Author Topic: Quiz3 TUT5101  (Read 473 times)

Jingjing Cui

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Quiz3 TUT5101
« on: February 10, 2020, 01:31:56 PM »
$$
u_{tt}-c^2u_{xx}=0\\
u|_{t=0}=0\\
u_{t}|_{t=0}=1\\
u_{x}|_{x=0}=0\\
$$
general solution: u(x,t)=f(x+ct)+g(x-ct)
when x+ct>0 and x-ct>0:
$$
u(x,0)=f(x)+g(x)=0\\
u_{t}(x,0)=cf'(x)-cg'(x)=1\\
f'(x)-g'(x)=\frac{1}{c}\\
f(x)-g(x)=\frac{x}{c}\\
$$
Let s>0
$$
f(s)=\frac{s}{2c}\\
g(s)=-\frac{s}{2c}\\
u(x,t)=\frac{x+ct}{2c}-\frac{x-ct}{2c}=t\\
$$
when x+ct>0 and x-ct<0:
$$
u_{x}(0,t)=f'(ct)+g'(-ct)=0\\
$$
Let s=-ct<0
$$
-f'(-s)=g'(s)\\
f(-s)=g(s)\\
u(x,t)=\frac{x+ct}{2c}+\frac{ct-x}{2c}=t\\
$$