What rationale am I missing with this simple boundary value problem?

Question

Consider $$u_{xy} = xy,\ u(0,\ y) = 0,\ u_x(x,\ 0) = 0$$

Antidifferentiating the PDE with respect to $y$ yields $u_x = \frac{xy^2}{2} + a(x)$, and antidifferentiating this equation with respect to $x$ yields $u = \frac{x^2 y^2}{4} + A(x) + b(y)$, where $A(x)$ is the antiderivative of $a(x)$ with respect to $x$.

Now the boundary conditions produce the system $\begin{cases}0 = A(0) + b(y) \\ 0 = a(x)\end{cases}$. Antidifferentiating the second equation, $A(x) = C$, where $C$ is constant with respect to both $x$ and $y$ (because $A(x) = C(y)$ makes no sense, except in a trivial way when $C$ is constant). Plugging this into the the first equation, $0 = C + b(y)$, or lumping the minus sign, $b(y) = C$. This suggests the solution $u = \frac{x^2 y^2}{4} + C$, which might be thought to require another boundary condition to pin down $C$.

However, only $C = 0$ seems to work when checking the solution. Where is the mistake in my reasoning which leads to the conclusion that any $C$ should solve the problem?

Answer 1

$$u(x,y) = \frac{x^2 y^2}{4} + A(x) + b(y) \quad\text{is OK.}$$

Condition $u(0,y) = 0$ : $$u(0,y) = A(0) + b(y) \quad\implies\quad b(y)=-A(0)=\text{constant}$$ $$u(x,y) = \frac{x^2 y^2}{4} + A(x) -A(0)$$ Condition $u_x(x, 0) = 0$ :

$u_x(x,y)=\frac{x y^2}{2} + A'(x)$ $$u_x(x,0)= A'(x)=0$$ $$A(x)=\text{constant}=C\quad\implies\quad A(0)=C$$ $$u(x,y) = \frac{x^2 y^2}{4} + C -C$$ $$\boxed{u(x,y) = \frac{x^2 y^2}{4}}$$

Answer 2

You have used only the condition $0=a(x)$. Now use also the other condition $0= C+b(y)$

Answer 3

After you have lumped the minus sign into $b(y)=C$, putting this back into the first equation gives $0=A(0)+C$, so $0=C+C \implies C=0$.