# 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?

• Commented Dec 11, 2018 at 2:51
• For b), answer is provided in this post Commented Dec 11, 2018 at 13:32

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