[tex]u_{a}+u_{t}=-\mu t_{u}[/tex](adsbygoogle = window.adsbygoogle || []).push({});

[tex]u(a,0)=u_{0}(a)[/tex]

[tex]u(0,t)=b\int_0^\infty \left u(a,t)da[/tex]

Solve [tex]u(a,t)[/tex] for the region [tex]a<t[/tex]

Got this question from assignment. My solution is incomplete though, need some inspirations! I have shown that the general solution is

[tex]u(a,t)=F(a-t)e^{-1/2{\mu}t^{2}}[/tex]

So for u(0, t) we have

[tex]u(0,t)=F(-t)^{-1/2{\mu}t^{2}}=b\int_0^\infty \left u(a,t)da[/tex]

Substitute the general solution u(a,t) we get

[tex]F(-t)=b\int_0^\infty \left F(a-t)da[/tex]

Now -t -> a-t, and we obtain F(a-t) to be

[tex]F(a-t)=b \int_0^\infty \left F(2a-t)da[/tex]

Hence our solution u(a,t) for a<t is

[tex]u(a,t)=be^{-1/2{\mu}t^{2}}\int_0^\infty\left F(2a-t)da[/tex]

However, when I talked to my lecturer, he told me I can simplify this further. I need some inspirations

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# Homework Help: Pde problem: inspiration needed

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