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--- math121a-f23:october_27_friday [2023/10/27 05:17]
pzhou created
+++ math121a-f23:october_27_friday [2026/02/21 14:41] (current)
@@ Line 39: / Line 39: @@
 ===== Inhomogenous term =====
-what if you had one
+If you had a equation of the form
+$$ D f(x) = 1 $$
+then it is not a homogeneous equation: the term on the right hand side is does not contain factor $f$. What does its solution space look like? We know it is of the form
+$$ f(x) = c + x. $$
+Note that, the solution space is not a vector space anymore. indeed, if you have $f_1(x)$ and $f_2(x)$ both satisfies the equation,
+$$ D f_1(x) = 1 $$
+$$ D f_2(x) = 1 $$
+then add the two equation up, we see
+$$ D (f_1(x) + f_2(x)) = 2 $$
+so $f_1(x) + f_2(x)$ is not a solution ($1 \neq 2$).
+Nonetheless, the solution space is a so called 'affine space' $V$, which means there is an associated vector space $V'$, and for any two elements $v_1, v_2 \in V$, we have their difference $v_1 - v_2 \in V'$.
+In our case, the associated vector space is the solution space for the homogenous equation
+$$ V' = \{f(x) \mid Df = 0 \}$$
+That means, if we pick a 'base point' $v_0 \in V$, and pick a basis $e_1, \cdots, e_n$ of $V'$, and then we can express any element $v \in V$ as
+$$ v = v_0 + (c_1 e_1 + \cdots + c_n e_n). $$
+for some coefficients $c_i$.
+Back to our problem here, any solution to the inhomogenous equation can be written as a 'particular solution' (playing the role of $v_0$ above), and plus a solution to the homogenous equation (an element of $V'$).

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