I obtained sum of unidirectional waves, and not the sum of a forward and backward wave. Where did I go wrong?
$$u_{tt}=c^2u_{xx}$$
$$\iff u_{tt}-c^2u_{xx}=0$$
$$\iff (\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x}) [ ( \dfrac{\partial}{\partial t} + c \dfrac{\partial }{\partial x})]u=0$$
$$\iff (\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x}) ( u_t+cu_x)=0$$
Assume $$v(x,t)=u_t+cu_x$$
$$(\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x})v=0$$
Then, $$cv_x+v_t=0.$$
This may be solved for $v(x,t)$ using the method of characteristics.
Lagrange-Charpit:
$$\dfrac{dx}{c}=\dfrac{dt}{1}=\dfrac{dv}{0}$$
$$\iff x=ct+A$$
$$x-ct=A$$
$$f(x-ct)=v(x,t)=cu_x+u_t$$
It remains to solve for $u(x,t).$
Lagrange-Charpit:
$$\dfrac{dx}{c}=\dfrac{dt}{1}=\dfrac{du}{f(x-ct)}$$
$$\dfrac{dx}{dt}=c \implies x=ct+A \implies x-ct=A$$
$$\dfrac{du}{dt}=f(x-ct)$$
$$u(x,t)=\int f(x-ct) dt$$
$$u(x,t)=F(x-ct) + B$$ where
$$\dfrac{\partial}{\partial t} F(x-ct)=f(x-ct).$$
$$u(x,t)-F(x-ct)=B$$
$$u(x,t)=g(x-ct)+F(x-ct)$$ as $$g(A)=B$$ gives the solution.
But this is incorrect as we should have $$u(x,t)=h(x-ct)+p(x+ct)$$
I obtained sum of unidirectional waves, and not the sum of a forward and backward wave. Where did I go wrong?
$$u_{tt}=c^2u_{xx}$$
$$\iff u_{tt}-c^2u_{xx}=0$$
$$\iff (\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x}) [ ( \dfrac{\partial}{\partial t} + c \dfrac{\partial }{\partial x})]u=0$$
$$\iff (\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x}) ( u_t+cu_x)=0$$
Assume $$v(x,t)=u_t+cu_x$$
$$(\dfrac{\partial}{\partial t} - c \dfrac{\partial }{\partial x})v=0$$
Then, $$***-cv_x+v_t=0.***$$
This may be solved for $v(x,t)$ using the method of characteristics.
Lagrange-Charpit:
$$\dfrac{dx}{**-c**}=\dfrac{dt}{1}=\dfrac{dv}{0}$$
$$\iff x=-ct+A$$
$$x+ct=A$$
$$f(x+ct)=v(x,t)=cu_x+u_t$$
It remains to solve for $u(x,t).$
Do the rest....