One dimensional wave equation and D-Alembert's solution of wave equation

Wave equation:

One dimensional wave equation :

Let us consider a string with mass ‘m’ and length ‘l’ such that its mass per unit length (\(\frac ml=\rho\) is constant. Also consider the string is fixed at two point ‘o’ and ‘c’. So that it undergoes transverse mode of vibration. Let A nd B be two reference point where tension \(T_1\) and \(T_2\) are exerted. Then \(T_1 cos\alpha\) and \(T_2 cos\beta\) are horizontal components and \(T_2 sin\beta\) and \(T_1 sin\alpha\) are vertical components.

Since there is no movement in horizontal direction, so

$$T_1 cos\alpha= T_2 cos\beta= T(let)$$

Now the vertical components

$$T_2 sin\beta- T_1 sin\alpha= m\frac{\delta ^2 u(x,t)}{\delta t^2}$$

Dividing both sides of above equation by \(T_1\), we get

$$\frac{T_2 sin\beta}{T}-\frac{T_1 sin\alpha}{T}=\frac{m}{T} \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or,\; \frac{T_2 sin\beta}{T_2 cos\beta}- \frac{T_1 sin\alpha}{T_1 cos\alpha}= \frac{m}{T}\frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or,\; Tan\beta-Tan\alpha=\frac mT \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or,\; \biggl(\frac{\delta u}{\delta x}\biggr)_{x+\delta x}-\biggl(\frac{\delta u}{\delta x}\biggr)_x=\frac mT \frac{\delta^2 u(x,t)}{\delta t^2}$$

In limiting case,

\(A\to B, \delta x=\delta x\to 0\)

$$or,\;\; \biggl(\frac{\delta u}{\delta x}\biggr)_{x+\delta x}-\biggl(\frac{\delta u}{\delta x}\biggr)_x= \frac{\rho \delta_s}{T} \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or,\;\;\frac{ lim_{\delta x\to 0}\biggl(\frac{\delta u}{\delta x}\biggr)_{x+\delta x}-\biggl(\frac{\delta u}{\delta x}\biggr)_x}{\delta x}= \frac{\rho}{T} \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or, \frac{\delta^2}{\delta x^2}=\biggl(\frac{\rho}{T}\biggr) \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or, \;\frac{\delta^2 u(x,t)}{\delta x^2}=\frac{1}{(\frac{T}{\rho)}} \frac{\delta^2 u(x,t)}{\delta t^2}$$

$$or,\; \frac{\delta^2 u(x,t)}{\delta x^2}=\frac{1}{c^2}\frac{\delta^2 u(x,t)}{\delta t^2}$$

Where, \(c^2=\frac{T}{\rho}\)

\(\therefore\) \(c=\sqrt{\frac{T}{\rho}}\) is called velocity of transverse wave in the string.

This is the required one dimensional wave equation.

Alembert’s solution of wave equation :

We know the one dimensional wave equation as,

$$\frac{\delta^2 u(x,t)}{\delta x^2}= \frac{1}{c^2} \frac{\delta^2 u(x,t)}{\delta t^2}\dotsm(1)$$

We see that from equation (1) U is a function of x and t. In this method we assume u as function of ‘v’ and ‘w’ so that,

$$v=x+ct$$

$$w=x-ct$$

Here,

$$\frac{\delta u}{\delta x}=\frac{\delta u}{\delta v}\cdot \frac{\delta v}{\delta x}+\frac{\delta u}{\delta w}\cdot \frac{\delta w}{\delta x}$$

$$= \frac{\delta u}{\delta v}+\frac{\delta u}{\delta w}$$

$$or,\;\; \frac{\delta^2 u}{\delta x^2}=\frac{\delta}{\delta x} \biggl(\frac{\delta u}{\delta x}\biggr)$$

$$=\frac{\delta}{\delta x}\biggl[\frac{\delta u}{\delta v}+\frac{\delta u}{\delta w}\biggr]$$

$$=\frac{\delta}{\delta v}\biggl(\frac{\delta \psi}{\delta v}+\frac{\delta u}{\delta w}\biggr)\frac{\delta v}{\delta x}+\frac{\delta}{\delta 2}\biggl(\frac{\delta u}{\delta v}+\frac{\delta u}{\delta w}\biggr)\frac{\delta w}{\delta x}$$

$$=\frac{\delta^2 u}{\delta v^2}+\frac{\delta^2 u}{\delta v\delta w}+\frac{\delta^2 u}{\delta w\delta v}+\frac{\delta^2 u}{\delta w^2}$$

$$=\frac{\delta^2 u}{\delta v^2}+2\frac{\delta^2 u}{\delta v\delta w}+\frac{\delta^2 u}{\delta w^2}$$

Now,

$$\frac{\delta u}{\delta t}=\frac{\delta u}{\delta v}\frac{\delta v}{\delta t}+\frac{\delta u}{\delta 2}\frac{\delta w}{\delta t}= c\frac{\delta u}{\delta v}-c\frac{\delta u}{\delta 2}$$

Again,

$$\frac{\delta^2}{\delta t^2}=\frac{\delta}{\delta t}(\frac{\delta u}{\delta t}=\frac{\delta}{\delta t}\biggl(c\frac{\delta u}{\delta v}-c\frac{\delta u}{\delta w}$$

$$or, \frac{\delta^2u}{\delta t^2}=\frac{\delta}{\delta v}\biggl(c\frac{\delta u}{\delta v}-c\frac{\delta u}{\delta w}\biggr)\frac{\delta v}{\delta t}+\frac{\delta }{\delta w}\biggl(c\frac{\delta u}{\delta v}-c^2\frac{\delta u}{\delta w}\biggr)\frac{\delta w}{\delta t}$$

$$=c^2\frac{\delta ^2 u}{\delta v^2}-c^2\frac{\delta^2 u}{\delta v\delta w}-c^2\frac{\delta^2 u}{\delta w \delta v}+c^2 \frac{\delta^2 u}{\delta w^2}$$

$$= c^2\frac{\delta^2u}{\delta v^2}-2c^2\frac{\delta^2 u}{\delta v\delta 2}+c^2\frac{\delta^2 u}{\delta w^2}$$

Substituting there values in equation (1)

$$\frac{\delta^2u}{\delta v^2}+2\frac{\delta^2 u}{\delta v \delta w}+\frac{\delta^2u}{\delta w^2}=\frac{1}{c^2} \biggl[c^2\frac{\delta^2u}{\delta v^2}-2c^2 \frac{\delta^2 u}{\delta v \delta 2}+\frac{c^2 \delta^2 u}{\delta w^2}\biggr]$$

$$or,\; \;\frac{\delta^2u}{\delta v^2}+2\frac{\delta^2 u}{\delta v \delta w}+\frac{\delta^2u}{\delta w^2}=\frac{\delta^2u}{\delta u^2}-2\frac{\delta^2 u}{\delta v\delta w}+\frac{\delta^2u}{\delta w^2}$$

$$or,\; 4\frac{\delta^2 u}{\delta v \delta w}=0$$

$$or,\; \frac{\delta^2 u}{\delta v \delta w}=0$$

Integrating with respect to v we get,

$$\int \frac{\delta^2u}{\delta v\delta w}dv= Constant\;on\;w$$

$$=f(w)$$

Now, \(\frac{\delta u}{\delta w}=f(w)\)

Again, Integrating with respect to w. we get,

$$\int \frac{\delta u}{\delta w} dw= \int f(w)dw+constant\;on\;v$$

$$\Rightarrow u(x,t)=h(w)+f(v)$$

$$\therefore\;\;\;\; u(x,t)=h(x-ct)+f(x+ct)$$

This is required D-Alembert’s solution of wave equation.

Reference:

1. Adhikari, Pitri Bhakta, and Dya Nidhi Chhatkuli.A Textbook of Physics. Third Revised Edition ed. Vol. III. Kathmandu: SUKUNDA PUSTAK BHAWAN, 2072.
2. Vaughn, Michael T.Introduction to Mathematical Physics. Weinheim: Wiley-VCH, 2007
3. Butkov, Eugene.Mathematical Physics. Reading, MA: Addison-Wesley Pub., 1968.
4. Carroll, Robert W.Mathematical Physics. Amsterdam: North-Holland, 1988.
5. Dass, H. K.Mathematical Physics. New Delhi: S. Chand, 2005.