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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?

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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.

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