Solve $\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$ with initial condition $u(0,y)=0.$

Question

Solve the PDE $\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$ with initial condition $u(0,y)=0.$

I solved the PDE, the solution is $u(x,y)=-y\phi(3x-y)+\frac{x^2y}{2}$, where $\phi$ is a real arbitrary function.

And I am having trouble with the initial condition:

$u(0,y)=-y\phi(-y)=0$.

The issue is the $-y$ factor outside the function $\phi$.

What can I do from here in order to get the complete answer $u(x,y)$?

Any kind of help is very appreciated

Answer 1

You were close to the solution you just did I guess some mistakes $$\frac{\partial u}{\partial x}+3\frac{\partial u}{\partial y}=xy$$ The differential equations are $$\frac {dx}1=\frac {dy}3=\frac {dz}{xy}$$ $$ \begin{cases} \int 3dx= \int dy \implies y=3x+K_1 \\ xydy=dz \implies \int x(3x+K_1)dx=z+K_2 \\ \implies K_2=x^3+\frac {x^2K_1}2-z \end{cases} $$ Therefore $$\phi((y-3x)=\frac {-x^3}2+\frac {x^2y}2-z \implies u(x,y)=-\phi((y-3x)-\frac {x^3}2+\frac {x^2y}2$$ Initital condition $u(0,y)=0 \implies \phi(y-3x)=0$ Therefore the solution is $$\boxed{ u(x,y)=\frac {x^2}2( y-x)}$$

Answer 2

I'm not sure how you solved it in the first case so I'll write out a full solution. We use the method of characteristics. Characteristics $(x(s),y(s))$ originating from points $(0,\eta)$ for some $\eta \in \mathbb R$ must obey \begin{align*} \dot x(s) &= 1, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, x(0) = 0, \\ \dot y(s) &= 3, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, y(0) = \eta, \\ \dot z(s) &= x(s)y(s), \,\,\,\, z(0) = 0, \end{align*} where $z(s) = u(x(s),y(s)$. Solving the first two equations gives $x(s) = s$, $y(s) = 3s + \eta$ and thus $$\dot z(s) = 3s^2 + \eta s \,\,\,\, \implies z(s) - \underbrace{z(0)}_{\,\,\,\,\,= 0} = s^3 + \frac{\eta s^2}{2}.$$ Now if we can invert the relationships (i.e., solve for $(s,\eta)$ in terms of $(x,y)$) we will have $$u(x,y) = z(s(x,y); \eta(x,y)).$$ This is very easy to do: we have $s = x$ and $\eta = y - 3x$. Thus $$\boxed{u(x,y) = x^3 + \frac{x^2(y-3x)}{2}= \frac{1}{2}(x^2y - x^3)}.$$ Indeed, we can check the answer: it's clear that $u(0,y) = 0$ and $$\frac{\partial u}{\partial x} + 3 \frac{\partial u}{\partial y} = xy - \frac{3}{2}x^2 + 3\cdot\frac{1}{2}x^2 = xy.$$