If the velocity distribution in a laminar boundary layer on a flat plate is

$\frac{\upsilon}{U}=a+b(\frac{y}{\delta})+c(\frac{y}{\delta})^2+d(\frac{y}{\delta})^3$

Determine the coefficient a,b,c and d. Here u is the velocity at the distance y from the surface of the flat plate and U be the free stream velocity at the boundary layer thickness $\delta$

Consider the velocity distribution in a laminar boundary layer over a flat plate,

We know that at the surface of the plate y=0 ⇒ u=0 due to no slip condition.

Substituting in the above equation,

$\frac{0}{U}=a+b(\frac{0}{δ})+c(\frac{0}{δ})^2+d(\frac{0}{δ})^3$

$0=a+b(0)+c(0)^2+d(0)^3$

∴a=0

∴$\frac{u}{U}=b(\frac{y}{δ})+c(\frac{y}{δ})^2+d(\frac{y}{δ}^3$

Now, At the boundary layer thickness, we know that $y=δ⇒u=U,\frac{du}{dy}$=0 and $\frac{d^2 u}{dy^2}$=0

Substituting $y=δ$⇒u=U

$\frac{U}{U}=b(\frac{δ}{δ})+c(\frac{δ}{δ})^2+d(\frac{δ}{δ})^3$

1=b+c+d … (1)

Substituting y=$δ⇒\frac{du}{dy}$=0

$\frac{du}{dy}=U[b(\frac{1}{δ})+c(\frac{2y}{δ^2})+d(\frac{3y^2}{δ^3})]$

∴$0=U[b(\frac{1}{δ})+c(\frac{2δ}{δ^2})+d(\frac{3δ^2}{δ^3})]$

∴0=b+2c+3d … (2)

Substituting y=$δ⇒\frac{d^2 u}{dy^2}=0$

$\frac{d^2 u}{dy^2}=U[c(\frac{2}{δ^2})+d(\frac{6y}{δ^3})]$

∴0=$U[c(\frac{2}{δ^2})+d(\frac{6}{δ^3})]$

∴0=2c+6d … (3)

Solving equations (1), (2) and (3) simultaneously,

b=3

c=-3

d=1

Hence the velocity distribution equation is given as,

$\frac{u}{U}=3(\frac{y}{δ})-3(\frac{y}{δ})^2+(\frac{y}{δ})^3$