Proof with Step-by-step explanation:
From the Newton's law of viscosity we have
![\tau =\mu (du)/(dy).........(i)](https://img.qammunity.org/2020/formulas/physics/college/mowxh2a46go6r656znkrowv3vmfvqm6qft.png)
where
is the wall shear stress
is the coefficient of dynamic viscosity
is the velocity gradient in the flow
Now from the principle of boundary layer condition we know that the velocity of the fluid that is in contact with a surface the velocity of the fluid is same as that of the boundary itself.
Hence from the attached figure we can infer that
Velocity at
![y=0=0](https://img.qammunity.org/2020/formulas/physics/college/ouxpz781bnts8udhf3185p202uqxyqiwd8.png)
Velocity at
![y=H=U](https://img.qammunity.org/2020/formulas/physics/college/fjm9ifwspdao5wjt20z5t706kli5cr1qbd.png)
Solving equation 'i' we get
![du=(\tau )/(\mu )dy\\\\\int du=\int (\tau )/(\mu )dy\\\\u(y)=(\tau )/(\mu )* y+c](https://img.qammunity.org/2020/formulas/physics/college/ua1iwnhctcclool7rp5jnh4q3sf6rgackq.png)
from the boundary conditions we obtain that c = 0 since
![v(0)=0](https://img.qammunity.org/2020/formulas/mathematics/college/486mt7nv923flfcpmjw8mvq8kiwr7m6a58.png)
Also we have
![U=(\tau )/(\mu )* H\\\\\therefore (\tau )/(\mu )=(U)/(H)\\\\\therefore u(y)=(U)/(H)* y](https://img.qammunity.org/2020/formulas/physics/college/2l2r973qqu3jsdgmuyegwdpbf8hcm218ld.png)