solve a PDE within a circle domain

Question

prove that the equation $$xu_x+yu_y = -u$$ in the domain $x^2+y^2 \le a^2$ has only one solution $u\equiv0$

i don't know how to proceed. I did figure out on the x and y axis the function u equals to 0.

Answer 1

Assuming the domain means the function must be defined on every point on the domain. Change the PDE into polar coordinates using

$$x\partial_x + y\partial_y = r\partial_r$$

So we get

$$ru_r = -u \implies u = \frac{f(\theta)}{r}$$

Since this function has a singularity at $r=0$, which is supposed to be in the domain, then the only solution is to set $f(\theta)=0$. But this is an interpretation I haven't seen given that in 2D, $\frac{1}{r}$ is a locally integrable function, so please check if there was more to the problem such as a boundary condition.

Answer 2

Given the equation

$xu_x + y u_y = -u \tag 1$

defined over the domain

$D = \{(x, y) \in \Bbb R^2, \; x^2 + y^2 \le a^2 \}, \tag 2$

we may assume $u(x, y) \in C^1(D)$, since the existence of the derivatives $u_x$ and $u_y$ of $u(x, y)$ is implicitly implied by (1), which we observe it may be written in the form

$(x, y) \cdot \nabla u = -u; \tag 3$

let $\mathbf e_r$ be the unit vector field in the radial direction; outward pointing along the radial lines; note $\mathbf e_r$ is not defined at $r = 0$; then

$r \ne 0 \Longrightarrow (x, y) = r\mathbf e_r, \tag 4$

whence

$r\mathbf e_r \cdot \nabla u = (x, y) \cdot \nabla u = - u; \tag 5$

thus

$r \dfrac{\partial u}{\partial r} = r u_r = r\mathbf e_r \cdot \nabla u = -u; \tag 6$

the equation

$r \dfrac{\partial u}{\partial r} = -u \tag 7$

is readily solved along radial lines, on each of which it may be treated as an ordinary differential equation since the parameter $\theta$ is fixed; furthermore, we may assume $u(r) \ne 0$ at any radial value $r$, for if $u(r) = 0$ for some $r$ along a given radius, then $u(r) = 0$ identically along this radius, by uniqueness of solutions of the ordinary differential equation (7); and since equations (1), (3), (5)-(7) are sign-invariant, we may take $u(r) > 0$; these observations then legitimize the following calculations, as long as $r \ne 0$:

$\dfrac{\partial \ln u}{\partial r} = \dfrac{1}{u} \dfrac{\partial u}{\partial r} = -\dfrac{1}{r} = -r^{-1}; \tag 8$

$\ln \left (\dfrac{u(r)}{u(r_0)}\right ) = \ln u(r) - \ln u(r_0) = \displaystyle \int_{r_0}^r \dfrac{\partial \ln u(s)}{\partial s} \; ds = -\int_{r_0}^r s^{-1} \; ds = \ln r_0 - \ln r = \ln \left (\dfrac{r_0}{r} \right ) = \ln (r_0r^{-1}); \tag 9$

$\dfrac{u(r)}{u(r_0)} = r_0r^{-1}, \tag{10}$

$u(r) = u(r_0)r_0r^{-1}, \tag{11}$

which shows that such $u(x, y)$ must become arbitrarily large as $r \to 0$; but since $u(x, y)$ is a differentiable, hence continuous, function on the compact set (2), it is bounded on this set; thus (11) is impossible an we conclude the only solution to (1) on (2) is

$u(x, y) = 0. \tag{12}$

$OE\Delta$.