How Can Separation of Variables Solve This Partial Differential Equation?

[sarahqwert]

Homework Statement

utt = uxx -(25/4)cos((5/2)x)
ux(0,t) =1
u(pi,t)= pi
u(x,0)=x
ut(x,0)=0

Homework Equations

u(x,t)=v(x) + w(x,t)

The Attempt at a Solution

This is what I did so far:

u(x,t)=v(x) + w(x,t)
u(x,0) = v(x) +w(x,0)

when t is large:
vxx - (25/4)cos((5/2)x) = 0
vx = (5/2)sin((5/2)x)
v(x) = -cos((5/2)x) +x

when t is not large
wtt = wxx
w(x,t)= x-v(x) = cos((5/2)x)

and I'm not sure what to do after this

Maybe: X''+ λX = 0
X'(0) = 1 and X(pi)= pi

 

[LCKurtz]

Homework Statement

utt = uxx -(25/4)cos((5/2)x)
ux(0,t) =1
u(pi,t)= pi
u(x,0)=x
ut(x,0)=0

Homework Equations

u(x,t)=v(x) + w(x,t)

The Attempt at a Solution

This is what I did so far:

u(x,t)=v(x) + w(x,t)
u(x,0) = v(x) +w(x,0)

when t is large:
vxx - (25/4)cos((5/2)x) = 0
vx = (5/2)sin((5/2)x)
v(x) = -cos((5/2)x) +x

when t is not large
wtt = wxx
w(x,t)= x-v(x) = cos((5/2)x)

What does ##t## being large have to do with anything?

and I'm not sure what to do after this

Maybe: X''+ λX = 0
X'(0) = 1 and X(pi)= pi

No. The point of your original substitution is to let ##v(x)## take care of the non-homogeneous terms. I will call ##f(x) = \frac {25}4\cos(\frac 5 2 x)## to save typing. What you want to do is substitute ##u(x,t) = v(x) + w(x,t)## into the equation ##u_{tt}=u_{xx}-f(x)## and its boundary conditions:

##u_{tt}=u_{xx}-f(x)## becomes ##w_{tt} = w_{xx}+v''(x) -f(x)## That will be satisfied if we take ##v''(x) = f(x)## and ##w_{tt} = w_{xx}##

Now ##u_x(0,t) = 1## becomes ##w_x(0,t) + v'(0) = 1##. That will be satisfied if we take ##v'(0)=1## and ##w_x(0,t)=0##.

##u(\pi,t)=\pi## becomes ##w(\pi,t) +v(\pi)=1##. That will be satisfied if we take ##v(\pi)=\pi## and ##w(\pi,t) = 0##.

Now with those conditions on ##v(x)## you should be able to solve for ##v(x)## and you have a homogeneous equation and boundary conditions in ##w##. Try to take it from there.

 

[sarahqwert]

Thank you for your help, I m almost done solving this problem and got:

u(x,0)= Ʃ (cos((2n+1)/2)x)*Dn = x

sigma from 0 to infinity and Dn is a constant.

How do I find Dn?

 

[LCKurtz]

Thank you for your help, I m almost done solving this problem and got:

u(x,0)= Ʃ (cos((2n+1)/2)x)*Dn = x

sigma from 0 to infinity and Dn is a constant.

How do I find Dn?

That isn't correct. You need to show your work so I can see where you fell off the tracks.

Remember, when you substitute in the initial condition you get$$
x=u(x,0) = v(x) + w(x,0)$$so your initial condition for ##w(x,t)## is ##w(x,0)=x - v(x)##. What did you get for ##v(x)##?

Also, did you remember to check whether the eigenvalue zero works?

[Edit: Added] Your focus should be on solving the ##w(x,t)## system completely. That's where I need to see your work. Once you have figured that out, all you have to do to solve the ##u## system is write ##u(x,t) = v(x) + w(x,t)## because you know both ##v## and ##w##.

 

[sarahqwert]

Sorry that was supposed to be w(x,t) not u(x,t). So:

Xn=C*cos((2n+1)/2)x
Tn=D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t)

w(x,t) = Ʃ Xn * Tn

w(x,t) = Ʃcos((2n+1)/2)x)*((D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t))

so

w(x,0)=Ʃcos((2n+1)/2)x)*D

and

w(x,0) = cos(5/2)x

 

[LCKurtz]

Sorry that was supposed to be w(x,t) not u(x,t). So:

Xn=C*cos((2n+1)/2)x
Tn=D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t)

w(x,t) = Ʃ Xn * Tn

w(x,t) = Ʃcos((2n+1)/2)x)*((D*cos((2n+1)/2)t)+(B*sin((2n+1)/2)t))

so

w(x,0)=Ʃcos((2n+1)/2)x)*D

You need subscripts on the constants, as ##D_n## and ##B_n##. You didn't answer some of my questions in post #4. What did you get for ##v(x)##? Did you check for zero eigenvalues? (This matters as to whether there is a constant term). What does the initial condition for ##w(x,0)## become?

[Edit: added] Also you need to show what your T(t) eigenvalue problem becomes, with boundary condition, and how you got ##T_n(t)##, because what you have isn't correct.