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I am stuck on the following problem that says:

a) Find a weak formulation for the partial differential equation $${partial uoverpartial t }+ c{partial u over partial x }=0$$
b) Show that $u=f(x-ct)$ is a generalized solution of $${partial uoverpartial t }+ c{partial u over partial x }=0$$ 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) $$int_{Omega } ( {partial uoverpartial t }+ c{partial u over partial x })*phi(x) dx= int_{Omega } {partial uover partial t}phi(x) dx - cint_{Omega } u(x,t)phi'(x)dx$$ 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.) 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