Compatible system of PDEs

Definition 3 (Compatible system of PDEs)   Two PDEs

$$f(x, y, u, u_x, u_y)=0~$$ and$$~ g(x, y, u, u_x, u_y)=0$$

are said to be compatible, if they have a common solution.

Theorem 2 (Necessary and sufficient condition for compatibility)   The necessary and sufficient condition for compatibility of the two PDEs $f(x, y, u, u_x, u_y)=0$ and $g(x, y, u, u_x, u_y)=0$ is

$$J=\frac{\partial(f, g)}{\partial(u_x, u_y)} \neq 0,$$    and $$[f, g]=0,$$

where $$[f, g] := \frac{\partial(f, g)}{\partial(x, u_x)} + \frac{\partia... ...artial(f, g)}{\partial(u, u_x)} + u_{y}\frac{\partial(f, g)}{\partial(u, u_y)}.$$

Proof Given

$$f(x, y, u, u_x, u_y) =$$ 0 (1.15)

$$g(x, y, u, u_x, u_y) = 0.$$ (1.16)

Solve () and () for $u_x$ and $u_y$ explicitly to get

$$u_x = \varphi(x, y, u)$$ (1.17)

$$u_y = \psi(x, y, u).$$ (1.18)

On the integral surface of the solution $u=$constant, the differential relation

$$du = u_x dx + u_y dy$$

should be integrable. Using () and ()

$$du = \varphi dx + \psi dy.$$

Rewrite it as

$$\varphi dx + \psi dy - du = 0.$$

This equation is in Pfaffian differential form for which one have the result.

Theorem 3 (Proof: See Chap. 1, Sec. 5 of Sneddon)   A necessary and sufficient condition that the Pfaffian differential equation $\vec{X} \cdot d\vec{r} = 0$ should be integrable is that $\vec{X} \cdot$   curl${\vec{X}} = 0$.

Using the above result

$$(\varphi, \psi, -1) \cdot$$   curl$$(\varphi, \psi, -1) = 0,$$

$$(\phi,\psi,-1). \begin{vmatrix} i & j & k \\ \partial_x & \partial_y & \partial_u\\ \psi & \phi & -1 \end{vmatrix}= 0,$$

$$\implies~\psi_x + \varphi \psi_u = \varphi_y + \psi \varphi_u.$$ (1.19)

Differentiate the given first PDE () w.r.t. $x$ and $u$ which gives

$$f_x + f_{u_x} (u_x)_x + f_{u_y}(u_y)_x = 0,$$

$$f_u + f_{u_x} (u_x)_u + f_{u_y}(u_y)_u = 0.$$

From () and (), express partial derivative of $u$ in terms of partial derivatives of $\varphi$ and $\psi$

$$(u_x)_x = \varphi_{x},~ (u_x)_u = \varphi_{u},~ (u_y)_x = \psi_{x}, ~(u_y)_u = \psi_{u}$$

Using these, one get

$$f_x + f_{u_x} \varphi_x + f_{u_y} \psi_x = 0,$$

$$f_u + f_{u_x} \varphi_u + f_{u_y} \psi_u = 0.$$

Multiply the second equation of the above pair by $\varphi$ and add to the first equation

$$(f_x + \varphi f_u ) + f_{u_x} ( \varphi_{x} + \varphi \varphi_u )+ f_{u_y} (\psi_x + \varphi \psi_u) = 0.$$

Following the similar procedure with the given second PDE () results

$$(g_x + \varphi g_u ) + g_{u_x} ( \varphi_{x} + \varphi \varphi_u )+ g_{u_y} (\psi_x + \varphi \psi_u) = 0.$$

Solving these two equations for $(\psi_x + \varphi \psi_u)$ gives

$$\psi_x + \varphi \psi_u = \frac{1}{J} \left \{ \frac{\partial(f, g)}{\partial(x, u_x)} + \varphi \frac{\partial(f, g)}{\partial(u, u_x)} \right \}$$ (1.20)

where $$J = \frac{\partial(f, g)}{\partial(u_x, u_y)}$$.

Differentiating given pair of equations w.r.t. $y$ and $u$ gives

$$\varphi_y + \psi \varphi_u = -\frac{1}{J} \left \{ \frac{\partial(f, g)}{\partial(y, u_y)} + \psi \frac{\partial(f, g)}{\partial(u, u_y)} \right \}$$ (1.21)

Put RHS of () and () in (), one get compatibility condition

$$[f, g]=0,$$

where $$[f, g] := \frac{\partial(f, g)}{\partial(x, u_x)} + \frac{\partia... ...artial(f, g)}{\partial(u, u_x)} + u_{y}\frac{\partial(f, g)}{\partial(u, u_y)}.$$

Hence proof.

Example 1.5.1   Show that the PDEs $u u_{x}^{2} + u_{y}^{2} - 4 = 0$ and $u _{x} - au_{y} = 0$ ($a \neq 0$, constant) are compatible equations.

Solution Here $f = u u_{x}^{2} + u_{y}^{2} - 4$, $g = u _{x} - a u_{y}$

$$\displaystyle [f, g] := \frac{\partial(f, g)}{\partial(x, u_x)} +... ...artial(f, g)}{\partial(u, u_x)} + u_{y}\frac{\partial(f, g)}{\partial(u, u_y)}.$$

Jacobian calculations

$$\frac{\partial(f, g)}{\partial(x, u_x)} = \left \vert \begin{matr... ...{matrix} u_{x}^{2} & 0 \\ 2 u u_x & 1 \end{matrix}\right \vert = u_{x}^{2}$$

$$\frac{\partial(f, g)}{\partial(y, u_y)} = \left \vert \begin{mat... ...u_{x}^{2} & 0 \\ 2 u_{y} & -a \\ \end{matrix} \right \vert = - a u_{x}^{2}$$

$\therefore$ The compatibility condition becomes

$$[f, g] = u_x (u_{x}^{2}) + u_{y} (-au_{x}^{2}) = u_{x}^{2}( u_{x} - au_y) = u_{x}^{2} \cdot 0 = 0$$

As $[f,g]=0$, the given two equations are compatible.