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Let $u(x,y)$ be a continuously differentiable function on the closed unit disc and is a solution to $$a(x,y)u_x+b(x,y)u_y=-u,$$ on the closed unit disc. Suppose $$a(x,y)x+b(x,y)y>0,$$ on the boundary of the closed unit disc. $a,b$ are given smooth functions.

Prove that $u$ vanishes identically zero.

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You can consult the document : The solution to your question corresponds to Week2/3 of this document

If the document can no longer be accessed, here is the idea:

  • first, parametrize the domain and its boundary: it is a unit disk. The boundary, $\partial \Omega$, can be parametrized by: $x = \cos(\phi)$, $y= \sin(\phi)$. The normal derivative is given by : $\frac{\partial u}{\partial r} = u_x \cos(\phi) + u_y \sin(\phi)$. The tangential derivative is given by: $\frac{\partial u}{\partial r} = - u_x \sin(\phi) + u_y \cos(\phi)$, where $\phi$ is the angle.

  • for the interior domain $\Omega$: if there is a local maximum or minimum at P, $(x_P, y_P)$, as the function $u$ is class $C¹$, its partial derivatives along $x$ and $y$ are equal to zero at this point: $u_x(x_P, y_P)=0$ and $u_y(x_P, y_P)=0$. As the PDE is also verified at P, we must have $u(x_P, y_P)=a(x,y)*0 + b(x,y)*0=0$. So, as the maximum and the minimum of $u$ are equal to zero on $\Omega$ and $u$ is continuous (class $C¹$) , $u$ has no choice but to be equal to zero also on $\Omega$.

  • for the boundary $\partial \Omega$, if there is a maximum, then the normal derivative is positive and the tangential derivative is equal to zero. After some calculations, one can show that: $(a \cos \phi + b \sin \phi)(u_x \cos \phi + u_y \sin \phi) = -u$. The hypothesis is $(a \cos \phi + b \sin \phi) \ge 0$ on $\partial \Omega$. As the normal derivative is positive, it implies $u \le 0$. So, if there is a maximum on the boundary, $u \le 0$. In the smae way, we can demonstrate that if there is a minimum on the boundary, $u \ge 0$. So, as $u$ is continuous, $u$ is constantly equal to $0$ on $\partial \Omega$.

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