In Evans' PDE textbook p. 136, 3.4.1 Shocks, entropy condition:

More precisely, assume $$ \left\lbrace \begin{aligned} &v : \Bbb R\times [0, \infty) \to \Bbb R \text{ is smooth, with}\\ &\text{compact support} \end{aligned} \right. \tag{2} $$

We call $v$ a test function. Now multiply the PDE $u_t + F(u)_x = 0$ by $v$ and integrate by parts: $$ \begin{aligned} 0&=\int_{0}^{\infty}\int_{-\infty}^{\infty}(u_t+F(u)_x)\, v\, dx dt& \\ &=-\int_{0}^{\infty}\int_{-\infty}^{\infty}uv_t\, dxdt - \int_{-\infty}^{\infty}uv\, dx|_{t=0} \\ &\qquad - \int_{0}^{\infty}\int_{-\infty}^{\infty}F(u)v_x\, dxdt . \end{aligned} \tag{3} $$

But my computation is $$\int_{0}^{\infty}\int_{-\infty}^{\infty}(u_t+F(u)_x)v dxdt=\\-\int_{0}^{\infty}\int_{-\infty}^{\infty}uv_t dxdt+\int_{-\infty}^{\infty}uv dx|_{t=0}^{\infty}\\+\int_{0}^{\infty}F(u)v dtdx|_{-\infty}^{\infty}-\int_{0}^{\infty}\int_{-\infty}^{\infty}F(u)v_x dxdt$$

How to get the answer?


1 Answer 1


Since $v$ is smooth with compact support, its values at $x = \pm\infty$ and $t = +\infty$ are zero. This is also true for $uv$, $F(u)v$, etc. Hence, the additional terms obtained in your (correct) computation vanish, and the formula $(3)$ of Evans' book is obtained.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.