Question

If $\mathit{u}\mathbf{=}\mathit{f}\mathbf{(}\mathit{x}\mathbf{-}\mathit{c}\mathit{t}\mathbf{)}\mathbf{+}\mathit{g}\mathbf{(}\mathit{x}\mathbf{+}\mathit{c}\mathit{t}\mathbf{)}$, show that $\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{c}}^{\mathbf{2}}}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{u}}{\mathbf{\partial }{\mathbf{t}}^{\mathbf{2}}}\mathbf{.}$

Short answer

It is proved that $\frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{x}}^2}=\frac{1}{{\mathrm{c}}^2}\frac{{\partial }^2\mathrm{u}}{\partial {\mathrm{t}}^2}.$

Step-by-step solution

Given information.

Given$u=f(x-ct)+g(x+ct)$

Definition of partial differentiation.

Partial differentiation is defined as the process, in which find the partial derivative of a function.

In Partial differentiation, the function has more than one variable and find the partial derivative of a function with respect to one variable and keeping the other variable constant.

Calculate ∂u2∂x2 and ∂2u∂t2.

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}$Find the partial differentiation of function,$u=f(x-ct)+g(x+ct)$ with respect to x.

$\frac{\partial u}{\partial x}=\frac{\partial f(x-ct)}{\partial x}+\frac{\partial g(x+ct)}{\partial x}.........\left(1\right)$

Find the partial differentiation of equation (1) with respect to x.

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{{\partial }^{2}f(x-ct)}{\partial x^2}+\frac{{\partial }^{2}g(x+ct)}{\partial x^2}.........\left(2\right)$

Find the partial differentiation of function,$u=f(x-ct)+g(x+ct)$ with respect to t.

$\frac{\partial u}{\partial t}=-c\frac{\partial f(x-ct)}{\partial t}+c\frac{\partial g(x+ct)}{\partial t}.........\left(3\right)$

Find the partial differentiation of equation (3) with respect to x.

$$\begin{gathered} \frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{\partial f(x-ct)}{\partial t}+c^2\frac{\partial g(x+ct)}{\partial t} \\ =c^2\left[\frac{{\partial }^{2}f(x-ct)}{\partial t^2}+\frac{{\partial }^{2}g(x+ct)}{\partial t^2}\right].........\left(4\right) \end{gathered}$$

From equation (2),

localid="1659151087874" $\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}$

Or it can be simplified as:

$\frac{{\partial }^{2}u}{\partial x^2}=\frac{1}{c^2}\frac{{\partial }^{2}u}{\partial t^2}$

Hence it is proved.