54 2. FOUR IMPORTANT LINEAR PDE Also, let us introduce the useful function (21) ψ := − n 2 log(−4πs) + |y|2 4s + n log r and observe ψ = 0 on ∂E(r) − {(0, 0)}, since Φ(y, −s) = r−n on ∂E(r). We utilize (21) to write B = 1 rn+1 E(r) 4us n i=1 yiψyi dyds = − 1 rn+1 E(r) 4nusψ + 4 n i=1 usyi yiψ dyds there is no boundary term since ψ = 0 on ∂E(r) − {(0, 0)}. Integrating by parts with respect to s, we discover B = 1 rn+1 E(r) −4nusψ + 4 n i=1 uyi yiψs dyds = 1 rn+1 E(r) −4nusψ + 4 n i=1 uyi yi − n 2s − |y|2 4s2 dyds = 1 rn+1 E(r) −4nusψ − 2n s n i=1 uyi yi dyds − A. Consequently, since u solves the heat equation, φ (r) = A + B = 1 rn+1 E(r) −4nΔuψ − 2n s n i=1 uyi yi dyds = n i=1 1 rn+1 E(r) 4nuyi ψyi − 2n s uyi yi dyds = 0, according to (21). Thus φ is constant, and therefore φ(r) = lim t→0 φ(t) = u(0, 0) ( lim t→0 1 tn E(t) |y|2 s2 dyds ) = 4u(0, 0), as 1 tn E(t) |y|2 s2 dyds = E(1) |y|2 s2 dyds = 4. We omit the details of this last computation.
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