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In Section 7.2.1 of Bergman's Fundamentals of Heat and Mass Transfer, there is a derivation of the Blasius equation $$ 2 \frac{\mathrm d^3 f}{\mathrm d \eta^3} + f \frac{\mathrm d^2 f}{\mathrm d \eta^2} = 0$$ through the definitions $$ \eta = y \sqrt{\frac{u_\infty}{\nu x}} \quad \text{and} \quad f(\eta) = \frac{\psi}{u_\infty \sqrt{\frac{\nu x}{u_\infty}}},$$ where $\psi$ is the velocity streamfunction satisfying $ u = \frac{\partial \psi}{\partial y} $ and $ v = - \frac{\partial \psi}{\partial x}$.

I have derived for myself the transformation from the momentum equation $$ u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = \nu \frac{\partial^2 u}{\partial y^2} $$ to the Blasius equation, which was mainly a lot of chain rule and algebraic manipulation.

I understand that the no-slip and freestream BCs are $$ u(x, 0) = 0, \quad v(x, 0) = 0, \quad u(x, \infty) = u_\infty.$$

Since the first two no-slip BCs are at $y = 0$, plugging this into the definition of $\eta$ yields that $\eta = 0$ at the wall. Similarly for the freestream, as $y \to \infty$, $\eta \to \infty$. But how do I derive the listed BCs? Unfortunately, Bergman's textbook simply states them without explanation. $$ f(\eta = 0) = 0, \quad \left.\frac{\mathrm df}{\mathrm d\eta}\right|_{\eta = 0} = 0, \quad \left.\frac{\mathrm df}{\mathrm d \eta}\right|_{\eta \to \infty} = 1.$$

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As you state, no-slip and matching to the outer Euler flow gives $$u(x, 0) = 0, \quad v(x,0) = 0, \quad u(x, \infty) = u_\infty.$$ The streamfunction is $$\psi = u_\infty \sqrt{\frac{\nu x}{u_\infty}} f(\eta) \quad \text{where} \quad \eta = \sqrt{\frac{u_\infty}{\nu x}} y.$$ This gives the velocity components $$u = \frac{\partial \psi}{\partial y} = u_\infty f'(\eta)\quad \text{and} \quad v = -\frac{\partial \psi}{\partial x} = \sqrt{\frac{u_\infty \nu}{x}} \left(-\frac 12 f + \frac 12 \eta f'\right)$$ after carefully applying the chain rule.

Thus, \begin{align*} u(x,0) = 0 &\implies u_\infty f'(0) = 0 \implies f'(0) = 0, \\ v(x,0) = 0 &\implies \sqrt{\frac{u_\infty \nu}{x}} \left(-\frac 12 f(0) + 0\right) =0 \implies f(0) = 0, \\ u(x, \infty) = u_\infty &\implies u_\infty f'(\infty) = u_\infty \implies f'(\infty) = 1. \end{align*} So we obtain the required boundary conditions $$f(0) = f'(0) = 0, \quad f'(\infty) = 1.$$

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