# Inhomogeneous First Order PDE with boundary condition

I'm having some trouble solving the following PDE:

$$u_x+u_y=1, u(y,\frac{y}{2})=y$$

I know I can use the total differential to find a general solution:

$$\frac{du}{dx}=u_x+u_y\frac{dy}{dx}\\$$

So \begin{align} \frac{du}{dx}&=1\\ \frac{dy}{dx}&=1\\ u&=x+C \end{align} where $C$ is a constant. How do I use the boundary data from here to find the solution though?

• You obtained $u(x,y)=x+C$ . According to the condition $u(y,\frac{y}{2})=y$ put $x=y$ in it. So, $y+C=y$ which implies $C=0$. May 17, 2015 at 7:45

## 1 Answer

a more general solution to $u_x + u_y = 1$ is $u(x,y)= Ax + (1-A)y+C$

the use of the variable $y$ in the expression of the boundary condition is unfortunate , we could say $u(p, \frac{p}{2}) = p$ for all real numbers $p$.

plug that into the general solution, keeping in mind that this is an identity in $p$, by equating coefficients you should get that $C=0$ and $A=1$

so $$u(x,y)=x$$

• how do you arrive at that general solution? May 17, 2015 at 4:52