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I'm stuck with the following problem.

$u_t+au_x=b(x,t),\ x \in \mathbb{R}, t>0 \\ u(x,0)=\Phi(x), x \in \mathbb{R}$

Here, $a$ is constant, and $b,\Phi$ are smooth functions. Now we assume some error in $\Phi$ and define

$v_t+av_x=b(x,t) \\ v(x,0)=\Phi(x)+\epsilon(x).$

I shall show that:

$sup_{x\in\mathbb{R},t>0} |u(x,t)-v(x,t)|=sup_{x\in\mathbb{R}}\epsilon(x).$

Can somebody give me a hint on the approach?

thank you very much! -marie

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The function $v-u$ solves a homogeneous hyperbolic equation with $\epsilon$ as initial function. Does it help? –  abatkai Feb 20 '12 at 20:34
    
Not quite, we haven't discussed hyperbolic equations yet, but I'll try to continue from there. But is there another way? –  Marie. P. Feb 20 '12 at 20:39
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1 Answer 1

up vote 1 down vote accepted

As I menitioned in my comment, consider the homogeneous equation $w_t + aw_x =0$.

Taking any differentiable function $f:\mathbb{R}\to\mathbb{R}$, $w(x,t):=f(t-ax)$ will be a solution of it (d'Alembert formula). Since it is a linear equation, the Cauchy problem will be unique if $w(x,0)=f(x)$ is bounded and differentiable. This should give you the answer.

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Thanks, now I got it! –  Marie. P. Feb 20 '12 at 21:35
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