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1. I start with simple differential inequality: find $u\in C^1[0,1]$ such that $u(0) = 0$ and $$ u'(t)\leq -u(t) $$ for all $t\in [0,1]$. Using Gronwall's lemma one can see that $u\leq 0$. On the other hand it seems to be the only solution, since this inequality keep $u$ non-decreasing for $u<0$ and non-increasing for $u>0$. Is it right that only $u=0$ satisfies this inequality?

2. With Gronwall's lemma one can see that any solution of $$ u'(t)\leq\beta(t)u(t) $$ is bounded from above by the solution of $$ u'(t) = \beta(t)u(t). $$

So there are two main results:

(i) there is a solution of $u'\leq \beta u$ which dominates any other solution.

(ii) this solution is attained on the correspondent equation.

Are there any similar results on PD inequalities of the type $$ u_t(t,x)\leq L_x u(t,x) $$ where $L_x$ is a differential operator in $x$ variable (of first or second order).

The main question for me if (i) is valid for such inequalities.

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2 Answers 2

up vote 5 down vote accepted

Related to your second question: I assume your functions are of the form $u: (a,b) \to L^2(\Omega)$. If you can get a bound on $(Lw, w)_2$ in terms of $\|w\|_2^2$ (as you can typically do by linearity, Holder's Inequality, Green's Identity, etc.) then assuming positivity you can do the following:

$$u_t(t, x) \le Lu(t, x)$$ $$1/2 u(t,x) u_t(t,x) \le 1/2 Lu(t,x) u(t,x)$$

$$\int_{\Omega} 1/2 u(t,x) u_t(t,x) dx \le 1/2(Lu(t), u(t))_2$$

$$d/dt \|u(t)\|_2^2 \le 1/2(Lu, u)_2 \le C\|u(t)\|_2^2$$

and now use Gronwall and the initial condition to get bounds on the $L^2$ norm of the solution. I don't know if that helps.

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Try $u(t)=\rm{e}^{-t}-1$.........................

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1  
Actually, $u(t)=-t$ will do. @Gortaur: I think $|u'(t)|\le u(t)$ makes for a more interesting problem -- is this perhaps what you had in mind? –  joriki Jun 10 '11 at 10:11
    
No, I was just confused and forget that sign of $u$ does not influence a sign of inequality. What about the second question? –  Ilya Jun 10 '11 at 12:18
1  
@Gortaur: I don't know anything about that, sorry. –  joriki Jun 11 '11 at 6:40

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