Solve $\frac{\partial u}{\partial t}=c^2\frac{\partial u}{\partial x \partial x}$ assuming we know the solutions of $\frac{\partial u}{\partial t}=\frac{\partial u}{\partial x \partial x}$
I don't know much about pde's. I've been reading about it but I'm not sure if I get it.
I think I could start changing slightly the form to $\frac{\partial u}{\partial t}-c^2\frac{\partial u}{\partial x \partial x}=0$. What I've read is about equations of the form $au_{xx}+bu_{xy}+cu_{yy}=0$.
Here I would like to use $\Delta=c^2-4(1)(0)=c^2.$ and then ince $\Delta >0$ it should be a hyperbolic solution: $u$ has the form $u=f(mx+y)$.
But does this makes sense? The first partial derivate with respect to $t$ bothers me, and it seems it what I wanted to do won't work here since I want $u_t=c^2u_{xx}$.
EDIT:
$v=u(a,b)$
With $a(x,t)=x/c, b(t)=t.$
$\displaystyle\frac{\partial v}{\partial t}=\displaystyle\frac{\partial u}{\partial a}\displaystyle\frac{\partial a}{\partial t}+\displaystyle\frac{\partial u}{\partial b}\displaystyle\frac{\partial b}{\partial t}$
Since $\displaystyle\frac{\partial a}{\partial t}=0, \displaystyle\frac{\partial b}{\partial t}=1$ then $\displaystyle\frac{\partial v}{\partial t}=\displaystyle\frac{\partial u}{\partial b}=\displaystyle\frac{\partial u}{\partial t}$
And $\displaystyle\frac{\partial v}{\partial x}=\displaystyle\frac{\partial u}{\partial a}\displaystyle\frac{\partial a}{\partial x}+\displaystyle\frac{\partial u}{\partial b}\displaystyle\frac{\partial b}{\partial x}$
since $\displaystyle\frac{\partial a}{\partial x} = 1/c$ and $\displaystyle\frac{\partial b}{\partial x}= 0$ it is $\displaystyle\frac{\partial v}{\partial t} =\displaystyle\frac{1}{c}\displaystyle\frac{\partial u}{\partial (x/c)}$