Question

Find the deflection of vibration of string of unit length having fixed ends with initial velocity zero and initial deflection f(x)=a (x-x2)

Answer 1

Let u(x,t) is a function, which describes vibration of string. So we can rewrite task in the next form:

$\dfrac{\partial^2 u}{\partial t^2}=b^2\dfrac{\partial^2 u}{\partial x^2}$

$u(0,t)=u(1,t)=0$

$u(x,0)=f(x)=a(x-x^2)$

$\dfrac{\partial u}{\partial t}|\begin{matrix} \\ t=0\end{matrix}=0$

To solve this task I use a Fourier method.

Let:

$u(x,t)=X(x)*T(t)$

Then,

$T''X=b^2TX''$

$\dfrac{T''}{b^2T}=\dfrac{X''}{X}=const=-\lambda^2$

So we have two tasks:

$1) X''+\lambda^2X=0$

$2)T''+(b\lambda)^2T=0$

For the task 1) we have 2 boundary conditions:

$u(0,t)=X(0)T(t)=0=>X(0)=0$

$u(1,t)=X(1)T(t)=0=>X(1)=0$

The first task solution is:

$X(x)=Acos(\lambda x)+Bsin(\lambda x)$

From the first boundary condition we have that A=0. The second give us:

$X(1)=Bsin(\lambda)=0=>\lambda=\pi n=>X(x)=Bsin(\pi n x)$

But we know that:

$||X(x)||=1=>$

$\begin{matrix} 1\\ \int \\ 0 \end{matrix} |X(x)|^2 dx=\begin{matrix} 1\\ \int \\ 0 \end{matrix} B^2sin^2(\pi nx)dx= B^2 * \dfrac{1}{2}=1$

$B=\sqrt 2$

For the task 2) we have next solution:

$T(t)=Ccos(\lambda bt) + Dsin(\lambda bt)=Ccos(\pi bn t)+Dsin(\pi b n t)$

The result of the main task is:

$u(x,t)=\sqrt 2\displaystyle\sum_{i=1}^{\infty} (Ccos(\pi n b t) +Dsin(\pi nbt)) sin(\pi n x)$

But we also have two initial conditions. The first one:

$u(x,0)=\sqrt 2 \displaystyle\sum_{i=1}^{\infty} C sin(\pi nx)=f(x)$

Multiply both side by $\sin(\pi kx)$:

$f(x)sin(\pi kx)=\sqrt 2\displaystyle\sum_{i=1}^{\infty} C sin(\pi nx)sin(\pi kx)$

Now integrate both sides with respect to x:

$\displaystyle\int_{0}^{1} f(x) sin(\pi kx)dx=\sqrt 2\displaystyle\int_{0}^{1} C sin(\pi nx)sin(\pi kx)dx$

Because trigonomical function are ortogonal, the right side give us:

$\begin{cases} 0 &\text {if } k {=}\mathllap{/\,} n \\ 1/2 &\text {if } k = n \end{cases}$

So

$C=\sqrt 2 \displaystyle\int_{0}^{1} f(x) sin(\pi nx)dx\\$

$C=\sqrt 2 a \displaystyle\int_{0}^{1} (x-x^2) sin(\pi nx)dx = -\dfrac{2a}{(\pi n)^3}cos(\pi nx)|\begin{matrix} 1 \\ 0 \end{matrix} =-\dfrac{2a}{(\pi n)^3}(cos(\pi n)-1)=\dfrac{4a}{(\pi (2n-1))^3}$

The second initial condition:

$\dfrac{\partial u}{\partial t}|\begin{matrix} \\ t=0\end{matrix} = \sqrt 2 \pi na\displaystyle\sum_{i=1}^{\infty} Dsin(\pi n x)=0=>D=0$

So the answer is:

$u(x,t)=\dfrac{4a\sqrt 2}{\pi ^3}\displaystyle\sum_{i=1}^{\infty} \dfrac{1}{(2n-1)^3}cos(\pi (2n-1) b t) sin(\pi (2n-1) x)$