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--- math121b:02-19 [2020/02/19 09:10]
pzhou created
+++ math121b:02-19 [2026/02/21 14:41] (current)
@@ Line 16: / Line 16: @@
 **Naive Approach**
-If for some $\lambda$, we can solve to get $A_{\lambda,i}(x)$ for $i=1,\cdots, n$, and $B_{\lambda, j}(y)$ for $j =1, \cdots, m$, then we get a collection of functions
+If for some $\lambda$, we can solve to get $A_{\lambda,i}(x)$ for $i=1,\cdots, n(\lambda)$, and $B_{\lambda, j}(y)$ for $j =1, \cdots, m(\lambda)$, then we get a collection of functions
 $$F_{\lambda, i, j} (x, y) = A_{\lambda,i}(x) B_{\lambda, j}(y) $$
 and we may hope to express any solution as
@@ Line 64: / Line 64: @@
 $$ C_n = \frac{2}{\sinh(n \pi)} \int_0^1 f(x) \sin(n \pi x) dx. $$
-I feel a bit lucky that this can be solved.
+** General Boundary Value? **: suppose the four sides of the square has non-zero values? Just break the problem to 4 smaller problems, where each one has only one side non-zero, then do this problem 4 times.
+===== Eigenvalue problem =====
+The equation
+$$ \d_x^2 X(x) = \lambda X(x), \quad X(0) = 1, \quad X(1)=0 $$
+is an eigenvalue problem. For a randomly chosen $\lambda$, say $0.721$, the solution is only $0$. Hence to have a non-zero solution, $\lambda$ can only be some special value.
+This is analogous to the problem of solving for eigenvalue and eigenvector for an $n \times n$ matrix $A$:
+$$ A v = \lambda v$$
+here, recall we solve for the roots of the  $P(\lambda) = \det(A - \lambda I) = 0$, then for each $\lambda$, we solve for $v$. (There might be several linearly independent $v$ corresponding to the same $\lambda$).
+Here we have the same problem, but, instead of having a matrix $A$ acting on a vector space $\R^n$, we have
+an operator $\d_x^2$ acting on (an infinite dimensional vector space) the function space on $[0,1]$.

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