含绝对值的方程化简求解咨询：d = abs(X²-vs²)/2*a1 + abs(vf²-X²)/2*a2

Alright, let's work through simplifying and solving this absolute value equation step by step. The core challenge with absolute values is identifying where the expressions inside them switch sign—once we map those critical points, we can break the problem into manageable intervals where we can drop the absolute value bars entirely.

First, let's restate your equation clearly for reference:

d = |X² - vs²|/(2*a1) + |vf² - X²|/(2*a2)

The key critical points here are where the expressions inside the absolute values equal zero:

• X² = vs² (so X = ±vs)
• X² = vf² (so X = ±vf)

We need to split our analysis based on whether vs² is greater than, less than, or equal to vf² (we'll cover equality as a special case at the end).

Case 1: vs² > vf² #

Here, the critical values of X² are ordered as vf² < vs². We'll analyze three intervals for X²:

Subcase 1a: X² ≥ vs² #

In this interval, X² - vs² ≥ 0 and vf² - X² ≤ 0, so we can rewrite the equation without absolute values:

d = (X² - vs²)/(2*a1) + (X² - vf²)/(2*a2)

Let's rearrange to solve for X²:

1. Multiply through by 2*a1*a2 to eliminate denominators:

   2*d*a1*a2 = a2*(X² - vs²) + a1*(X² - vf²)
   

2. Expand and collect like terms for X²:

   2*d*a1*a2 = X²*(a1 + a2) - a2*vs² - a1*vf²
   

3. Isolate X²:

   X² = [2*d*a1*a2 + a2*vs² + a1*vf²] / (a1 + a2)
   

4. Finally, take the square root (remember X can be positive or negative):

   X = ±√( [2*d*a1*a2 + a2*vs² + a1*vf²] / (a1 + a2) )
   

Important: Verify that the resulting X² is indeed ≥ vs²—if not, this solution is invalid for this interval and can be discarded.

Subcase 1b: vf² < X² < vs² #

Here, X² - vs² < 0 and vf² - X² < 0, so the equation becomes:

d = (vs² - X²)/(2*a1) + (X² - vf²)/(2*a2)

Rearrange to solve for X²:

1. Multiply through by 2*a1*a2:

   2*d*a1*a2 = a2*(vs² - X²) + a1*(X² - vf²)
   

2. Expand and collect like terms:

   2*d*a1*a2 = X²*(a1 - a2) + a2*vs² - a1*vf²
   

3. Isolate X²:

   X² = [2*d*a1*a2 - a2*vs² + a1*vf²] / (a1 - a2)
   

4. Take the square root:

   X = ±√( [2*d*a1*a2 - a2*vs² + a1*vf²] / (a1 - a2) )
   

Check: Ensure the calculated X² falls strictly between vf² and vs²—if not, discard this solution.

Subcase 1c: X² ≤ vf² #

Here, X² - vs² ≤ 0 and vf² - X² ≥ 0, so the equation simplifies to:

d = (vs² - X²)/(2*a1) + (vf² - X²)/(2*a2)

Solve for X²:

1. Multiply through by 2*a1*a2:

   2*d*a1*a2 = a2*(vs² - X²) + a1*(vf² - X²)
   

2. Expand and collect like terms:

   2*d*a1*a2 = -X²*(a1 + a2) + a2*vs² + a1*vf²
   

3. Isolate X²:

   X² = [a2*vs² + a1*vf² - 2*d*a1*a2] / (a1 + a2)
   

4. Take the square root:

   X = ±√( [a2*vs² + a1*vf² - 2*d*a1*a2] / (a1 + a2) )
   

Verify: Confirm X² ≤ vf² and that the numerator is non-negative (since X² can't be negative)—if either condition fails, this solution is invalid.

Case 2: vs² < vf² #

The ordered critical values here are vs² < vf². The subcases follow the same logic as Case 1, just with the interval bounds swapped:

Subcase 2a: X² ≥ vf² #

Same equation as Subcase 1a—follow the same steps to solve, then verify X² ≥ vf².

Subcase 2b: vs² < X² < vf² #

Here, X² - vs² > 0 and vf² - X² > 0, so the equation becomes:

d = (X² - vs²)/(2*a1) + (vf² - X²)/(2*a2)

Rearrange to solve for X²:

1. Multiply through by 2*a1*a2:

   2*d*a1*a2 = a2*(X² - vs²) + a1*(vf² - X²)
   

2. Expand and collect like terms:

   2*d*a1*a2 = X²*(a2 - a1) - a2*vs² + a1*vf²
   

3. Isolate X²:

   X² = [2*d*a1*a2 + a2*vs² - a1*vf²] / (a2 - a1)
   

4. Take the square root:

   X = ±√( [2*d*a1*a2 + a2*vs² - a1*vf²] / (a2 - a1) )
   

Check: Ensure vs² < X² < vf² and the numerator is non-negative.

Subcase 2c: X² ≤ vs² #

Same equation as Subcase 1c—solve and verify X² ≤ vs².

Special Case: vs² = vf² #

If vs = ±vf, then both absolute value expressions simplify to |X² - vs²|, so the equation becomes:

d = |X² - vs²|*(1/(2*a1) + 1/(2*a2))

Solve for |X² - vs²|:

|X² - vs²| = 2*d*a1*a2/(a1 + a2)

This splits into two equations:

1. X² - vs² = 2*d*a1*a2/(a1 + a2) → X = ±√( vs² + 2*d*a1*a2/(a1 + a2) )
2. X² - vs² = -2*d*a1*a2/(a1 + a2) → X = ±√( vs² - 2*d*a1*a2/(a1 + a2) )
   For the second equation, ensure the expression under the square root is non-negative (otherwise, no real solution here).

Final Notes #

• Always double-check that your solutions fall within the interval you solved for—absolute value equations often produce extraneous solutions that don't satisfy the original equation's absolute value conditions.
• Ensure denominators (a1 + a2, a1 - a2, etc.) are not zero. If any denominator is zero, you'll need to handle that as an additional special case (e.g., if a1 = a2, the middle interval equations simplify differently).

内容的提问来源于stack exchange，提问作者Silouane