### 7 Homogeneous Wave PDE in 2D

#### 7.1 Rectangular membrane

Using Cartesian coordinates. Wave displacement is (out of page).

Boundary conditions on

And boundary conditions on

Initial conditions

Let . Plug in the PDE

Hence

Hence time ODE becomes

And the space ODE is

Separating this again

Using new separation variable , we obtain two new ODE’s

Or

Solving for ODE, and knowing that from earlier, we obtain

Applying B.C. at

Hence . Applying B.C. at

Hence

Therefore the solution is

Solving the ODE

The solution is

Applying first B.C.

Hence

Applying second B.C.

Hence

Hence the solution is

Now we solve the time ode

Combining all solution , and merging all constants into two, we find

We now use initial conditions to find . At

Applying first initial condition to (1)  gives

Applying 2D orthogonality gives

Taking time derivative of (1)

AT the above becomes

Applying 2D orthogonality gives

Summary of solution

Here are few animations, all using . For different modes. The mode is given below each animation. These run for seconds. Initial conditions are and . Boundary conditions zero on all 4 edges.

#### 7.2 circular membrane

##### 7.2.1 not circularly symmetric

Using polar coordinates. Disk has radius . Wave displacement is (out of page).

Boundary conditions on

And boundary conditions on

Initial conditions

Let . Plug in the PDE

Dividing by

Hence

The time ODE is

Now we separate again the space ODE’s (remember to move the with the and not the )

Let separation constant be , therefore

With periodic boundary conditions and

Now it is in SL form, where . This is singular SL. Can be written as

Before we solve the above ODE, we solve the to find Eigenvalues. The solution is

With B.C  and . From first B.C. we obtain

Looking at second B.C.

Hence

From (1,2), we see that both are satisfied if

Hence

There is another solution for which is constant (that is why one of the sums below starts from ). We can combine the zero eigenvalue with the above and write

Since at the above reduces to constant .

Now that we know , from solving the part, we go and solve the ODE. For each , the solution to the (Bessel) ode

The solution turns out to be

Where is found from roots of giving the eigenvalues. Now the time ODE is solved

Hence the solution is

We now break this sum as follows

Or

Then we break the above into 4 sums

Finally, we merge constants in the above as follows

Hence the final solution now becomes

Now initial conditions is used to find using orthogonality. At the solution simplifies to (all terms with vanish giving

Hence

 (4)

When iterating over index, the terms and will be constant. So for each , we have and . So orthogonality is carried out on the index on the Bessel functions, as in . But in this problem (from the Bessel ODE above). Multiplying (4) by and integrating

Or

Replacing back with , the above becomes

We now apply orthogonality on using the . Hence

Or

The term goes to zero, and we are left with (also, changed back to since we are doing with summation)

Hence

We now need to apply orthogonality on using the . to find . Using (5)

Hence

Therefore

Now we will look at the second initial conditions Taking derivative w.r.t. time of the solution in (3) gives

At time the above becomes (all terms with vanish).

Now orthogonality is used. At the above becomes

Similarly to the above we now find and . The only difference, is that now we have extra terms that show up. The final result will be

And

Summary of solution

With being the solutions for . For each , we find , which are the zeros of the Bessel function. So for each , we have infinite number of zeros. This generates all the needed . Hence , therefore

The following animations run for 80 seconds. They are for different modes. All have zero for initial velocity, which means . Radius used is and , initial position used is .

Cases for

Cases for

Cases for

Cases for

##### 7.2.2 case circularly symmetric

In this case, there is no dependency in boundary conditions or in initial conditions. Using polar coordinates. Disk has radius . Wave displacement is (out of page).

Boundary conditions on

Initial conditions

Let . Plug in the PDE

Dividing by

Hence

The time ODE is

And the ODE is (Sturm-Liouville)

Where . This is singular SL.  The solution turns out to be

Where is found from roots of giving the eigenvalues. Now the time ODE is solved

Hence the solution is

Now initial conditions is used to find using orthogonality. At the solution simplifies to

Hence

Now we will look at the second initial conditions Taking derivative w.r.t. time of the solution in (1) gives

At time the above becomes

Now orthogonality is used. The above becomes

Summary of solution

With being the solutions for . We have infinite number of zeros. This generates all the needed . Hence , therefore

This animation runs for 40 seconds. Using radius and zero initial velocity. Initial position is And . Number of terms in the sum used was . Which means . The terms are all zero since initial velocity was zero.