Fie \( u, v:I=[a, b]\to\mathbb{R} \) doua functii continue, \( v\geq 0 \) si exista \( M\geq 0 \) astfel incat \( u(t)\leq M+\int_0^t u(s)v(s)ds \) pentru orice \( t\in [a,b] \).
Sa se arate ca \( u(t)\leq Me^{\int_0^t v(s)ds} \).
Lema Bellman-Gronwall
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
Cezar Lupu
- Site Admin
- Posts: 612
- Joined: Wed Sep 26, 2007 2:04 pm
- Location: Bucuresti sau Constanta
- Contact:
Lema Bellman-Gronwall
Iata, am sa postez o lema foarte utila in teoria ecuatiilor diferentiale ordinare si nu numai.
Fie \( u, v:I=[a, b]\to\mathbb{R} \) doua functii continue, \( v\geq 0 \) si exista \( M\geq 0 \) astfel incat \( u(t)\leq M+\int_0^t u(s)v(s)ds \) pentru orice \( t\in [a,b] \).
Sa se arate ca \( u(t)\leq Me^{\int_0^t v(s)ds} \).
Fie \( u, v:I=[a, b]\to\mathbb{R} \) doua functii continue, \( v\geq 0 \) si exista \( M\geq 0 \) astfel incat \( u(t)\leq M+\int_0^t u(s)v(s)ds \) pentru orice \( t\in [a,b] \).
Sa se arate ca \( u(t)\leq Me^{\int_0^t v(s)ds} \).
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
O solutie posibila
Inmultim cu \( v(t)>0 \) si obtinem \( u(t)v(t)\leq v(t) \left(M+\int_{0}^{t}u(s)v(s)ds \right) \)
Notam \( F(t)=\int_{0}^{t}u(s)v(s)ds \), atunci \( \frac{F^{,}(t)} {M+F(t)} \leq v(t) \Leftrightarrow ln \left( M+F(t) \right) ^{,} \leq v(t). \) Acum integram pe un interval \( [0,x] \).
\( ln(M+F(x))-ln M \leq \int_{0}^{x} v(t)dt \Leftrightarrow M+F(x) \leq Me^{\int_{0}^{x}v(t)dt} \) si tinand seama de ipoteza obtinem ca \( u(t)\leq M e^{\int_{0}^{x} v(t)dt}\ \)
Notam \( F(t)=\int_{0}^{t}u(s)v(s)ds \), atunci \( \frac{F^{,}(t)} {M+F(t)} \leq v(t) \Leftrightarrow ln \left( M+F(t) \right) ^{,} \leq v(t). \) Acum integram pe un interval \( [0,x] \).
\( ln(M+F(x))-ln M \leq \int_{0}^{x} v(t)dt \Leftrightarrow M+F(x) \leq Me^{\int_{0}^{x}v(t)dt} \) si tinand seama de ipoteza obtinem ca \( u(t)\leq M e^{\int_{0}^{x} v(t)dt}\ \)