# A question on integration by parts

From PDE Evans, 2nd edition, page 148...

Observe \begin{align} \int_0^\infty \int_{-\infty}^\infty w_t v_x \, dx dt &= -\int_0^\infty \int_{-\infty}^\infty wv_{tx} \, dx dt - \int_{-\infty}^\infty wv_x \, dx \vert_{t=0} \\ &= \int_0^\infty \int_{-\infty}^\infty w_x v_{t} \, dx dt + \int_{-\infty}^\infty w_x v \, dx \vert_{t=0} \end{align}

For the first line, I believe the formula to the integration by parts was employed. (Please correct me if I'm mistaken.) But for the last step, may I ask how can we just move the subscript of $x$ from $v$ to $w$, in both the terms, and expect to remove the negative?

(Note: This step is part of proving the Lax-Oleinik formula as an integral solution to a certain PDE with initial conditions.)

• Are we to assume compact support on $w$? – DanZimm Jun 23 '14 at 8:52
• Yes. Perhaps I should note also that the mapping x↦w(x,t) is Lipschitz continuous from earlier proofs (thus, absolutely continuous). Similar with t↦w(x,t) -- absolutely continuous for each x∈R. – Cookie Jun 23 '14 at 9:05

Assuming $w,v$ are in the necessary spaces you are correct about the first line. Remember that the general idea for IBP is $$\int_S u v' \, \mathrm{d} \mu = \int_{\partial S} uv \, \mathrm{d} \mu - \int_S u'v \, \mathrm{d} \mu$$ and that the first integral vanishes as long as either $v$ or $u$ has compact support (i.e. the functions are in the correct space).
So now consider $$\begin{eqnarray} \int_0^\infty \int_{-\infty}^\infty w_t v_x \, \mathrm{d} x \, \mathrm{d} t & = & \int_{-\infty}^\infty \left. w v_x \right|_{t=0}^{t\to\infty} \,\mathrm{d} x - \int_0^{\infty} \int_{-\infty}^\infty w v_{xt} \, \mathrm{d} x \, \mathrm{d} t \\ & = & - \int_{-\infty}^\infty \left. w v_x \right|_{t=0} \, \mathrm{d}x - \int_0^{\infty} \int_{-\infty}^\infty w v_{xt} \, \mathrm{d} x \, \mathrm{d} t \end{eqnarray}$$ Here we are integrating by parts in the $t$, and the second equality comes from an assumption that $w$ has compact support in $t$. Now let's do this again but this time in the $x$ integral: $$- \int_{-\infty}^\infty \left. w v_x \right|_{t=0} \, \mathrm{d}x - \int_0^{\infty} \int_{-\infty}^\infty w v_{xt} \, \mathrm{d} x \, \mathrm{d} t \\ = - \left. \left. wv \right|_{t=0} \right|_{x\to -\infty}^{x \to \infty} + \int_{-\infty}^{\infty} \left. w_x v \right|_{t=0} \, \mathrm{d} x - \int_0^{\infty} \left. w v_t \right|_{x\to-\infty}^{x\to\infty} \, \mathrm{d}t + \int_0^{\infty} \int_{-\infty}^\infty w_x v_t \, \mathrm{d} x \, \mathrm{d} t \\ = \int_{-\infty}^\infty \left. w_x v \right|_{t=0} \, \mathrm{d} x + \int_0^{\infty} \int_{-\infty}^\infty w_x v_t \, \mathrm{d} x \, \mathrm{d} t$$ where the last equality follows from the assumption that $w$ has compact support in $x$.
• Note that you'll also need some derivatives of some sort here, i.e. $w,v \in \mathcal{C}^1$ or $w,v \in H^1$ – DanZimm Jun 23 '14 at 9:14
• For the work of $\int_0^\infty \int_{-\infty}^{\infty} w_t v_x \, dx dt$, the first step is using the integration by parts formula which I understand. In between the next two steps, you asserted that $$\int_{-\infty}^\infty \left. w v_x \right|_{t=0}^{t\to\infty}=\int_{-\infty}^\infty \left. w v_x \right|_{t=0}$$ That is what I still have trouble understanding; can you explain how the negative sign comes out when you simply remove the "$t \rightarrow \infty$"? – Cookie Jun 23 '14 at 17:42
• @glace well note that I assert $$\int_{-\infty}^\infty \left. w v_x \right|_{t=0}^{t \to \infty} = - \int_{-\infty}^{\infty} \left. w v_x \right|_{t=0}$$ and this follows since the side becomes $$\lim_{b \to \infty} w(x,b) v_x(x,b) - w(x,0) v_x(x,0)$$ but this first term drops out due to compact support, thus leaving $$\left. w v_x \right|_{t=0}$$ which is then inside the integral – DanZimm Jun 23 '14 at 18:32