Solving inhomogenous wave equation

[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

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

 

[gabbagabbahey]

Well... that means that

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

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

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

 

[gabbagabbahey]

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

$$A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})sin(\frac{(2n-1)\pi}{2}x)dx$$
, 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...

$$A_{n}D_{n}=2[(\frac{x^{3}}{6}-\frac{x}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}<br /> <br /> +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx$$
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:
$$A_{n}D_{n}=\frac{32}{(2n-1)^{4}\pi^{4}}*-(-1)^{n}$$ ?
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.
$$A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx =$$
$$=-\frac{16}{(2n+1)^{2}\pi{2}}\int^{1}_{0}(x*sin(\frac{(2n-1)\pi}{2}x)dx = \frac{32}{(2n+1)^{3}\pi{3}}\int^{1}_{0}(cos(\frac{(2n-1)\pi}{2}x)dx=$$
$$=\frac{32}{(2n+1)^{3}\pi{3}}[(sin(\frac{(2n-1)\pi}{2}x)]=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}$$
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
$$w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)$$ 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.
$$A_{n}D_{n}=+\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx =$$
$$=-\frac{16}{(2n+1)^{2}\pi{2}}\int^{1}_{0}(x*sin(\frac{(2n-1)\pi}{2}x)dx = \frac{32}{(2n+1)^{3}\pi{3}}\int^{1}_{0}(cos(\frac{(2n-1)\pi}{2}x)dx=$$
$$=\frac{32}{(2n+1)^{3}\pi{3}}[(sin(\frac{(2n-1)\pi}{2}x)]=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}$$
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
$$w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)$$ 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
$$u(1,0)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}*-(-1)^{n}*1-\frac{1}{6}+\frac{1}{2}+1$$
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
$$u(1,0)= -\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n-1)^{4}}*-(-1)^{n}*1-\frac{1}{6}+\frac{1}{2}+1$$
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?

 

[wahoo2000]

Aha.. but then
$$u(1,0)= -\frac{32}{\pi^{4}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{4}}*-(-1)^{m}*1-\frac{1}{6}+\frac{1}{2}+1=1<br /> \Rightarrow \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=(1-\frac{4}{3})*(-\frac{\pi^{2}}{32})=\frac{\pi^{4}}{96}<br />$$

 

[gabbagabbahey]

Yep, and

$$\sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} =\frac{\pi^{4}}{96}$$

can be used to answer your question by substituting n=m+1 :

$$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} = \sum_{m+1=1}^{\infty} \frac{1}{(2(m+1)-1)^{4}}<br /> = \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=\frac{\pi^{4}}{96}$$

 

[wahoo2000]

Thank you SO much for your patience. Tomorrow I am going through this problem from the start again to make sure I get each step.

 

[gabbagabbahey]

Thank you SO much for your patience. Tomorrow I am going through this problem from the start again to make sure I get each step.

Your welcome. The critical point of the problem was the part about the boundary conditions for w(x,t). Choosing the right boundary conditions (by choosing boundary conditions for S) makes the solution much easier. In general, you should try to set the conditions on S such that w(x,t) vanishes at one endpoint (x=0 in this case).