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I need to find a solution to the I.V.P:

$$ u_y + uu_x = -au, a > 0 \\ u(x,0) = f(x), x \in \Bbb{R} $$

After that i need to find conditions in which $f(x)$ is singular, for :

$$ f(x) = \left\{\begin{matrix} 1 & x<0\\ 1-x & 0<x<1\\ 0 & x>1 \end{matrix}\right. $$

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I used the mehod of characteristics to solve this PDE, getting: $$ u = e^{-at}f(s) \\ y = t \\ x = -\frac{e^{-at}}{a}f(s) + s + \frac{f(s)}{a} $$

Now, i try to isolate $s$ from $x$ equation in order to substitute it in $f(s)$, yet i dont succeed, it stays dependent on $f(s)$..

$$ s = x + \frac{e^{-at}}{a}f(s) - \frac{f(s)}{a} $$

If i can’t substitute $s$, how can i proceed? Maybe i did something wrong?

Thank you.

K.defaoite
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Alon
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1 Answers1

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This looks correct. Using $u = e^{-at}f(s)$ we can write the solution in implicit form $$ u = e^{-at}\, f\!\left(x - \frac{e^{at}-1}{at} ut\right) . $$ Thus,

  • where $f(s) = 0$ we know $u=0$ for $s=x>1$;
  • where $f(s) = 1$ we know $u=e^{-at}$ for $s=x-\frac{1-e^{-at}}{a} < 0$;
  • between the curves $x=1$ and $x = \frac{1-e^{-at}}{a}$, we have $$ u = e^{-at}\left[ 1 - \left(x - \frac{e^{at}-1}{at} ut\right) \right] $$ which gives us an explicit expression $$ u = e^{-at}\frac{1-x}{1-\frac{1-e^{-at}}{a}}, \qquad \tfrac{1-e^{-at}}{a} < x < 1 $$ upon solving with respect to $u$.

Now ask yourself where this piecewise solution is well-defined (sub-domain bounds, vanishing denominator, etc.). You will find useful complements for other $f$ in this post.

EditPiAf
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