Side conditions don't make sense

Question

I have a function of two variables $u(x,y)$. The side conditions are $u(x,1) = 0$ and $u(0,y) = 0$. The differential equation is $$yu_{xy} + 2u_x = x.$$ I solved it to get a solution of the form $$u = \frac{x^2}{4} + \frac{F(y)}{y} - \frac{G(y)}{y^2},$$
where $ F(y)$ is the derivative of $G(y)$. I first integrate with respect to $x$ and get $$yu_y + 2u = \frac{x^2}{2} + f(y),$$ where $f(y)$ is an arbitrary constant as a function of $y$. then I solve by treating $x$ as a constant and solve it using ode techniques, which gives $$u = \frac{x^2}{4} + \frac{F(y)}{y} - \frac{G(y)}{y^2}. $$ I can't figure out if there is any other way to do it as according to the side conditions, this is wrong.

Answer 1

Let me replicate your computation, but in a way that is aware of the boundary conditions.

Integrating both sides with respect to $x$ from $0$ to $x$,

$$ y (u_y(x, y) - u_y(0, y)) + 2(u(x, y) - u(0, y)) = \frac{x^2}{2}. $$

Invoking the boundary condition $u(0, y) = 0$ and assuming that $u$ is $C^1$ near $x = 0$, we get $u_y(0, y) = \frac{\partial}{\partial y} u(0, y) = 0$. Hence, the above equality reduces to

$$ y u_y + 2u = \frac{x^2}{2}. $$

Now multiplying the integrating factor $y$ to both sides and integrating with respect to $y$ from $1$ to $y$,

$$ y^2 u(x, y) - u(x, 1) = \frac{(y^2 - 1)x^2}{4}. $$

Again, plugging the boundary condition $u(x, 1) = 0$ into the above equality,

$$ u = \frac{(1 - y^{-2})x^2}{4}. \tag{*} $$

Conversely, if $u$ is given by $\text{(*)}$, then $u$ satisfies the differential equation $ yu_{xy} + 2u_x = x$.

Answer 2

$$yu_y + 2u = \frac{x^2}{2} + f(y)\quad \text{is OK.}$$ You made a mistake in integrating wrt $y$. The result is not $u = \frac{x^2}{4} + \frac{F(y)}{y} - \frac{G(y)}{y^2}$ but is : $$u = \frac{x^2}{4} + \frac{F(y)}{y} - \frac{G(x)}{y^2}$$ or equivalently : $$u = \frac{x^2}{4} + \Phi(y) + \frac{\Psi(x)}{y^2}$$ Condition $u(x,1)=0=\frac{x^2}{4} + \Phi(1) + \Psi(x)\quad\implies\quad \Psi(x)=-\frac{x^2}{4} -\Phi(1)$

Condition $u(0,y)=0=\Phi(y) + \frac{\Psi(0)}{y^2}\quad\implies\quad \Phi(y)=-\frac{\Psi(0)}{y^2}$ $$u = \frac{x^2}{4} -\frac{\Psi(0)}{y^2} + \frac{-\frac{x^2}{4} -\Phi(1)}{y^2}=\frac{x^2}{4}\left(1-\frac{1}{y^2} \right)-\frac{\Psi(0)+\Phi(1)}{y^2}$$ $u(0,y)=0=-\frac{\Psi(0)+\Phi(1)}{y^2}\quad\implies\quad \Psi(0)+\Phi(1)=0$. $$u(x,y)=\frac{x^2}{4}\left(1-\frac{1}{y^2}\right)$$