# Math Help - Help on solving first-order NONhomogeneous PDE

1. ## Help on solving first-order NONhomogeneous PDE

Hello everybody. I'm trying to solve this first-order NONhomogenous PDE. But because my awfully bad textbook only gave examples on how to solve homogeneous PDEs, I'm not sure what to do. Here's my scratch work.

I also know the theorem that if two solutions solve linear NONhomogeneous PDE, then their difference solve linear homogeneous PDE. But I don't think that's going to help here.

Problem: Solve $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$

Scratch work:
I know how to solve the HOMOgeneous PDE: $\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=0$

The LHS is just the directional derivative of $u$ in the direction of $[1, 1]$.

So if a curve in 3D has $[1,1]$ as a tangent vector, the curve's equation is just $\frac{dy}{dx} = 1$

so $y = x + C_1$ so $x - y = -C_1 = C$. So $u(x,y) = f(x - y)$, where $f$ is differentiable.

But how do I solve the NONhomogeneous question from this?

Thanks everybody.

2. ## Re: Help on solving first-order NONhomogeneous PDE

One solution is u = x. If you let $u = x + v$ you'll end up with $v_x + v_y = 0$.

BTW - do you know the method of characteristics?

3. ## Re: Help on solving first-order NONhomogeneous PDE

Originally Posted by Danny
One solution is u = x. If you let $u = x + v$ uou'll end up with $v_x + v+y = 0$.

BTW - do you know the method of characteristics?
Hello and thanks Danny. So $u(x,y) = x$ could've been just guessed?

But if $u(x,y) = x + v$, wouldn't you get $u_x + u_y = (1 + v_x) + 0$? Since $(x + v)_y = 0$ because there is no y in here.

And I don't think so since I'm taking a first course in PDEs.

4. ## Re: Help on solving first-order NONhomogeneous PDE

Well, you assme that $u(x,y) = v(x,y) + x$. The method of characteristics will probably come up. It's a way to solve

$a(x,y,u)u_x + b(x,y,u) u_y = c(x,y,u)$.

5. ## Re: Help on solving first-order NONhomogeneous PDE

Hello and thanks Danny.

Originally Posted by Danny
Well, you assme that $u(x,y) = v(x,y) + x$.
But when you say $v(x,y)$ here, do you mean $v(x,y) = f(x - y)$ and not just any $v(x,y)$ which I found in my original post for the homogeneous PDE?

And did I get $u_x + u_y = (1 + v_x) + 0$ right?

6. ## Re: Help on solving first-order NONhomogeneous PDE

Well, you would end up with that $v$ in the end. Also, if you let $u = v + x$ then

$u_x + u_y = 1$ gives $\left( v_x + 1\right) + \left( v_y + 0\right) = 1$ gives $v_x + v_y = 0$.