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.
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