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I'm trying to solve the following question:

Prove that $\mathcal{J} = \frac{\partial^3}{\partial z^3}$ is formally skew-adjoint on the space $C^ \infty(I, \mathbb{R}) $ of smooth real-valued functions on $I=[a,b]\subset\mathbb{R}$.

I'm trying to do this calculating the $L_2$ norm of $\mathcal{J}$:

\begin{align} (e(z), \mathcal{J}e(z))_{L_2} & = & \int_a^b e^T(z)\mathcal{J}e(z)\mathrm{d}z &\\ & = & \int_a^b \frac{\partial^3}{\partial z^3} (e(z)^2) & \mathrm{d}z \quad(e(z)\in\mathbb{R}) \\ & = & \int_a^b \frac{\partial^2}{\partial z^2} \left(\frac{\partial e^2}{\partial e}\frac{\partial e}{\partial z}\right)\mathrm{d}z & \\ & = & \int_a^b \frac{\partial^2}{\partial z^2} \left(2ee'\right)\mathrm{d}z &\\ & = & 2\int_a^b \frac{\partial}{\partial z} \left(e'^2+ee''\right)\mathrm{d}z &\\ & = & 2(e'^2(b) + e(b)e''(b) - e'^2(a) - e(a)e''(a)) &\\ \end{align}

At this part I don't know exactly how to proceed in the proof. Is it done already? Do I need to perhaps find a connection between the derivatives? I appreciate the help!

olenscki
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1 Answers1

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A differential operator $L^*$ is said to be the "formal adjoint" of the differential operator $L$, if for all test functions $u,v \in C_c^\infty(I)$, i.e. smooth functions with compact support we have $$\int_I u \cdot Lv \ dx = \int_I L^*u \cdot v \ dx$$ Those are basically exactly the $L^2$-norms you were calculating, but you made some mistakes when you wrote $e(z)^T \mathcal J e(z) = \frac{\partial^3}{\partial z^3}(e(z)^2)$ in your first step. This is wrong because the operator $\mathcal J$ operates only on the second function and you can't interchange it with the first $e$. The second little mistake is that you have to show the result not for $u=e, v=e$ but for two different functions.

The term "formal" means that we only show the result for test functions and not on the whole space that we are working with. The goal is now $\mathcal J^*=-\mathcal J$ because the exercise says that $\mathcal J$ should be skew-adjoint. Let's see why this is the case:

Let $u,v \in C_c^\infty(I)$. Then \begin{align} (u, \mathcal Jv)_{L^2} = \int_I u \cdot \mathcal Jv \ dx = \int_I u \cdot v''' \ dx\\ = - \int_I u' \cdot v'' \ dx = \int_I u'' \cdot v' \ dx \\ =-\int_Iu''' \cdot v \ dx = \int_I (-\mathcal J)(u) \cdot v \ dx \end{align} which was to be shown. In the second, third and fourth equality I used the partial integration formula: $$\int_a^b u\cdot v' \ dx = u\cdot v \bigg |_a^b - \int_a^b u' \cdot v$$ where when using test functions (functions with compact support) the boundary terms vanish in each step: $$u\cdot v \bigg |_a^b = 0$$

This is how we see $\mathcal J^* = - \mathcal J$.

Lukas
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