Given a solution $u = f(x-c_1t)+g(x-c_2t)$ to $Au_{tt}+2Bu_{tx}+Cu_{xx}=0$ what equation should $c_1$ and $c_2$ satisfy

Question

We find a solution to the equation $$Au_{tt}+2Bu_{tx}+Cu_{xx}=0$$ as $$u = f(x-c_1t)+g(x-c_2t)$$ with aribitrary $f,g,$ and real $c_1 < c_2.$

1. What equation should satisfy $c_1$ and $c_2$
2. When does this equation have such roots?

So far I have used the solution to rewrite the PDE as $$f''(x-c_1t)\left[Ac_1^2 - 2Bc_1 + C\right] + g''(x-c_2t)\left[Ac_2^2 - 2Bc_2 +C\right]=0.$$ I'm stuck on what I ought to do next. I recognize that within each bracket is a quadratic in the respective $c.$ I'm just not sure where to go from here.

Answer 1

Calling

$$ M = \left( \begin{array}{cc} A & B \\ B & C \\ \end{array} \right) $$

we have

$$ \left( \begin{array}{c} \partial t \\ \partial x \\ \end{array} \right)^{\dagger}M\left( \begin{array}{c} \partial t \\ \partial x \\ \end{array} \right)u=0 $$

or

$$ \left( \begin{array}{c} \partial t \\ \partial x \\ \end{array} \right)^{\dagger}T^{-1}\Lambda T\left( \begin{array}{c} \partial t \\ \partial x \\ \end{array} \right)u=0 $$

with

$$ T = \left( \begin{array}{cc} \frac{A-C-\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\ \frac{A-C+\sqrt{4 B^2+(A-C)^2}}{2 B} & 1 \\ \end{array} \right)\\ \Lambda=\left( \begin{array}{cc} \frac{1}{2} \left(A+C-\sqrt{4 B^2+(A-C)^2}\right) & 0 \\ 0 & \frac{1}{2} \left(A+C+\sqrt{4 B^2+(A-C)^2}\right) \\ \end{array} \right) = \left( \begin{array}{cc} \lambda_1 & 0 \\ 0 & \lambda_2 \\ \end{array} \right) $$

and now calling

$$ \left( \begin{array}{c} \partial \eta \\ \partial \xi \\ \end{array} \right) = T\left( \begin{array}{c} \partial t \\ \partial x \\ \end{array} \right) $$

the PDE reduces to

$$ \lambda_1 u_{\eta\eta}+\lambda_2 u_{\xi\xi}=0 $$

and also we can verify by substitution that

$$ u(\eta,\xi) = f(\eta)+g(\xi) $$

is a solution, or changing variables again

$$ u(t,x) = f\left(\frac{\lambda_1}{B} t + x\right) + g\left(\frac{\lambda_2}{B} t+x\right) $$

Answer 2

You almost got it. If $c_1, c_2$ are zeros of $A x^2 + B x + C$, both terms are zero, whatever $f''(), g''()$ may be. Thus in the proposed solution $f, g$ are arbitrary (but have second derivatives).