Question

Method of separation of variables: pic attached

Answer 1

not sure entirely where to start

Answer 2

I'd listen to their advice, start out with the standard separation of variables. \[\Theta (x,t) = X(x)T(t)\] plug this into the wave equation. \[\frac{\partial^2 \Theta}{\partial t^2}=4\frac{\partial^2 \Theta}{\partial x^2}\]\[\frac{\partial^2 (XT)}{\partial t^2}=4\frac{\partial^2 (XT)}{\partial x^2}\]since X doesn't depend on t and T doesn't depend on x, we pull them out of the derivatives\[X\frac{\partial^2 T}{\partial t^2}=4T\frac{\partial^2 X}{\partial x^2}\]Since we know that the only derivative of T is of t and of X is also of x, we just replace them to make it easier on the eyes to look at \[XT''=4TX''\]Now we start the trick by putting all x and t parts on each side of the equation \[\frac{T''}{T}=4\frac{X}{X''}\]This is interesting. Since X and T are both functions, that means we can plug in anything. But then that would mean one function of x is really dependent upon t and the other way around because they're equal! That must mean then that these are equal to constants to maintain that they're not dependent on each other. \[\frac{T''}{T}=4\frac{X}{X''}=-\lambda^2\] Here I've just chosen negative lambda squared because I'm anticipating what we did last time with the sines and cosines.

I hope that can get you started. Ask me anything if you need more help.

Answer 3

It would be appropriate to solve around T first?

Answer 4

So \[T ^{''}+\lambda^2 T=0\]

Answer 5

\[T ^{''}+\mu^2 T=0\]

Answer 6

and we know
\[\mu=\frac{ \pi(2n-1) }{ 6 }\]

Answer 7

nup im still unsure

Answer 8

Actually you can solve both at the same time as long as you notice that the constant they're equal to is slightly different, and plug it in for each one at the end if that makes sense.

Answer 9

where\[T _{n}=\sin(\lambda x)\] assuming we are indexing from n=0

Answer 10

now to solve for the X equation.

\[\lambda^2 X ^{''}+4\lambda=0\]

Answer 11

4X instead of 4L

Answer 12

where do i go from here @Kainui

Answer 13

Plug in X=e^{ax} and solve it like a normal ODE for a.

Answer 14

Yep! i found the general solution for X

Answer 15

what do i do with that? Solve for initial conditions?

Answer 16

@mathslover

Answer 17

@dan815
HELP A BRO OUT

Answer 18

BRO too long!

Answer 19

where have u been

Answer 20

@Bswan

Answer 21

its crrrraaaaa right now. been stuyding hey. exams soon. Last semeter for engineering maths as well! So sad.

Answer 22

haha im on summer :P

Answer 23

wow. thats fked

Answer 24

omg

Answer 25

okay so u only learnt the separation of var method for solving PDE right

Answer 26

thinking of going to toronto to study 1st sem next yr aye

Answer 27

yep. thats right

Answer 28

Kainui has started it

Answer 29

thats sick

Answer 30

didn't say ur in canada?

Answer 31

yeah

Answer 32

okay now do the 3 cases for lambda

Answer 33

for T?

Answer 34

lemme check something

Answer 35

whats ur final solution for x?
initial condition is used to give the constant that u would have

Answer 36

ive done the lamda cases for T''=-L^2T

Answer 37

think is that im not sure what i'm doing - got so much on my plate right now.

Answer 38

thing*

Answer 39

show me theta (x,t)
or u dint solve it yet ?

Answer 40

haven't solved yet thats what you want to find out though?

Answer 41

ohh , i thought u asking specific qn
u wanna the whole sol ?

Answer 42

yea!! if u look up the top, kainui has started it for me

Answer 43

providing a equation with both T and X

Answer 44

for
\[T''=-\lambda ^2 T\]

Answer 45

i've already solved for the eigenvalues and eigen function for them in a previous question - not sure if you need them though

Answer 46

no need i think we gotta just express in fourier serie right

Answer 47

wait
at
\(\lambda ^2 x'' + u \lambda=0\)
untill here its good ?

Answer 48

i'm not sure if thats right though it should be 4X''/X

Answer 49

are we refering to kainui's initial workings?

Answer 50

yes kianui work is perfect

Answer 51

if you re-arrange the second to last step though it should be 4X''/X though?

Answer 52

only if u chick from last equation with lambda u and x xD
plz check it xD

Answer 53

whereas he wrote 4X/X''

Answer 54

where did you get 'u' from?

Answer 55

thats what im calling the theta

Answer 56

don't we just want to deal with
\[\frac{T''}{T}=4\frac{X}{X''}=-\lambda^2\]

Answer 57

yeah lets do it from here agian

Answer 58

@rational

Answer 59

and forget the lambda^2 call it lambda

Answer 60

ahk\[\frac{T''}{T}=4\frac{X}{X''}=\lambda\]

Answer 61

T''+lamT=0, what the boundary conditions given for x or t?

Answer 62

and shud that be 4X''/X=lambda

Answer 63

boundary conditions are 0

Answer 64

X(0)=0, X(L)=0
X''+lambaX=0 <--- lets change your equations to
T''/4T = X''/x =-Lamba

Answer 65

so the 3 possibilties lamba =0, lamba = -k^2 , lamba =k^2

Answer 66

correct

Answer 67

we would only get eigenvalues of 0 if we look at the first two possibilites right?

Answer 68

so only look at lamba=k^2?

Answer 69

X(x)=c1+c2x
X(x)=c1coshax+c2sinhax
X(X)=c1cos ax + c2 sin ax
X(4pi)=0, x(0) =0

Answer 70

yeah so u left with the cos solutions only

Answer 71

correct!

Answer 72

X(x)=c2sin npix/L, n=1,2,3...
L=4pi

Answer 73

Now we need T(t) to find X(t)T(t) for some n

Answer 74

i mean X(x)T(t)

Answer 75

right

Answer 76

as we said theta = X(x)T(t)

Answer 77

yep

Answer 78

what did u get for T(t)

Answer 79

c3cos(...)+c4sin(..)?

Answer 80

is the boundary condition for T t(0)=0?

Answer 81

Do we need that to find that?

Answer 82

ah this is fkd.

do we use what they give us for T''-lamdaT=0?

Answer 83

where eigen function is cos(nx)/4

Answer 84

we rearrange it like remember
T''/4T = X''/x =-Lamba

Answer 85

so solve T''+4LambaT=0

Answer 86

ah

T''+4lamdaT=0

Answer 87

so now find general solution of T?

Answer 88

given that T'(0)=0?

Answer 89

wait im lemme check something is our general solution for X(X) right

Answer 90

whats a?

Answer 91

u are only left with c2 sin nx/4 right?

Answer 92

with boundary condition for X(0)=0?

Answer 93

and X(4pi)=0

Answer 94

isn't it X'(0)=0? and X'(4pi)=0?

Answer 95

oh are u serious

Answer 96

ye cuz