I am presently working on a problem in fluid dynamics where our group is investigating the behavior of temperature and velocity at the leading edge of a flat plate when fluid flows past it. The momentum equation, after introducing stream functions and cross-differentiation, reads $$\rho\left(\psi_{y}\nabla^{2}\psi_{x} - \psi_{x}\nabla^{2}\psi_{y} \right) = \mu\nabla^{4}\psi$$

I have made use of the self-similar transformation $\eta=\frac{y}{x}$ with $f(\eta,R)=\frac{\psi}{Ux}$ to obtain the following PDE (It checks out with this paper which I have referred (pg 156); note: there is one error in the paper, it has $2\eta ff_{\eta}$ instead of $2\eta ff_{\eta\eta}$ in the $R[...]$ terms)

$$\left[(1+\eta^{2})^{2}f_{\eta\eta\eta\eta}+8(1+\eta^{2})f_{\eta\eta\eta}+4(1+3\eta^{2})f_{\eta\eta}\right]+\left[2\eta ff_{\eta\eta}+(1+\eta^{2})(ff_{\eta\eta}+f_{\eta}f_{\eta\eta}) - 4(1+3\eta^{2})f_{\eta\eta R}-4\eta(1+\eta^{2})f_{\eta\eta\eta R}\right]R+\left[2\eta(f_{R}f_{\eta\eta}-ff_{\eta\eta R})-(1+\eta^{2})(f_{\eta}f_{\eta\eta R}-f_{R}f_{\eta\eta\eta})+2(1+3\eta^{2})f_{\eta\eta RR}\right]R^{2}+\left[2\eta(f_{\eta}f_{\eta RR}-f_{R}f_{\eta \eta R})+ff_{\eta RR}-3f_{\eta}f_{RR}+4f_{RRR}-4\eta f_{\eta RRR}\right]R^{3}+\left[f_{RRRR}+f_{R}f_{\eta RR}-f_{\eta}f_{RRR}\right]R^{4}=0$$

Boundary conditions: At the plate, $\eta=0$ and we have $f_{\eta}=0$ & $f+f_{R}R=0$. At the leading edge, $\eta \rightarrow \infty$ and we have $f_{\eta}\rightarrow 1$ & $f+f_{R}R\rightarrow \eta$.

Since this problem is at the leading edge, we consider Reynolds numbers of the order $10^{-3}$ and hence, we can neglect all $\mathcal{O}(R^{2})$ and higher order terms of the PDE, but still, solving the resulting PDE is daunting. Any help is appreciated.

The paper uses an asymptotic expansion in $R$ to solve the PDE but I require a solution in $\eta$ and $R$ since I need to use it in the energy equation. I did try to solve the PDE neglecting the terms of $R^{2}$ and higher powers and later trying to compare it to a standard form mentioned in Polyanin, but couldn't find a solution.

Edit: As per Tomas Kojar’s suggestion in the comment section, consider the following two equations Now, setting $u=f_{\eta\eta}$, we obtain the following second-order PDE $$(1+η^2 )^2 u_{ηη}+8η(1+η^2 )u_{η}+4(1+3η^2 )u=0 $$ Which is solved with boundary conditions to obtain the solution $f_{1}(\eta,R)=\frac{2}{\pi}tan^{-1}(η)+\frac{2η}{π(1+η^2)}$. Now, the second equation reads $$[2ηff_{ηη}+(1+η^2 )(ff_{ηηη}+f_η f_{ηη} )-4(1+3η^2 ) f_{ηηR}-4η(1+η^2 ) f_{ηηηR}]R=0,$$ and since $R\neq 0$, we obtain a non-linear PDE which upon setting $v(\eta,R)=f_{\eta\eta\eta}$ yields the following PDE $$η^6 v^2+η^2 v^3 (1+η^2 )-4η(1+η^2 ) v_R=0.$$ But, this is still complicated to solve.

  • 1
    $\begingroup$ Since we are ignoring R^2 terms, how about setting the coefficient of the R term to zero to get one equation? Then in the constant (in R) term we can set u=f'' to get a second order equation. So you will have two equations to solve. Then you can try building a common solution. $\endgroup$ – OOESCoupling Jun 21 '19 at 1:13
  • $\begingroup$ @ThomasKojar Thank you for the comment, the second-order PDE is solvable but the second part of the equation (the one with the R term) is still pretty non-linear and Mathematica is unable to churn out a solution. I have added this as an edit. Any suggestions? $\endgroup$ – Spoilt Milk Jun 22 '19 at 7:05
  • $\begingroup$ You should try a series $f=f_0+R\cdot f_1+R^2\cdot f_2+O(R^3)$, choosing for $f_0$ the solution of the leading term set to 0. Then, non-linearities in the higher order terms will disappear. This is standard perturbation theory. You are not granted to get closed form solutions for the higher order corrections but you are working with linear ODE at least. $\endgroup$ – Jon Jun 22 '19 at 17:04
  • $\begingroup$ @Jon an asymptotic expansion was used to solve the PDE in the paper I have linked. I ask this question since I need a solution in terms of $\eta$ and $R$ since both play an important role when I solve for the heat transfer coefficient via the energy equation. $\endgroup$ – Spoilt Milk Jun 22 '19 at 21:19
  • $\begingroup$ Dear Naveen: you might not be aware, but each edit bumps your question to the top of the queue and thus bumps another question vying for attention off the top page. Thus very many small edits in a short time span might be resented by other members of the community. $\endgroup$ – Todd Trimble Jun 23 '19 at 11:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.