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I am currently working on solving the p.d.e.

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

with the initial condition

$$ u = 0 \text{ on } x + y = 0 $$

using the method of characteristics. Thus, so far, I have obtained a system of characteristic equations

$$ \begin{cases} \frac{dx}{dt} = 1 \implies x = t + x_0\\ \frac{dy}{dt} = 1 \implies y = t + y_0\\ \frac{du}{dt} = 1 \implies u = t + u_0 \end{cases} $$ However, I'm unsure how to proceed to encorporate the initial conditions or solve for $u(x, y)$ explicitly. Any help is appreciated.

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  • $\begingroup$ Eliminate $t$ from each of the equations i.e $$x-x_{0} = y-y_{0} = u-u_{0}$$ which implies $y = x + c_{1}$ and $u = x + f(c_{1}) = x + f(y-x)$. Now apply your boundary data. $\endgroup$ – Mattos Mar 10 at 1:36
  • $\begingroup$ How do you deduce $u = x + f(c_1)$? All that seems clear to me is $u = x - x_0 + u_0$. $\endgroup$ – Scriniary Mar 10 at 1:59
  • $\begingroup$ Because the implicit solution to the PDE problem is given by $$\phi(c_{1}, c_{2}) = 0 \implies \phi(y-x, u-x) = 0$$ which, in a different form, is $$u-x = f(y-x)$$ i.e $c_{2} = f(c_{1})$. $\endgroup$ – Mattos Mar 10 at 2:52

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