Multivariable Chain Rule – Proving an equation of partial derivatives – Exercise 6511

Exercise

Given the differentiable function (with parameter a)

$y=f(x-at)+g(x+at)$

Prove the equation

$y''_{tt}=a^2y''_{xx}$

Proof

Define

$u=x-at$

$v=x+at$

We get the function

$y=f(u)+g(v)$

And the internal functions

$u(x,t)=x-at$

$v(y,t)=x+at$

We will use the chain rule to calculate the first and second order partial derivatives of y.

$y'_t=f'_u\cdot u'_t+g'_v\cdot v'_t=$

$=f'_u\cdot (-a)+g'_v\cdot a$

$=-af'_u+ag'_v$

$y''_{tt}=-a\cdot f''_u\cdot u'_t+ag''_v\cdot v'_t=$

$=-af''_u\cdot (-a)+ag''_v\cdot a=$

$=a^2f''_u+a^2g''_v$

$y'_x=f'_u\cdot u'_x+g'_v\cdot v'_x=$

$=f'_u\cdot 1+g'_v\cdot 1=$

$=f'_u+g'_v$

$y''_{xx}=f''_u\cdot u'_x+g''_v\cdot v'_x$

$f''_u\cdot 1+g''_v\cdot 1=$

$=f''_u+g''_v$

We will put the second order partial derivatives in the left side of the equation we need to prove.

$y''_{tt}=a^2f''_u+a^2g''_v=$

$=a^2(f''_u+g''_v)=$

$=a^2y''_{xx}$

We got to the right side of the equation as required.

Have a question? Found a mistake? – Write a comment below!
Was it helpful? You can buy me a cup of coffee here, which will make me very happy and will help me upload more solutions!