# Solving inhomogenous wave equation

by wahoo2000
Tags: equation, inhomogenous, solving, wave
HW Helper
P: 5,003
 Quote by wahoo2000 Hmm, does that mean we are looking for Fourier sinus series of f(x)=x^3/6-x/2? Is it $$A_{n}D_{n}=2\int^{1}_{0}(\frac{x^{3}}{6}-\frac{x}{2})(\frac{(2n-1)\pi}{2}x)dx$$
Yes, you can evaluate the integral by using integration by parts 3 times.
 P: 37 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 :)
HW Helper
P: 5,003
 Quote by 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 :)
Yes, I just assumed you meant to put the sin in there.
 P: 37 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: $$A_{n}D_{n}=-\frac{2}{(2n+1)^{2}\pi^{2}}(-(-1)^{n})-\frac{16}{(2n+1)^{2}\pi^{2}}-\frac{64}{(2n+1)^{4}\pi^{4}}$$
HW Helper
P: 5,003
 Quote by 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: $$A_{n}D_{n}=-\frac{2}{(2n+1)^{2}\pi^{2}}(-(-1)^{n})-\frac{16}{(2n+1)^{2}\pi^{2}}-\frac{64}{(2n+1)^{4}\pi^{4}}$$
This is incorrect. Show me what you did when you applied integration by parts the first time.
 P: 37 first time was $$2[(\frac{x^{3}}{6}-\frac{x^{2}}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0} +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{2x}{2})(cos(\frac{(2n-1)\pi}{2}x)dx$$ But I saw now when writing this that I used x^2/2 instead of x/2 in my calculations.. :( :(
HW Helper
P: 5,003
 Quote by wahoo2000 first time was $$2[(\frac{x^{3}}{6}-\frac{x^{2}}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0} +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{2x}{2})(cos(\frac{(2n-1)\pi}{2}x)dx$$ 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.
 P: 37 Here is the first step.. if you are online, just tell me if that step i correct before I do it all again... $$A_{n}D_{n}=2[(\frac{x^{3}}{6}-\frac{x}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0} +\frac{4}{(2n+1)\pi}\int^{1}_{0}(\frac{3x^{2}}{6}-\frac{1}{2})(cos(\frac{(2n-1)\pi}{2}x)dx$$ where the first term, $$2[(\frac{x^{3}}{6}-\frac{x}{2})(cos(\frac{(2n-1)\pi}{2}x)]^{1}_{0}=0$$
 P: 37 I worked it through anyway... all terms except the last became zero. I now got $$A_{n}D_{n}=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}$$
HW Helper
P: 5,003
 Quote by wahoo2000 I worked it through anyway... all terms except the last became zero. I now got $$A_{n}D_{n}=\frac{32}{(2n-1)^{3}\pi^{3}}*-(-1)^{n}$$
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?
 P: 37 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 $$u(x,t)=w(x,t)+S(x)= w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1$$
HW Helper
P: 5,003
 Quote by 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 $$u(x,t)=w(x,t)+S(x)= w(x,t)=-\frac{32}{\pi^{4}} \sum_{n=1}^{\infty} \frac{(-1)^{n}}{(2n+1)^{4}}sin(\frac{(2n-1)\pi}{2}x)cos(\frac{(2n-1)\pi}{2}t)-\frac{x^{3}}{6}+\frac{x}{2}+1$$
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?
 P: 37 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 .....
HW Helper
P: 5,003
 Quote by 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 .....
You have u(x,0)=1 don't you? Why wouldn't u(1,0)=1 too?
 P: 37 Aha.. but then $$u(1,0)= -\frac{32}{\pi^{4}} \sum_{m=0}^{\infty} \frac{(-1)^{m}}{(2m+1)^{4}}*-(-1)^{m}*1-\frac{1}{6}+\frac{1}{2}+1=1 \Rightarrow \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=(1-\frac{4}{3})*(-\frac{\pi^{2}}{32})=\frac{\pi^{4}}{96}$$
 HW Helper P: 5,003 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 : $$\Rightarrow \sum_{n=1}^{\infty} \frac{1}{(2n-1)^{4}} = \sum_{m+1=1}^{\infty} \frac{1}{(2(m+1)-1)^{4}} = \sum_{m=0}^{\infty} \frac{1}{(2m+1)^{4}}=\frac{\pi^{4}}{96}$$
 P: 37 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.
HW Helper
P: 5,003
 Quote by 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.
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).

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