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loesung

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Hi! I am trying to understand the MP for the heat eqn. in [tex]\mathbb{R}^n[/tex] (see attached jpegs below). I don't understand why we are working with the function

[tex] v(x,t):=u(x,t)-\frac{\mu}{(T+\epsilon-t)^{n/2}}e^{\frac{|x-y|^2}{4(T+\epsilon-t)}}[/tex]

Why not just work with u(x,t)? I understand that, eventually we want [tex]\epsilon\to 0[/tex] and get [tex]u[/tex] back, but I guess I really don't understand the proof (as I can't convince myself that v(x,t) is necessary)!

My second issue is in Part (1) of the proof, we are working in the cylindrical domain [tex]U_T=B^0(y,r)\times(0,T]

[/tex].

In Part (2), though, we somehow jump from this domain to [tex]\mathbb{R}^n[/tex]. Why do we need to start with the cylindrical domain in order to show the MP for [tex]\mathbb{R}^n[/tex]?

Thanks for your time. I look forward to any helpful replies!

los

[tex] v(x,t):=u(x,t)-\frac{\mu}{(T+\epsilon-t)^{n/2}}e^{\frac{|x-y|^2}{4(T+\epsilon-t)}}[/tex]

Why not just work with u(x,t)? I understand that, eventually we want [tex]\epsilon\to 0[/tex] and get [tex]u[/tex] back, but I guess I really don't understand the proof (as I can't convince myself that v(x,t) is necessary)!

My second issue is in Part (1) of the proof, we are working in the cylindrical domain [tex]U_T=B^0(y,r)\times(0,T]

[/tex].

In Part (2), though, we somehow jump from this domain to [tex]\mathbb{R}^n[/tex]. Why do we need to start with the cylindrical domain in order to show the MP for [tex]\mathbb{R}^n[/tex]?

Thanks for your time. I look forward to any helpful replies!

los

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