text: Author Topic: FE Problem 3  (Read 3853 times)
  
  Victor Ivrii
  
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  FE Problem 3
  « on: April 14, 2015, 07:41:13 PM »
  Solve by the method of separation of variables
  \begin{equation*}
  \left\{\begin{aligned}
  & u_x (0,t)= u _x (1,t)=0,\\[2pt]
  & u(x,0)=f(x),\\[2pt]
  & u_t(x,0)=g(x)\end{aligned}\right.
  \end{equation*}
  with $f(x)=x(1 -x)$, $g(x)=0$.  Write the answer in terms of  Fourier series.
  
  Zacharie Leger
  
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  Re: FE Problem 3
  « Reply #1 on: April 16, 2015, 02:16:09 AM »
  Hopeful solution
  By letting $u(x,t)=X(x)T(t)$ and plug into the wave equation we can get
  $$\frac{X''(x)}{X(x)}-\frac{T''(t)}{4T(t)}=0 \Rightarrow \frac{X''(x)}{X(x)}=\frac{T''(t)}{4T(T)}=-\lambda$$
  We know that $\lambda\geq 0$, so let of first develop the the case where $\lambda =0$. Hence,
  $$X''(x)=0 \Rightarrow X(x)=Ax+B \Rightarrow X'(0)=A=0 \Rightarrow X(x)=B$$
  Now if $\lambda =\omega ^2,\,\omega >0$ we get the following solution
  $$X(x)=C\sin(\omega X)+D\cos(\omega X)$$
  Applying the boundry conditions on the solution we see that
  $$X'(0)=C\omega\cos(\omega (0))+D\omega\sin(\omega (0))=C=0\,\mathrm{and}\, X(1)=D\omega\sin(\omega (1))=0\Rightarrow \omega=n\pi$$
  $$\Rightarrow X_n (x)=D_n \cos(n\pi x)$$
  As for the solution for $T(t)$, we again consider first when $\lambda=0$
  $$T''(t)=0 \Rightarrow T(t)=Et+F \Rightarrow T'(0)=E=0 \Rightarrow T(t)=F$$
  Now when $\lambda =(n\pi)^2$
  $$T(t)=G\cos(2n\pi x)+H\sin(2n\pi x)$$
  Applying the boundary conditions we see that
  $$T'(0)=2n\pi G\sin(n(0))+2n\pi H\cos(n(0))=2n\pi F=0$$
  Hence,
  $$T_n (t)=G_n \cos(2n\pi t)$$
  So the full solution can be expressed as
  $$u(x,y)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)\cos(2n\pi t)$$
  where we let $\alpha _0 =B+F$ and $\alpha _n = D_n G_n$. We have one last boundary conditions to apply to get our final answer:
  $$u(x,0)=x(1-x)=\alpha _0 +\sum\limits_{n=1}^\infty \alpha _n \cos(n\pi x)$$
  Fourier tells us that the coefficiants will be given by
  $$\alpha _0 =\int_0^1 x-x^2\mathrm{d}x=[\frac{x^2}{2}-\frac{x^3}{3}]_0^1=\frac{1}{6}$$
  $$\alpha _n =2\int_0^1 (x-x^2)\cos(n\pi x)\mathrm{d}x=2[(-n\pi(-1+2 x)\cos(n\pi x)+(2-n^2\pi^2 (-1+x)x)\sin(n\pi x))/(n^3\pi^3)]_0^1=\frac{(1+(-1)^n)}{\pi^2 n^2}$$
  Therefore, our final solution is:
  $$u(x,y)=\frac{1}{6} +\sum\limits_{n=1}^\infty \frac{2(1+(-1)^n)}{\pi^2 n^2} \cos(n\pi x)\cos(2n\pi t)$$
  
  Victor Ivrii
  
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  Re: FE Problem 3
  « Reply #2 on: April 17, 2015, 08:17:08 AM »
  Separation of variables results in $X''+\lambda X=0$, $X'(0)=X'(1)=0$ and thus $\lambda_0=0$, $X_0= \frac{1}{2}$ and $\lambda_n=\pi^2n^2$, $X_n=\cos (\pi n x)$ with $n=1,2,\ldots$; also $T''+4\pi^2 T=0$ and thus $T_0=A_0+B_0t$, $T_n= A_n \cos (2\pi n t)+B_n \sin (2\pi n t)$, and
  
  u=\frac{1}{2}(A_0+B_0t) + \sum_{n=1}^\infty \Bigl(A_n \cos (2\pi n t)+B_n \sin (2\pi n t)\Bigr)\cos (\pi n x).
  
  The initial conditions result in
  \begin{equation*}
  \frac{1}{2}A_0+ \sum_{n=1}^\infty A_n \cos (\pi n x)=x(1-x),\qquad
  \frac{1}{2}B_0 + \sum_{n=1}^\infty 2\pi n B_n \cos (\pi n x)=0
  \end{equation*}
  and $B_n=0$ ($n=0,1,2,\ldots$) and
  \begin{multline*}
  A_n =2 \int_0^1 x(1-x)\cos (\pi n x)\,dx=- \frac{2}{\pi n} \int_0^1 (1-2x) \sin (\pi nx)=\\
  -\frac{2}{\pi ^2n^2 }  (1-2x) \cos (\pi nx)\Bigr|_{x=0}^{x=1} +
  \frac{4}{\pi ^2n^2 } \int_0^1 \cos (\pi nx)\,dx =\left\{\begin{aligned}  -\frac{1}{\pi ^2m^2 } & && n=2m,\\
  0& &&n=2m+1\end{aligned}\right.
  \end{multline*}
  $m=1,2, \ldots$. Meanwhile $A_0=\frac{1}{3}$. Then
  \begin{equation*}
  u=\frac{1}{12}- \sum_{m=1}^\infty \frac{1}{\pi^2m^2} A_n \cos (4\pi m t)\cos (2\pi m x).
  \end{equation*}
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