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Problem Statement

Prompt

provide me the solution to Question 21.

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Question 21
**Sol.** The vibrations of the square membrane are governed by two dimensional wave equation

\[\frac{\partial^2 u}{\partial t^2} = c^2 \left( \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} \right)\]

... (1)

Here the boundary conditions are \( u(0, y, t) = 0, \, u(1, y, t) = 0, \, u(x, 0, t) = 0, \, u(x, 1, t) = 0 \) and the initial conditions are \( u(x, y, 0) = f(x, y) = A \sin \pi x \sin 2\pi y \) and

\[\left. \frac{\partial u}{\partial t} \right|_{t=0} = 0\]

To solve equation (1),
Let

\[u = XY T\]

where, \( X \) is a function of \( x \) only, \( Y \) is a function of \( y \) only, and \( T \) is a function of \( t \) only.

\[\frac{\partial^2 u}{\partial t^2} = \frac{\partial^2}{\partial t^2} (XYT) = XY T''\]

\[\frac{\partial^2 u}{\partial x^2} = X'' Y T, \, \frac{\partial^2 u}{\partial y^2} = XY'' T\]

Substituting these values in equation (1), we have

\[XY T'' = c^2 (X'' Y T + XY'' T')\]

\[\Rightarrow \frac{1}{c^2} \cdot \frac{T''}{T} = \frac{X''}{X} + \frac{Y''}{Y}\]

This will be true only when each member is a constant. Choosing the constants suitably, we have

\[\frac{d^2X}{dx^2} + k^2X = 0, \quad \frac{d^2Y}{dy^2} + l^2Y = 0, \quad \frac{d^2T}{dt^2} + (k^2 + l^2)c^2T = 0\]

The solutions of these equations are respectively,

\[X = c_1 \cos kx + c_2 \sin kx, \quad Y = c_3 \cos ly + c_4 \sin ly\]

\[T = c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct.\]

\[u(x, y, t) = (c_1 \cos kx + c_2 \sin kx)(c_3 \cos ly + c_4 \sin ly)\]

\[(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct) \tag{3}\]

From (3),

\[u(0, y, t) = 0 = c_1(c_3 \cos ly + c_4 \sin ly)(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct)\]

\[\Rightarrow c_1 = 0\]

From (3),

\[u(x, y, t) = c_2 \sin kx(c_3 \cos ly + c_4 \sin ly)(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct) \tag{4}\]

\[u(1, y, t) = 0 = c_2 \sin k(c_3 \cos ly + c_4 \sin ly)(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct)\]

\[\Rightarrow \sin k = 0 = \sin m\pi(m \in I) \text{ or } k = m\pi\]

\[u(x, y, t) = c_2 \sin m\pi x(c_3 \cos ly + c_4 \sin ly)(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct) \tag{5}\]

\[u(x, 0, t) = 0 = c_2 \sin m\pi x(c_3 \cos ly + c_4 \sin ly)(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct) \Rightarrow c_3 = 0\]

From (5),

\[u(x, y, t) = c_2c_4 \sin m\pi x \sin ly(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct) \tag{6}\]

\[u(x, 1, t) = 0 = c_2c_4 \sin m\pi x \sin l(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct)\]

\[\Rightarrow \sin l = 0 = \sin n\pi(n \in I) \Rightarrow l = n\pi\]

From (6),

\[u(x, y, t) = c_2c_4 \sin m\pi x \sin n\pi y(c_5 \cos \sqrt{k^2 + l^2} ct + c_6 \sin \sqrt{k^2 + l^2} ct)\]

\[u(x, y, t) = \sin m\pi x \sin n\pi y(A_{mn} \cos pt + B_{mn} \sin pt) \tag{7}\]

where

\[p = \pi c \sqrt{m^2 + n^2}\]

The most general solution is

\[u(x, y, t) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \sin m\pi x \sin n\pi y \, (A_{mn} \cos pt + B_{mn} \sin pt) \tag{8}\]

Applying the condition \(\frac{\partial u}{\partial t} = 0\) at \(t = 0\) we get \(B_{mn} = 0\) (9)

Also, using the condition \(u = f(x, y) = A \sin \pi x \sin 2\pi y\)

When \[(t = 0), we get (A \sin \pi x \sin 2\pi y = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} A_{mn} \sin m\pi x \sin n\pi y\]

This is a double Fourier Series.

\[A_{mn} = \frac{2}{1} \cdot \frac{2}{1} \int_0^1 \int_0^1 A \sin \pi x \sin 2\pi y \sin m\pi x \sin n\pi y \, dxdy\]

Obviously

\[A_{m_1} = A_{m_3} = A_{m_4} = \ldots = 0\]

But,
\[A_{m_2} = 4A \int_0^1 \int_0^1 \sin \pi x \sin m\pi x \sin^2 2\pi y \cdot dxdy\]
\[= 2A \int_0^1 \int_0^1 \sin \pi x \sin m\pi x (1 - \cos 4\pi y) dxdy\]
\[= 2A \int_0^1 \sin \pi x \sin m\pi x \left( y - \frac{\sin 4\pi y}{4\pi} \right)_0^1 dx\]
\[= 2A \int_0^1 \sin \pi x \sin m\pi x dx\]

Again obviously,
\[A_{22} = A_{32} = A_{42} = ... = 0\]

But,
\[A_{12} = 2A \int_0^1 \sin^2 \pi x \ dx = A \int_0^1 (1 - \cos 2\pi x) dx\]
\[= A \left( x - \frac{\sin 2\pi x}{2\pi} \right)_0^1 = A\]
\[\therefore \text{From (8), (9) and (10), we get}\]
\[u(x, y, t) = A \sin \pi x \sin 2\pi y \cos pt\]
\[\text{where, } p = \pi c \sqrt{m^2 + n^2} = \pi (1) \sqrt{1 + 4} = \pi \sqrt{5}\]

\[\therefore \text{From (11), } u(x, y, t) = A \cos \pi \sqrt{5} t \sin \pi x \sin 2\pi y \text{ which is the required deflection.} \]

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Problem #10