# Thread: 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 $\displaystyle \frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}=1$

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

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

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

so $\displaystyle y = x + C_1$ so $\displaystyle x - y = -C_1 = C$. So $\displaystyle u(x,y) = f(x - y)$, where $\displaystyle f$ 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 $\displaystyle u = x + v$ you'll end up with $\displaystyle v_x + v_y = 0$.

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 $\displaystyle u = x + v$ uou'll end up with $\displaystyle v_x + v+y = 0$.

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

But if $\displaystyle u(x,y) = x + v$, wouldn't you get $\displaystyle u_x + u_y = (1 + v_x) + 0$? Since $\displaystyle (x + v)_y = 0$ 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 $\displaystyle u(x,y) = v(x,y) + x$. The method of characteristics will probably come up. It's a way to solve

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

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

Hello and thanks Danny.

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

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

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

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

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