Solve $3u_y+u_{xy}=0$

Question

Solve $3u_y+u_{xy}=0$, does a solution exist with conditions $u(x,0)=0$ and $u_y(x,0)=0$

I made substitution $v=u_y$

To get $3v+v_x=0$

Which is solved by $v={3}e^{-3x}$

So $u_y= e^{-3x}$

then $e^{-3x}+u_{xy}=0$

$u_y=\frac{1}{3}e^{-3x}+g(y)$

$u=\frac{1}{3}e^{-3x}y+xg(y)+f(x)$

I'm not sure what to do with the initial conditions, I get $u_y=\frac{1}{3}e^{-3x}+g(0)=0$

So $g(0) = \frac{1}{3}e^{-3x}$

and $u(x,0)=e^{-3x}=0-\frac{1}{3}xe^{-3x}+f(x)$ so $f(x)=\frac{1}{3}xe^{-3x}-e^{-3x}$

But I still need to find a solution of $g(y)$to find $u$.

Answer 1

The general solution to $v_x = -3v$ is $v(x,y) = \phi(y)e^{-3x}$ where $\phi$ some function. Thus $u_y(x,y) = \phi(y)e^{-3x}$ leads to $\Phi(y)e^{-3x} + C$ where $\Phi$ is an antiderivative of $\phi$ and $C$ is a constant. Now, $$u(x,0) = \Phi(0) e^{-3x} + C$$ and $$u_y(x,0) = -3\phi(0) e^{-3x}.$$ so $u_y(x,0) = 0$ forces $\phi(0) = 0$ and $\Phi(0) e^{-3x} + C = 0$ (for all $x$) forces $\Phi(0) = 0$ and $C = 0$.

Thus any function of the form $u(x,y) = \Phi(y)e^{-3x}$ is a solution provided that $\Phi$ is differentiable, $\Phi(0) = 0$, and $\Phi'(0) = 0$.

Answer 2

Let $v = u_y$. Then:

$$3u_y+u_{xy}=0 \Rightarrow 3v + v_x = 0 \Rightarrow v_x = -3v \Rightarrow u_y = Ae^{-3x}.$$

Additionally, given an arbitrary point $(x_1, y_1)$, we have that:

$$u(x,y)-u(x_1,y_1) = \int_{y_1}^{y}u_y(x,s) ds = \int_{y_1}^{y} A e^{-3x} ds = Ae^{-3x}(y-y_1).$$

By choosing $y_1 = 0$, we obtain:

$$\begin{cases} u(x,y) &= Ae^{-3x}y+u(x_1,0)\\ u_y(x,y) &= Ae^{-3x} \end{cases}.$$

Using the boundary condition $u(x,0) = 0$, we get:

$$\begin{cases} u(x,y) &= Ae^{-3x}y\\ u_y(x,y) &= Ae^{-3x} \end{cases}.$$

Finally, $u_y(x,0)=0$ yields to:

$$u_y(x,0) = Ae^{-3x} = 0 \Rightarrow A = 0 \Rightarrow \begin{cases} u(x,y) &= 0\\ u_y(x,y) &=0 \end{cases}.$$