# Integrating by parts, where did I go wrong?

Suppose u satisfies $$\left(\partial_{t}^{2}-\partial_{x}^{2}\right) u(x, t)=0,(x, t) \in \mathbb{R} \times(0, \infty)$$

 I'm trying to show: $$\int_{\mathbb{R}} \int_{0}^{\infty} u(x, t)\left(\partial_{t}^{2}-\partial_{x}^{2}\right) \phi(x, t) \mathrm{d} t \mathrm{~d} x=\int_{\mathbb{R}} \phi(x, 0) h(x)-g(x) \partial_{t} \phi(x, 0) \mathrm{d} x$$

Where $$u(x, 0)=g(x), \partial_{t} u(x, 0)=h(x), x \in \mathbb{R}$$

I have: Multiply the wave equation by $$\phi$$ and integrate over $$\mathbb{R}\times(0,\infty)$$ $$\begin{gathered} \left.\int_{\mathbb{R}} \phi(x, t) u_{t}(x, t)\right|_{0} ^{\infty}-\int_{\mathbb{R}} \int_{0}^{\infty} u_{t}(x, t) \phi_{t}-u_{x x}(x, t) \phi(x, t) d t d x \\ =\int_{\mathbb{R}} \phi(x, t) h(x) d x-\int_{\mathbb{R}} \int_{0}^{\infty} u_{t}(x, t) \phi_{t}-u_{x x}(x, t) \phi(x, t) d t d x \\ =\int_{\mathbb{R}} \phi(x, 0) h(x)-\phi_{t}(x, 0) g(x) d x+\int_{\mathbb{R}} \int_{0}^{\infty} u(x, t) \phi_{t t}-u_{x x}(x, t) \phi(x, t) d t d x \\ =\int_{\mathbb{R}} \phi(x, 0) h(x)-\phi_{t}(x, 0) g(x) d x+\int_{\mathbb{R}} \int_{0}^{\infty} u(x, t) \phi_{t t}-u(x, t) \phi_{x x}(x, t) d t d x-\int_{0}^{\infty}( \phi u_{x}-\phi_{x}u)|_\mathbb{R}dt \end{gathered}$$ Where did I go wrong?

• Are you possibly supposed to use that u satisfies a PDE?
– Jeff
Dec 20, 2021 at 13:53
• Yes, u satisfies $\left(\partial_{t}^{2}-\partial_{x}^{2}\right) u(x, t)=0,(x, t) \in \mathbb{R} \times(0, \infty)$ Dec 20, 2021 at 13:54
• How did you get the first line of the equation from the given assumptions? Also, how did you get the second line of the equation without $u_t(x,\infty)$ given?
– I H
Dec 20, 2021 at 14:20
• This is because u goes to 0 as t goes to $\infty$ (you can prove this using Kircchoff's formula) Dec 20, 2021 at 14:21
• What are you doing? Dec 27, 2021 at 20:01

So if you are trying to show $$\int_{\mathbb{R}} \int_{0}^{\infty} u(x, t)\left(\partial_{t}^{2}-\partial_{x}^{2}\right) \phi(x, t) \mathrm{d} t \mathrm{~d} x=\int_{\mathbb{R}} \phi(x, 0) h(x)-g(x) \partial_{t} \phi(x, 0) \mathrm{d} x$$
\begin{align}&\int_\mathbb{R} \int_0^\infty u \phi_{tt} -u \phi_{xx} \mathrm d t \mathrm d x \\ \overset{\text{Fubini}}{\Rightarrow}=& \int_\mathbb{R} \int_0^\infty u \phi_{tt} \mathrm d t \mathrm d x - \int_0^\infty \int_\mathbb{R} u \phi_{xx} \mathrm d x \mathrm d t \\ \overset{\text{Int. by parts}}{\Rightarrow} = &\int_\mathbb{R} u(t)\phi_t(t) \Big \vert_0^\infty - \int_0^\infty u_t \phi_{t} \mathrm d t \mathrm d x - \int_0^\infty u(x)\phi_x(x) \Big \vert_\mathbb{R} - \int_\mathbb{R} u_x \phi_{x} \mathrm d x \mathrm d t\end{align} Typically test functions are chosen such that they have compact support. Here this could be something like $$\phi \in H_c^2(\mathbb{R} \times [0, \infty ) )$$ which implies $$\lim_{x \to \pm \infty} \phi(x,t) = 0, \lim_{t \to \infty} \phi(x, t)= 0$$. Same holds for $$\phi_x, \phi_t$$. Invoking this gives \begin{align}&\int_\mathbb{R} -u(x, 0) \phi_t(x, 0) - \int_0^\infty u_t \phi_{t} \mathrm d t \mathrm d x + \int_0^\infty \int_\mathbb{R} u_x \phi_{x} \mathrm d x \mathrm d t \\ \overset{\text{Int. by parts}}{\Rightarrow} =& \int_\mathbb{R} -g(x) \phi_t(x, 0) - u_t \phi \big \vert_0^\infty + \int_0^\infty u_{tt} \phi \mathrm d t \mathrm d x + \int_0^\infty u_x(x) \phi(x) \big \vert_\mathbb{R} - \int_\mathbb{R} u_{xx} \phi \mathrm d x \mathrm d t \end{align} Again, using the compact support of $$\phi$$ gives \begin{align} &\int_\mathbb{R} -g(x) \phi_t(x, 0) + u_t(x, 0) \phi(x, 0) \mathrm d x + \int_\mathbb{R} \int_0^\infty u_{tt} \phi \mathrm d t \mathrm dx - \int_0^\infty \int_\mathbb{R} u_{xx} \phi \mathrm d x \mathrm d t \\ \overset{\text{Fubini}}{\Rightarrow}=& \int_\mathbb{R} h(x) \phi(x, 0) - g(x) \phi_t(x, 0) \mathrm d x + \int_\mathbb{R} \int_0^\infty (u_{tt} - u_ {xx}) \phi \mathrm d t \mathrm d x\end{align} which is the desired outcome for $$u$$ satisfying the wave equation.
Basically you can do all the steps in reverse order (if starting from $$\int_\mathbb{R} \int_0^\infty \phi (u_{xx} - u_{tt} ) \mathrm d t \mathrm d x$$) to get the same result.