Question about 2D PDE

Question

Suppose I have this PDE $$ \frac{\partial A(r,\theta)}{\partial r} + \frac{\partial B(r,\theta)}{\partial \theta} = 0 $$ Is it ok to conclude (without losing any generality) that \begin{align} A(r,\theta) &= \frac{\partial \psi(r,\theta)}{\partial \theta} \\ B(r,\theta) &= -\frac{\partial \psi(r,\theta)}{\partial r} \end{align} for some unknown function $\psi$? This would allow me to write the PDE in terms of one unknown instead of 2. Of course I am assuming all functions are smooth, differentiable, etc.

Answer 1

It's OK locally, that is, as long as $A(r, \theta)$ and $B(r, \theta)$ are considered to be defined in some open ball $B(p, \epsilon) = \{ q \in \Bbb R^2 \mid \vert q - p \vert \ < \epsilon \}$ about some point $p = (r_0, \theta_0) \in \Bbb R^2$.

I will use subscript notation for partial derivatives throughout; thus

$A_r = \dfrac{\partial \psi}{\partial r}, \; B_\theta = \dfrac{\partial B}{\partial \theta}, \tag 0$

and so forth.

I will exploit the language of differential forms.

Consider the one form defined on $B(p, \epsilon)$,

$\omega(r, \theta) = -B(r, \theta) \; dr + A(r, \theta) \; d\theta; \tag 1$

we have

$d\omega= -dB \wedge dr + dA \wedge d \theta$ $= -(B_r \; dr \wedge dr + B_\theta \; d \theta \wedge dr) + A_r \; dr \wedge d\theta + A_\theta \; d\theta \wedge d\theta; \tag 2$

since

$dr \wedge dr = d\theta \wedge d\theta = 0, \tag 3$

(2) becomes

$d\omega = -B_\theta \; d\theta \wedge dr + A_r \; dr \wedge d\theta$ $= B_\theta \; dr \wedge d\theta + A_r \; dr \wedge d\theta = (B_\theta + A_r) \; dr \wedge d\theta = 0; \tag 4$

thus $\omega$ is a closed one-form; as such it is also exact, by the Poincare Lemma; thus there is a zero-form, or function, $\psi$ on $B(p, \epsilon)$ such that

$-Bd\theta + A dr = \omega = d\psi = \psi_r \; dr + \psi_\theta \; d\theta; \tag 5$

then

$A = \psi_r \tag 6$

and

$-B = \psi_\theta, \tag 7$

or

$B = -\psi_\theta. \tag 8$

Of course, all of this may be stated in a more old-school style by using the notation of vector calculus; the reader may easily check that the condition

$A_r + B_\theta = 0 \tag{9}$

is sufficient to conclude that

$\nabla \times (-B, A) = (\partial/\partial r, \partial/\partial \theta) \times (-B, A) = 0; \tag{10}$

thus $(-B, A)$ must be a gradient, that is,

$(-B, A) = (\psi_\theta, \psi_r), \tag{11}$

which is the requisite result.

Answer 2

That would be nice, but it's not ok, I'm afraid. The problem is that writing $A$ and $B$ that way forces a relationship between $A$ and $B$ that the original PDE does not assume.

Now, if you have additional information, like you're supposed to have harmonic functions, that would be another matter.