Learning PDEs from Scratch in 24 Hours

[imsleepy]

ehh my head's not working, i should go to bed soon.

question 4. no clue.
H and L are just arbitrary points on the y and x-axis respectively, i presume..
what do each of those initial constraints mean, please?
and how do i know what is insulated?

 

[imsleepy]

ok found some stuff in my notes.
i think the du/dx (0,y) = 0 line tells you that something is insulated.

i've done some question in my book (obviously lecturer has done it and i was merely copying it, lol), and i have written down:

u_xx + u_yy = 0
u(0,y) = 0
u(a,y) = 0
u(x,0) = x (maintain temp-profile 'x')
du/dy (x,0) = 0 -> temp gradient perpendicular to side on bottom is zero, ie. no heat can flow across side => INSULATED

i might've missed something because my graph has a 'b', and i see no b's written in the question, but main thing is that from this i can gather that with question 4, the du/dx (0,y) = 0 line tells me that something is insulated :)

now to find out what side is insulated...

 

[imsleepy]

du/dx (0,y) = 0
so, gradient at 0,y is equal to 0?
would this be just the y-axis (up to point H) that is insulated?

 

[imsleepy]

from there onwards, i simply cannot understand what worked solutions are doing >_<

 

[lanedance]

in you question i think a=L, b=H as you have
0<x<L, 0<y<H

the insulated side is the one where the derivative is held at zero,