Distributional solution of $frac12(ah)''-(bh)'=0$
Let $b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous and at most of linear growth. Moreover, let $a:=sigma^2$ and $$s:=int_0^{;cdot;}expleft(-2int_0^yfrac ba(x):{rm d}xright){rm d}y.$$ Assume $s(mathbb R)=mathbb R$ (are there nice sufficient conditions on $b,sigma$ that ensure this?). Now, let $g:mathbb Rto[0,infty)$ be Borel measurable with $$int g(x):{rm d}x=1.tag1$$
If $bin C^1(mathbb R)$ and $sigmain C^2(mathbb R)$, then $$L^ast h:=frac12(ah)''-(bh)';;;text{for }hin C^2(mathbb R).$$ is well-defined. Assuming $gin C^2(mathbb R)$, it's straightforward to deduce that $$L^ast g=0Leftrightarrow g=frac caexpleft(2int_0^{;cdot;}frac ba(x):{rm d}xright)tag2.$$ However, the expression found for $g$ would even be well-defined without the assumptions $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$ and $gin C^2(mathbb R)$. We simply need to note that linear growth assumptions yield local integrability of $asigma$ and $bsigma$. Further, we need to assume that $b/a$ is locally integrable. $L^ast g=0$ has then to be understood in the distributional sense. Is it possible to show $(2)$ in that setting too?
integration functional-analysis differential-equations distribution-theory
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
add a comment |
Let $b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous and at most of linear growth. Moreover, let $a:=sigma^2$ and $$s:=int_0^{;cdot;}expleft(-2int_0^yfrac ba(x):{rm d}xright){rm d}y.$$ Assume $s(mathbb R)=mathbb R$ (are there nice sufficient conditions on $b,sigma$ that ensure this?). Now, let $g:mathbb Rto[0,infty)$ be Borel measurable with $$int g(x):{rm d}x=1.tag1$$
If $bin C^1(mathbb R)$ and $sigmain C^2(mathbb R)$, then $$L^ast h:=frac12(ah)''-(bh)';;;text{for }hin C^2(mathbb R).$$ is well-defined. Assuming $gin C^2(mathbb R)$, it's straightforward to deduce that $$L^ast g=0Leftrightarrow g=frac caexpleft(2int_0^{;cdot;}frac ba(x):{rm d}xright)tag2.$$ However, the expression found for $g$ would even be well-defined without the assumptions $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$ and $gin C^2(mathbb R)$. We simply need to note that linear growth assumptions yield local integrability of $asigma$ and $bsigma$. Further, we need to assume that $b/a$ is locally integrable. $L^ast g=0$ has then to be understood in the distributional sense. Is it possible to show $(2)$ in that setting too?
integration functional-analysis differential-equations distribution-theory
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53
add a comment |
Let $b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous and at most of linear growth. Moreover, let $a:=sigma^2$ and $$s:=int_0^{;cdot;}expleft(-2int_0^yfrac ba(x):{rm d}xright){rm d}y.$$ Assume $s(mathbb R)=mathbb R$ (are there nice sufficient conditions on $b,sigma$ that ensure this?). Now, let $g:mathbb Rto[0,infty)$ be Borel measurable with $$int g(x):{rm d}x=1.tag1$$
If $bin C^1(mathbb R)$ and $sigmain C^2(mathbb R)$, then $$L^ast h:=frac12(ah)''-(bh)';;;text{for }hin C^2(mathbb R).$$ is well-defined. Assuming $gin C^2(mathbb R)$, it's straightforward to deduce that $$L^ast g=0Leftrightarrow g=frac caexpleft(2int_0^{;cdot;}frac ba(x):{rm d}xright)tag2.$$ However, the expression found for $g$ would even be well-defined without the assumptions $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$ and $gin C^2(mathbb R)$. We simply need to note that linear growth assumptions yield local integrability of $asigma$ and $bsigma$. Further, we need to assume that $b/a$ is locally integrable. $L^ast g=0$ has then to be understood in the distributional sense. Is it possible to show $(2)$ in that setting too?
integration functional-analysis differential-equations distribution-theory
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
Let $b,sigma:mathbb Rtomathbb R$ be Lipschitz continuous and at most of linear growth. Moreover, let $a:=sigma^2$ and $$s:=int_0^{;cdot;}expleft(-2int_0^yfrac ba(x):{rm d}xright){rm d}y.$$ Assume $s(mathbb R)=mathbb R$ (are there nice sufficient conditions on $b,sigma$ that ensure this?). Now, let $g:mathbb Rto[0,infty)$ be Borel measurable with $$int g(x):{rm d}x=1.tag1$$
If $bin C^1(mathbb R)$ and $sigmain C^2(mathbb R)$, then $$L^ast h:=frac12(ah)''-(bh)';;;text{for }hin C^2(mathbb R).$$ is well-defined. Assuming $gin C^2(mathbb R)$, it's straightforward to deduce that $$L^ast g=0Leftrightarrow g=frac caexpleft(2int_0^{;cdot;}frac ba(x):{rm d}xright)tag2.$$ However, the expression found for $g$ would even be well-defined without the assumptions $bin C^1(mathbb R)$, $sigmain C^2(mathbb R)$ and $gin C^2(mathbb R)$. We simply need to note that linear growth assumptions yield local integrability of $asigma$ and $bsigma$. Further, we need to assume that $b/a$ is locally integrable. $L^ast g=0$ has then to be understood in the distributional sense. Is it possible to show $(2)$ in that setting too?
integration functional-analysis differential-equations distribution-theory
integration functional-analysis differential-equations distribution-theory
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
edited Dec 28 '18 at 10:36
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
asked Dec 28 '18 at 0:28
0xbadf00d
1,75441430
1,75441430
When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53
add a comment |
When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53
When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53
When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53
add a comment |
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When I try to solve the equation I don't get the same solution as you have but $$g = e^{-F} int frac{c}{a} e^F,$$ where $$F = -2 int frac{b}{a}.$$ Here $int f$ denotes a primitive function of $f.$
– md2perpe
Dec 29 '18 at 23:53