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# 数学代写|偏微分方程代考Partial Differential Equations代写|The Origin of Second-order Equations

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## 数学代写|偏微分方程代考Partial Differential Equations代写|The Origin of Second-order Equations

Suppose that the function $z$ is given by an expression of the type
$$z=f(u)+g(v)+w$$
where $f$ and $g$ are arbitrary functions of $u$ and $v$, respectively, and $u, v$, and $w$ are prescribed functions of $x$ and $y$. Then writing
$$p=\frac{\partial z}{\partial x}, \quad q=\frac{\partial z}{\partial y}, \quad r=\frac{\partial^2 z}{\partial x^2}, \quad s=\frac{\partial^2 z}{\partial x \partial y}, \quad t=\frac{\partial^2 z}{\partial y^2}$$
we find, on differentiating both sides of (1) with respect to $x$ and $y$, respectively, that
\begin{aligned} & p=f^{\prime}(u) u_x+g^{\prime}(v) v_x+w_x \ & q=f^{\prime}(u) u_y+g^{\prime}(v) v_y+w_y \end{aligned}
and hence that
\begin{aligned} & r=f^{\prime \prime}(u) u_x^2+g^{\prime \prime}(v) v_x^2+f^{\prime}(u) u_{x x}+g^{\prime}(v) v_{x x}+w_{x x} \ & s=f^{\prime \prime}(u) u_x u_y+g^{\prime \prime}(v) v_x u_y+f^{\prime}(u) u_{x v}+g^{\prime}(v) v_{x v}+w_{x y} \ & t=f^{\prime \prime}(u) u_y^2+g^{\prime \prime}(v) v_y^2+f^{\prime}(u) u_{y v}+g^{\prime}(v) v_{y y}+w_{y v} \end{aligned}
We now have five equations involving the four arbitrary quantities $f^{\prime}$, $f^{\prime \prime}, g^{\prime}, g^{\prime \prime}$. If we eliminate these four quantities from the five equations, we obtain the relation
$$\left|\begin{array}{cllcc} p-w_x & u_x & v_x & 0 & 0 \ q-w_y & u_y & v_y & 0 & 0 \ r-w_{x x} & u_{x x} & v_{x x} & u_x^2 & v_x^2 \ s-w_{x y} & u_{x v} & v_{x v} & u_x u_v & v_x v_y \ t-w_{v y} & u_{y v} & v_{y v} & u_y^2 & v_y^2 \end{array}\right|=0$$
which involves only the derivatives $p, q, r, s, t$ and known functions of $x$ and $y$. It is therefore a partial differential equation of the second order. Furthermore if we expand the determinant on the left-hand side of equation (3) in terms of the elements of the first column, we obtain an equation of the form
$$R r+S s+T t+P p+Q q=W$$
where $R, S, T, P, Q, W$ are known functions of $x$ and $y$. Therefore the relation (1) is a solution of the second-order linear partial differential equation (4). It should be noticed that the equation (4) is of a particular type: the dependent variable $z$ does not occur in it.

## 数学代写|偏微分方程代考Partial Differential Equations代写|Equations with Variable Coefficients

We shall now consider equations of the type
$$R r+S s+T t+f(x, y, z, p, q)=0$$

which may be written in the form
$$\mathrm{L}(z)+f(x, y, z, p, q)=0$$
where $L$ is the differential operator defined by the equation
$$\mathrm{L}=R \frac{\partial^2}{\partial x^2}+S \frac{\partial^2}{\partial x \partial y}+T \frac{\partial^2}{\partial y^2}$$
in which $R, S, T$ are continuous functions of $x$ and $y$ possessing continuous partial derivatives of as high an order as necessary. By a suitable change of the independent variables we shall show that any equation of the type (2) can be reduced to one of three canonical forms. Suppose we change the independent variables from $x, y$ to $\xi, \eta$, where
$$\xi=\xi(x, y), \quad \eta=\eta(x, y)$$
and we write $z(x, y)$ as $\zeta(\xi, \eta)$; then it is readily shown that equation (1) takes the form
\begin{aligned} A\left(\xi_x, \xi_y\right) \frac{\partial^2 \zeta}{\partial \xi^2}+2 B\left(\xi_x, \xi_y ; \eta_x, \eta_y\right) & \frac{\partial^2 \zeta}{\partial \xi \partial \eta} \ & +A\left(\eta_x, \eta_y\right) \frac{\partial^2 \zeta}{\partial \eta^2}=F\left(\xi, \eta, \zeta, \zeta_{\xi}, \zeta_\eta\right) \end{aligned}
where
$$\begin{gathered} A(u, v)=R u^2+S u v+T v^2 \ B\left(u_1, v_1 ; u_2, v_2\right)=R u_1 u_2+\frac{1}{2} S\left(u_1 v_2+u_2 v_1\right)+T v_1 v_2 \end{gathered}$$
and the function $F$ is readily derived from the given function $f$.
The problem now is to determine $\xi$ and $\eta$ so that equation (4) takes the simplest possible form. The procedure is simple when the discriminant $S^2-4 R T$ of the quadratic form (5) is everywhere either positive, negative, or zero, and we shall discuss these three cases separately.

# 偏微分方程代写

## 数学代写|偏微分方程代考Partial Differential Equations代写|The Origin of Second-order Equations

$$z=f(u)+g(v)+w$$

$$p=\frac{\partial z}{\partial x}, \quad q=\frac{\partial z}{\partial y}, \quad r=\frac{\partial^2 z}{\partial x^2}, \quad s=\frac{\partial^2 z}{\partial x \partial y}, \quad t=\frac{\partial^2 z}{\partial y^2}$$

\begin{aligned} & p=f^{\prime}(u) u_x+g^{\prime}(v) v_x+w_x \ & q=f^{\prime}(u) u_y+g^{\prime}(v) v_y+w_y \end{aligned}

\begin{aligned} & r=f^{\prime \prime}(u) u_x^2+g^{\prime \prime}(v) v_x^2+f^{\prime}(u) u_{x x}+g^{\prime}(v) v_{x x}+w_{x x} \ & s=f^{\prime \prime}(u) u_x u_y+g^{\prime \prime}(v) v_x u_y+f^{\prime}(u) u_{x v}+g^{\prime}(v) v_{x v}+w_{x y} \ & t=f^{\prime \prime}(u) u_y^2+g^{\prime \prime}(v) v_y^2+f^{\prime}(u) u_{y v}+g^{\prime}(v) v_{y y}+w_{y v} \end{aligned}

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