Lektion 11

Extrema unter Nebenbedingungen



In [1]:

    
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline
from sympy import *
init_printing()



In [2]:

    
x = Symbol('x')
y = Symbol('y')
lam = Symbol('lambda')



In [3]:

    
g = x**4 + y**4 - 1



In [4]:

    
f = x**2 + y**2

Gesucht: Maximum von $f$ unter der Nebenbedingung $g=0$



In [5]:

    
grf = Matrix([f]).jacobian([x,y])
grf









    Out[5]:





$$\left[\begin{matrix}2 x & 2 y\end{matrix}\right]$$



In [6]:

    
grg = Matrix([g]).jacobian([x,y])
grg









    Out[6]:





$$\left[\begin{matrix}4 x^{3} & 4 y^{3}\end{matrix}\right]$$



In [7]:

    
gls = grf - lam*grg
gls









    Out[7]:





$$\left[\begin{matrix}- 4 \lambda x^{3} + 2 x & - 4 \lambda y^{3} + 2 y\end{matrix}\right]$$



In [8]:

    
glsmenge = set(gls)



In [9]:

    
glsmenge.add(g)
glsmenge









    Out[9]:





$$\left\{- 4 \lambda x^{3} + 2 x, - 4 \lambda y^{3} + 2 y, x^{4} + y^{4} - 1\right\}$$



In [10]:

    
Lsg = solve(glsmenge)
Lsg









    Out[10]:





$$\left [ \left \{ \lambda : - \frac{1}{2}, \quad x : 0, \quad y : - i\right \}, \quad \left \{ \lambda : - \frac{1}{2}, \quad x : 0, \quad y : i\right \}, \quad \left \{ \lambda : - \frac{1}{2}, \quad x : - i, \quad y : 0\right \}, \quad \left \{ \lambda : - \frac{1}{2}, \quad x : i, \quad y : 0\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : -1, \quad y : 0\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 0, \quad y : -1\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 0, \quad y : 1\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 1, \quad y : 0\right \}, \quad \left \{ \lambda : - \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}} i}{2}, \quad y : - \frac{2^{\frac{3}{4}} i}{2}\right \}, \quad \left \{ \lambda : - \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}} i}{2}, \quad y : \frac{2^{\frac{3}{4}} i}{2}\right \}, \quad \left \{ \lambda : - \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}} i}{2}, \quad y : - \frac{2^{\frac{3}{4}} i}{2}\right \}, \quad \left \{ \lambda : - \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}} i}{2}, \quad y : \frac{2^{\frac{3}{4}} i}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}}}{2}, \quad y : - \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}}}{2}, \quad y : \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}}}{2}, \quad y : - \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}}}{2}, \quad y : \frac{2^{\frac{3}{4}}}{2}\right \}\right ]$$



In [11]:

    
def test_real(l):
    x0 = x.subs(l)
    y0 = y.subs(l)
    return x0.is_real and y0.is_real



In [12]:

    
reelle_lsg = [l for l in Lsg if test_real(l)]
reelle_lsg









    Out[12]:





$$\left [ \left \{ \lambda : \frac{1}{2}, \quad x : -1, \quad y : 0\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 0, \quad y : -1\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 0, \quad y : 1\right \}, \quad \left \{ \lambda : \frac{1}{2}, \quad x : 1, \quad y : 0\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}}}{2}, \quad y : - \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : - \frac{2^{\frac{3}{4}}}{2}, \quad y : \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}}}{2}, \quad y : - \frac{2^{\frac{3}{4}}}{2}\right \}, \quad \left \{ \lambda : \frac{\sqrt{2}}{2}, \quad x : \frac{2^{\frac{3}{4}}}{2}, \quad y : \frac{2^{\frac{3}{4}}}{2}\right \}\right ]$$



In [13]:

    
werte = [f.subs(l) for l in reelle_lsg]
werte









    Out[13]:





$$\left [ 1, \quad 1, \quad 1, \quad 1, \quad \sqrt{2}, \quad \sqrt{2}, \quad \sqrt{2}, \quad \sqrt{2}\right ]$$



In [14]:

    
fn = lambdify((x,y), f, 'numpy')
gn = lambdify((x,y), g, 'numpy')



In [15]:

    
x = np.linspace(-1.5, 1.5)
y = np.linspace(-1.5, 1.5)
X, Y = np.meshgrid(x, y)
Wf = fn(X, Y)
Wg = gn(X, Y)



In [16]:

    
plt.contour(X, Y, Wg, levels=[0])
plt.contour(X, Y, Wf, levels=[1, np.sqrt(2)], colors=['red', 'green'])
plt.axis('image');

Plotverschönerung



In [17]:

    
x = Symbol('x')
f = 1/(1+x**2)
f









    Out[17]:





$$\frac{1}{x^{2} + 1}$$



In [18]:

    
wendestellen = solve(f.diff(x,2))
wendestellen









    Out[18]:





$$\left [ - \frac{\sqrt{3}}{3}, \quad \frac{\sqrt{3}}{3}\right ]$$



In [19]:

    
werte = [f.subs(x, w) for w in wendestellen]
werte









    Out[19]:





$$\left [ \frac{3}{4}, \quad \frac{3}{4}\right ]$$



In [20]:

    
xn = np.linspace(-3.5, 3.5, 100)
fn = lambdify(x, f)
wn = fn(xn)



In [21]:

    
props = {}
props['arrowstyle'] = '-|>'



In [22]:

    
plt.plot(xn, wn)
wx = wendestellen[0].n()
wy = fn(wx)
plt.annotate("Wendepunkt", (wx, wy), (wx-2.5, wy+.03),
            arrowprops=props);

Differentialgleichungen



In [23]:

    
x = Symbol('x')
y = Function('y')
dgl = Eq(y(x).diff(x), 2*y(x)+1)
dgl









    Out[23]:





$$\frac{d}{d x} y{\left (x \right )} = 2 y{\left (x \right )} + 1$$



In [24]:

    
Lsg = dsolve(dgl)
Lsg









    Out[24]:





$$y{\left (x \right )} = \frac{C_{1}}{2} e^{2 x} - \frac{1}{2}$$



In [25]:

    
print(Lsg)









    



Eq(y(x), C1*exp(2*x)/2 - 1/2)



In [26]:

    
Lsg.subs(C1, 3)   # NameError









    



---------------------------------------------------------------------------
NameError                                 Traceback (most recent call last)
<ipython-input-26-40e6ce7cd214> in <module>()
----> 1 Lsg.subs(C1, 3)   # NameError

NameError: name 'C1' is not defined



In [27]:

    
C1 = Symbol('C1')



In [28]:

    
Lsg.subs(C1, 3)









    Out[28]:





$$y{\left (x \right )} = \frac{3}{2} e^{2 x} - \frac{1}{2}$$



In [29]:

    
f = Lsg.rhs



In [30]:

    
y0 = 0



In [31]:

    
ab = Eq(f.subs(x,0), y0)
ab









    Out[31]:





$$\frac{C_{1}}{2} - \frac{1}{2} = 0$$



In [32]:

    
Lsg_AWA = solve(ab)
f.subs(C1, Lsg_AWA[0])









    Out[32]:





$$\frac{e^{2 x}}{2} - \frac{1}{2}$$



In [33]:

    
dgl = Eq(y(x).diff(x), y(x)-x**3+3*x-2)
dgl









    Out[33]:





$$\frac{d}{d x} y{\left (x \right )} = - x^{3} + 3 x + y{\left (x \right )} - 2$$



In [34]:

    
Lsg = dsolve(dgl).expand()
Lsg









    Out[34]:





$$y{\left (x \right )} = C_{1} e^{x} + x^{3} + 3 x^{2} + 3 x + 5$$



In [35]:

    
f = Lsg.rhs



In [36]:

    
xn = np.linspace(-1.5,2)
for y0 in [S(1)/2, 1, S(3)/2]:
    ab = Eq(f.subs(x, 0), y0)
    Lsg_AWA = solve(ab)
    fn = lambdify(x, f.subs(C1, Lsg_AWA[0]), 'numpy')
    plt.plot(xn, fn(xn), label=y0)
plt.legend(loc='lower left');



In [37]:

    
xq = np.linspace(-1.4, 1.8, 13)
yq = np.linspace(-2.5, 5.5, 11)
X, Y = np.meshgrid(xq, yq)
vf = np.array([dgl.rhs.subs({x: xx, y(x): yy})
               for yy in yq for xx in xq]).reshape(11, 13).astype(float)
X.shape, vf.shape









    Out[37]:





$$\left ( \left ( 11, \quad 13\right ), \quad \left ( 11, \quad 13\right )\right )$$



In [38]:

    
plt.quiver(X, Y, np.ones_like(X), vf, angles='xy')
for y0 in [S(1)/2, 1, S(3)/2]:
    ab = Eq(f.subs(x, 0), y0)
    Lsg_AWA = solve(ab)
    fn = lambdify(x, f.subs(C1, Lsg_AWA[0]), 'numpy')
    plt.plot(xn, fn(xn), label=y0)
plt.legend(loc='lower left');

Definitionsbereiche



In [39]:

    
y = Function('y')
y0 = Symbol('y_0', real=True)
x = Symbol('x')



In [40]:

    
dgl = Eq(y(x).diff(x), exp(y(x))*sin(x))
dgl









    Out[40]:





$$\frac{d}{d x} y{\left (x \right )} = e^{y{\left (x \right )}} \sin{\left (x \right )}$$



In [41]:

    
Lsg = dsolve(dgl)
f = Lsg.rhs
f









    Out[41]:





$$\log{\left (- \frac{1}{C_{1} - \cos{\left (x \right )}} \right )}$$



In [42]:

    
xq = np.linspace(-3.5, 9.5, 13)
yq = np.linspace(-.8, 1.95, 12)
X, Y = np.meshgrid(xq, yq)
vf = np.array([dgl.rhs.subs({x: xx, y(x): yy})
               for yy in yq for xx in xq]).reshape(12, 13).astype(float)



In [43]:

    
xn = np.linspace(-2.2, 9, 200)
ab = Eq(f.subs(x, 0), y0)
Lsg_AWA = solve(ab, C1)
Lsg_AWA









    Out[43]:





$$\left [ 1 - e^{- y_{0}}\right ]$$



In [44]:

    
f_y0 = f.subs(C1, Lsg_AWA[0])
f_y0









    Out[44]:





$$\log{\left (- \frac{1}{- \cos{\left (x \right )} + 1 - e^{- y_{0}}} \right )}$$

Probe



In [45]:

    
dgl.subs(y(x), f_y0).doit().simplify()









    Out[45]:





$$\mathrm{True}$$



In [46]:

    
f_y0.subs(x, 0) == y0









    Out[46]:





True



In [47]:

    
f_y0.subs(x, 0).simplify()









    Out[47]:





$$y_{0}$$



In [48]:

    
fn = lambdify(x, f_y0.subs(y0, -.5), 'numpy')
wn = fn(xn)
wn[wn>2] = np.nan
plt.plot(xn, wn)
plt.quiver(X, Y, np.ones_like(X), vf, angles='xy');









    



/home/braun/miniconda3/envs/WS1617/lib/python3.5/site-packages/numpy/__init__.py:1: RuntimeWarning: invalid value encountered in log
  """

Wo ist die Lösung definiert?



In [49]:

    
f









    Out[49]:





$$\log{\left (- \frac{1}{C_{1} - \cos{\left (x \right )}} \right )}$$



In [50]:

    
denom(exp(f_y0))









    Out[50]:





$$- \cos{\left (x \right )} + 1 - e^{- y_{0}}$$



In [51]:

    
solve(denom(exp(f_y0)), x)









    Out[51]:





$$\left [ - \operatorname{acos}{\left (1 - e^{- y_{0}} \right )} + 2 \pi, \quad \operatorname{acos}{\left (1 - e^{- y_{0}} \right )}\right ]$$



In [52]:

    
solve(Eq(1-exp(-y0), -1))









    Out[52]:





$$\left [ - \log{\left (2 \right )}\right ]$$



In [53]:

    
plt.quiver(X, Y, np.ones_like(X), vf, angles='xy')
for yy in [-1, -log(2), -S(1)/2]:
    fn = lambdify(x, f_y0.subs(y0, yy), 'numpy')
    wn = fn(xn)
    wn[wn>2] = np.nan
    plt.plot(xn, wn, label=yy)
plt.legend(loc='lower left');









    



/home/braun/miniconda3/envs/WS1617/lib/python3.5/site-packages/numpy/__init__.py:1: RuntimeWarning: invalid value encountered in log
  """

Die Lösung der Anfangswertaufgabe $$ y' = e^y \sin(x), y(0) = y_0 $$ ist für $y_0 < -\ln(2)$ auf ganz $\mathbb R$ definiert, sonst nur auf dem Intervall $]-\arccos(1-e^{-y_0}), \arccos(1-e^{-y_0})[ $.

Höhere Ordnung



In [54]:

    
a = Symbol('a')
lam = Symbol('lambda')



In [55]:

    
dgl = Eq(y(x).diff(x,2) + a*diff(y(x),x) + y(x), 0)
dgl









    Out[55]:





$$a \frac{d}{d x} y{\left (x \right )} + y{\left (x \right )} + \frac{d^{2}}{d x^{2}}  y{\left (x \right )} = 0$$



In [56]:

    
dsolve(dgl)









    Out[56]:





$$y{\left (x \right )} = C_{1} e^{\frac{x}{2} \left(- a - \sqrt{a^{2} - 4}\right)} + C_{2} e^{\frac{x}{2} \left(- a + \sqrt{a^{2} - 4}\right)}$$



In [57]:

    
tmp = dgl.lhs.subs(y(x), exp(lam*x)).doit()
tmp









    Out[57]:





$$a \lambda e^{\lambda x} + \lambda^{2} e^{\lambda x} + e^{\lambda x}$$



In [58]:

    
chi = (tmp * exp(-lam*x)).expand()
chi









    Out[58]:





$$a \lambda + \lambda^{2} + 1$$

$\chi$ ist das charakteristische Polynom



In [59]:

    
Lsg = solve(chi, lam)
Lsg









    Out[59]:





$$\left [ - \frac{a}{2} - \frac{1}{2} \sqrt{a^{2} - 4}, \quad - \frac{a}{2} + \frac{1}{2} \sqrt{a^{2} - 4}\right ]$$



In [ ]:



In [ ]: