Hyperreality
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u_{a}+u_{t}=-\mu t_{u}
u(a,0)=u_{0}(a)
u(0,t)=b\int_0^\infty \left u(a,t)da
Solve u(a,t) for the region a<t
Got this question from assignment. My solution is incomplete though, need some inspirations! I have shown that the general solution is
u(a,t)=F(a-t)e^{-1/2{\mu}t^{2}}
So for u(0, t) we have
u(0,t)=F(-t)^{-1/2{\mu}t^{2}}=b\int_0^\infty \left u(a,t)da
Substitute the general solution u(a,t) we get
F(-t)=b\int_0^\infty \left F(a-t)da
Now -t -> a-t, and we obtain F(a-t) to be
F(a-t)=b \int_0^\infty \left F(2a-t)da
Hence our solution u(a,t) for a<t is
u(a,t)=be^{-1/2{\mu}t^{2}}\int_0^\infty\left F(2a-t)da
However, when I talked to my lecturer, he told me I can simplify this further. I need some inspirations
u(a,0)=u_{0}(a)
u(0,t)=b\int_0^\infty \left u(a,t)da
Solve u(a,t) for the region a<t
Got this question from assignment. My solution is incomplete though, need some inspirations! I have shown that the general solution is
u(a,t)=F(a-t)e^{-1/2{\mu}t^{2}}
So for u(0, t) we have
u(0,t)=F(-t)^{-1/2{\mu}t^{2}}=b\int_0^\infty \left u(a,t)da
Substitute the general solution u(a,t) we get
F(-t)=b\int_0^\infty \left F(a-t)da
Now -t -> a-t, and we obtain F(a-t) to be
F(a-t)=b \int_0^\infty \left F(2a-t)da
Hence our solution u(a,t) for a<t is
u(a,t)=be^{-1/2{\mu}t^{2}}\int_0^\infty\left F(2a-t)da
However, when I talked to my lecturer, he told me I can simplify this further. I need some inspirations

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