I have come across a question which asks the following of me:

By defining $$ \alpha = \frac{\partial \phi}{\partial x} \hspace{10mm} \beta = \frac{\partial \phi}{\partial y} $$ express the partial differential equation $$ \frac{\partial^2 \phi}{\partial x^2} - \frac{\partial^2 \phi}{\partial y^2} = 0 $$ as a system of 2 first order partial differential equations in $\alpha$ and $\beta$.

I am not sure how I am to go about this one. Clearly, I can see that the PDE may be expressed in terms of $\alpha$ and $\beta$ as $$ \frac{\partial \alpha}{\partial x} - \frac{\partial \beta}{\partial y} = 0 $$ and (I believe) we may write this as the system $$ \gamma(x,y) = \frac{\partial \alpha}{\partial x} \hspace{10mm} \gamma (x,y) = \frac{\partial \beta}{\partial y} $$ but I'm unsure of where to go from here.

Note: I have a check sheet for this question which states the answer as $$ U_x + U_y = 0 \hspace{10mm} V_x - V_y = 0 $$


Lauds to John Doe for an accurate and accepted answer. In the following, I have tried to explain why both the relations

$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y}, \; \dfrac{\partial \alpha}{\partial y} = \dfrac{\partial \beta}{\partial x} \tag 0$

are necessary and sufficient.

Given that

$\dfrac{\partial^2 \phi}{\partial x^2} - \dfrac{\partial^2 \phi}{\partial y^2} = 0, \tag 1$


$\dfrac{\partial^2 \phi}{\partial x^2} = \dfrac{\partial^2 \phi}{\partial y^2}, \tag 2$

then clearly by setting

$\alpha = \dfrac{\partial \phi}{\partial x} \tag 3$


$\beta = \dfrac{\partial \phi}{\partial y}, \tag 4$

we obtain the first-order equation

$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial^2 \phi}{\partial x^2} = \dfrac{\partial^2 \phi}{\partial y^2} = \dfrac{\partial \beta}{\partial y}, \tag 5$


$\dfrac{\partial \alpha}{\partial x} - \dfrac{\partial \beta}{\partial y} = 0. \tag 6$

However, (6) by itself is not yet a system of first-order equations equivalent to (1), (2); indeed, so far all (6) tells us is that we have a vector field $(\alpha, \beta)^T$ on some region of $\Bbb R^2$ the components of which satisfy this equation. What is not evident from (6), which we need to affirm the equivalence of (6) and (1), (2), is the existence of a scalar function $\phi$ such that (3) and (4) apply; that is, $(\alpha, \beta)^T$ is the gradient some $\phi$; we thus need $\phi$ such that

$(\alpha, \beta)^T = \nabla \phi. \tag 7$

It is well-known that such a $\phi$ exists if and only if

$\nabla \times (\alpha, \beta)^T = 0, \tag 8$


$\dfrac{\partial \alpha}{\partial y} - \dfrac{\partial \beta}{\partial x} = 0; \tag 9$

thus the equation (1), (2) is equivalent to the system (6), (9); that is, to

$\dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y}, \tag{10}$

$\dfrac{\partial \alpha}{\partial y} = \dfrac{\partial \beta}{\partial x}. \tag{11}$

To re-iterate, taking $\alpha$ and $\beta$ as in (3), (4), we see via (5) that (1)-(2) imply (6), (10); (11) follows from the fact that, for sufficiently differentiable $\phi$, the mixed partials satisfy

$\dfrac{\partial^2 \phi}{\partial x \partial y} = \dfrac{\partial^2 \phi}{\partial y \partial x}, \tag{12}$


$\dfrac{\partial \alpha}{\partial y} = \dfrac{\partial}{\partial y} \dfrac{\partial \phi}{\partial x} = \dfrac{\partial^2 \phi}{\partial y \partial x} = \dfrac{\partial^2 \phi}{\partial x \partial y} = \dfrac{\partial}{\partial x} \dfrac{\partial \phi}{\partial y} = \dfrac{\partial \beta}{\partial x}; \tag{13}$

having arrived at (10), (11) starting from (1), (2), we may walk in the other direction: for sufficiently differentiable functions $\alpha$, $\beta$, (11) implies that the vector field $(\alpha, \beta)^T$ is curl free on its region of definition in $\Bbb R^2$, hence must be the gradient of some function $\phi$, that is, (7) binds, and then we have

$\alpha = \dfrac{\partial \phi}{\partial x}, \tag{14}$


$\beta = \dfrac{\partial \phi}{\partial y}; \tag{15}$

thus, using (10),

$\dfrac{\partial^2 \phi}{\partial x^2} = \dfrac{\partial}{\partial x}\dfrac{\partial \phi}{\partial x} = \dfrac{\partial \alpha}{\partial x} = \dfrac{\partial \beta}{\partial y} = \dfrac{\partial}{\partial y}\dfrac{\partial \phi}{\partial y} = \dfrac{\partial^2 \phi}{\partial y^2}, \tag{16}$

whence follows (1).

I can't speak to our OP J Chapman's introduction of $U$ and $V$ with

$U_x + U_y = 0 = V_x - V_y \tag{17}$

since these functions are left undefined in the text of the question.

If we are give a sufficiently differentiable function $\phi$ of $x$ and $y$, then the symmetry relation of the mixed partials (12) immediately follows; thus (11) comes "for free", so to speak, when we define $(\alpha, \beta)^T = \nabla \phi$; it is in the attempt to define $\phi$ from $\alpha$ and $\beta$ satisfying (10), however, we discover that the assumption (11) is essential to ensure that the requisite $\phi$ exists.

Finally, I couldn't help but observe that the system (10)-(11) bears a strong resemblance to the Cauchy-Riemann equations of complex analysis; indeed, if instead of (11) we take

$\dfrac{\partial \alpha}{\partial y} = -\dfrac{\partial \beta}{\partial x}. \tag{18}$

then (10) and (18) are in fact a Cauchy-Riemann system for $\alpha + i \beta$, which is then holomorphic. Indeed, it follows from (10) and (18) that $\alpha$ and $\beta$ are harmonic:

$\dfrac{\partial}{\partial x} \dfrac{\partial \alpha}{\partial x} = \dfrac{\partial}{\partial x} \dfrac{\partial \beta}{\partial y} = \dfrac{\partial}{\partial y} \dfrac{\partial \beta}{\partial x} = -\dfrac{\partial}{\partial y} \dfrac{\partial \alpha}{\partial y}, \tag{19}$

that is,

$\nabla^2 \alpha = 0, \tag{20}$

with a similar derivation holding to show that $\beta$ obeys

$\nabla^2 \beta = 0. \tag{21}$

However, it the absence of the "$-$" sign occurring in (18), i.e. when (10) holds instead, there seems to be no analog of these facts for the resulting (hyperbolic) PDEs.

Finally, we note that $\alpha$ and $\beta$ also obey the equation (2); to see this, simply take $\partial / \partial x$ or $\partial / \partial y$ of that equation.

  • 1
    $\begingroup$ Excellent answer, +1. The Cauchy Riemann equations can be derived as a consequence of Stokes’ theorem which applies here. Problem 4-33 in Calculus on Manifolds (Spivak) explains how. $\endgroup$ – Mathemagical Jan 16 '18 at 2:57
  • $\begingroup$ @Mathemagical: yes I'm aware of how Stokes' leads to Cauchy-Riemann. Are you saying that Stoke's may be used in a similar way here? $\endgroup$ – Robert Lewis Jan 16 '18 at 7:31
  • 1
    $\begingroup$ No. I said it because I thought you may not have been aware. Because of your comment noting a strong resemblance without mentioning one can be derived from another. $\endgroup$ – Mathemagical Jan 16 '18 at 7:46
  • $\begingroup$ @Mathemagical: Got it. I had to stop typing someplace . . . ;) $\endgroup$ – Robert Lewis Jan 16 '18 at 7:56

I believe the second equation required is $$\frac{\partial \alpha}{\partial y}=\frac{\partial \beta}{\partial x}$$by symmetry of mixed partials.


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