# Small Oscillations

Formulation of Small Oscillations only consider systems that satisfy the following:

• The system has $$n-$$ independent generalized coordinates, $$q$$
• $$V = V(q)$$
• $$\vec{r}_i = \vec{r}_i(q)$$, alternatively $$x_i = x_i(q)$$

Assume that the system undergoes small oscillations around a static equilibrium, $$q_\sigma = q_\sigma^0$$, $$\sigma=1,\cdots,n$$ Then, $$\dot{q}_\sigma|_{q_\sigma^0} = \ddot{q}_\sigma|_{q_\sigma^0} = 0$$. Hence, $$-\left(\frac{\partial V}{\partial q_\sigma}\right)_{q_\sigma^0} = 0$$. We can rewrite: $$q_\sigma = q_\sigma^0 + \eta_\sigma$$.

$$\mathcal{L} = T - V = \sum_j \frac{1}{2}m_j\frac{dx_j}{dt}^2 - V(q) = \sum_j\frac{1}{2}m_j\left(\sum_\sigma \frac{\partial x_j}{\partial q_\sigma}\dot{q}_\sigma\right)\left(\sum_\lambda \frac{\partial x_j}{\partial q_\lambda}\dot{q}_\lambda\right) - V(q) = \frac{1}{2}\sum_{j\sigma\lambda}m_j\left(\frac{\partial x_j}{\partial q_\sigma}\right)\left(\frac{\partial x_j}{\partial q_\lambda}\right)\dot{q}_\sigma\dot{q}_\lambda - V(q)$$. We can think of the inner sum term as the $(σ,λ)-$th element of a symmetric, real, square matrix $$M$$. Then, $$\mathcal{L} = T - V = \frac{1}{2}M_\sigma^\lambda\dot{\eta}_\sigma\dot{\eta}_\lambda$$. Note that $$M_{\sigma\lambda} = M_{\lambda\sigma} = (M_{\sigma\lambda})^*$$.

We can also write: $$V = V(q) = V(q^0 + \eta)$$. Taylor expanding, $$V = V(q^0) + \sum_\sigma \frac{\partial V}{\partial q_\sigma}|_{q_\sigma^0}\eta_\sigma + \frac{1}{2}\sum_\sigma\sum_\lambda\left(\frac{\partial^2 V}{\partial q_\sigma q_\lambda}q_\sigma q_\lambda\right) + \mathcal{O}(q^3) = V(q^0) + \frac{1}{2}\sum_\sigma\sum_\lambda\left(\frac{\partial^2 V}{\partial q_\sigma q_\lambda}\eta_\sigma \eta_\lambda\right)$$. Let $$V_{\sigma}^\lambda = \frac{\partial^2 V}{\partial q_\sigma q_\lambda}$$ be the potential square, real, symmetric matrix. So, $$V = V(q^0) + \frac{1}{2}\sum_{\sigma\lambda}V_\sigma^\lambda q_\sigma q_\lambda$$. Note that $$V_{\sigma\lambda} = V_{\lambda\sigma} = (V_{\sigma\lambda})^*$$.

Therefore, $$\mathcal{L} = \frac{1}{2}\sum_{\sigma\lambda}[M_\sigma^\lambda\dot{\eta}_\sigma\dot{\eta}_\lambda - V_\sigma^\lambda \eta_\sigma \eta_\lambda] - V(q^0)$$.

\begin{align*} 0 &= \frac{d}{dt}\frac{\partial\mathcal{L}}{\partial \dot{\eta}_k} - \frac{\partial\mathcal{L}}{\partial \eta_k} \\ &= \frac{d}{dt}\left(\frac{1}{2}\sum_{\sigma\lambda}[M_\sigma^\lambda](\ddot{q}_\sigma \dot{q}_\lambda\delta_{\sigma k} + \dot{q}_\sigma\ddot{q}_\lambda\delta_{\lambda k})\right) - \left(\frac{1}{2}\sum_{\sigma\lambda}[M_\sigma^\lambda](\dot{q}_\sigma q_\lambda\delta_{\sigma k} + q_\sigma\dot{q}_\lambda\delta_{\lambda k})\right) \\ \end{align*}

$$\frac{\partial\mathcal{L}}{\partial\dot{\eta}_k} = \sum_\lambda M_k^\lambda\dot{\eta}_\lambda$$. $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\eta}_k} = \sum_\lambda M_k^\lambda\ddot{\eta}_\lambda$$. $$\frac{\partial\mathcal{L}}{\partial\eta_k} = -\sum_\lambda V_k^\lambda\eta_\lambda$$. So, for each $$\sigma$$, the E-L of the Second Kind are,

\begin{align} 0 &= \sum_\lambda (M_\sigma^\lambda\ddot{\eta}_\lambda + V_\sigma^\lambda\eta_\lambda). \end{align}

Let $$z_\sigma$$ where $$(\sigma=1,\cdots,n)$$ satisfy the same system of equations, that is complex.

\begin{align*} 0 &= \sum_\lambda (M_\sigma^\lambda\ddot{z}_\lambda + V_\sigma^\lambda z_\lambda). \end{align*}

Hence, $$\eta_\sigma = \Re\{z_\sigma\}$$.

## Solve for Normal Mode Solutions

Hence, $$z_\sigma = z_\sigma^0\exp(i\omega_0 t)$$ for $$\sigma\in\{1,\cdots,n\}$$. Then,

\begin{align*} 0 &= \sum_\lambda(-\omega_0^2 m_\sigma^\lambda + V_\sigma^\lambda)z_\lambda^0\exp(i\omega t) \\ &= \sum_\lambda(V_\sigma^\lambda-\omega_0^2 m_\sigma^\lambda)z_\lambda^0 \end{align*}

Then,

\begin{align*} [V_0^\lambda-\omega_0^2M_\sigma^\lambda]\begin{pmatrix}| \\ z^0 \\ |\end{pmatrix} = 0 \end{align*}

$$\text{det}\left|V - \omega_0^2M\right| = 0$$. We will have $$n$$ frequencies, $$\omega_0^2 = \omega_s^2$$, $$s\in\{1,\cdots,n\}$$. Which gives us, $$\sum_\lambda(V_\sigma^\lambda - \omega_s^2M_\sigma^\lambda)z_\lambda^{(s)} = 0$$. It can be shown that $$\omega_s^2 = (\omega_s^2)^*$$ and $zλ(s)/zp(s) =$real. So, $$z_\lambda^{(s)} = \rho_\lambda(s)\exp(i\phi_s)$$, for $$\rho_\lambda(s)$$ real and $$\phi_s$$ is a common phase.

$$\sum_\sigma z_\sigma^{(s)*}\sum_\lambda(V_\sigma^\lambda-\omega_S^2M_\sigma^\lambda)z_\lambda^{(s)} = 0$$.

\begin{align*} \omega_S^2\sum_{\sigma,\lambda}z_\sigma^{(s)}M_\sigma^\lambda z_\lambda^{(s)} &= \sum z_\sigma^{(s)*}V_\sigma^\lambda z_\lambda^{(s)} \\ \omega_S^2 &= \frac{\sum z_\sigma^{(s)*}V_\sigma^\lambda z_\lambda^{(s)}}{\sum_{\sigma,\lambda}z_\sigma^{(s)*}M_\sigma^\lambda z_\lambda^{(s)}} \\ (\omega_S^2)^* &= \frac{\sum z_\sigma^{(s)}V_\sigma^\lambda z_\lambda^{(s)*}}{\sum_{\sigma,\lambda}z_\sigma^{(s)}M_\sigma^\lambda z_\lambda^{(s)*}} \\ &= \frac{\sum z_\sigma^{(s)*}V_\sigma^\lambda z_\lambda^{(s)}}{\sum_{\sigma,\lambda}z_\sigma^{(s)*}M_\sigma^\lambda z_\lambda^{(s)}} \\ &= \omega_S^2. \end{align*}

## Recall from last time

1. $$\sum_\lambda (M_\sigma^\lambda\ddot{\eta}_\lambda + V_\sigma^\lambda\eta_\lambda) = 0,\forall\sigma\in\{1,\cdots,n\}$$ $$\eta_\lambda = \Re\{z_\lambda\}$$
2. $$\sum_\lambda(M_\sigma^\lambda\ddot{z}_\lambda + V_\sigma^\lambda z_\lambda) = 0,\forall\sigma\in\{1,\cdots,n\}$$
3. (Normal mode solutions) $$z_\lambda = z_\lambda^0\exp(i\omega_0 t),\lambda = 1,\cdots$$. $$\sum_\lambda(V_\sigma^\lambda-\omega_0^2M_\sigma^\lambda)z_\lambda^0 = 0,\forall\sigma\in\{1,\cdots,n\}$$.
4. $$\left|V_\sigma^\lambda - \omega_0M_\sigma^\lambda\right| = 0$$. The characteristic frequencies are real and $$z_\lambda^{(s)},z_\pi^{(s)},\cdots$$ all have the same phase
5. For each frequency, $$\sum_\lambda(V_\sigma^\lambda-\omega_S^2M_\sigma^\lambda)z_\lambda^{(S)} = 0,\forall\sigma\in\{1,\cdots,n\}$$.
6. Plugging in this shared complex phase: $$\sum_\lambda(V_\sigma^\lambda-\omega_S^2M_\sigma^\lambda)\rho_\lambda^{(s)}\exp(i\phi_S),\forall\sigma\in\{1,\cdots,n\}$$. $$\sum_\lambda(V_\sigma^\lambda-\omega_S^2M_\sigma^\lambda)\rho_\lambda^{(s)},\forall\sigma\in\{1,\cdots,n\}$$. Boundary conditions/ initial conditions can be used to find the phase. Writing this in vector form, $$\rho_\lambda^{(s)}\Rightarrow \rho^{(s)}$$. One can show that $$\sum_{\lambda,\sigma}\rho_\sigma^{(t)}M_\sigma^\lambda\rho_\lambda^{(s)} = \rho^{(t)T}M\rho^{(s)} = \delta_{s,t}\rho^{(t)T}M\rho^{(s)}$$ Thus, we can make them orthonormal to the mass matrix. At that point: $$\rho^{(t)T}M\rho^{(s)} = \delta_{s,t}$$. $$z_\sigma = z_\sigma^{(s)}\exp(i\omega_St) = \rho_\sigma^{(s)}\exp(i(\phi_S + \omega_St))$$. Hence, our solution is $$\eta_\sigma = \rho_\sigma^{(s)}\cos(\phi_S + \omega_S t)$$. Our general solution is then, $$Z_\sigma = \sum_S C^{(S)}\rho_\sigma^{(S)}\exp(i(\phi_S\omega_St))$$, $$H_\sigma = \sum_S \Re\{C^{(S)}\}\rho_\sigma^{(S)}\cos(\phi_S + \omega_S t)$$.

## Example

Consider the coupled pendulum, each hanging from a celing by a string of length $$l$$ and the (equal $$m$$) masses are connected to each other by a spring with constant $$k$$ and length $$d_0$$ apart. Consider a perturbation.

Let $$\eta_n$$ be the horizontal displacement of mass n. Then, $$\eta_n/\ell = \sin\theta_i\approx\theta_i$$. $$1-\cos\theta \approx \frac{1}{2}\theta^2$$. So, $$\dot{\eta}_i = \ell\dot{\theta}_i$$.

Let $$\phi_1$$ be the angle of the first mass and $$\phi_2$$ be the angle of the second mass.

\begin{align*} \mathcal{L} &= \frac{1}{2}m\ell^2(\dot{\phi}_1^2 + \dot{\phi}_2^2) - (-mg(\ell\cos\phi_1 + \ell\cos\phi_2)) + \cdots \\ &= \frac{1}{2}m(\dot{\eta}_1^2 + \dot{\eta}_2^2) - \left(\frac{1}{2}k(\eta_1-\eta_2)^2\right) - \left(\frac{1}{2\ell}mg(\eta_1^2 + \eta_2^2)\right) \end{align*}

Step 2: $$m\ddot{\eta}_1 + \left(\frac{mg}{\ell} + k\right)\eta_1 - k\eta_2 = 0$$. $$m\ddot{\eta}_2 + \left(\frac{mg}{\ell} + k\right)\eta_2 - k\eta_1 = 0$$.

Thus,

\begin{align*} M &= \begin{pmatrix}m & 0 \\ 0 & m\end{pmatrix}. \\ V &= \begin{pmatrix}\frac{mg}{\ell}+k & -k \\ -k & \frac{mg}{\ell}+k\end{pmatrix}. \end{align*}

Step 3: $$\left|V-\omega_S^2M\right| = \begin{vmatrix}\frac{mg}{\ell}+k-\omega_S^2m & -k \\ -k & \frac{mg}{\ell}+k-\omega_S^2m\end{vmatrix} = 0$$. So, $$\left(\frac{mg}{\ell}+k-\omega_S^2m\right)^2 - k^2 = 0$$. This gives solutions, $$\omega_1 = \sqrt{\frac{g}{\ell}},\omega_2 = \sqrt{\frac{g}{\ell} + \frac{2k}{m}}$$. Now, get the eigenvectors! $$\begin{pmatrix}\frac{mg}{\ell}+k-\frac{g}{\ell}m & -k \\ -k & \frac{mg}{\ell}+k-\frac{g}{\ell}m\end{pmatrix}\rho^{(1)} k\rho_1^{(1)} - k\rho_2^{(2)}= 0$$

$$\begin{pmatrix}\frac{mg}{\ell}+k-\left(\frac{g}{\ell} + \frac{2k}{m}\right)m & -k \\ -k & \frac{mg}{\ell}+k-\left(\frac{g}{\ell} + \frac{2k}{m}\right)m\end{pmatrix}\rho^{(2)} = 0$$ $$\begin{pmatrix}\frac{mg}{\ell}+k-\left(\frac{g}{\ell} + \frac{2k}{m}\right)m & -k \\ -k & \frac{mg}{\ell}+k-\left(\frac{g}{\ell} + \frac{2k}{m}\right)m\end{pmatrix}\rho^{(2)} = -k\rho_1^{(2)} -k\rho_2^{(2)}= 0$$ Thus, we get the eigenvectors, $$\rho^{(1)} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ 1\end{pmatrix}$$ and $$\rho^{(2)} = \frac{1}{\sqrt{2}}\begin{pmatrix}1 \\ -1\end{pmatrix}$$. Normalizing with respect to the mass matrix: we get the eigenvectors, $$\rho^{(1)} = \frac{1}{\sqrt{2m}}\begin{pmatrix}1 \\ 1\end{pmatrix}$$ and $$\rho^{(2)} = \frac{1}{\sqrt{2m}}\begin{pmatrix}1 \\ -1\end{pmatrix}$$. The general solution is then, $$\eta = \sum_S C^{(S)}\rho^{(S)}\cos\left(\phi_S + \omega_S t\right)$$.

Note that $$\eta_1$$ is the first row and $$\eta_2$$ is the second row in this vector form.

Suppose $$\dot{\eta}(0) = 0$$ and $$\eta(0) = \begin{pmatrix}\alpha \\ 0\end{pmatrix}$$… Then, $$C_1 = \frac{\alpha\sqrt{m}}{\sqrt{2}}$$, $$C_2 = \frac{\alpha\sqrt{m}}{\sqrt{2}}$$, $$\phi_1=\phi_2=0$$.

For each ωS4 we have $$\rho^{(S)}$$ a column vector. The Modal matrix, $$A = \begin{pmatrix} | & | & \cdots & | \\\rho^{(1)} & \rho^{(2)} & \cdots & \rho^{(n)}\\ | & | & \cdots & |\end{pmatrix}$$. Then, $$A^TMA = \mathbb{I}$$. But also, $$A^TVA = \begin{pmatrix}\omega_1^2 & & &\\ &\omega_2^2 & & \\ & & \cdots & \\ & & & \omega_n^2\end{pmatrix} = \Omega$$. Normal Coordinates: $$\eta(t) = A\xi(t)$$, $$\xi(t)$$ is called the normal coordinate. $$A^Tm\eta(t) = A^TmA\xi(t) = \xi(t)$$. Recall the Lagrangian: \begin{align*} \mathcal{L} &= T - V \\ &= \frac{1}{2}\sum_\sigma\sum_\lambda M_\sigma^\lambda\dot{\eta}_\sigma\dot{\eta}_\lambda - V_\sigma^\lambda\eta_\sigma\eta_\lambda \\ &= \frac{1}{2}\dot{\eta}^T M\dot{\eta} - \frac{1}{2}\eta^T V\eta \\ &= \frac{1}{2}\left((A\dot{\xi}(t))^T m(A\dot{\xi}(t)) - (A\xi(t))^TV(A\xi(t))\right) \\ &= \frac{1}{2}\left(\dot{\xi}^2(t) - \xi^T(t)\Omega\xi(t)\right). \end{align*} So, we get $$\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{\xi}_i} - \frac{\partial\mathcal{L}}{\partial\xi_i} \ddot{\xi}_i(t) - \omega_i^2\xi_i(t)= 0$$ for each coordinate. Thus, $$\xi_\sigma(t) = C^{(\sigma)}\cos(\omega_\sigma t + \phi_\sigma)$$. We can then get, $$\eta(t) = A\xi(t)$$. ## Equation of Motion for Continuous String ### Newtonian Approach Consider a horizontal string anchored on both ends. Now, we perturb it vertically. The mass of the string per length is $$\sigma(x)$$. Consider a $$dx$$ piece of the string. The vertical displacement at that point at a given time is going to be given by $$u(x,t)$$. So, the end of the segment is $$u(x+dx,t)$$. The tension is then $$T_L = T(x)$$ and $$T_R = T(x+dx)$$. The angle at the bottom, $$x$$, is $$\theta$$ and the angle at $$x+dx$$ is $$\phi$$. So, $$\sin\theta\approx\theta\approx\tan\theta = \frac{\partial u(x,t)}{\partial x}$$. Similarly, $$\sin\phi\approx\phi\approx\tan\theta = \frac{\partial u(x+dx,t)}{\partial x}$$. So, $$ma_y = [\sigma(x) dx]\left[\frac{\partial^2 u(x,t)}{\partial t^2}\right] = T(x+dx)\sin\phi - T(x)\sin\theta = T(x+dx)\frac{\partial u(x+dx,t)}{\partial x} - T(x)\frac{\partial u(x,t)}{\partial x}$$. Let $$f(x,t) = T(x)\frac{\partial u(x,t)}{\partial x}$$. Then, $$[\sigma(x)dx]\left[\frac{\partial^2 u(x,t)}{\partial t^2}\right] = f(x+dx,t) - f(x,t)$$ So, $$[\sigma(x)dx]\left[\frac{\partial^2 u(x,t)}{\partial t^2}\right] = T(x)\frac{\partial u(x,t)}{\partial x} + \frac{\partial}{\partial x}\left(T(x)\frac{\partial u(x,t)}{\partial x}\right)dx + \cdots - T(x)\frac{\partial u(x,t)}{\partial x} = \frac{\partial}{\partial x}\left(T(x)\frac{\partial u(x,t)}{\partial x}\right)dx$$. So, $$\sigma(x)\frac{\partial^2 u(x,t)}{\partial t^2} = \frac{\partial}{\partial x}\left(T(x)\frac{\partial u(x,t)}{\partial x}\right)$$. Hence, $$\sigma(x)\frac{\partial^2 u(x,t)}{\partial t^2} = T(x)\frac{\partial^2 u(x,t)}{\partial x^2} + T'(x)\frac{\partial u(x,t)}{\partial x}$$. For mass density and tension position independent, $$\frac{\partial^2 u(x,t)}{\partial t^2} = \frac{T}{\sigma}\frac{\partial^2 u(x, t)}{\partial x^2}$$ which can be rewritten as $$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2} - \frac{\partial^2}{\partial x^2}\right)u(x,t) = 0$$. ### Hamilton’s Principle in a Continuous Medium Consider breaking it up into $$N$$ masses–then take the limit to $$N\to\infty$$. Let the position of then-\$th mass be $$x_n$$ and the vertical displacement $$u_n$$. Consider the spacing between the masses as $$a$$.

Writing the lagrangian, with a tension of $$\tau$$, so $$k = \frac{\tau}{a}$$.

\begin{align*} \mathcal{L} &= \sum\limits_{i=1}^N\frac{1}{2}m_i\dot{u}_i^2 - \sum\limits_{i=0}^{N} \frac{1}{2}k(u_{i+1}-u_i)^2 \\ &= \sum\limits_{i=1}^N\frac{1}{2}\frac{m_i}{a}\dot{u}_i^2a - \sum\limits_{i=0}^{N} \frac{1}{2}\tau\left(\frac{u_{i+1}-u_i}{a}\right)^2a. \end{align*}

Let $$N\to\infty$$. So, $$a\to dx$$, $$\frac{m}{a}=\sigma(x)$$, $$u_i(x,t)\to u(x,t)$$.

\begin{align*} \mathcal{L} &= \int_0^\ell\frac{1}{2}\sigma(x)\left(\frac{\partial u(x)}{\partial t}\right)^2 dx - \int_0^\ell\frac{1}{2}\tau(x)\left(\frac{\partial u(x,t)}{\partial x}\right)^2 dx \\ &= \frac{1}{2}\int_0^\ell \sigma(x)\left(\frac{\partial u(x)}{\partial t}\right)^2 - \tau(x)\left(\frac{\partial u(x,t)}{\partial x}\right)^2 dx. \end{align*}

Then, the lagrangian is an integral of the lagrangian density,

\begin{align*} \mathcal{L} &= \int_0^\ell L\left(u(x,t),\frac{\partial u(x,t)}{\partial x},\frac{\partial u(x,t)}{\partial t},x ,t\right)dx. \end{align*}

So,

\begin{align*} I &= \int_{t_1}^{t_2}\mathcal{L}dt \\ &= \int_{t_1}^{t_2}\int_0^\ell L\left(u(x,t),\frac{\partial u(x,t)}{\partial x},\frac{\partial u(x,t)}{\partial t},x ,t\right)dxdt. \\ \delta I &= 0. \end{align*}

Notice, $$\delta u(x,t_1) = \delta u(x,t_2) = 0 = \delta u(0,t) = \delta u(\ell,t)$$.

\begin{align*} \delta I &= \delta \int_{t_1}^{t_2}\int_0^\ell L\left(u(x,t),\frac{\partial u(x,t)}{\partial x},\frac{\partial u(x,t)}{\partial t},x ,t\right)dxdt \\ &= \int_{t_1}^{t_2}\int_0^\ell \delta L\left(u(x,t),\frac{\partial u(x,t)}{\partial x},\frac{\partial u(x,t)}{\partial t},x ,t\right)dxdt \\ &= \int_{t_1}^{t_2}\int_0^\ell \frac{\partial L}{\partial u}\delta u + \frac{\partial L}{\partial u'}\frac{\partial}{\partial x}\delta u + \frac{\partial L}{\partial\dot{u}}\frac{\partial}{\partial t}\delta udxdt \\ &= \int_{t_1}^{t_2}\int_0^\ell \left(\frac{\partial L}{\partial u} + \frac{\partial L}{\partial u'}\frac{\partial}{\partial x} + \frac{\partial L}{\partial\dot{u}}\frac{\partial}{\partial t}\right)\delta udxdt \\ &=\cdots (IBP) \\ &= \int_{t_1}^{t_2}\int_0^\ell \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} - \frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{u}}\right)\delta udxdt \\ 0 &= \left(\frac{\partial L}{\partial u} - \frac{\partial}{\partial x}\frac{\partial L}{\partial u'} - \frac{\partial}{\partial t}\frac{\partial L}{\partial\dot{u}}\right) \end{align*}

Note that I abbreviated $$u' = \frac{\partial u}{\partial x}$$ and $$\dot{u} = \frac{\partial u}{\partial t}$$. Recall,

\begin{align*} L &= \sigma(x)\left(\frac{\partial u(x)}{\partial t}\right)^2 - \tau(x)\left(\frac{\partial u(x,t)}{\partial x}\right)^2 \end{align*}

If $$\sigma(x) = \sigma$$ and $$\tau(x) = \tau$$. Then,

\begin{align*} 0 &= 0 - \left(-2\tau\frac{\partial^2 u(x,t)}{\partial x^2}\right) - \left(2\sigma\frac{\partial^2u(x,t)}{\partial t^2}\right), \\ \left(\tau\frac{\partial^2 u(x,t)}{\partial x^2}\right) &= \left(\sigma\frac{\partial^2u(x,t)}{\partial t^2}\right). \end{align*}

Giving us our wave equation again.

Created: 2024-05-30 Thu 21:20

Validate