That would result in:
A_{jilk} + A_{jkil} = A_{lkji} + A_{likj}
The only thing it reminds me is of the third symmetry again, but if I use it I end up with a meaningless result:
A_{jlki} = A_{ljik}
Which translates in the first two symmetries.
New attempt, got further but still missing something, hope this was what you meant.
From the third property:
A_{ijkl} + A_{iklj} + A_{iljk} = 0
A_{klij} + A_{kijl} + A_{kjli} = 0
Therefore:
A_{ijkl} + A_{iklj} + A_{iljk} = A_{klij} + A_{kijl} + A_{kjli}
Since the first two properties...
Homework Statement
Let Aijkl be a rank 4 square tensor with the following symmetries:
A_{ijkl} = -A_{jikl}, \qquad A_{ijkl} = - A_{ijlk}, \qquad A_{ijkl} + A_{iklj} + A_{iljk} = 0,
Prove that
A_{ijkl} = A_{klij}
Homework EquationsThe Attempt at a Solution
From the first two properties...
I think I do, because when taking the limit, the only 2 possibilities that satisfy that relation are:
- the limits do not exist (which would be the case of a sine or cosine)
- the limit is 0, which is the point that prevents the function from oscillating
That's a nice example, but taking...
Homework Statement
Given the differential equation : \frac{dy}{dt} + a(t)*y = f(t)
in which a and f are continuous functions in ℝ and verify:
a(t) > c > 0 \forallt , lim f(t) = 0
Show that any solution of the differential equation...