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I apologise, since this question does already exist. But I cannot for the life of me understand how it is solved. Could anyone provide me with a simplified version of the solution?

ktktktkt
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    That is exactly the question that I saw that exists. Unfortunately it does not. – ktktktkt Feb 13 '20 at 19:26
  • What do you know about solving partial differential equations? – Andrew Chin Feb 13 '20 at 19:28
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    What is the first part you don't understand? – Robert Israel Feb 13 '20 at 19:29
  • I am trying to understand the method of characteristics, and I understand that the solution remains a constant on the same curve. My tutor suggested solving the equations using such a curve but I don't really seem to understand him that well. – ktktktkt Feb 13 '20 at 19:30
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    What do you mean by a simplified version of the solution? The equation $u_x+u_y=1$ is specific example of the general equation $au_x+bu_y=c$. The theory for these equations is one of the topics covered at the beginning of a partial differential equations course. What are you confused about in regards to the method of characteristics? I could write a different solution to your problem but it might not help you if you haven't done at least one example on your own. – Axion004 Feb 13 '20 at 19:59
  • In my textbook, the method for solving $au_x + bu_y = 0$ is covered. However, the method for the inhomogeneous equation wasn't very well explained to me. I was hoping to get a more detailed explanation. For example, in this solution, https://stemjock.com/STEM%20Books/Strauss%20PDEs%202e/Chapter%201/Section%202/StraussPDEch1s2p09.pdf why does $y(\xi, 0)$ have parameters? – ktktktkt Feb 13 '20 at 20:01

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If you're using the method of characteristics: the charactistics here are the lines $x+y=$constant. It's not true that the solutions are constant on such a curve. Rather, the PDE tells you how the solution changes on each characteristic curve. In this case, the characteristic curves are $y = x + c$ for arbitrary constant $c$. If we write $u(x, x+c) = g(x)$, the PDE (together with the chain rule) says $$ \dfrac{d}{dx} g(x) = u_x(x,x+c) + u_y(x,x+c) = 1 $$ You can integrate that to get $$g(x) = x + \text{constant}$$ where the constant can depend on $c$, i.e. $$ u(x,x+c) = x + K(c)$$ where $K$ is arbitrary. Writing $x+c=y$, this says $$ u(x,y) = x + K(y-x)$$ $K$ is an arbitrary function, but should be differentiable for the pde to be defined.

Robert Israel
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  • Thank you, this makes sense to me. However, I would like to clarify your statement that "It's not true that the solutions are constant on such a curve." If the equation were homogeneous, i.e. for $u_x + u_y = 0$, would the solution $u(x, y)$ then be a constant on the characteristic curve? – ktktktkt Feb 13 '20 at 20:15
  • Yes, the solutions of $u_x + u_y = 0$ are constant on the characteristic curves. – Robert Israel Feb 13 '20 at 20:56
  • Perfect, thanks. – ktktktkt Feb 13 '20 at 20:57