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I have used the steam functions $u = \psi_{y}$ and $v = -\psi_{x}$ to transform the momentum equations to the following form $$\rho\left(\psi_{y}\psi_{xy} - \psi_{x}\psi_{yy}\right)=-p_{x}+\mu\left(\psi_{xxy}+\psi_{yyy} \right) $$ $$\rho\left(-\psi_{y}\psi_{xx} + \psi_{x}\psi_{xy}\right)=-p_{y}+\mu\left(\psi_{xxx}+\psi_{xyy} \right) $$

I eliminated the pressure term via cross-differentiation and have obtained the final form of the equation to be as follows $$\rho\left(\psi_{y}\left(\psi_{xyy}+\psi_{xxx} \right) - \psi_{x}\left(\psi_{yyy}+\psi_{xxy} \right) \right) = \mu\left(\psi_{xxxx} +\psi_{yyyy} + 2\psi_{xxyy} \right)\tag{1}, $$ which can be rewritten with $\nabla = \left<\partial_{x},\partial_{y} \right>$ as follows $$\rho\left(\psi_{y}\nabla^{2}\psi_{x} - \psi_{x}\nabla^{2}\psi_{y} \right) = \mu\nabla^{4}\psi$$

Now, this is a leading edge problem of a fluid flowing over a flat plate with viscous forces dominating. The self-similar variable was found to be $$\eta = \frac{y}{x}\tag{2},$$ and the self-similar stream function has the following form $$f(\eta) = \frac{\psi}{Ux}.\tag{3} $$

Substituting $(2)$ and $(3)$ into $(1)$ transforms the PDE to an ODE a follows $$\left(1+\eta^{2}\right)^{2}f_{\eta\eta\eta\eta}+8\eta\left(1+\eta^{2}\right)f_{\eta\eta\eta} + 4\left(1+3\eta^{2}\right)f_{\eta\eta} + Re\left[2\eta ff_{\eta}+\left(1+\eta^{2}\right)\left(ff_{\eta\eta\eta} + f_{\eta}f_{\eta\eta}\right)\right]=0,\tag{4}$$ where $Re=\frac{\rho Ux}{\mu}$

Could someone please show me how this transformation is done. I tried proceeding via the chain rule and was able to compute all the derivatives but failed to obtain the final form as shown in $(4)$. Any help is appreciated.

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  • $\begingroup$ I guess your $\psi_{y} \nabla^{2} \psi_{x}-\psi_{x} \nabla^{2}$ should be $\psi_{y} \nabla^{2} \psi_{x}-\psi_{x} \nabla^{2}\psi_y$, and I guess $\nabla^2$ is the Laplacian(not the Hessian). Is your question about using $(2)$ and $(3)$ to get $(4)$? $\endgroup$ – Calvin Khor May 22 at 11:25
  • $\begingroup$ @CalvinKhor Thank you for pointing out that, have edited the question. Yeah, it's about using (2) and (3) in (1) to get (4) $\endgroup$ – Naveen Balaji May 22 at 11:41
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It seems that I have been able to obtain the answer. After terribly failing to manually compute the derivatives (since it's terribly taxing), I turned to mathematica and was successful in churning out the answer via a small program.

eta = y/x;

psi = f[[$\eta$][x, y]]Ox; \\Using 'O' instead of velocity 'U' since mathematica realizes 'U' as a function

a = D[psi, {x, 3}] + D[psi, x, {y, 2}]; \\Computation of $\psi_{xxx}+\psi_{xyy}$

a2 = D[psi, y]*a; \\Computation of $\psi_{y}\left(\psi_{xxx}+\psi_{xyy}\right)$

b = D[psi, {y, 3}] + D[psi, y, {x, 2}]; \\Computation of $\psi_{yyy}+\psi_{yxx}$

b2 = D[psi, x]*b; \\Computation of $\psi_{x}\left(\psi_{yyy}+\psi_{yxx}\right)$

c = D[psi, {x, 2}, {y, 2}]; \\Computation of $\psi_{xxyy}$

H = D[psi, {y, 2}, {x, 2}]; \\Computation of $\psi_{yyxx}$

P = $\nu$[D[psi, {x, 4}] + D[psi, {y, 4}]]; \\Computation of $\psi_{xxxx}+\psi_{yyyy}$

L = $\nu$[c + H]; \\Computation of $\psi_{xxyy}+\psi_{yyxx}$

K = P + L; \\Computation of $\psi_{xxxx}+\psi_{yyyy}+\psi_{xxyy}+\psi_{yyxx}$

m = a2-b2; \\Computation of $\psi_{y}\nabla^{2}\psi_{x}-\psi_{x}\nabla^{2}\psi_{y}$

Derivative[n_, 1][$\eta$][x, y] := D[eta, y]; \Replacing the functional forms with the actual derivatives of eta.

Derivative[1, n_][$\eta$][x, y] := D[eta, x];

Derivative[1, 1][$\eta$][x, y] := D[eta, x, y];

Derivative[2, n_][$\eta$][x, y] := D[eta, {x, 2}];

Derivative[n_, 2][$\eta$][x, y] := D[eta, {y, 2}];

Derivative[2, 2][$\eta$][x, y] := D[eta, {x, 2}, {y, 2}];

Derivative[3, n_][$\eta$][x, y] := D[eta, {x, 3}];

Derivative[n_, 3][$\eta$][x, y] := D[eta, {y, 3}];

Derivative[3, 3][$\eta$][x, y] := D[eta, {x, 3}, {y, 3}];

Derivative[4, n_][$\eta$][x, y] := D[eta, {x, 4}];

Derivative[n_, 4][$\eta$][x, y] := D[eta, {y, 4}];

Simplify[m - K]

This program churns-out an output in terms of $x's$ and $y's$ and realizing that $y=\eta x$ & $Re=Ux/\nu$ yields the correct final expression.

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