First-Order PDE from f(phi, psi)=0

The first-order PDE from

f(\phi,\psi)

is obtained by differentiating the relation

f(\phi,\psi)=0

and eliminating the derivatives of the arbitrary function. Here

f

is an arbitrary function. The functions

\phi

and

\psi

are known functions of

x,y,z

. The aim is to find the partial differential equation satisfied by

z=z(x,y)

Let

\begin{equation*} \phi=\phi(x,y,z), \qquad \psi=\psi(x,y,z). \end{equation*}

Suppose that

z=z(x,y)

is connected with

x

and

y

through the relation

\begin{equation*} f(\phi,\psi)=0. \quad \text{(1)} \end{equation*}

Since

z

depends on

x

and

y

, we write

\begin{equation*} p=\frac{\partial z}{\partial x}, \qquad q=\frac{\partial z}{\partial y}. \end{equation*}

The required first-order PDE is obtained by differentiating (1) with respect to

x

and

y

, and then eliminating the derivatives of the arbitrary function

f

The first-order PDE from

f(\phi,\psi)

is obtained by differentiating the relation

f(\phi,\psi)=0

and eliminating the derivatives of the arbitrary function. Here

f

is an arbitrary function. The functions

\phi

and

\psi

are known functions of

x,y,z

. The aim is to find the partial differential equation satisfied by

z=z(x,y)

Let

\begin{equation*} \phi=\phi(x,y,z), \qquad \psi=\psi(x,y,z). \end{equation*}

Suppose that

z=z(x,y)

is connected with

x

and

y

through the relation

\begin{equation*} f(\phi,\psi)=0. \quad \text{(1)} \end{equation*}

Since

z

depends on

x

and

y

, we write

\begin{equation*} p=\frac{\partial z}{\partial x}, \qquad q=\frac{\partial z}{\partial y}. \end{equation*}

The required first-order PDE is obtained by differentiating (1) with respect to

x

and

y

, and then eliminating the derivatives of the arbitrary function

f

The relation

f(\phi,\psi)=0

contains an arbitrary function, so the construction removes the derivatives of that function rather than removing constants. Choose a stage and click Show Stage to see how the relation changes during the construction. Notice that

f_\phi

and

f_\psi

appear after differentiation but disappear after the determinant condition is used. The final equation contains only

x,y,z,p,q

through Jacobians of

\phi

and

\psi

\phi=\phi(x,y,z)

and

\psi=\psi(x,y,z)

are two given functions, and if

\begin{equation*} f(\phi,\psi)=0 \end{equation*}

where

f

is an arbitrary function, then

z=z(x,y)

satisfies

\begin{equation*} p\frac{\partial(\phi,\psi)}{\partial(y,z)}+q\frac{\partial(\phi,\psi)}{\partial(z,x)}=\frac{\partial(\phi,\psi)}{\partial(x,y)}. \quad \text{(2)} \end{equation*}

Here

\begin{equation*} \frac{\partial(\phi,\psi)}{\partial(x,y)}= \begin{vmatrix} \phi_x & \phi_y \\ \psi_x & \psi_y \end{vmatrix}. \end{equation*}

Differentiate (1) with respect to

x

. By the chain rule,

\begin{equation*} f_\phi(\phi_x+p\phi_z)+f_\psi(\psi_x+p\psi_z)=0. \quad \text{(3)} \end{equation*}

Differentiate (1) with respect to

y

. Again by the chain rule,

\begin{equation*} f_\phi(\phi_y+q\phi_z)+f_\psi(\psi_y+q\psi_z)=0. \quad \text{(4)} \end{equation*}

Equations (3) and (4) are two homogeneous linear equations in

f_\phi

and

f_\psi

. For a non-trivial solution, the determinant of coefficients must vanish. Hence

\begin{equation*} \begin{vmatrix} \phi_x+p\phi_z & \psi_x+p\psi_z \\ \phi_y+q\phi_z & \psi_y+q\psi_z \end{vmatrix}=0. \quad \text{(5)} \end{equation*}

Now expand (5):

\begin{align*} (\phi_x+p\phi_z)(\psi_y+q\psi_z)&-(\psi_x+p\psi_z)(\phi_y+q\phi_z)=0 \nonumber \\ \phi_x\psi_y-\psi_x\phi_y+p(\phi_z\psi_y-\psi_z\phi_y)&+q(\phi_x\psi_z-\psi_x\phi_z)=0. \quad \text{(6)} \end{align*}

Using Jacobian notation, (6) becomes

\begin{equation*} \frac{\partial(\phi,\psi)}{\partial(x,y)}-p\frac{\partial(\phi,\psi)}{\partial(y,z)}-q\frac{\partial(\phi,\psi)}{\partial(z,x)}=0. \quad \text{(7)} \end{equation*}

Therefore

\begin{equation*} p\frac{\partial(\phi,\psi)}{\partial(y,z)}+q\frac{\partial(\phi,\psi)}{\partial(z,x)}=\frac{\partial(\phi,\psi)}{\partial(x,y)}. \end{equation*}

This is the required first-order partial differential equation. \hfill

\square

The chain rule must include the hidden dependence

z=z(x,y)

. Choose differentiation with respect to

x

y

, then click Build Terms to see how

\phi

and

\psi

change. For differentiation with respect to

x

, the extra factor is

p=\partial z/\partial x

; for differentiation with respect to

y

, the extra factor is

q=\partial z/\partial y

. This explains why the differentiated equations contain

\phi_x+p\phi_z

and

\phi_y+q\phi_z

After differentiation, the unknown quantities are

f_\phi

and

f_\psi

. The two equations are homogeneous and linear in these two quantities, so a non-trivial solution requires the coefficient determinant to vanish. Click Show Determinant to see the coefficient matrix and the determinant condition. This step is what removes the derivatives of the arbitrary function from the final PDE.

The Jacobian

\begin{equation*} \frac{\partial(\phi,\psi)}{\partial(x,y)} \end{equation*}

means that we differentiate

\phi

and

\psi

with respect to

x

and

y

and then form the determinant. Similarly, the Jacobians with respect to

(y,z)

and

(z,x)

are formed by using those pairs of variables.

The theorem becomes practical once the three Jacobians are computed. Choose simple functions

\phi

and

\psi

, then click Compute PDE to calculate

J_{xy}

J_{yz}

, and

J_{zx}

. The calculator then assembles the result as

pJ_{yz}+qJ_{zx}=J_{xy}

. This helps you see how the compact Jacobian formula becomes an explicit first-order PDE.

The equation in (2) is linear in the first derivatives

p

and

q

. The coefficients may depend on

x,y,z

. Therefore it is a first-order quasi-linear partial differential equation.

The final Jacobian equation has the same structure as a quasi-linear first-order PDE. Click Highlight Parts to identify the coefficient of

p

, the coefficient of

q

, and the right side. Notice that

p

and

q

appear only to the first power. The coefficients may depend on

x,y,z

through the Jacobians, which is allowed in a quasi-linear equation.

Some books describe this result as the Jacobian form of a first-order PDE. Some also connect it with the general integral

f(\phi,\psi)=0

. In this page, we use the title First-Order PDE from f(phi, psi)=0 because it states the construction directly.

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