Help on how to tackle second order PDE

Question

I don’t really have a background in differential equations and am studying functional analysis right now. In the textbook we follow I came across this second order partial differential equation. I have no idea how to tackle it.

$$\frac{\partial^2y}{\partial t^2}=a^2\frac{\partial^2y}{\partial x^2}$$ With boundary condition: $y(t,0)=y(t,1)=0$. And initial condition: $\frac{\partial y}{\partial t}(0,x)=0$. Can someone explain me how to tackle such a problem. I don’t really want a full solution just a push or two in the right direction.

Answer 1

The equation can be rearranged as the following (because partial derivatives commute):

(∂/∂t - a∂/∂x)(∂/∂t + a∂/∂x)y = 0

we know that f(x+at) is nulled by (∂/∂t - a∂/∂x) and g(x-at) is nulled by (∂/∂t + a∂/∂x)y

Therefore (because the operators commute because partial derivatives commute), the general form of the solution is :

y(x,t) = f(x+at) + g(x-at)

Now you can find substitute in your initial conditions to find the answer. I'll leave that to you, unless you want more help

Edit: in case you don't know how to do this, I'll briefly explain how to solve by method of characteristics.

Take the equation:

∂y/∂t + a∂y/∂x =0 (*)

We want to find a path x(t) such that y does not vary along this path. In other words, y(x(t),t)=constant

By the multivariate chain rule,

dy/dt = (∂y/∂t)(dt/dt) + (∂y/∂x)(dx/dt)

Thus dy/dt = (∂y/∂t) + (∂y/∂x)(dx/dt)

we can rearrange (*) to get: ∂y/∂t = -a∂y/∂x

Thus dy/dt = -a(∂y/∂x) + (∂y/∂x)(dx/dt)

Since y is constant on this path, we have dy/dt =0

Thus 0= (∂y/∂x) (dx/dt -a)

Thus if x = at +B where B is some other constant, then this equality is satisfied. Thus y doesn't change along the path: x= at + B

Thus y(x,t) = y(x-at,0)

Thus the general solution is y(x,t) = f(x-at) for some function f

Answer 2

Use the function $\Phi:[0,\pi]\to [0,1]$ such that $\Phi(x)=\dfrac{1}{\pi}x$ and define $z(t,x):=y(t,\Phi(x))$.
Obserbve that $z(t,0)=y(t,\Phi(0))=y(t,0)=0=y(t,1)=y(t,\Phi(\pi))=z(t,\pi).$ We can now consider the problem on the functional space $L^2[0,\pi]$ with periodic conditions.
$\partial_{tt}y=\partial_{tt}z$ and $\partial_{xx}y=\pi^2\partial_{xx}z$ so we have a "new" problem to solve, which is $$\begin{cases}\partial_{tt}z(t,x)=c^2\pi^2\partial_{xx} z(t,x)\\ z(t,0)=0\\ z(t,\pi)=0\end{cases}.$$ If you now express $z(t,x)$ as $\sum_{n\ge 1}z_n(t)\sin(nx)$, where $z_n(t):=(\sin(nx),z(t,x))_{L^2[0,\pi]}$, the wave equation is now $$\sum_{n\ge 1}(\ddot z_n(t)+\pi^2n^2c^2z_n(t))\sin(nx)=0\iff \ddot z_n(t)+\pi^2n^2c^2z_n(t)=0$$ and the second equation is an harmonic oscillator!
Once you have determined $z_n(t)$ for $n\ge 1$, you can write the general solution on $[0,1]$ $$y(t,x)=\sum_{n\ge 1}y_n(t)\sin(n\pi x).$$