Prove the differential equation for the Weierstrass elliptic function by showing a derived function is holomorphic and constant | Step-by-Step Solution

Hello! This is a problem that connects the theory of elliptic functions with complex analysis.

What We're Solving:

We need to prove that the function g(z) = ℘'(z)² - 4℘(z)³ - g₂℘(z) - g₃ is identically zero by showing it's an elliptic function with no poles, which means it must be constant (and therefore zero).

The Approach:

We'll use Liouville's theorem for elliptic functions! This powerful theorem tells us that any elliptic function without poles must be constant. Since g(z) involves the Weierstrass function and its derivative, we expect it to be doubly periodic (elliptic). If we can show it has no poles, then it's constant. And since we can show g(z) = 0 at certain points, it must be zero everywhere!

We're proving the differential equation ℘'(z)² = 4℘(z)³ + g₂℘(z) + g₃ by showing the difference between left and right sides is always zero.

Step-by-Step Solution:

Step 1: Verify g(z) is elliptic (doubly periodic)

• ℘(z) is elliptic with periods 2ω₁ and 2ω₂
• ℘'(z) is also elliptic with the same periods (derivative of elliptic function)
• Since g(z) is built from ℘(z) and ℘'(z), it inherits the same periods
• Therefore, g(z) is elliptic! ✓

Step 2: Analyze the poles of g(z) This is the crucial part! Let's examine what happens at the poles of ℘(z):

• ℘(z) has double poles at lattice points (z = 2mω₁ + 2nω₂)
• Near a pole, ℘(z) ≈ 1/z² and ℘'(z) ≈ -2/z³

Step 3: Check the behavior near poles Near z = 0 (a typical pole):

• ℘'(z)² behaves like 4/z⁶
• 4℘(z)³ behaves like 4/z⁶
• The terms g₂℘(z) and g₃ are less singular

The key insight: The most singular terms (℘'(z)² and 4℘(z)³) have the same leading behavior and cancel each other out!

Step 4: Precise pole cancellation The Laurent expansion shows that:

• The 1/z⁶ terms from ℘'(z)² and -4℘(z)³ cancel exactly
• The 1/z⁴ and 1/z² terms also cancel due to the specific choice of g₂ and g₃
• What remains has no negative powers of z

Step 5: Apply Liouville's theorem

• g(z) is elliptic (doubly periodic)
• g(z) has no poles (we just proved this!)
• By Liouville's theorem: g(z) must be constant

Step 6: Determine the constant Since we can compute that g(z) approaches 0 in certain limits (or by direct calculation at specific points), the constant must be 0.

The Answer:

Therefore, g(z) = ℘'(z)² - 4℘(z)³ - g₂℘(z) - g₃ ≡ 0

This gives us the fundamental differential equation: ℘'(z)² = 4℘(z)³ + g₂℘(z) + g₃

Memory Tip:

Remember this pattern: "Elliptic + No Poles = Constant"! This is a powerful technique in complex analysis. Whenever you have a function that's doubly periodic but has no singularities, Liouville's theorem tells you it must be boring (constant). The art is in showing the pole cancellation - that's where the magic happens in elliptic function theory!

The constants g₂ and g₃ aren't arbitrary - they're precisely chosen to make this cancellation work perfectly!