Partial Differential.Equation

4. note ∂p/∂x=x/p, ∂p/∂y=y/p, ∂p/∂z=z/p

ux=(du/dp)(∂p/∂x)=(du/dp)(x/p)
uxx=(d^2 u /dp^2)(x/p)^2 + (du/dp)(1/p - x^2 /p^3)
= (d^2 u /dp^2)(x/p)^2 + (1/p)(du/dp)(1-x^2 / p^2)

similarly,
uyy=(d^2 u /dp^2)(y/p)^2 + (1/p)(du/dp)(1-y^2 / p^2)
uzz=(d^2 u /dp^2)(z/p)^2 + (1/p)(du/dp)(1-z^2 / p^2)

Therefore
∆u = (d^2 u /dp^2)((x^2 + y^2 + z^2)/p^2) + (1/p)(du/dp)(3-(x^2 + y^2 + z^2)/p^2)
=(d^2 u /dp^2) + (2/p)(du/dp)

∆u = 0
(d^2 u /dp^2) + (2/p)(du/dp) = 0
d(u')/dp = -2/p u'
d(u')/u' = -2dp / p
ln (u') = -2ln(p) +c
u' = Cp^(-2)
u(p) = -Cp^(-1) + B = Ap^(-1)+B
where A, B are constants.

5. U(x, t) = V(x, t)e^(ax−bt),
Ut = Vt e^(ax−bt) -bV e^(ax−bt)
Ux = Vx e^(ax−bt) + aV e^(ax−bt)
Uxx = Vxx e^(ax−bt) +2aVx e^(ax−bt) + a^2 V e^(ax−bt)

Ut − φUxx − cUx −λU =0

(Vt e^(ax−bt) -bV e^(ax−bt))-φ(Vxx e^(ax−bt) +2aVx e^(ax−bt) + a^2 V e^(ax−bt))
-c( Vx e^(ax−bt) + aV e^(ax−bt)) - λU = 0

e^(ax−bt) (Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V)=0.

Vt -φVxx -(2aφ+c)Vx -(b+φa^2+ac+λ)V=0

We need (2aφ+c)=0 and (b+φa^2+ac+λ)=0
Hence a= -c/2φ
b=-(φa^2+ac+λ)=-(c^2/4φ - c^2/2φ + λ) = c^2/4φ - λ.