Question

Steady state solution of a first order partial differential equation

Answer 1

From

12ddσ(b2⋅p)=a⋅p12ddσ(b2⋅p)=a⋅p

you concluded

1pdp=2ab2dσ1pdp=2ab2dσ

But this is not correct: Note that bbdoes depend on σσ. You have to apply the chain rule:

12ddσ(b2⋅p)⇒b⋅dbdσ⋅p+12b2⋅dpdσ⇒1pdp=a⋅p=a⋅p=2(a−b⋅dbdσ)b2dσ=2ab2dσ−21b⋅dbdσdσ12ddσ(b2⋅p)=a⋅p⇒b⋅dbdσ⋅p+12b2⋅dpdσ=a⋅p⇒1pdp=2(a−b⋅dbdσ)b2dσ=2ab2dσ−21b⋅dbdσdσ

From the second term on the righthand-side you obtain the factor 1b21b2 in the given solution.

Concerning dσ′dσ′: Since p=p(σ)p=p(σ)you shouldn't use σσ as variable of integration at the same time. That's why they called it σ′σ′ instead (has nothing to do with the derivative!)