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--- math121b:04-01 [2020/04/01 06:55]
pzhou created
+++ math121b:04-01 [2026/02/21 14:41] (current)
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 Hence, we have the following solution to the heat equation (ignoring the initial condition for now)
-$1, \z{ and } e^{-n^2 t} \sin(n \theta), e^{-n^2 t} \cos(n \theta) $$
+$$1, \z{ and } e^{-n^2 t} \sin(n \theta), e^{-n^2 t} \cos(n \theta) $$
 Thus, if we decompose the initial condition $u_0$ as
 $$ u_0(\theta) = c + \sum_{n = 1}^\infty a_n \cos(n \theta) + b_n \sin(n \theta)$$
@@ Line 36: / Line 36: @@
 To fix the coefficients $v_{n,m}$, we use the initial conditions
 $$ v(0,x,y) = \sum_{n,m} a_{n,m}  v_{n,m}(0,x,y) $$
-so
+so multiply both sides by $v_{n,m}(0,x,y)$ and integrate, only one term on the RHS contribute, and we get
 $$a_{n,m} = \frac{\int_{[0,1]^2} v(0,x,y) v_{n,m}(0,x,y) dx dy}{\int_{[0,1]^2}  v^2_{n,m}(0,x,y) dx dy}.$$
+Remark: if the boundary temperature is not constant $T$ (but still time-independent), we may still find a special solution first, a function $U(x,y)$ that satisfies the boundary condition, and $\Delta U(x,y) = 0$. Such function $U(x,y)$ exists and is unique, it is called the harmonic extension of the boundary value to the interior.  Then we can still get rid of the boundary condition by setting
+$$ u(t,x,y) = U(x,y) + v(t,x,y) $$
+where $v(t,x,y)$ now has boundary condition $0$, and initial condition $v(0,x,y) = u(0,x,y) - U(x,y)$.
+==== Schroedinger Equation (without potential) ====
+$$ i \d_t u = - \Delta u $$
+We may reuse the analysis for the heat equation, except replacing $t$ in heat equation to $it$. Thus, exponential decay now become oscillation.
+==== Wave equation ====
+$$ \d_t^2 u = \Delta u$$
+Suppose $u$ lives on a domain $D$ with boundary value zero, or $u$ lives on a space without boundary, e.g $S^1$ or a torus. We may then consider eigenvalue of $\Delta$, $\lambda_1 \leq \lambda_2 \leq \cdots, $ with $\lambda_n \geq 0$, (repeated with multiplicity), with eigenfunction $u_1(x), u_2(x), \cdots,$ then we may write the general solution
+$$ u(t,x) = \sum_{n=1}^\infty (a_n \cos(\sqrt{\lambda_n} t)   + b_n \sin(\sqrt{\lambda_n} t) ) u_n(x) $$
+(if $\lambda_n=0$,  then we may set $b_n=0$.) To fix the coefficient, we use initial condition $u(0, x)$ and $\dot u(0, x)$:
+$$ \int_D u(0,x) u_n(x) dx = a_n \int u_n^2 dx $$
+$$ \int_D \dot u(0,x) u_n(x) dx = \sqrt{\lambda_n} b_n \int u_n^2 dx $$
+Example: 1-dim string vibration on an interval.

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