}+\sum_{j=1}^n\frac{\partial V}{\partial y_j}\text{det}\frac{\partial(V_1,\cdots,V_n)}{\partial (y_1,\cdots,y_{k-1},x,y_{k+1},\cdots,y_n)}\right]\\
&=\frac{1}{J}\text{det}\frac{\partial (V,V_1,\cdots,V_n)}{\partial (x,y_1,\cdots,y_n)}
\end{align*}
\]
While for the right side matrix, its corresponding linear system could be
\[
\left[\begin{array}{ccccc}
\displaystyle \frac{\partial V}{\partial x} &
\displaystyle \frac{\partial V}{\partial y_1} & \displaystyle \frac{\partial V}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V}{\partial y_n} \\
\displaystyle \frac{\partial V_1}{\partial x} &
\displaystyle \frac{\partial V_1}{\partial y_1} & \displaystyle \frac{\partial V_1}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_1}{\partial y_n} \\
\displaystyle \frac{\partial V_2}{\partial x} &
\displaystyle \frac{\partial V_2}{\partial y_1} & \displaystyle \frac{\partial V_2}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_2}{\partial y_n}\\
\displaystyle \vdots& \vdots& \vdots & \ddots &\vdots\\
\displaystyle \frac{\partial V_n}{\partial x} &
\displaystyle \frac{\partial V_n}{\partial y_1} & \displaystyle \frac{\partial V_n}{\partial y_2} &\cdots &
\displaystyle \frac{\partial V_n}{\partial y_n}
\end{array}\right]_{(n+1)\times(n+1)} \cdot \left[\begin{array}{c}
1 \\
\displaystyle \frac{\partial y_1}{\partial x}\\
\displaystyle \frac{\partial y_2}{\partial x}\\ \displaystyle \vdots\\
\displaystyle \frac{\partial y_n}{\partial x}
\end{array}\right]_{(n+1)\times 1}=\pmb{0}
\]
which has non-zero solution (by property of first integral for \(V,V_1,\cdots,V_n\) ), meaning its determinant equals zero.
And we are done.
Note: the above conclusion holds locally. we could not guarantee the existence of first integrals in a large region.
Homogeneous linear first order PDE
The form of homogeneoud first order PDE can be
\[
\begin{align}
\sum_{j=1}^n A_j(x_1,x_2,\cdots,x_n)\frac{\partial u}{\partial x_j}=0 \label{PDE}
\end{align}
\]
where \(u=u(x_1,x_2,\cdots,x_n)\) is unknown function, coefficient function
\[
A_j(x_1,x_2,\cdots,x_n)\in C^1,\quad j=1,2,\cdots,n,\quad \forall(x_1,x_2,\cdots, n)\in D
\]
and
\[
\sum_{j=1}^n|A_j(x_1,x_2,\cdots,x_n)|>0.
\]
Transformation of first order PDE
If \(A_1(x_1,x_2,\cdots,x_n)\neq 0\) , then equation \(\ref{PDE}\) could be written as
\[
\frac{\partial u}{\partial x_1}+\frac{\partial u}{\partial x_2}\frac{A_2}{A_1}+\cdots+\frac{\partial u}{\partial x_n}\frac{A_n}{A_1}=0
\]
Compared to equation \(\ref{Property}\) we could deduce that \(u\) is a first integral of ODEs
\[
\frac{d x_j}{dx_1}=\frac{A_j}{A_1},\quad j=2,3,\cdots,n.
\]
We could transfer a PDE problem into a ODEs problem.
So here comes the following theorem.
General solution to First Order PDE
We define characteristic equation of PDE \(\ref{PDE}\)
\[
\begin{equation}
\frac{dx_1}{A_1(x_1,x_2,\cdots,x_n)}=\frac{dx_2}{A_2(x_1,x_2,\cdots,x_n)}=\cdots=\frac{dx_n}{A_n(x_1,x_2,\cdots,x_n)}.\label{Characteristic}
\end{equation}
\]
which has \(n-1\) independent first integrals
\[
\phi_j(x_1,x_2,\cdots,x_n)=c_j,\quad j=1,2,\cdots, n.
\]
Then the general solution to PDE \(\ref{PDE}\) is
\[
u=\varPsi(\phi_1(x_1,x_2,\cdots,x_n),\phi_2(x_1,x_2,\cdots,x_n),\cdots,\phi_n(x_1,x_2,\cdots,x_n))
\]
where \(\varPsi\) is an arbitary continuously differentiable function.
Proof
Assume \(\phi(x_1,x_2,\cdots,x_n)\) is a first integral of characteristic equation \(\ref{Characteristic}\) . Cause \(A_1,A_2,\cdots,A_n\) do not equal zero, assume there exists a neighborhood, in which \(A_1(x_1,x_2,\cdots,x_n)\neq 0\) . So characteristic equation \(\ref{Characteristic}\) is equivalent to the following ODEs
\[
\frac{dx_j}{dx_i}=\frac{A_j(x_1,x_2,\cdots,x_n)}{A_1(x_1,x_2,\cdots,x_n)}, \quad j=2,3,\cdots,n
\]
Thus \(\phi(x_1,x_2,\cdots,x_n)\) is also a first integral of this ODEs. So
\[
\frac{\partial\phi}{\partial x_1}+\sum_{j=2}^n\frac{A_j(x_1,x_2,\cdots,x_n)}{A_1(x_1,x_2,\cdots,x_n)}\frac{\partial \phi}{\partial x_j}=0
\]
that is,
\[
\sum_{j=1}^n A_j(x_1,x_2,\cdots,x_n)\frac{\partial \phi}{\partial x_j}=0
\]
Due to \(\phi_1,\phi_2,\cdots,\phi_{n-1}\) are \(n-1\) independent first integrals, so for an arbitrary \(C^1\) function \(\varPsi\) ,
\[
\varPsi(\phi_1,\phi_2,\cdots,\phi_n)
\]
is also a first integral of characteristic equation \(\ref{Characteristic}\) . By the theory of expression with first integrals , for an arbitrary first integral \(\psi(x_1,x_2,\cdots,x_n)\) , there exists a \(C^1\) function \(\varPsi_0\) such that
\[
\psi(x_1,x_2,\cdots,x_n)=\varPsi_0(\phi_0(x_1,x_2,\cdots,x_n),\phi_2(x_1,x_2,\cdots,x_n),\cdots,\phi_{n-1}(x_1,x_2,\cdots,x_n)).
\]
Quasi-linear first order PDE
We call the following PDE Quasi-linear first order PDE
\[
\begin{align}
\sum_{j=1}^n A_j(x_1,\cdots,x_n,u)\frac{\partial u}{\partial x_j}=B(x_1,\cdots,x_n,u) \label{ql-PDE}
\end{align}
\]
where \(A_1,\cdots,A_n,B\in C^1(G)\) . This PDE is characterized that it is linear with respect to the partial direvative of \(u\) .
Transformation of Quasi-linear first order PDE
Assume \(u=\phi(x_1,\cdots,x_n)\) is solution to PDE \(\ref{ql-PDE}\) , if we let
\[
\varPhi(x_1,\cdots,x_n)=\phi(x_1,\cdots,x_n)-u
\]
which follows
\[
\varPhi(x_1,\cdots,x_n,u)\equiv 0 \Leftrightarrow u=\phi(x_1,\cdots,x_n)
\]
and
\[
\frac{\partial \varPhi}{\partial x_j}=\frac{\partial \phi}{\partial x_j} (j=1,2,\cdots,n),\quad \frac{\partial \varPhi}{\partial u}=-1
\]
So \(\varPhi\) satisfies a homogeneous first order PDE
\[
\begin{align}
\sum_{j=1}^nA_j(x_1,\cdots,x_n,u)\frac{\partial \varPhi}{\partial x_j}+B(x_1,\cdots,x_n,u)\frac{\partial \varPhi}{\partial u}=0. \label{alter-PDE}
\end{align}
\]
Here we transform a Quasi-linear PDE into a homogeneous PDE. Then we only have to show that the solution to PDE \(\ref{alter-PDE}\) is also a solution to PDE \(\ref{ql-PDE}\) .
Theorem for Quasi-linear first order PDE
Assume general solution to PDE \(\ref{alter-PDE}\) is
\[
\varPhi=\varPhi(\psi_1(x_1,\cdots,x_n,u),\cdots,\psi_n(x_1,\cdots,x_n,u))
\]
where
\[
\psi_j(x_1,\cdots,x_n,u)=c_j,\quad j=1,2\cdots,n
\]
are \(n\) independent first integrals of the corresponding characteristic equations
\[
\frac{dx_1}{A_1(x_1,\cdots,x_n,u)}=\cdots=\frac{dx_n}{A_1(x_1,\cdots,x_n,u)}=\frac{du}{B(x_1,\cdots,x_n,u)}
\]
with \(\varPhi\) is a continuously differentiable function. Then the general solution to PDE \(\ref{ql-PDE}\) is
\[
\begin{align}
\varPhi(\psi_1(x_1,\cdots,x_n,u),\cdots,\psi_n(x_1,\cdots,x_n,u))=0 \label{solu-ql}
\end{align}
\]
where \(\frac{\partial}{\partial u}\varPhi(x_1,\cdots,x_n,u)\neq 0\) .
Proof
Similar to the previous proof, we have to show that all solution to PDE \(\ref{ql-PDE}\) could be expressed by formula \(\ref{solu-ql}\) . That is, for an arbitrary solution \(u=f(x_1,\cdots,x_n)\) of PDE \(\ref{ql-PDE}\) , there exists \(\varPhi_0\) such that
\[
\varPhi_0(\psi_1(x_1,\cdots,x_n,f(x_1,\cdots,x_n)),\cdots,\psi_n(x_1,\cdots,x_n,f(x_1,\cdots,x_n)))\equiv 0
\]
which is equivalent to that
\[
\psi_j(x_1,\cdots,x_n,f(x_1,\cdots,x_n))\overset{\Delta}{=}\phi_j(x_1,\cdots,x_n),\quad j=1,2,\cdots,n
\]
are relevant. Cause
\[
\frac{\partial \phi_j}{\partial x_k}=\frac{\partial \psi_j}{\partial x_k}+\frac{\partial \psi_j}{\partial u}\frac{\partial f}{\partial x_k},\quad j,k=1,2,\cdots,n
\]
by using \(f(x_1,\cdots,x_n)\) is solution to PDE \(\ref{ql-PDE}\) , \(\psi_j(x_1,\cdots,x_n,f(x_1,\cdots,x_n))(j=1,2,\cdots,n)\) is solution to PDE \(\ref{alter-PDE}\) , we have
\[
\begin{align*}
\sum_{k=1}^nA_k(x_1,\cdots,x_n,f(x_1,\cdots,x_n))\frac{\partial \phi_j}{\partial x_k}&=\sum_{k=1}^n\left[A_k\frac{\partial \psi_j}{\partial x_k}+A_k\frac{\partial \psi_j}{\partial u}\frac{\partial f}{\partial x_k}\right]\\
&=\sum_{k=1}^n \left[A_k\frac{\partial \psi_j}{\partial x_k}+\frac{\partial \psi_j}{\partial u}A_k\frac{\partial f}{\partial x_k}\right]\\
&=\sum_{k=1}^n \frac{\partial \psi_j}{\partial x_k}A_k+\frac{\partial \psi_j}{\partial u}B_k\\
&=0,\quad j=1,2,\cdots,n
\end{align*}
\]
which means the following matrix multiplication
\[
\left[\begin{array}{cccc}
\displaystyle \frac{\partial \phi_1}{\partial x_1} & \displaystyle \frac{\partial \phi_1}{\partial x_2} &\cdots &
\displaystyle \frac{\partial \phi_1}{\partial x_n} \\
\displaystyle \frac{\partial \phi_2}{\partial x_1} & \displaystyle \frac{\partial \phi_2}{\partial x_2} &\cdots &
\displaystyle \frac{\partial \phi_2}{\partial x_n}\\
\displaystyle \vdots& \vdots& \ddots &\vdots\\
\displaystyle \frac{\partial \phi_n}{\partial x_1} & \displaystyle \frac{\partial \phi_n}{\partial x_2} &\cdots &
\displaystyle \frac{\partial \phi_n}{\partial x_n}
\end{array}\right] \cdot \left[\begin{array}{c}
\displaystyle A_1\\
\displaystyle A_2\\ \displaystyle \vdots\\
\displaystyle A_n
\end{array}\right]=\pmb{0}
\]
\(A_1,A_2,\cdots,A_n\) do not equal zeto at the same time, so
\[
\text{det}\frac{\partial (\phi_1,\cdots,\phi_n)}{\partial(x_1,\cdots,x_n)}=0
\]
which means \(\phi_1,\cdots,\phi_n\) are relevent.
Solve the general solution to PDE
\[
xz\frac{\partial z}{\partial x}+yz\frac{\partial z}{\partial y} = -xy
\]
and its solution plane which passes curve \(y=x^2\) , \(z=x^3\) .
Answer
The corresponding characteristic equation
\[
\frac{dx}{xz}=\frac{dy}{yz}=\frac{dz}{-xy}
\]
which has two independent first integrals
\[
\frac{x}{y}=c_1,\quad z^2+xy=c_2
\]
so the general solution to PDE is
\[
V\left(\frac{x}{y},z^2+xy\right)=0
\]
where \(V(\xi,eta)\) is an arbitrary continuously differentiable function.
From above we could solve
\[
f\left(\frac{x}{y}\right)=z^2+xy
\]
substitute initial condition, we have
\[
f\left(\frac{1}{x}\right)=x^6+x^3 \Rightarrow f(x)=1/x^6+1/x^3
\]
so the solution with initial condition is
\[
(z^2+xy)=\left(\frac{y}{x}\right)^6+\left(\frac{y}{x}\right)^3
\]