Find weak form of linear transport equation I am stuck on the following problem that says:  

a) Find a weak formulation for the partial differential equation $${\partial u\over\partial t\ }+ c{\partial u \over \partial x\ }=0$$ 
  b) Show that $u=f(x-ct)$ is a generalized solution of $${\partial u\over\partial t\ }+ c{\partial u \over \partial x\ }=0$$ for any distribution $f$ 

My Attempt: 
I know that in order to find a weak form of a pde, we need to multiply it by a test function, then integrate it. Also, to find a generalized solution, we need to find a weak solution and just multiply it by the Heaviside function.
Let's take any test function $\phi $, then we have (integrating by parts second part of the integral) $$\int_{\Omega } ( {\partial u\over\partial t\ }+ c{\partial u \over \partial x\ })*\phi(x) dx= \int_{\Omega } {\partial u\over \partial t}\phi(x) dx - c\int_{\Omega } u(x,t)\phi'(x)dx$$ where $\phi $ vanishes at boundaries. So, is it the final form or can we proceed further? And how am I supposed to find a generalized solution?
Can someone help me out? Thanks in advance for your time.
 A: a.) The idea of integral solutions is a little more complicated than just integration with a test function. It's in fact integration with a variety of test functions, all of which yield the same result. But you are on the right track. We multiply our equation by a test function $v$ in the integrand: 
$$\int_0^{\infty} \int_{\mathbb{R}} (u_t + c u_x) v dx dt$$
And then using integration by parts: 
$$\int_0^{\infty} \int_{\mathbb{R}} u_t v dx dt = - \int_0^{\infty} \int_{\mathbb{R}} u v_t dx dt - \int_{\mathbb{R}} u(x,0) v(x,0) dx $$
Similarly, 
$$\int_0^{\infty} \int_{\mathbb{R}} c u_x v dx dt = - \int_0^{\infty} \int_{\mathbb{R}} c u v_x dx dt$$
So the entire equation becomes: 
$$\int_0^{\infty} \int_{\mathbb{R}} (u_t + c u_x) v dx dt = - \int_0^{\infty} \int_{\mathbb{R}} u ( v_t + c v_x ) dx dt - \int_{\mathbb{R}} u(x,0) v(x,0) dx = 0$$
Or simplified: 
$$\int_0^{\infty} \int_{\mathbb{R}} u ( v_t + c v_x ) dx dt + \int_{\mathbb{R}} u(x,0) v(x,0) dx = 0$$
We notice that $u$ is not necessarily differentiable. So if such a $u$ were to satisfy this equation for every test function $v$, then $u$ is considered to be a weak solution, because it doesn't necessarily satisfy the original equation. 
b.) Use the weak formulation to integrate $u = f(x - ct)$. You're just showing it's a formula, not the formula. But you could also choose to show uniqueness via derivation. Consider the changing of variables: 
$$s = x - ct \implies s \in \mathbb{R}$$
Performing the integration via change in coordinates: 
$$\int_0^{\infty} \int_{\mathbb{R}} f(x - ct) ( v_t + c v_x ) dx dt = \int_0^{\infty} \int_{\mathbb{R}} f(s) v_t ds dt$$
Now, integration by parts tells us: 
$$\int_0^{\infty} \int_{\mathbb{R}} f(s) v_t ds dt = \int_{\mathbb{R}} [f(s) v]_{t = 0}^{t = \infty} ds - \int_{0}^{\infty}\int_{\mathbb{R}} \frac{d}{dt}[f(s)] v ds dt$$
Differentiating $f(s)$ by the independent variable $t$ must result in zero. Also, assuming $v$ has compact support, $v(\infty) = 0$. Therefore:
$$\int_0^{\infty} \int_{\mathbb{R}} f(s) v_t ds dt = - \int_{\mathbb{R}} [f(s) v]_{t = 0} ds$$
That is, 
$$\int_0^{\infty} \int_{\mathbb{R}} f(s) v_t ds dt = - \int_{\mathbb{R}} f(x) v(x,0) dx$$
Then placing this into the weak formulation: 
$$\int_0^{\infty} \int_{\mathbb{R}} u ( v_t + c v_x ) dx dt + \int_{\mathbb{R}} u(x,0) v(x,0) dx = - \int_{\mathbb{R}} f(x) v(x,0) dx + \int_{\mathbb{R}} u(x,0) v(x,0) dx = 0$$
Which shows that $u = f(x - ct)$ is a weak (general) solution. 
Citations: Partial Differential Equations in Action, Sandro Salsa, pg 178
