← Advanced Engineering Mathematics

Orthogonal Trajectories

Real-World Scenario: Heat Flow in a Biofilm

In whole cell modeling, biofilms grow on surfaces where temperature gradients drive metabolic activity. The isotherms (curves of constant temperature) and the heat flow lines (curves along which heat flows) form two families of curves that are everywhere perpendicular to each other.

This is an example of orthogonal trajectories: given one family of curves, find a second family that intersects the first at right angles everywhere.

Section 1.6 of AEM develops the method:

1. Start with a one-parameter family of curves $G(x, y, c) = 0$
2. Differentiate to find the ODE of the family (eliminating the parameter $c$)
3. Replace $y'$ with $-1/y'$ to get the ODE of the orthogonal trajectories
4. Solve the new ODE

import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
from scipy.integrate import solve_ivp

plt.style.use('seaborn-v0_8-whitegrid')
mpl.rcParams['font.family'] = 'DejaVu Sans'

1. The Method — Ellipses and Parabolas

Consider the family of ellipses from the textbook (Eq. 1):

$$\tfrac{1}{2}x^2 + y^2 = c \qquad (c > 0)$$

These could represent equipotential curves (surfaces of equal voltage) between two elliptic cylinders in an electric field.

Step 1 — Find the ODE of the given family

Differentiate implicitly with respect to $x$ to eliminate $c$:

$$x + 2y\,y' = 0 \quad \Longrightarrow \quad y' = f(x,y) = -\frac{x}{2y}$$

Step 2 — ODE of the orthogonal trajectories

At any intersection point, the given curve has slope $f$. The orthogonal trajectory must have slope $-1/f$ (since perpendicular lines satisfy $m_1 \cdot m_2 = -1$):

$$\tilde{y}' = -\frac{1}{f(x,\tilde{y})} = +\frac{2\tilde{y}}{x}$$

Step 3 — Solve by separating variables

$$\frac{d\tilde{y}}{\tilde{y}} = \frac{2\,dx}{x} \quad \Longrightarrow \quad \ln|\tilde{y}| = 2\ln|x| + C \quad \Longrightarrow \quad \boxed{\tilde{y} = c^* x^2}$$

The orthogonal trajectories are parabolas — the paths along which charged particles would move in the electric field.

fig, ax = plt.subplots(1, 1, figsize=(7, 7))

theta = np.linspace(0, 2 * np.pi, 300)
x_dense = np.linspace(-3.5, 3.5, 400)

# Ellipses: x²/2 + y² = c
for c in [0.3, 0.8, 1.5, 2.5, 4.0, 6.0]:
    a = np.sqrt(2 * c)  # semi-axis along x
    b = np.sqrt(c)       # semi-axis along y
    xe = a * np.cos(theta)
    ye = b * np.sin(theta)
    label = 'Ellipses  $x^2/2 + y^2 = c$' if c == 0.3 else None
    ax.plot(xe, ye, color='steelblue', lw=1.5, label=label)

# Parabolas: y = c* x²
for cstar in [-0.6, -0.3, -0.12, 0.12, 0.3, 0.6]:
    yp = cstar * x_dense**2
    mask = np.abs(yp) < 4
    label = 'Parabolas  $\\tilde{y} = c^* x^2$' if cstar == -0.6 else None
    ax.plot(x_dense[mask], yp[mask], color='coral', lw=1.5, label=label)

ax.set_xlim(-4, 4)
ax.set_ylim(-4, 4)
ax.set_aspect('equal')
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_title('Orthogonal Trajectories: Ellipses and Parabolas')
ax.legend(loc='upper left')
plt.tight_layout()
plt.show()

2. Verifying Orthogonality Numerically

At any point $(x_0, y_0)$, the slopes of the ellipse and the parabola passing through that point should satisfy:

$$m_{\text{ellipse}} \cdot m_{\text{parabola}} = -1$$

• Ellipse slope: $m_1 = -x_0 / (2y_0)$
• Parabola slope: $m_2 = 2y_0 / x_0$

Product: $m_1 \cdot m_2 = \dfrac{-x_0}{2y_0} \cdot \dfrac{2y_0}{x_0} = -1 \;\checkmark$

# Verify at several random points
rng = np.random.default_rng(42)
test_points = rng.uniform(-3, 3, size=(6, 2))
# Avoid points too close to axes
test_points = test_points[np.abs(test_points).min(axis=1) > 0.3]

print(f'{"Point (x, y)":>20s}   {"m_ellipse":>10s}   {"m_parab":>10s}   {"product":>10s}')
print('-' * 60)
for x0, y0 in test_points:
    m1 = -x0 / (2 * y0)
    m2 = 2 * y0 / x0
    print(f'  ({x0:+5.2f}, {y0:+5.2f})   {m1:10.4f}   {m2:10.4f}   {m1 * m2:10.4f}')

        Point (x, y)    m_ellipse      m_parab      product
------------------------------------------------------------
  (+1.64, -0.37)       2.2411      -0.4462      -1.0000
  (+2.15, +1.18)      -0.9084       1.1008      -1.0000
  (-2.43, +2.85)       0.4266      -2.3440      -1.0000
  (+1.57, +1.72)      -0.4564       2.1909      -1.0000
  (-0.78, +2.56)       0.1514      -6.6062      -1.0000

3. Biotech Application: Isotherms and Heat Flow in a Biofilm

Consider a simplified model where the temperature distribution in a biofilm near a heated surface is described by concentric circles:

$$x^2 + y^2 = c \qquad (\text{isotherms})$$

Step 1 — Differentiate: $2x + 2y\,y' = 0 \Longrightarrow y' = -x/y$

Step 2 — Orthogonal trajectories ODE: $\tilde{y}' = y/x$

Step 3 — Solve: $d\tilde{y}/\tilde{y} = dx/x \Longrightarrow \tilde{y} = c^* x$ (straight lines through the origin)

The heat flows radially — along straight lines through the origin — perpendicular to the circular isotherms. This is exactly what we expect from physical intuition.

fig, axes = plt.subplots(1, 2, figsize=(13, 5.5))

# --- Left panel: circles and radial lines ---
ax = axes[0]
theta = np.linspace(0, 2 * np.pi, 300)

for r in [0.5, 1.0, 1.5, 2.0, 2.5, 3.0]:
    label = 'Isotherms $x^2+y^2=c$' if r == 0.5 else None
    ax.plot(r * np.cos(theta), r * np.sin(theta), 'steelblue', lw=1.5, label=label)

for slope in np.tan(np.linspace(0, np.pi, 7)[:-1]):
    xs = np.linspace(-3.2, 3.2, 100)
    ys = slope * xs
    mask = xs**2 + ys**2 <= 10
    label = 'Heat flow $y=c^*x$' if slope == np.tan(0) else None
    ax.plot(xs[mask], ys[mask], 'coral', lw=1.5, label=label)

ax.set_xlim(-3.5, 3.5)
ax.set_ylim(-3.5, 3.5)
ax.set_aspect('equal')
ax.set_title('Circular Isotherms + Radial Heat Flow')
ax.legend(loc='upper left', fontsize=9)

# --- Right panel: direction field of the OT ODE ---
ax = axes[1]
xg = np.linspace(-3, 3, 16)
yg = np.linspace(-3, 3, 16)
X, Y = np.meshgrid(xg, yg)

# ODE of orthogonal trajectories: y' = y/x
U = np.ones_like(X)
V = np.where(np.abs(X) > 0.1, Y / X, 0)
norm = np.sqrt(U**2 + V**2)
U /= norm
V /= norm

ax.quiver(X, Y, U, V, color='gray', alpha=0.6, scale=30)

# Overlay some solution curves (trajectories)
for slope in [-2, -1, -0.5, 0, 0.5, 1, 2]:
    xs = np.linspace(-3, 3, 100)
    ys = slope * xs
    ax.plot(xs, ys, 'coral', lw=1.2, alpha=0.8)

ax.set_xlim(-3.5, 3.5)
ax.set_ylim(-3.5, 3.5)
ax.set_aspect('equal')
ax.set_title("Direction Field of $y' = y/x$ (heat flow ODE)")

plt.tight_layout()
plt.show()

4. A Richer Example — Exponential Curves

Consider the family $y = c\,e^{-x^2}$ (Gaussian-shaped curves, common in diffusion problems).

Step 1 — Eliminate $c$. Since $c = y\,e^{x^2}$, differentiate $y = c\,e^{-x^2}$:

$$y' = -2x\,c\,e^{-x^2} = -2xy$$

Step 2 — Orthogonal trajectories ODE:

$$\tilde{y}' = \frac{1}{2x\tilde{y}}$$

Step 3 — Separate and integrate:

$$2\tilde{y}\,d\tilde{y} = \frac{dx}{x} \quad \Longrightarrow \quad \tilde{y}^2 = \ln|x| + C$$

The orthogonal trajectories are curves of the form $\tilde{y} = \pm\sqrt{\ln|x| + C}$.

fig, ax = plt.subplots(figsize=(8, 6))

x = np.linspace(-2.5, 2.5, 500)

# Given family: y = c * exp(-x²)
for c in [-2, -1.2, -0.5, 0.5, 1.2, 2.0]:
    y = c * np.exp(-x**2)
    label = 'Given: $y = c\, e^{-x^2}$' if c == -2 else None
    ax.plot(x, y, 'steelblue', lw=1.5, label=label)

# Orthogonal trajectories: y² = ln|x| + C
# Use solve_ivp to trace them numerically
def ot_ode(t, y):
    if abs(t) < 1e-10 or abs(y[0]) < 1e-10:
        return [0.0]
    return [1.0 / (2 * t * y[0])]

colors_ot = 'coral'
first = True
for y0 in [0.3, 0.7, 1.1, 1.5]:
    # Trace forward from x = 0.05
    sol_fwd = solve_ivp(ot_ode, [0.05, 2.5], [y0], max_step=0.02, dense_output=True)
    sol_bwd = solve_ivp(ot_ode, [0.05, 2.5], [-y0], max_step=0.02, dense_output=True)
    
    t_plot = np.linspace(0.05, 2.5, 300)
    label = 'OT: $y^2 = \\ln|x| + C$' if first else None
    ax.plot(t_plot, sol_fwd.sol(t_plot)[0], colors_ot, lw=1.5, label=label)
    ax.plot(t_plot, sol_bwd.sol(t_plot)[0], colors_ot, lw=1.5)
    # Mirror for negative x
    ax.plot(-t_plot, sol_fwd.sol(t_plot)[0], colors_ot, lw=1.5)
    ax.plot(-t_plot, sol_bwd.sol(t_plot)[0], colors_ot, lw=1.5)
    first = False

ax.set_xlim(-2.8, 2.8)
ax.set_ylim(-2.5, 2.5)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_title('Gaussian Curves and Their Orthogonal Trajectories')
ax.legend(loc='upper right', fontsize=9)
plt.tight_layout()
plt.show()

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/var/folders/34/4mb6rzb52l76jcqm_pjx3fph0000gn/T/ipykernel_49818/848487223.py:8: SyntaxWarning: invalid escape sequence '\,'
  label = 'Given: $y = c\, e^{-x^2}$' if c == -2 else None

Summary

Step	Action	Key Formula
1	Given family $G(x,y,c)=0$ → find its ODE $y' = f(x,y)$	Differentiate, eliminate $c$
2	ODE of orthogonal trajectories	$\tilde{y}' = -1/f(x,\tilde{y})$
3	Solve the new ODE	Separation, exact, integrating factor, etc.

Physical intuition: orthogonal trajectories arise naturally whenever two related physical quantities have gradients that are perpendicular — temperature vs. heat flow, electric potential vs. electric field, pressure vs. fluid streamlines.