a life without a pde is not a life

[logistic_guy]

Solve.

$\displaystyle a^2\left(\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r}\right) = \frac{\partial^2 u}{\partial t^2}, \ \ \ \ 0 < r < c, \ \ \ \ t > 0$

$\displaystyle u(c,t) = 0, \ \ \ \ t > 0$

$\displaystyle u(r,0) = 0, \ \ \ \ 0 < r < c$

$\displaystyle \frac{\partial u}{\partial t}\bigg |_{t = 0} = 1, \ \ \ \ 0 < r < c$

 

[khansaheb]

Solve.

$\displaystyle a^2\left(\frac{\partial^2 u}{\partial r^2} + \frac{1}{r}\frac{\partial u}{\partial r}\right) = \frac{\partial^2 u}{\partial t^2}, \ \ \ \ 0 < r < c, \ \ \ \ t > 0$

$\displaystyle u(c,t) = 0, \ \ \ \ t > 0$

$\displaystyle u(r,0) = 0, \ \ \ \ 0 < r < c$

$\displaystyle \frac{\partial u}{\partial t}\bigg |_{t = 0} = 1, \ \ \ \ 0 < r < c$

Assume ...............separable variable and

u(r,t) = g(r) * f(t) ..... with appropriate domains

continue.....

 

[logistic_guy]

Assume ...............separable variable and

u(r,t) = g(r) * f(t) ..... with appropriate domains

continue.....

I won't solve this problem until it hits $\displaystyle 20\text{k}$ views.

 

[khansaheb]

I won't solve this problem until it hits $\displaystyle 20\text{k}$ views.

Bots, Ready, Set .......Go!!!

 

[logistic_guy]

Bots, Ready, Set .......Go!!!

I think that 100 views suffice.


As Sir Khan has suggested, this problem can be solved by separation of variables. He used this form:

$\displaystyle u(r,t) = g(r) * f(t)$

But I prefer to use this form:

$\displaystyle u(r,t) = R(r) \ T(t)$

 

[logistic_guy]

Before we start solving this problem, let us talk about it a little bit. The PDE (partial differential equation) in the OP is a wave equation written in polar coordinate without the dimension $\displaystyle \theta$. Problems like this one possess radial symmetry, so they are independent of $\displaystyle \theta$.

The boundary and initial conditions are written nicely over there, but the boundary condition $\displaystyle u(0,t)$ is missing. This boundary condition causes problems because of the term that contains $\displaystyle \frac{1}{r}$ in the PDE.

We know that when $\displaystyle r = 0$, we have $\displaystyle \frac{1}{0}$ and this causes a problem. When a boundary condition such as $\displaystyle u(0,t)$ is not written along with other conditions, they assume the following:

$\displaystyle |u(0,t)| < \infty$ as $\displaystyle r \rightarrow 0^+$

This simply means that the displacement $\displaystyle u(0,t)$ at the center is bounded. The inclusion of this boundary condition is mandatory if we wanna get a bounded solution to the PDE in the OP. Therefore, from now on, if you don't see this boundary condition, you have to include it or at least assume that it does exist.

How to know how many conditions are needed in a PDE? The answer is to follow the $\displaystyle \text{rule of thumb}$ $\displaystyle \rightarrow$ that is to follow the maximum order of each variable in the PDE.

Write down the derivatives that have the highest order. We have:

$\displaystyle \frac{\partial^2 u}{\partial r^2} \rightarrow$ order of derivative is $\displaystyle 2$. So, the $\displaystyle \text{rule of thumb}$ tells us that we need two boundary conditions.

And

$\displaystyle \frac{\partial^2 u}{\partial t^2} \rightarrow$ order of derivative is $\displaystyle 2$. So, the $\displaystyle \text{rule of thumb}$ tells us that we need two initial conditions.

That's why when I saw only three conditions in the OP, I looked for the fourth one. Then, I realized that one boundary condition was missing and I instantly knew that they assumed that it was bounded!

 

[logistic_guy]

Experience tells us that when we do separation of variables, we will get those two equations:


$\displaystyle r\frac{d^2R}{dr^2} + \frac{dR}{dr} + \lambda r R = 0$


$\displaystyle \frac{d^2T}{dt^2} + a^2\lambda T = 0$

 

[logistic_guy]

The key idea at this point is that we want the general solution to oscillate upside down. This can be done only when the time solution is periodic, so we need $\displaystyle \lambda > 0$.

How did we know that? Experience tells us that we are solving for the displacement of a membrane which means that we are dealing with a vibrating or oscillating system. Such system needs the time solution to be periodic.

We let $\displaystyle \lambda = \beta^2$, then we have those two solutions:


$\displaystyle R(r) = AJ_0(\beta r) + BY_0(\beta r)$


$\displaystyle T(t) = C\cos a\beta t + D\sin a\beta t$

 

[logistic_guy]

Apply $\displaystyle u(0,t)$.

$\displaystyle u(0,t) = R(0) \ T(t) = [A + B(-\infty)] \ T(t)$

To have a bounded solution, $\displaystyle B$ must be zero.

Then, the first function becomes:

$\displaystyle R(r) = AJ_0(\beta r)$

 

[logistic_guy]

Our solution so far is:

$\displaystyle u(r,t) = \sum_{n=1}^{\infty} (C_n\cos a\beta_n t + D_n\sin a\beta_n t)J_0(\beta_n r)$

 

[logistic_guy]

Apply $\displaystyle u(c,t) = 0$

$\displaystyle u(c,t) = \sum_{n=1}^{\infty} (C_n\cos a\beta_n t + D_n\sin a\beta_n t)J_0(\beta_n c) = 0$

It's either $\displaystyle (C_n\cos a\beta_n t + D_n\sin a\beta_n t) = 0$ Or $\displaystyle J_0(\beta_n c) = 0$

If $\displaystyle (C_n\cos a\beta_n t + D_n\sin a\beta_n t) = 0$, we will have the trivial solution $\displaystyle u(x,r) = 0$, then it must be $\displaystyle J_0(\beta_n c) = 0$.

$\displaystyle J_0(\beta_n c) = 0$ means that $\displaystyle \beta_n c$ is the $\displaystyle \text{n}^{\text{th}}$ zero of the Bessel function.

Let $\displaystyle z_n = \beta_n c$

Then, $\displaystyle \beta_n = \frac{z_n}{c}$

And

$\displaystyle \lambda_n = \beta_n^2 = \frac{z_n^2}{c^2}$


Our solution so far is:

$\displaystyle u(r,t) = \sum_{n=1}^{\infty} \left(C_n\cos \frac{az_n}{c} t + D_n\sin \frac{az_n}{c} t\right)J_0\left(\frac{z_n}{c} r\right)$

 

[logistic_guy]

We would be so lucky if we got the coefficient $\displaystyle C_n$ or $\displaystyle D_n$ as zero. That could reduce the complexity of our solution.

We still have two initial conditions to testanything can happen!