Solving inhomogenous wave equation

[wahoo2000]

I have made an attempt on this one, but I'm not quite sure that I have done it correctly so far..? I am now heading a (for me massive partial integration, and therefore I think it's better to ask before I start.

Homework Statement

Find the solution u(x,t) of the inhomogenous wave equation

u_{tt}-u_{xx}=x, 0<x<1, t>0
u(0,t)=1, u_{x}(1,t)=0
u(x,0)=1, u_{t}(x,0)=0

The Attempt at a Solution

Let u(x,t)=S(x)+w(x,t) where S''(x)=x, S(0)=S'(1)=0
then
S'(x)= X^2+C
S(x)=x^3/6+C*x+D
S(0)=0 ==> D=0
S'(1)=0 ==> C=-1/2
==>
S(x)=x^3/6-x/2

Then the equation for w is:
<br /> w_{tt}-w_{xx}=0, 0&lt;x&lt;1, t&gt;0
w(0,t)=1, w_{x}(1,t)=0
w(x,0)=u(x,0)-S(x)=1-\frac{x^{3}}{6}+\frac{x}{2}, w_{t}(x,0)=0<br />

I will try to send the rest as a reply to this post... I get database error when I try to send more than this.. :/

 

[wahoo2000]

Now I use separation of variables
w(x,t)=X(x)T(t)=/0
==>

\frac{X&#039;&#039;(x)}{X(x)}=\frac{T&#039;&#039;(t)}{T(t)}=\lambda&lt;0
==>

X(x)=Asin(\sqrt{-\lambda}x+Bcos(\sqrt{-\lambda}x)
And since X(0)=1 ==> A=0
and therefore \lambda=-n^{2}\pi^{2}
And then X_{n}x=cos(nx\pi)

 

[wahoo2000]

same thing for t ==> T_{n}t=P_{n}sin(nt\pi)+Q_{n}cos(nt\pi)
And since w_{t}(x,0)=0, T_{n}&#039;t=P_{n}cos(nt\pi)-Q_{n}sin(nt\pi)
==>Q_{n}=0
This means that w(x,t)=\sum^{n=1}_{\infty}P_{n}sin(nx\pi)cos(nx\pi)

 

[wahoo2000]

And to identify P_{n} let t=0 (to use w(x,0))

w(x,0)=\sum^{\infty}_{n=1}P_{n}sin(nt\pi)=1-\frac{x^{3}}{6}+\frac{x}{2}
Multiply both sides with sin(kx\pi) and integrate over (0,1)

==>
\sum^{\infty}_{n=1}P_{n}\int^{0}_{1}sin(nt\pi)sin(kx\pi)dx = \int^{0}_{1}(1-\frac{x^{3}}{6}+\frac{x}{2})sin(kx\pi)dx

And now it is time for the integration by parts... If it was correct so far!

 

[gabbagabbahey]

S'(x)= X^2+C
S(x)=x^3/6+C*x^2/2+D

==>
S(x)=x^3/6-x^2/4

Your solution for S(x) is incorrect:

\int Cdx = Cx+D

NOT

\frac{Cx^2}{2} + D

 

[wahoo2000]

Yes, of course... S(x)=x^3/6-x/2

 

[gabbagabbahey]

Now I use separation of variables
w(x,t)=X(x)T(t)=/0
==>

\frac{X&#039;&#039;(x)}{X(x)}=\frac{T&#039;&#039;(t)}{T(t)}=\lambda&lt;0
==>

X(x)=Asin(\sqrt{-\lambda}x+Bcos(\sqrt{-\lambda}x)
And since X(0)=1 ==> A=0
and therefore \lambda=-n^{2}\pi^{2}
And then X_{n}x=cos(nx\pi)

Why are you assuming \lambda &lt; 0 here? Don't you generally have to treat all three cases:
\lambda &lt; 0, \lambda = 0, and \lambda &gt; 0 ?

 

[wahoo2000]

Hmm, I might have misunderstood this. In all examples I have seen, \lambda&lt;0 is the one used. Therfore I thought it might be ok to assume that this is the case, if I find a non-trivial solution using that \lambda

 

[gabbagabbahey]

Hmm, I might have misunderstood this. In all examples I have seen, \lambda&lt;0 is the one used. Therfore I thought it might be ok to assume that this is the case, if I find a non-trivial solution using that \lambda

You might also find non-trivial solutions for the other two cases. If you do, then the complete solution will be the product of your solutions for each of the 3 cases. I think you need to make sure that there are no non-trivial solutions for the other two cases.

 

[gabbagabbahey]

Now I use separation of variables
w(x,t)=X(x)T(t)=/0
==>

\frac{X&#039;&#039;(x)}{X(x)}=\frac{T&#039;&#039;(t)}{T(t)}=\lambda&lt;0
==>

X(x)=Asin(\sqrt{-\lambda}x+Bcos(\sqrt{-\lambda}x)
And since X(0)=1 ==> A=0
and therefore \lambda=-n^{2}\pi^{2}
And then X_{n}x=cos(nx\pi)

Anyways, for the \lambda&lt;0 case,

X(0)=1 \Rightarrow Asin(0) + Bcos(0) = B =1

NOT A=0

 

[wahoo2000]

Ah, okey! Then I have the check the other two cases as well. Thanks so far!

 

[gabbagabbahey]

\lambda=-n^{2}\pi^{2}

Where did you get that from?

 

[wahoo2000]

Ok, I have studied the two other cases now...

For X the boundary conditions given are X(0)=1 and X'(1)=0
And X''(x)=\lambda*X(x)
The characteristic equation is r^2=lambda and hence r_{1,2}=+-\sqrt{\lambda}, \lambda&gt;0
and
r_{1,2}=+-i\sqrt{-\lambda}, \lambda&lt;0.

For case 1, lambda=0
X(x)=Ax+B
X'(x)=A

X(0)=B=1
X'(1)=A=0 ==> X(x)=0 and that is not what we are looking for... so now to the case where lambda>0:

X(x)=Ae^{\sqrt{\lambda}x}+Be^{-\sqrt{\lambda}x}
and
X&#039;(x)=\sqrt{\lambda}*(Ae^{\sqrt{\lambda}x}-Be^{-\sqrt{\lambda}x})
then
X(0)=A+B=1
and
X'(1)= \sqrt{\lambda}*(Ae^{\sqrt{\lambda}}-Be^{-\sqrt{\lambda}})=0
then I am not sure.. sqrt(lambda) obviously can't be zero, since lambda is larger than zero in this case.. Then the other part might be zero? But I am not sure if I can say it is.. If it is zero, then the case where lambda<0 alone gives the solution.

There is no need to do the same calculations for T since the equations and boundary conditions are the same, right?

 

[wahoo2000]

Where did you get that from?

From the boundary condition u_{x}(1,t)=0
X' is -Bsin(\sqrt{-\lambda}x)=0 for x=1
and since B is not allowed to be 0, sin(\sqrt{-\lambda}) has to be zero. And that is true if sqrt(-lamba)=n*Pi, which menas that \lambda=-n^{2}\pi^{2}

 

[gabbagabbahey]

From the boundary condition u_{x}(1,t)=0
X' is -Bsin(\sqrt{-\lambda}x)=0 for x=1
and since B is not allowed to be 0, sin(\sqrt{-\lambda}) has to be zero. And that is true if sqrt(-lamba)=n*Pi, which menas that \lambda=-n^{2}\pi^{2}

This is incorrect.

Anyways, for the \lambda&lt;0 case,

X(0)=1 \Rightarrow Asin(0) + Bcos(0) = B =1

NOT A=0

If X(0)=1 , then B=1 \Rightarrow X(x)=Asin(kx) +cos(kx) \Rightarrow X&#039;(x) = Akcos(kx) -ksin(kx) where k \equiv \sqrt{- \lambda}

So,

X&#039;(1)=0 \Rightarrow Akcos(k) -ksin(k)=0 \Rightarrow A = tan(k) \Rightarrow X(x)=tan(k)sin(kx)+cos(kx)

But(!) these are not the boundary conditions you were given; you only have conditions for w(x,t) NOT for X(x) and T(t) individually.

Here is what you actually know:

w(x,t)=(Asin(kx)+Bcos(kx))(Csin(kt) + Dcos(kt))

And so, for the first boundary condition:

w(0,t)=1 \Rightarrow (Asin(0)+Bcos(0))(Csin(kt)+Dcos(kt))=B(Csin(kt)+Dcos(kt))=1

\Rightarrow (Csin(kt)+Dcos(kt))= \frac{1}{B}

\Rightarrow w(x,t) = \frac{A}{B}sin(kx)+cos(kx)

What do the other 3 boundary conditions say about w(x,t) ?

 

[wahoo2000]

"\lambda=-n^{2}\\pi^{2}
This is incorrect. "

Hmm.. Tried again and saw an error, but still get the same result?
X'= <br /> -B\sqrt{-\lambda}sin(\sqrt{-\lambda}*1)=0<br /> But B or lambda can't be zero so then sqrt(-lambda)=n*Pi?

Then I made an attempt with the boundary conditions but I am getting confused... Is the solution really t-independent?
From the first BC you got w(x,t)=(A/B)sin(kx)+cos(kx), but since B=1, w(x,t)=Asin(kx)+Bcos(kx).

Then I uesd the second one, w_{x}(1,t)=0=(Ak cos(k)- k sin(k))=0<br /> ==&gt;<br /> A=\frac{k cos(k)}{k sin(k)} ==&gt; \frac{1}{A}=tan(k) ==&gt; A=\frac{1}{tan(k)}

Then the third, w_t(x,0)=0
Is true because the soulution is independent of t, (because C and D = 0) ?

Fourth
w(x,0)=Asin(kx)+cos(kx)=1-x^3/6-x/2 I don't know what to do with it? True since there is no t..?

Then w(x,t)=(1/tan(k))sin(kx)+cos(kx)-1+x^3/6-x/2

But I don't know what to do now, this can't be it..? Seems suspiciously simple.. Need to get to u(x,t), and also it has to be written in some convenient form that can help me to calculate \sum^{\infty}_{k=0}\frac{1}{(2k+1)^{4}}

 

[gabbagabbahey]

Are you sure you've posted the correct boundary conditions for u(x,t)?

i.e. are these correct:
u(0,t)=1, u_{x}(1,t)=0
u(x,0)=1, u_{t}(x,0)=0

?

 

[wahoo2000]

Yes... that is what it says.

 

[gabbagabbahey]

hmmm...okay, I think I found the problem:

(1) first, for w(x,t)+S(x) to satisfy the given wave equation, S''(x) should be negative x, not +x. This gives a general solution of S(x)=-(x^3)/6 +Ax +B

(2) second, S(0)=0 leaves w(x,t) with too harsh restrictions at x=0; it is better to set S(0)=1 so that w(0,t)=0. S'(1)=0 is fine though. This gives S(x)=-(x^3)/6 +x/2 +1.

So the boundary conditions for w(x,t) are:
(i) \quad w(0,t)=0
(ii) \quad w_x (1,t)=0
(iii) \quad w(x,0)=\frac{x^3}{6} - \frac{x}{2}
(iv) \quad w_t (x,0)=0

Now you can get a meaningful solution for w(x,t)

Start with the \lambda &lt;0 and show me what you get.

 

[wahoo2000]

Hmm..

k=n*Pi=sqrt(-lambda) if my lambda was correct.

then
(i) w(0,t)=0=B(Csin(kt)+Dcos(kt)), and since B=1, Csin(kt)+Dcos(kt) must be =0

(ii) w_x(1,t)=0=(Akcos(k)-Bksin(k))(Csin(kt)+Dcos(kt))
And I know from (i) that Csin(kt)+Dcos(kt)=0 and hence Akcos(k)-ksin(k)=0 ==> A=(kcos(k))/ksin(k)=tan(k)

(iii) w(x,0)=x^3/6-x/2=(Asin(kx)+Bcos(kx))(Csin(0)+Dcos(0))=D(Asin(kx)+(Bcos(kx)=x^3/6-x/2
==> D(tan(k)sin(kx)+cos(kx))=x^3/6-x/2

(iv) w_t(x,0)=(Asin(kx)+Bcos(kx))(Ckcos(0)-Dksin(0))=Ck(Asin(kx)+Bcos(kx))=0
==> Ck=(tan(k)sin(kx)+cos(kx))=0

But I don't know how to continue from here?

 

[gabbagabbahey]

Hmm..

k=n*Pi=sqrt(-lambda) if my lambda was correct.

then
(i) w(0,t)=0=B(Csin(kt)+Dcos(kt)), and since B=1, Csin(kt)+Dcos(kt) must be =0

Why would B=1 here? doesn't (i) tell you that B=0?

 

[wahoo2000]

Why would B=1 here? doesn't (i) tell you that B=0?

Sorry, got that from <br /> X(0)=1 \Rightarrow Asin(0) + Bcos(0) = B =1 <br /> but that is history now..
Hmm.. But why must B=0 and not Csin(kt)+Dcos(kt)=0?
Just because we _want_ to find solutions including t?

 

[gabbagabbahey]

Sorry, got that from <br /> X(0)=1 \Rightarrow Asin(0) + Bcos(0) = B =1 <br /> but that is history now..
Hmm.. But why must B=0 and not Csin(kt)+Dcos(kt)=0?
Just because we _want_ to find solutions including t?

<br /> w(x,t)=(Asin(kx)+Bcos(kx))(Csin(kt) + Dcos(kt)) <br />

If

<br /> (Csin(kt) + Dcos(kt))=0 <br />

then

<br /> w(x,t)=(Asin(kx)+Bcos(kx))(0) =0 <br />

which is the trivial solution

 

[wahoo2000]

Ah, I see..

So then all Bcos(kx) disappears
and then, from
(iii)=> D(tan(k)sin(kx))=x^3/6-x/2
and
(iv)=> Ck(tan(k)sin(kx))=0

 

[gabbagabbahey]

Ah, I see..

So then all Bcos(kx) disappears
and then, from
(iii)=> D(tan(k)sin(kx))=x^3/6-x/2
and
(iv)=> Ck(tan(k)sin(kx))=0

What happened to A?

 

[wahoo2000]

A=tan(k) ?

from (ii)
Akcos(k)-ksin(k)=0
A=(ksin(k))/(kcos(k))=tan(k)

 

[gabbagabbahey]

A=tan(k) ?

from (ii)
Akcos(k)-ksin(k)=0
A=(ksin(k))/(kcos(k))=tan(k)

But your no longer using the same w(x,t) as before; remember (i) gives you B=0 now

\Rightarrow w(x,t)=Asin(kx)(Csin(kt)+Dcos(kt))

or absorbing A into C and D,

w(x,t)=sin(kx)(Csin(kt)+Dcos(kt))

What does (ii) say about this w(x,t)?

 

[wahoo2000]

But your no longer using the same w(x,t) as before; remember (i) gives you B=0 now

\Rightarrow w(x,t)=Asin(kx)(Csin(kt)+Dcos(kt))

or absorbing A into C and D,

w(x,t)=sin(kx)(Csin(kt)+Dcos(kt))

What does (ii) say about this w(x,t)?

Hmm, absorbing? Never heard of that.. If w(x,t)=Asin(kx)(Csin(kt)+Dcos(kt)) and then after absorbing:
w(x,t)=sin(kx)(Csin(kt)+Dcos(kt)), have you not said that A=1 then?

 

[gabbagabbahey]

Hmm, absorbing? Never heard of that.. If w(x,t)=Asin(kx)(Csin(kt)+Dcos(kt)) and then after absorbing:
w(x,t)=sin(kx)(Csin(kt)+Dcos(kt)), have you not said that A=1 then?

Not really, I've just taken advantage of the fact that AC and AD are still constants and called them C and D. You can call them E and F if it makes you feel better, but it makes no real difference.

 

[wahoo2000]

Anyway.. if I keep the A
(ii) \Rightarrow w_{x}(1,t)=Akcos(kx)(Csin(kt)+Dcos(kt))=0 then cos(kx) should be zero because k and (Csin(kt)+Dcos(kt) are not zero, (and neither is A..), because if A=0 we have only trivial solution X(x)T(t)=0, and same if (Csin(kt)+Dcos(kt)=0, then per definition k is not zero since lambda<0.


then
(iii)\Rightarrow w(x,0)=D(Asin(kx))=\frac{x^3}{6}-\frac{x}{2}

(iv)\Rightarrow w_{t}(x,0)=Ck(Asin(kx))=0

Hmmm...and then, from (iv) Asin(kx) can't be zero (since A is not allowed to be zero), and k is not zero.
Then C must be zero.

so w(x,t)=Asin(kx)*Dcos(kt)
then this, together with (iii) says that
Asin(kx)*Dcos(0)=ADsin(kx)=x^3/6-x/2
then (ii)=>
w_x(1,t)=kAcos(k)*Dcos(kt)=0

 

[gabbagabbahey]

Anyway.. if I keep the A
(ii) \Rightarrow w_{x}(1,t)=Akcos(kx)(Csin(kt)+Dcos(kt))=0 then cos(kx) should be zero because k and (Csin(kt)+Dcos(kt) are not zero, (and neither is A..), because if A=0 we have only trivial solution X(x)T(t)=0, and same if (Csin(kt)+Dcos(kt)=0, then per definition k is not zero since lambda<0.

Right, so what does cos(kx)=0 tell you about k?

 

[wahoo2000]

that k=(n*Pi)/2...
so \lambda=-\frac{n^{2}\pi^{2}}{4}

 

[gabbagabbahey]

that k=(n*Pi)/2...
so \lambda=-\frac{n^{2}\pi^{2}}{4}

Close, but only true if n is odd. The best way to represent that is :

(ii) \Rightarrow k= \frac{(2n-1) \pi}{2} \quad \ni n=1,2,3 \ldots

\Rightarrow w(x,t)= \sum_{n=1}^{\infty} A_nsin \left( \frac{(2n-1) \pi}{2} x \right)(C_nsin \left(\frac{(2n-1) \pi}{2} t \right)+D_n cos \left( \frac{(2n-1) \pi}{2} t \right) )

And (iv) still implies that C=0,

\Rightarrow w(x,t)= \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right) cos \left( \frac{(2n-1) \pi}{2} t \right)

So, what do you get from (iii) now?

 

[wahoo2000]

Well... that means that

<br /> \Rightarrow w(x,0)= \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right) cos \left( 0\right) \Rightarrow \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right)=\frac{x^{3}}{6}-\frac{x}{2}<br />

 

[gabbagabbahey]

Well... that means that

<br /> \Rightarrow w(x,0)= \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right) cos \left( 0\right) \Rightarrow \sum_{n=1}^{\infty} A_n D_n sin \left( \frac{(2n-1) \pi}{2} x \right)=\frac{x^{3}}{6}-\frac{x}{2}<br />

Yes, and from this you can use Fourier's method to determine A_nD_n. Give it a shot; what do you get?

 

[wahoo2000]

Hmm, does that mean we are looking for Fourier sinus series of f(x)=x^3/6-x/2?
Is it A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})(\frac{(2n-1)\pi}{2}x)dx

 

[gabbagabbahey]

Hmm, does that mean we are looking for Fourier sinus series of f(x)=x^3/6-x/2?
Is it A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})(\frac{(2n-1)\pi}{2}x)dx

Yes, you can evaluate the integral by using integration by parts 3 times.

 

[wahoo2000]

I think I made a typing error there, of course it should be

<br /> A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})sin(\frac{(2n-1)\pi}{2}x)dx<br />
, right? I'll attack it right now :)

 

[gabbagabbahey]

I think I made a typing error there, of course it should be

<br /> A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})sin(\frac{(2n-1)\pi}{2}x)dx<br />
, right? I'll attack it right now :)

Yes, I just assumed you meant to put the sin in there.

 

[wahoo2000]

Ok..puh, now I have tried.. I am not too confident it is OK though :( so much to write if I write everything, but I can show how I did and thought if this is completely wrong..
This is what I got:

A_{n}D_{n}=-\frac{2}{(2n+1)^{2}\pi^{2}}(-(-1)^{n})-\frac{16}{(2n+1)^{2}\pi^{2}}-\frac{64}{(2n+1)^{4}\pi^{4}}

 

[gabbagabbahey]

Ok..puh, now I have tried.. I am not too confident it is OK though :( so much to write if I write everything, but I can show how I did and thought if this is completely wrong..
This is what I got:

A_{n}D_{n}=-\frac{2}{(2n+1)^{2}\pi^{2}}(-(-1)^{n})-\frac{16}{(2n+1)^{2}\pi^{2}}-\frac{64}{(2n+1)^{4}\pi^{4}}

This is incorrect. Show me what you did when you applied integration by parts the first time.

 

[wahoo2000]

first time was 2[(\frac{x^{3}}{6}-\frac{x^{2}}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}<br /> <br /> +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{2x}{2})(cos(\frac{(2n-1)\pi}{2}x)dx

But I saw now when writing this that I used x^2/2 instead of x/2 in my calculations.. :( :(

 

[gabbagabbahey]

first time was 2[(\frac{x^{3}}{6}-\frac{x^{2}}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}<br /> <br /> +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{2x}{2})(cos(\frac{(2n-1)\pi}{2}x)dx

But I saw now when writing this that I used x^2/2 instead of x/2 in my calculations.. :( :(

Try it with x/2 and show me what you get.

 

[wahoo2000]

Here is the first step.. if you are online, just tell me if that step i correct before I do it all again...

A_{n}D_{n}=2[(\frac{x^{3}}{6}-\frac{x}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}<br /> <br /> +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx<br />
where the first term, 2[(\frac{x^{3}}{6}-\frac{x}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}=0

 

[wahoo2000]

I worked it through anyway... all terms except the last became zero.
I now got A_{n}D_{n}=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}

 

[gabbagabbahey]

I worked it through anyway... all terms except the last became zero.
I now got A_{n}D_{n}=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}

Do you mean:
<br /> A_{n}D_{n}=\frac{32}{(2n-1)^{4}\pi^{4}}*-(-1)^{n}<br /> ?
If so, then your right. What are w(x,t) and u(x,t) then?

 

[wahoo2000]

Hmm.. no, actually i got ^3... I will show you how I did and then you can see where it went wrong. I don't write out the terms that went zero.
<br /> A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx =
=-\frac{16}{(2n+1)^{2}\pi{2}}\int^{1}_{0}(x*sin(\frac{(2n-1)\pi}{2}x)dx = \frac{32}{(2n+1)^{3}\pi{3}}\int^{1}_{0}(cos(\frac{(2n-1)\pi}{2}x)dx=
=\frac{32}{(2n+1)^{3}\pi{3}}[(sin(\frac{(2n-1)\pi}{2}x)]=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}
edit: I think I lost one inner derivative there at the end...but it doesn't make sense to me anyway..in that case it would be 64/((2n+1)^4*pi^4)
:/


Hope we can find the mistake. I go on with "your" AnDn meanwhile.

Then it means that
w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)<br /> and
u(x,t)=w(x,t)+S(x)= w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1

 

[gabbagabbahey]

Hmm.. no, actually i got ^3... I will show you how I did and then you can see where it went wrong. I don't write out the terms that went zero.
<br /> A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx =
=-\frac{16}{(2n+1)^{2}\pi{2}}\int^{1}_{0}(x*sin(\frac{(2n-1)\pi}{2}x)dx = \frac{32}{(2n+1)^{3}\pi{3}}\int^{1}_{0}(cos(\frac{(2n-1)\pi}{2}x)dx=
=\frac{32}{(2n+1)^{3}\pi{3}}[(sin(\frac{(2n-1)\pi}{2}x)]=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}
edit: I think I lost one inner derivative there at the end...but it doesn't make sense to me anyway..in that case it would be 64/((2n+1)^4*pi^4)
:/


Hope we can find the mistake. I go on with "your" AnDn meanwhile.

Then it means that
w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)<br /> and
u(x,t)=w(x,t)+S(x)= w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1

In your second line you multiplied by 4 instead of 2 and in your 3rd line you missed a factor of 2/(2n-1)pi also, they should all be (2n-1)'s not(2n+1)'s. This gives:

u(x,y)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1

or if you make the substitution n=m+1 you get:

u(x,y)= \frac{32}{\pi^{4}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{4}}sin(\frac{(2m+1)\pi}{2}x)cos(\frac{(2m+1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1

Now, what is u(1,0) according to the boundary conditions? What is it according to the above formula?

 

[wahoo2000]

Ah, I see.. The missed inner derivative I found myself, but I squared once instead of multiplying by 2.. Time to go to bed soon
<br /> u(1,0)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}*-(-1)^{n}*1-\frac{1}{6}+\frac{1}{2}+1<br />
but I don't have u(1,0) in my BC?

EDIT: and (-1)^n*-(-1)^n = 1 ...

 

[gabbagabbahey]

Ah, I see.. The missed inner derivative I found myself, but I squared once instead of multiplying by 2.. Time to go to bed soon
<br /> u(1,0)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}*-(-1)^{n}*1-\frac{1}{6}+\frac{1}{2}+1<br />
but I don't have u(1,0) in my BC?

EDIT: and (-1)^n*-(-1)^n = 1 ...

You have u(x,0)=1 don't you? Why wouldn't u(1,0)=1 too?