Operation on a vector with a special property
I have a vector $vec u$ with the following property: $$vec u (tx, ty, tz) = t^n vec u (tx, ty, tz)$$ Now I have to prove that: $$(vec r.vec nabla)vec u = n.vec u$$ How can I do this? I tried to do it in the following way: $$(vec r.vec nabla)vec u = (x.partial /partial x + y.partial/partial y + z.partial/partial z) vec u(tx, ty, tz)$$
$$ = (x.partial/partial x + y.partial/partial y + z.partial/partial z)[t^n vec u(x, y, z)]$$
$$ = t^n(xpartialvec u(x)/partial x + y.partial vec u(y)/partial y + z.partial vec u(z)/partial z)$$ since under partial differentiation $vec u(x, y, z)$ will get separated into $vec u(x), vec u(y), vec u(z)$ for $partial/partial x, partial/partial y, partial/partial z$ respectively, and $t$ is a scalar. But I can't go any further. What is/are the mistake(s) in this approach, if any? Also, how can it be proven?
vector-analysis divergence
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
add a comment |
I have a vector $vec u$ with the following property: $$vec u (tx, ty, tz) = t^n vec u (tx, ty, tz)$$ Now I have to prove that: $$(vec r.vec nabla)vec u = n.vec u$$ How can I do this? I tried to do it in the following way: $$(vec r.vec nabla)vec u = (x.partial /partial x + y.partial/partial y + z.partial/partial z) vec u(tx, ty, tz)$$
$$ = (x.partial/partial x + y.partial/partial y + z.partial/partial z)[t^n vec u(x, y, z)]$$
$$ = t^n(xpartialvec u(x)/partial x + y.partial vec u(y)/partial y + z.partial vec u(z)/partial z)$$ since under partial differentiation $vec u(x, y, z)$ will get separated into $vec u(x), vec u(y), vec u(z)$ for $partial/partial x, partial/partial y, partial/partial z$ respectively, and $t$ is a scalar. But I can't go any further. What is/are the mistake(s) in this approach, if any? Also, how can it be proven?
vector-analysis divergence
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
1
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23
add a comment |
I have a vector $vec u$ with the following property: $$vec u (tx, ty, tz) = t^n vec u (tx, ty, tz)$$ Now I have to prove that: $$(vec r.vec nabla)vec u = n.vec u$$ How can I do this? I tried to do it in the following way: $$(vec r.vec nabla)vec u = (x.partial /partial x + y.partial/partial y + z.partial/partial z) vec u(tx, ty, tz)$$
$$ = (x.partial/partial x + y.partial/partial y + z.partial/partial z)[t^n vec u(x, y, z)]$$
$$ = t^n(xpartialvec u(x)/partial x + y.partial vec u(y)/partial y + z.partial vec u(z)/partial z)$$ since under partial differentiation $vec u(x, y, z)$ will get separated into $vec u(x), vec u(y), vec u(z)$ for $partial/partial x, partial/partial y, partial/partial z$ respectively, and $t$ is a scalar. But I can't go any further. What is/are the mistake(s) in this approach, if any? Also, how can it be proven?
vector-analysis divergence
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
I have a vector $vec u$ with the following property: $$vec u (tx, ty, tz) = t^n vec u (tx, ty, tz)$$ Now I have to prove that: $$(vec r.vec nabla)vec u = n.vec u$$ How can I do this? I tried to do it in the following way: $$(vec r.vec nabla)vec u = (x.partial /partial x + y.partial/partial y + z.partial/partial z) vec u(tx, ty, tz)$$
$$ = (x.partial/partial x + y.partial/partial y + z.partial/partial z)[t^n vec u(x, y, z)]$$
$$ = t^n(xpartialvec u(x)/partial x + y.partial vec u(y)/partial y + z.partial vec u(z)/partial z)$$ since under partial differentiation $vec u(x, y, z)$ will get separated into $vec u(x), vec u(y), vec u(z)$ for $partial/partial x, partial/partial y, partial/partial z$ respectively, and $t$ is a scalar. But I can't go any further. What is/are the mistake(s) in this approach, if any? Also, how can it be proven?
vector-analysis divergence
vector-analysis divergence
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
asked Nov 30 '18 at 12:07
darthasterdarthaster
11
11
Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
1
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23
add a comment |
Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
1
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23
Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
1
1
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23
add a comment |
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Is there a mistake in the first equation? Did you mean homogeneous function: $u(t mathbf r) = t^nu(mathbf r)$?
– Vasily Mitch
Nov 30 '18 at 12:18
1
Hint: Differentiate both sides of $u(tx,ty,tz) = t^nu(x,y,z)$ with respect to $t$.
– Timothy Hedgeworth
Nov 30 '18 at 12:23