Question

partial derivative

Answer 1

\[u=e^{-\alpha^2k^2t}sinkx\]

Answer 2

with respect to t

Answer 3

I took log on both sides \[lnu= -\alpha^2k^2t sinkx\]

Answer 4

\[\frac{ 1 }{ u }=-\sin(kx)\alpha^2k^2\]
\[u_t=-\sin(kx)\alpha^2k^2\]

Answer 5

*u

Answer 6

so \[-\sin(kx)\alpha^2k^2 e^{-\alpha^2k^2t}sinkx\]

Answer 7

@AccessDenied

Answer 8

they asked to prove u=.... same as \[u_t=\alpha^2_{xx}\]

Answer 9

so I took with respect to u_xx

Answer 10

I got \[-k^2e^{\alpha^2k^2t}\sin(kx)\]

Answer 11

so finally my ans is \[sin(kx)alpha^2u_{xx}\]

Answer 12

I am not following the extraneous work...
From \( u = e^{- \alpha^2 k^2 t} \sin (kx) \)
Taking the partial derivative with respect to t, we consider every other variable a constant.
\( u = e^{- \alpha^2 k^2 \color{red}t} \sin (k x) \)
Everything other than the t marked in red is like a constant, so the partial derivative is the equivalent to taking this derivative with respect to t:
\( u = e^{A t} \times B \)
Where A and B are constants.

Answer 13

lol I typed those for nothing. Anyways ^= \[e^{At} AB\]

Answer 14

Yep. Then we just replace A and B with what was originally there in its place.
\( A = -\alpha^2 k^2 \), \(B = \sin (kx) \)

Answer 15

okay, will this work if I use logs

Answer 16

As in,
\( \ln u = \ln \left( e^{-\alpha^2 k^2 t} \sin (kx) \right) \)
and then take the derivative?

Answer 17

ya ln and e cancels

Answer 18

ohhh but ln for sin stays right?

Answer 19

The product property would apply.
\( \ln u = \ln \left( e^{-\alpha^2 k^2 t} \right) + \ln \left( \sin (kx) \right) \)
\( \ln u = -\alpha^2 k^2 t + \ln \sin (kx) \)
Which would work, you take the partial derivative and the ln(sin (kx)) cancels out to 0.

Answer 20

okay got u thx

Answer 21

Then we just have to watch for chain rule in the end.
\( \ln u = -\alpha^2 k^2 t + \ln \sin (kx) \)
\( \dfrac{1}{u} u_t = - \alpha^2 k^2 \)
\( \dfrac{1}{e^{-\alpha^2 k^2 t} \sin (kt) } u_t = - \alpha^2 k^2 \)