question on one of the steps in a weak solution proof

**hatsoff** · November 7th 2011, 04:06 AM

My textbook has the following problem:

Show that for a continuous function $f$ the expression $u=f(x-ct)$ is a weak solution of the partial differential equation

$u_t+cu_x=0.$

[Hint: Transform for $\phi\in C_0^1(\mathbb{R}^2)$ the integral

$\int\int(\phi_t+c\phi_x)u\;dx\;dt$

to the coordinates $y_1=x-ct,y_2=x$ . Use $\phi=\psi(y_1)X(y_2)$ .]

They apparently mean in $\mathbb{R}^2$ . Recall that a "weak solution" (in this case) is a solution $u$ which satisfies

(1) $0=\int\int(\phi_t+c\phi_x)u\;dx\;dt$

for every test function $\phi\in C^1_0(\mathbb{R}^2)$ , i.e. for every continuously differentiable function with compact support in $\mathbb{R}^2$ . However it is (supposedly) sufficient to show that (1) holds for all $\phi\in C^\infty_0(\mathbb{R}^2)$ , i.e. all smooth functions with compact support in $\mathbb{R}^2$ .

My question is this: Why can we assume that a smooth (or continuously differentiable) function of two variables $\phi:\mathbb{R}^2\to\mathbb{R}^2$ can be written as the product of functions of a single variable $\phi(y_1,y_2)=\psi(y_1)X(y_2)$ ? Or am I missing something here ?

Thanks !

**Jose27** · November 20th 2011, 07:36 PM

I don't have an answer, but for example your conditions imply that the identity works for $\phi \in H^1(\mathbb{R}^2)=W^{1,2}(\mathbb{R}^2)$ . Now, it's well known that $L^2(X)\hat{\otimes} L^2(Y) \cong L^2( X\times Y)$ so "separated" simple functions are dense in the product (alternatively "separated" linear combinations of test functions are dense). One could then suspect that something along the lines of $H^1(\mathbb{R}) \hat{\otimes} H^1(\mathbb{R}) \cong H^1(\mathbb{R}^2)$ which would justify the argument. I don't know if the last identity holds.

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