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$u_t+cu_x=-{\lambda}u : x, t > 0, u(x, 0) = 0 \forall x \in \Re$ and $u(0, t) = g(t) \forall t > 0$

I am confused on a particular step for $x

For $x

$e^{-\lambda{t}}\phi(0-ct)=g(t)$, so $\phi(-ct)=g(t)e^{-\lambda{t}}$.

We let $s=-ct$, so $\phi(s) = g(\dfrac{-s}{c})e^{\lambda{\dfrac{s}{c}}}$.

This is where I am confused. My professor said by an algebraic trick with $s$, we can replace s with $x-ct$ and get the solution:

$u(x, t)=e^{-\lambda{t}}e^{-\lambda(\dfrac{x-ct}{c})}g(-\dfrac{x-ct}{c})=e^{-\lambda\dfrac{x}{c}}g(-\dfrac{x-ct}{c})$

Hopefully I typed the solution correctly. Let me know if I messed up or missed any negative signs.

What is the algebraic trick? I did ask my professor, but he seemed frustrated with why I didn't know what the trick was.

  • 1
    What is $\phi$? – 2017-02-11

1 Answers 1

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I follow a slightly different approach when solving linear first order PDE's. So this may look a bit more involved to you. Skip to 'Summarizing' to see a somewhat more concise explanation of what I think was going on in your professor steps.

To solve this kind of equation I find extremely useful to use the change of variable $w(x,t)=e^{\lambda t} u(x,t)$. It is easy to verify that $w(x,t)$ satisfies

$$ w_t+cw_x=0,\quad w(x,0)=0,\quad w(0,t)=e^{-\lambda t}g(t). $$

Now, this is the simple transport equation, and the solution for $w$ is constant along the characteristics $x-ct=s$, so you can travel back on the characteristics until you hit either the initial or the boundary condition. If $x

$$ w(x,t)=w(0,t_0) $$ where $t_0$ is the point where the characteristic hits the boundary condition, and depends on both $x$ and $t$. Its value is easily computed:

$$ x-ct=s \Rightarrow 0-ct_0 = s = x-ct\Rightarrow t_0 -\frac{x-ct}{c}. $$

Therefore, for $x

Summarizing: I think your professor used a somewhat confusing notation. The first equation containing $\phi$ looked somewhat "given from above". Yes, $w$ is constant on lines of the form $x-ct$, and he basically called $\phi(x-ct)=w(x,t)$. But then from there he took it to a somewhat mystic route, hard to follow. In particular, when he writes $0-ct$, that $t$ is not a variable anymore, but it is the time at which the characteristic $x-ct$ hits the $t$-axis (what I called $t_0$). Letting $s=-ct$ is misleading then, cause it's actually $s=-ct_0$. The starting point should always be $s=x-ct$, with $s$ constant along the characteristic (the general equation of a characteristic), and when you're on the $t$-axis, then $x=0$ and $t=t_0$, leaving you with $s=-ct_0$. Then you can say

$$ e^{-\lambda t_0} \phi (-ct_0)=g(t_0) $$ and since $s=x-ct$ is constant over the characteristic, then $s=-ct_0$, so you can write

$$ e^{-\lambda t_0} \phi (x-ct) = g(t_0). $$ Then use $-ct_0=s=x-ct$ to solve for $t_0$ in terms of $x$ and $t$ to give a complete expression to the solution.