# Toronto Math Forum

## APM346-2015S => APM346--Tests => Final Exam => Topic started by: Victor Ivrii on April 14, 2015, 07:41:13 PM

Title: FE Problem 3
Post by: Victor Ivrii on April 14, 2015, 07:41:13 PM
Solve by the method of separation of variables
\begin{equation*}
\left\{\begin{aligned}
&u_{tt}-  4u_{xx}=0,\qquad 0<x<1, \; t>0,\\[2pt]
& 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.
Title: Re: FE Problem 3
Post by: Zacharie Leger 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:
$$u(x,y)=\frac{1}{6} +\sum\limits_{n=1}^\infty \frac{2(1+(-1)^n)}{\pi^2 n^2} \cos(n\pi x)\cos(2n\pi t)$$
Title: Re: FE Problem 3
Post by: Victor Ivrii 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*}