Evans PDE Solutions for Ch2 and Ch3

1. Consider the function z : < → < for fixed x ∈ < and t ∈ (0,∞) z(s) = u(x+ bs, t+ s)e Then ż(s) := ∂z ∂s = e(b ·Dxu(x+ sb, t+ s) + ut(x+ sb, t+ s) + cu(x+ sb, t+ s)) = 0 by the condition given by the problem. Therefore, z is a constant function with respect to s. Finally, by using the fact that u = g on < × {t = 0}, we conclude that u(x, t) = z(0) = z(−t) = u(x− tb, 0)e−ct = g(x− tb)e−ct 2. Let O = (aij) and φ(x) = O · x. Then it’s clear that Dφ(x) = O. Since v is defined to be v(x) = u(O · x) = (u ◦ φ)(x), we calculate Dv(x) = Du(φ(x)) ·Dφ(x) = Du(O · x) ·O Then vxi(x) = ∑n j=1 uxj (φ(x))aji. Note that by the same calculations, we have D(uxj ◦ φ)(x) = Duxj (φ(x)) ·O Therefore