Proof: $F(x,t) := t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t})$ is solution of $Delta F - frac{partial F}{partial...
How can one prove that the function
$F: mathbb{R}^n times mathbb{R}^+ text{ {0}} to mathbb{R}$ with
$$F(x,t) := t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t})$$
is a solution of the partial differential equation
$$Delta F - frac{partial F}{partial t} = 0$$
I know that this can be done using the Laplace operator, which is given by $Delta := sum_{i=1}^n frac{partial^2}{partial x_i^2}$, but I don't know what the partial derivative of $Delta F - frac{partial F}{partial t} = 0$ looks like.
analysis functions partial-derivative
add a comment |
How can one prove that the function
$F: mathbb{R}^n times mathbb{R}^+ text{ {0}} to mathbb{R}$ with
$$F(x,t) := t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t})$$
is a solution of the partial differential equation
$$Delta F - frac{partial F}{partial t} = 0$$
I know that this can be done using the Laplace operator, which is given by $Delta := sum_{i=1}^n frac{partial^2}{partial x_i^2}$, but I don't know what the partial derivative of $Delta F - frac{partial F}{partial t} = 0$ looks like.
analysis functions partial-derivative
Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33
add a comment |
How can one prove that the function
$F: mathbb{R}^n times mathbb{R}^+ text{ {0}} to mathbb{R}$ with
$$F(x,t) := t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t})$$
is a solution of the partial differential equation
$$Delta F - frac{partial F}{partial t} = 0$$
I know that this can be done using the Laplace operator, which is given by $Delta := sum_{i=1}^n frac{partial^2}{partial x_i^2}$, but I don't know what the partial derivative of $Delta F - frac{partial F}{partial t} = 0$ looks like.
analysis functions partial-derivative
How can one prove that the function
$F: mathbb{R}^n times mathbb{R}^+ text{ {0}} to mathbb{R}$ with
$$F(x,t) := t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t})$$
is a solution of the partial differential equation
$$Delta F - frac{partial F}{partial t} = 0$$
I know that this can be done using the Laplace operator, which is given by $Delta := sum_{i=1}^n frac{partial^2}{partial x_i^2}$, but I don't know what the partial derivative of $Delta F - frac{partial F}{partial t} = 0$ looks like.
analysis functions partial-derivative
analysis functions partial-derivative
asked Nov 22 '18 at 23:09
Bad At MathBad At Math
203
203
Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33
add a comment |
Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33
Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33
Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33
add a comment |
1 Answer
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It's very long, so let me help you with some expressions
$$Vert x Vert _2^{2} = x_1^2 + x_2^2 + cdots + x_n^2, $$
$$ dfrac{partial}{partial x_i}(x_1^2 + x_2^2 + cdots + x_n^2) = 2x_i, $$
$$dfrac{partial}{partial x_i}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i, $$
$$dfrac{partial^2}{partial x_i^2}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = dfrac{-1}{4t} Bigg[ exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i cdot 2 x_i + exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot 2Bigg] $$
$$dfrac{partial}{partial t} Big( t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t}) Big) = dfrac{-n t^{n/2-1}}{2} exp !Big(-dfrac{Vert x Vert_2^2}{4t} Big) + t^{-n/2} exp! Big(-frac{Vert x Vert_2^2}{4t} Big) cdot dfrac{Vert x Vert_2^2}{4t^2}.$$
Can you continue with the calculations?
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
votes
It's very long, so let me help you with some expressions
$$Vert x Vert _2^{2} = x_1^2 + x_2^2 + cdots + x_n^2, $$
$$ dfrac{partial}{partial x_i}(x_1^2 + x_2^2 + cdots + x_n^2) = 2x_i, $$
$$dfrac{partial}{partial x_i}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i, $$
$$dfrac{partial^2}{partial x_i^2}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = dfrac{-1}{4t} Bigg[ exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i cdot 2 x_i + exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot 2Bigg] $$
$$dfrac{partial}{partial t} Big( t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t}) Big) = dfrac{-n t^{n/2-1}}{2} exp !Big(-dfrac{Vert x Vert_2^2}{4t} Big) + t^{-n/2} exp! Big(-frac{Vert x Vert_2^2}{4t} Big) cdot dfrac{Vert x Vert_2^2}{4t^2}.$$
Can you continue with the calculations?
add a comment |
It's very long, so let me help you with some expressions
$$Vert x Vert _2^{2} = x_1^2 + x_2^2 + cdots + x_n^2, $$
$$ dfrac{partial}{partial x_i}(x_1^2 + x_2^2 + cdots + x_n^2) = 2x_i, $$
$$dfrac{partial}{partial x_i}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i, $$
$$dfrac{partial^2}{partial x_i^2}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = dfrac{-1}{4t} Bigg[ exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i cdot 2 x_i + exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot 2Bigg] $$
$$dfrac{partial}{partial t} Big( t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t}) Big) = dfrac{-n t^{n/2-1}}{2} exp !Big(-dfrac{Vert x Vert_2^2}{4t} Big) + t^{-n/2} exp! Big(-frac{Vert x Vert_2^2}{4t} Big) cdot dfrac{Vert x Vert_2^2}{4t^2}.$$
Can you continue with the calculations?
add a comment |
It's very long, so let me help you with some expressions
$$Vert x Vert _2^{2} = x_1^2 + x_2^2 + cdots + x_n^2, $$
$$ dfrac{partial}{partial x_i}(x_1^2 + x_2^2 + cdots + x_n^2) = 2x_i, $$
$$dfrac{partial}{partial x_i}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i, $$
$$dfrac{partial^2}{partial x_i^2}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = dfrac{-1}{4t} Bigg[ exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i cdot 2 x_i + exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot 2Bigg] $$
$$dfrac{partial}{partial t} Big( t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t}) Big) = dfrac{-n t^{n/2-1}}{2} exp !Big(-dfrac{Vert x Vert_2^2}{4t} Big) + t^{-n/2} exp! Big(-frac{Vert x Vert_2^2}{4t} Big) cdot dfrac{Vert x Vert_2^2}{4t^2}.$$
Can you continue with the calculations?
It's very long, so let me help you with some expressions
$$Vert x Vert _2^{2} = x_1^2 + x_2^2 + cdots + x_n^2, $$
$$ dfrac{partial}{partial x_i}(x_1^2 + x_2^2 + cdots + x_n^2) = 2x_i, $$
$$dfrac{partial}{partial x_i}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i, $$
$$dfrac{partial^2}{partial x_i^2}, exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) = dfrac{-1}{4t} Bigg[ exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot dfrac{-1}{4t} cdot 2x_i cdot 2 x_i + exp! Big(-dfrac{Vert x Vert_2^2}{4t} Big) cdot 2Bigg] $$
$$dfrac{partial}{partial t} Big( t^{-n/2} exp(-frac{Vert x Vert_2^2}{4t}) Big) = dfrac{-n t^{n/2-1}}{2} exp !Big(-dfrac{Vert x Vert_2^2}{4t} Big) + t^{-n/2} exp! Big(-frac{Vert x Vert_2^2}{4t} Big) cdot dfrac{Vert x Vert_2^2}{4t^2}.$$
Can you continue with the calculations?
edited Nov 23 '18 at 2:23
answered Nov 23 '18 at 2:11
DavidDavid
784410
784410
add a comment |
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Try with $n=2$, $F(x_1,x_2,t) = t^{-1} exp(-frac{x_1^2+x_2^2}{4t})$. What are $partial_{x_i}F$ and $partial_{x_i}^2F$ and $partial_t F$ and $Delta F$ ?
– reuns
Nov 23 '18 at 1:33