Verify the PDE has the following solution

[docnet]

Homework Statement

Verify the PDE has the following solution

Relevant Equations

$u(t,x)=f(x-t)+g(x+t)$

Hello, please lend give me your wisdom.

I suspect this problem is about the wave equation $\partial_t^2-\partial_x^2=0$ commonly encountered in physics. I tried a search for information but I could not find help.

Attempt at arriving at solution:

So I took the partial derivatives of $u(t,x)=f(x-t)+g(x+t)$

$\partial_t u=-\partial_tf(x-t)+\partial_tg(x+t)$
$\partial_t^2 u=\partial_t^2f(x-t)+\partial_t^2g(x+t)$
$\partial_x u=\partial_xf(x-t)+\partial_xg(x+t)$
$\partial_x^2 u=\partial_x^2f(x-t)+\partial_x^2g(x+t)$

Giving
$\partial_t^2-\partial_x^2 =0$
$\partial_t^2f(x-t)+\partial_t^2g(x+t) -\partial_x^2f(x-t)-\partial_x^2g(x+t)=0$
$\partial_t^2f(x-t) -\partial_x^2f(x-t)+\partial_t^2g(x+t)-\partial_x^2g(x+t)=0$

I try to reason if the last line is true, then so is $\partial_t^2-\partial_x^2=0$ because it shows
$\partial_t^2f(x-t) -\partial_x^2f(x-t)=0, \partial_t^2g(x+t)-\partial_x^2g(x+t)=0$.
Although it does not feel sufficient.

For $u=f+g$ I can set $t=0$ in $u(t,x)=f(x-t)+g(x+t)$ and just get back $u=f+g$.

For $\partial_t u=-f+g$ I can take $\partial_t u=-\partial_tf(x-t)+\partial_tg(x+t)$ and set $t=0$ to get back $\partial_t u=-\partial_tf+\partial_tg$. So the last condition is not satisfied unless setting t=0 somehow means removing $\partial_t$ from both terms.

 

[lurflurf]

If that from a book?
Unless I missed something I am confused why the second condition is not
$$\partial_tu=-f^\prime+g^\prime \textrm{ on } \{t=0\}\times\mathbb{R}$$

 

[FactChecker]

It seems like you have proven it and then doubt your own proof. I see nothing wrong with your proof and do not understand why you doubt it.