Let there's a function $ f: \mathbb{R}\rightarrow\mathbb{R}$ 
Find all functions $ f$ that satisfies:
a) $ f(2u)\equal{}f(u\plus{}v)f(v\minus{}u)\plus{}f(u\minus{}v)f(\minus{}u\minus{}v)$
b) $ f(u)\geq0$
1. Given the function \( f: \mathbb{R} \rightarrow \mathbb{R} \) that satisfies:
   \[
   f(2u) = f(u + v)f(v - u) + f(u - v)f(-u - v)
   \]
   and
   \[
   f(u) \geq 0
   \]

2. First, we substitute \( u \to -u \) and \( v \to -v \) into the given equation:
   \[
   f(-2u) = f(-u - v)f(u - v) + f(v - u)f(u + v)
   \]
   Since \( f \) is non-negative, we can compare this with the original equation:
   \[
   f(2u) = f(u + v)f(v - u) + f(u - v)f(-u - v)
   \]
   By comparing both equations, we get:
   \[
   f(-2u) = f(2u)
   \]
   This implies that \( f \) is an even function:
   \[
   f(x) = f(-x) \quad \forall x \in \mathbb{R}
   \]

3. Using the fact that \( f \) is even, we can rewrite the original equation:
   \[
   f(2u) = f(u + v)f(v - u) + f(u - v)f(-u - v)
   \]
   Since \( f \) is even, \( f(-u - v) = f(u + v) \) and \( f(v - u) = f(u - v) \). Thus, the equation simplifies to:
   \[
   f(2u) = f(u + v)f(u - v) + f(u - v)f(u + v)
   \]
   \[
   f(2u) = 2f(u + v)f(u - v)
   \]

4. Let \( x = u + v \) and \( y = u - v \). Then the equation becomes:
   \[
   f(x + y) = 2f(x)f(y)
   \]

5. Now, consider substituting \( x \to -x \):
   \[
   f(y - x) = 2f(-x)f(y) = 2f(x)f(y)
   \]
   This gives:
   \[
   f(y - x) = f(x + y)
   \]

6. Since \( f \) is non-negative and \( f(y - x) = f(x + y) \), we can conclude that:
   \[
   f(x) = 0
   \]
   for all \( x \in \mathbb{R} \).

7. It is easy to check that \( f(x) \equiv 0 \) is indeed a solution to the given functional equation.

\(\blacksquare\)

The final answer is \( \boxed{ f(x) \equiv 0 } \).