```python
%matplotlib inline
from sympy import *
from sympy.utilities.lambdify import implemented_function
from sympy.abc import x, y, z
import numpy as np
import matplotlib.pyplot as plt
init_printing(use_unicode=True)
```


```python
r, u, v, c, r_c, u_c, v_c, E, p, r_p, u_p, v_p, e, a, b, q, b_0, b_1, b_2, b_3, q_0, q_1, q_2, q_3, q_4, q_5  = symbols('r u v c r_c u_c v_c E p r_p u_p v_p e a b q b_0 b_1 b_2 b_3 q_0 q_1 q_2 q_3 q_4 q_5')
```


```python
gamma = symbols('gamma',positive=True)
```

#### $f_{1}(c,p) = \dfrac{1}{2}r_{c}c^{2}+\dfrac{1}{4}u_{c}c^{4}+\dfrac{1}{6}v_{c}c^{6}+\dfrac{1}{2}r_{p}p^{2}-\gamma cp-Ep$


```python
f1 = (1/2)*r_c*c**2+(1/4)*u_c*c**4+(1/6)*v_c*c**6-E*p+(1/2)*r_p*p**2-gamma*c*p
```

### $\dfrac{\partial f_{1}(c,p)}{\partial p} = 0 = $


```python
pmin = solve(f1.diff(c),p)[0]
pmin
```


```python
E_cp = solve(f1.diff(p),E)[0]
E_cp
```


```python
expand(E_cp.subs(p,pmin))
```


```python

```

259bde4f381841dbe645239370f086b2871a012f
ipynb
Jupyter Notebook
Smectic/SimplePol.ipynb
brettavedisian/Liquid-Crystals
c7c6eaec594e0de8966408264ca7ee06c2fdb5d3
MIT
Smectic/SimplePol.ipynb
brettavedisian/Liquid-Crystals
c7c6eaec594e0de8966408264ca7ee06c2fdb5d3
MIT
Smectic/SimplePol.ipynb
brettavedisian/Liquid-Crystals
c7c6eaec594e0de8966408264ca7ee06c2fdb5d3
MIT
Qwen/Qwen-72B
1. YES
2. YES