Mixing
Integrability of bounded functions
This question arises from my studies concerning first order partial differential equations, in particular from the definition of weak solution for a conservation law.
Let $q in C^1(mathbb{R})$ and $u=u(t,x)$ be a bounded function on $[0,+infty)times mathbb{R}$. Let $v=v(t,x)in C^1([0,+infty)timesmathbb{R})$ with compact support.
What are sufficient conditions for the function $[uv_t+ (q(u))v_x]$ to be integrable on $(0,+infty)times mathbb{R}$, ie the (improper) integral $$int_{0}^{+infty}dtint_{-infty}^{+infty}[uv_t+ (q(u))v_x] ,dx$$ exists?
Note
With $v_t,v_x$ I mean partial derivative of $v$ with respect to $t,x$.
Thanks a lot in Advance.
real-analysis integration improper-integrals riemann-integration
asked Nov 21 '18 at 7:15
eleguitar
122114
add a comment |
This question arises from my studies concerning first order partial differential equations, in particular from the definition of weak solution for a conservation law.
Let $q in C^1(mathbb{R})$ and $u=u(t,x)$ be a bounded function on $[0,+infty)times mathbb{R}$. Let $v=v(t,x)in C^1([0,+infty)timesmathbb{R})$ with compact support.
What are sufficient conditions for the function $[uv_t+ (q(u))v_x]$ to be integrable on $(0,+infty)times mathbb{R}$, ie the (improper) integral $$int_{0}^{+infty}dtint_{-infty}^{+infty}[uv_t+ (q(u))v_x] ,dx$$ exists?
Note
With $v_t,v_x$ I mean partial derivative of $v$ with respect to $t,x$.
Thanks a lot in Advance.
real-analysis integration improper-integrals riemann-integration
asked Nov 21 '18 at 7:15
eleguitar
122114
add a comment |
This question arises from my studies concerning first order partial differential equations, in particular from the definition of weak solution for a conservation law.
Let $q in C^1(mathbb{R})$ and $u=u(t,x)$ be a bounded function on $[0,+infty)times mathbb{R}$. Let $v=v(t,x)in C^1([0,+infty)timesmathbb{R})$ with compact support.
What are sufficient conditions for the function $[uv_t+ (q(u))v_x]$ to be integrable on $(0,+infty)times mathbb{R}$, ie the (improper) integral $$int_{0}^{+infty}dtint_{-infty}^{+infty}[uv_t+ (q(u))v_x] ,dx$$ exists?
Note
With $v_t,v_x$ I mean partial derivative of $v$ with respect to $t,x$.
Thanks a lot in Advance.
real-analysis integration improper-integrals riemann-integration
asked Nov 21 '18 at 7:15
eleguitar
122114
This question arises from my studies concerning first order partial differential equations, in particular from the definition of weak solution for a conservation law.
Let $q in C^1(mathbb{R})$ and $u=u(t,x)$ be a bounded function on $[0,+infty)times mathbb{R}$. Let $v=v(t,x)in C^1([0,+infty)timesmathbb{R})$ with compact support.
What are sufficient conditions for the function $[uv_t+ (q(u))v_x]$ to be integrable on $(0,+infty)times mathbb{R}$, ie the (improper) integral $$int_{0}^{+infty}dtint_{-infty}^{+infty}[uv_t+ (q(u))v_x] ,dx$$ exists?
Note
With $v_t,v_x$ I mean partial derivative of $v$ with respect to $t,x$.
Thanks a lot in Advance.
real-analysis integration improper-integrals riemann-integration
real-analysis integration improper-integrals riemann-integration
asked Nov 21 '18 at 7:15
eleguitar
122114
asked Nov 21 '18 at 7:15
eleguitar
122114
asked Nov 21 '18 at 7:15
eleguitar
122114
asked Nov 21 '18 at 7:15
eleguitar
122114
asked Nov 21 '18 at 7:15
eleguitar
122114
122114
add a comment |
add a comment |
1 Answer
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The integral exists in Lebsgue sense without any further hypothesis. Since $int_0^{N}int_{-N}^{N}$ in the Lebesgue sense coincides with the Riemann integral it follows that the improper Riemann integral also exists.
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
add a comment |
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1 Answer
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1 Answer
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active
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The integral exists in Lebsgue sense without any further hypothesis. Since $int_0^{N}int_{-N}^{N}$ in the Lebesgue sense coincides with the Riemann integral it follows that the improper Riemann integral also exists.
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
add a comment |
The integral exists in Lebsgue sense without any further hypothesis. Since $int_0^{N}int_{-N}^{N}$ in the Lebesgue sense coincides with the Riemann integral it follows that the improper Riemann integral also exists.
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
add a comment |
The integral exists in Lebsgue sense without any further hypothesis. Since $int_0^{N}int_{-N}^{N}$ in the Lebesgue sense coincides with the Riemann integral it follows that the improper Riemann integral also exists.
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
The integral exists in Lebsgue sense without any further hypothesis. Since $int_0^{N}int_{-N}^{N}$ in the Lebesgue sense coincides with the Riemann integral it follows that the improper Riemann integral also exists.
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
answered Nov 21 '18 at 7:54
Kavi Rama Murthy
51.1k31854
51.1k31854
add a comment |
add a comment |
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