This post primarily introduces The Variational Approach and demonstrates how to employ the method to derive the Euler-Lagrange equations for the water wave equations
We consider first the variational principle \[ \delta J=\delta \iint_R L\left(\varphi_t, \varphi_x, \varphi\right) d t d \mathbf{x}=0 \]
where \(\varphi(\mathbf{x}, t)\), the integral \(J[\phi]\) should be unchanged to small \(\delta \phi\), the smallness \(\Delta \varphi\) is (Let \(\left.\Delta \varphi=h, x \rightarrow\left(x_1, x_2, \cdots\right)\right)\) \[ \|h\|=\max |h|+\max \left|h_t\right|+\max \left|h_x\right| \] \(L\) has second derivatives. Then using taylor's expansion \[ \begin{aligned} J[\varphi+h]-J[\varphi&]\\ &\begin{aligned}=\iint_R\left[\frac{\partial L}{\partial \varphi_t} \Delta \varphi_t+\frac{\partial L}{\partial \varphi_i} \Delta \varphi_{x_i}\right. &\left.+\frac{\partial L}{\partial \varphi} \Delta \varphi\right] d t d x \\ &+\mathcal{O}\left(\|h\|^2\right)\\ \end{aligned}\\ &=\iint_R\left[L \varphi_t h_t+L_{\varphi_i} h_{x_i}+L_{\varphi} h\right] d t d x+\mathcal{O}\left(\|h\|^2\right) \end{aligned} \] This expression is the first variation, \(\delta J[\varphi, h]=0\) for all \(h\). Integrating the preceding equation by parts we have \[ \delta J\left[\varphi, h\right]=\iint_R\left[-\frac{\partial}{\partial t} L_{\varphi_t}-\frac{\partial}{\partial x_i} L_{\varphi, i}+L_{\varphi}\right] h d t d x \]
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\[\begin{aligned}\iint_R L_{\varphi_t} h_t d t d x & =\iint_R\left[\frac{\partial}{\partial t}\left(L_{\varphi_t} h\right)-\left(\frac{\partial}{\partial t} L_{\varphi_t}\right) h\right] d t d x \\& =\iint_R \frac{\partial}{\partial t}\left(L_{\varphi_t} h\right) d t d x-\iint\left(\frac{\partial}{\partial t} L_{\varphi_t}\right) h d t d x\end{aligned}\] similarly \[\begin{aligned}& \delta J=\iint_R[\underbrace{\left[\frac{\partial}{\partial t}\left(L_{\varphi_t} h\right)+\frac{\partial}{\partial x_i}\left(L_{\varphi_i} h\right)\right]}_{=0} d t d x+ \\& \iint_R\left[-\frac{\partial}{\partial t} L_{\varphi_t}-\frac{\partial}{\partial x_i} L_{\varphi_i}+L_{\varphi}\right] h d t d x \\&\end{aligned}\]
Why is the first terms equal to 0 . still unsolved
\[
\delta J=J[\varphi+h]-J[\varphi]
\] \(h\) vanish on boundary \(R\). for all such \(R\) \[
\frac{\partial}{\partial t} L_t+\frac{\partial}{\partial x_i} L_{\varphi_i}-L_{\varphi}=0
\] for higher derivatives in \(\varphi\), the variational equation is \[
L_{\varphi}-\frac{\partial}{\partial t} L_{\varphi_t}-\frac{\partial}{\partial x_i} L_{\varphi_i}+\frac{\partial^2}{\partial t^2} L_{\varphi_{t t}}+\frac{\partial^2}{\partial t \partial x_i} L_{\varphi_{ti}}+\frac{\partial^2}{\partial x_i \partial x_j} L_{\varphi_{i j}}+\cdots=0\tag{E1}
\] \[L\left[\varphi, \varphi_t, \varphi_i, \varphi_{t t}, \varphi_{t i}, \varphi_{i j}\right]\] expand \(L\) around \(\varphi\) as taylor series \[\begin{aligned}& L[\varphi+h]=L[\varphi]+L_{\varphi_t} \varphi_t+L_{\varphi_i} h_i+L_{\varphi} h+L_{\varphi_{t t}} h_{t t}+L_{\varphi_{ij}} h_{i j}+L_{\varphi_{t i}} h_{t i} \\& +\mathcal{O}\left(\|h\|^3\right) \\&\end{aligned}\] integrate the preceding equation then easy to get equ (E1) Questions: How to get this lagrangian** \[\begin{array}{ccc}\varphi_{t t} & -\alpha^2 \nabla^2 \varphi & \beta^2 \varphi \\\downarrow & \downarrow & \downarrow \\\varphi_t & \varphi_i & \varphi \\-1\text { (1-order differential) }&-1 & 1 \text { (0-order differential) }\end{array}\] sometimes we cant determine the sign of the symbols, but by substituting them into the Euler-Lagrangian Equ. we get it by comparisonnote
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Euler–Lagrange equation
for single function of several variables with single derivative. the Euler - Lagrangian equation in terms of \(\mathcal{L}\left(x_1, x_2, \ldots, x_n, f_1, f_2, \ldots, f_n\right)\) is \[ \frac{\partial \mathcal{L}}{\partial f}-\sum_{j=1}^n \frac{\partial}{\partial x_j}\left(\frac{\partial \mathcal{L}}{\partial f_j}\right)=0 \]
Several functions of several variables with higher derivatives, Let \[ \begin{gathered} \mathcal{L}\left(x_1, x_2, \ldots, x_m ; f_1, f_2, \ldots, f_m ; f_{1,1}, \ldots, f_{p, m} ; f_{1,11}, \ldots, f_{p, m m} ;\right. \\ \left.f_{p, 1 \ldots 1}, \ldots, f_{p, m \ldots m}\right) \end{gathered} \] where \[ f_{i, \mu}=\frac{\partial f_i}{\partial x_\mu}, \quad f_{i, \mu_1, \mu_2}=\frac{\partial^2 f_i}{\partial x_{\mu_1} \partial x_{\mu_2}}, \ldots \]
Then the Euler - Lagrangian (motion) is \[ \frac{\partial \mathcal{L}}{\partial f_i}+\sum_{j=1}^n \sum_{\mu_1 \leq \cdots \leq \mu_j}(-1)^j \frac{\partial f}{\partial x_{\mu_1} \ldots \partial x_{\mu_j}}\left(\frac{\partial \mathcal{L}}{\partial f_{i_i, \mu_1 \cdots \mu_j}}\right)=0 \] The aforementioned equation is one of a set of equations indexed by \(i\), with there being a total of \(p\) (\(f_1,f_2,..,f_p\)) such equations in reality.
Another example. for motion: \[ \varphi_{t t}-\alpha^2 \nabla^2 \varphi=\beta^2 \nabla^2 \varphi_{t t} \quad \mathcal{L}\left[p, \varphi_t, \varphi_i\right] \] then the corresponding Lagrangian is \[ \mathcal{L}=\underbrace{\frac{1}{2}\left(\varphi_t\right)^2-\frac{\alpha^2}{2}\left(\varphi_i\right)^2}_{j=1}+\underbrace{\frac{\beta^2}{2}\left(\varphi_{t i}\right)^2}_{j=2} \] note that \[ \begin{aligned} & \varphi_{i i t t} \rightarrow \varphi_{i t}^2 \rightarrow j=2 \\ & \varphi_{t t} \rightarrow \varphi_t^2 \rightarrow j=1 \\ & \varphi_{t x} \rightarrow \varphi_t \varphi_x \rightarrow j=1 \end{aligned} \] why '\(+\)', \((-1)^j\) from the defintion of Euler - Lagrangian, \(j=1\), 1st order differential.
Note: Euler - Lagrangian equation is motion equation!
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we give a brief proof of the preceding equation \[\begin{aligned}& \frac{\partial \mathcal{L}}{\partial \varphi}-\frac{\partial}{\partial t}\left(\frac{\partial \mathcal{L}}{\partial \varphi_t}\right)-\frac{\partial}{\partial x_i}\left(\frac{\partial \mathcal{L}}{\partial \varphi_i}\right)+\frac{\partial}{\partial t \partial x_i}\left(\frac{\partial \mathcal{L}}{\partial \varphi_{t i}}\right)\quad\text{(E2)} \\& \frac{\partial \mathcal{L}}{\partial \varphi}=0, \frac{\partial \mathcal{L}}{\partial \varphi_t}=\varphi_t, \frac{\partial \mathcal{L}}{\partial \varphi_{t i}}=\beta^2 \varphi_{t i}, \frac{\partial \mathcal{L}}{\partial \varphi_i}=-\alpha^2 \varphi_{i i}\end{aligned}\] therefore (E2) becume \[-\varphi_{t t}+\alpha^2 \varphi_{i i}+\beta^2 \varphi_{t t i i}=0\]
Another unique examples, when we have equation with odd order of differentiation. \[ \varphi_t+\alpha \varphi_x+\beta \varphi_{x x x}=0 \] we put \(\varphi=\psi_x\) then we have \[ \psi_{x t}+\alpha\psi_{x x}+\beta \psi_{x x x x}=0 \] the corresponding Lagrangian is \[ \mathcal{L}=-\frac{1}{2} \psi_x \psi_t-\frac{1}{2} \alpha \psi_x^2+\frac{1}{2} \beta \psi_x^2 \] now we calculate \(\mathcal{L}\) as energy density and flux, and let \(a(a \rightarrow A\).), \(\eta, w, \vec{k}\) small (can be neglected), where \(a=|A|, \eta= \arg A\) for these examples before we have \[ \begin{aligned} & \mathcal{L}=\frac{1}{4}\left(\omega^2-\alpha^2 k^2-\beta^2\right) a^2 \\ & \mathcal{L}=\frac{1}{4}\left(\omega^2-\alpha^2 k^2+\beta^2 \omega^2 k^2\right) a^2 \end{aligned} \] propose the "average variational principle" \[ \delta \iint \mathcal{L}\left(-\theta_t, \theta_x, a\right) d x d t \] where \(a(x, t), \theta(x, t), x(x_1, x_2, \ldots)\)
variation in \(a\) is \[
\delta a: \quad \mathcal{L}_a=0
\] variation in \(\theta\) is \[
\delta \theta: \quad \frac{\partial}{\partial t} \mathcal{L}_{\theta_t}+\frac{\partial}{\partial x_i} \mathcal{L}_{\theta_i}=0
\] why \(\delta a: \mathcal{L}_a=0\) firstly we must mention that it's two functions, \(a, \theta\), with two variables for each function, we have Euler -Lagrangian equation \[\delta a: \frac{\partial \mathcal{L}}{\partial a}-\frac{\partial}{\partial t} \mathcal{L}_{a_t}-\frac{\partial}{\partial x_i} \mathcal{L}_{a_i}=0\] since \(a\) don't have derivatives. then: \[\mathcal{L}_a=0\] Similarly \[\delta \theta: \quad \frac{\partial \mathcal{L}}{\partial \theta}-\frac{\partial}{\partial t} \mathcal{L}_{\theta_t}-\frac{\partial}{\partial x_i} \mathcal{L}_{\theta_i}=0\] and importantly: \(\mathcal{L}_\theta=0\) ! sui no \(\theta\) terms in \(\mathcal{L}\) like \(\mathcal{L}\left(\theta, \theta_t, \theta_x\right) \cdots \text { so } \mathcal{L}_\theta=0\)notes
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we have \(\theta=k x-\omega t\), then it follows that \[\begin{aligned}& \frac{\partial \theta}{\partial x}=k \\& \frac{\partial \theta}{\partial t}=-\omega\\\end{aligned} \] therefore \[ \begin{aligned}& \frac{\partial^2 \theta}{\partial x \partial t}=\frac{\partial k_i}{\partial t}=\frac{\partial^2 \theta}{\partial t \partial x}=-\frac{\partial \omega}{\partial x_i} \\& \Rightarrow \frac{\partial k_i}{\partial t}+\frac{\partial \omega}{\partial x_i}=0 \\&\end{aligned}\] for \(\forall \vec{k}=\left(k_1, k_2, k_3\right), \quad \exists\vec{k}=\nabla \theta\) thus \(\nabla \times(\nabla \theta)=0\), rewrite it in tensor form \[\frac{\partial}{\partial x_i} \vec{e}_i \times k_j \vec{e}_j=0 \text{ (Einstein sum)}\] then \[\frac{\partial k_j}{\partial x_i} \varepsilon_{i j k} \vec{e}_k=0\] combine same \(\vec{e}_k\) \[\begin{gathered}\left[\frac{\partial k_j}{\partial x_i} \varepsilon_{i j k}+\frac{\partial k_i}{\partial x_j} \varepsilon_{j ik}\right] \vec{e}_k=0, \quad \varepsilon_{i j k}=-\varepsilon_{j i k} \\\Rightarrow\frac{\partial k_j}{\partial x_i}-\frac{\partial k_i}{\partial x_j}=0\end{gathered}\]
we let dispersion solution: \(G(\omega, k) \quad(=0)\) \[
\mathcal{L}=G(\omega, k) a^2
\] therefore \[
\frac{\partial}{\partial t}\left(G_\omega a^2\right)-\frac{\partial}{\partial x_i}\left(G_{k_i} a^2\right)=0
\] there is a solution: \(\omega=W(k)\) so that \[
G(\omega, k)=G\{W(k), k\}=0
\] thus \[
G_\omega \frac{\partial W}{\partial k_i}+G_{k_i}=0
\] for \(G\{W(k), k\}=0\). take derivative with respect to \(k\) in both side \[\begin{aligned}\Rightarrow & \frac{D G}{D k_i}=0 \\\Rightarrow & \frac{\partial G}{\partial W} \frac{\partial W}{\partial k_i}+\frac{\partial G}{\partial k_i}=0 \\& G_\omega \frac{\partial W}{\partial k_i}+G_{k i}=0\end{aligned}\]notes
According to Norther's theorem, the corresponding energy equation is \[
\frac{\partial}{\partial t}\left(\omega \mathcal{L}_\omega-\mathcal{L}\right)+\frac{\partial}{\partial x_j}\left(-\omega \mathcal{L}_{k_j}\right)=0\tag{1}
\] we don't attempt derive eq(1), but rather simplify it \[\begin{aligned}& \frac{\partial}{\partial t}(\omega \mathcal{L} \omega-\mathcal{L}) \\= & \frac{\partial \omega}{\partial t} \mathcal{L}_\omega+\frac{\partial \mathcal{L}_\omega}{\partial \omega} \frac{\partial \omega}{\partial t} \omega+\frac{\partial \mathcal{L}_\omega}{\partial k_i} \frac{\partial k_i}{\partial t} \omega-\frac{\partial \mathcal{L}}{\partial \omega} \frac{\partial \omega}{\partial t}-\frac{\partial \mathcal{L}}{\partial k_i} \frac{\partial k_i}{\partial t} \\= & \frac{\partial \mathcal{L}_\omega}{\partial \omega} \frac{\partial \omega}{\partial t} \omega+\frac{\partial \mathcal{L}_\omega}{\partial k_i} \frac{\partial k_i}{\partial t} \omega-\frac{\partial \mathcal{L}}{\partial k_i} \frac{\partial k_i}{\partial t}\end{aligned} \tag{2}\] Since \[\frac{\partial k_i}{\partial t}+\frac{\partial \omega}{\partial x_i}=0 ; \frac{\partial k_i}{\partial x_j}=\frac{\partial k_j}{\partial x_i} ;\] rewrite eq(2) we have \[\frac{\partial \mathcal{L}_\omega}{\partial \omega} \frac{\partial \omega}{\partial t} \omega+\frac{\partial \mathcal{L}_\omega}{\partial k_i} \frac{\partial k_i}{\partial t} \omega+\frac{\partial \mathcal{L}}{\partial k_i} \frac{\partial \omega}{\partial x_i}\quad \text { (3) }\] similarly \[\begin{aligned}& \frac{\partial}{\partial x_j}\left(-\omega \mathcal{L}_{k_j}\right) \\& =-\frac{\partial \omega}{\partial x_j} \mathcal{L}_{k_j}-\omega \frac{\partial \mathcal{L}_{k_j}}{\partial \omega} \frac{\partial \omega}{\partial x_j}-\omega \frac{\partial \mathcal{L}_{k_j}}{\partial k_l} \frac{\partial k_l}{\partial x_j} \quad \text { (4) } \end{aligned}\] (3)+(4) \[\omega\left[\frac{\partial \mathcal{L}_\omega}{\partial \omega} \frac{\partial \omega}{\partial t}+\frac{\partial \mathcal{L}_\omega}{\partial k_i} \frac{\partial k_i}{\partial t}-\left(\frac{\partial \mathcal{L}_{k_j}}{\partial \omega} \frac{\partial \omega}{\partial x_j}+\frac{\partial \mathcal{L}_{k_j}}{\partial k_l} \frac{\partial k_l}{\partial x_j}\right)\right]=0\] Hence \[\frac{\partial}{\partial t} \mathcal{L}_\omega-\frac{\partial}{\partial x_j} \mathcal{L}_{k_j}=0 \tag{5}\]notes