PDEMOC16

Consider the problem $u x + u y + u = e x + 2 y$ with initial conditions: $u (x,0) = 0$

We may write this as $(1,1)\cdot\nabla u = e^{x+2y} - u$ , with the initial curve as $Γ(s 1,0,0)$

So... $\partial_t x = 1$ and $\partial_t y = 1$

Note: $\partial_t \equiv \frac{\partial}{\partial t}$

This implies $x = t + C 1$ and $y = t + C 2$

Applying initial conditions: $x (0) = C 1 = s 1$ and $y (0) = C 2 = 0$

So, now $x = t + s 1$ and $y = t$

The characteristic curves are given by the following equation: $x - y = s 1$

Now, $\partial_t u + u = e^{3t + s_1}$ , applying the integrating factor $e t$

$e^t\partial_t u + e^t u = \partial_t\left(e^t u\right) = e^{4t +s_1}$

$e^t u = \frac{1}{4}e^{4t +s_1} + C$

And now applying initial conditions...

$u(0) = \frac{1}{4}e^{s_1} + C = 0$

So finally,

$u = \frac{1}{4}\left(e^{3t + s_1} - e^{s_1 - t}\right)$

or $u(x,y) = \frac{1}{4}\left(e^{x + 2y} - e^{x - 2y}\right)$