Compuational Fluid Dynamics for Incompressible Flows

Week-02 (Live Session)

Durga Prasad Pydi

2026-01-31

Accessing Live Session Materials

Live Session timings: 05:00 PM - 07:00 PM IST on every Saturday Meeting Link: Will be circulated to enrolled students on their registered email id

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Recorded Lectures Session Written Notes Session Slides

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Course Contents

  • Applications of CFD and brief recap of governing equations and boundary conditions
  • Classification of PDEs - elliptic, parabolic and hyperbolic - highlighting the relevance to CFD
  • Key qualifications for solvers and Finite difference method (FDM) as a tool to solve PDEs
  • Application of FDM
    • elliptic
    • parabolic
    • hyperbolic
  • Evaluating the learned strategies using techniques presented in 3
  • Alternate formulations of governing equation and their solutions
  • MAC Algorithm
  • Finite Volume Method using SIMPLE (last 4-5 weeks)

Concepts

Classification of Phenomena

For a general 2nd2^{nd} order PDE, defined as aϕxx+bϕxy+cϕyy=h a \phi_{x x} + b \phi_{x y} + c \phi_{yy}=h

The characteristics are given by dydx=b±b2−4ac2a\frac{dy}{dx}=\frac{b \pm \sqrt{ b^{2}-4ac }}{2a}

Discriminant: D=b2−4acD=\sqrt{ b^{2}-4ac } determines the classes of PDEs

  • D>0D>0
  • D=0D=0
  • D<0D<0

Water Ripples

Which PDE can describe this phenomenon?

Dye in Agar

Which PDE can describe this phenomenon?

Can you give an example of an elliptic phenomenon?

Problems

Question 1

The lines along which a partial differential equation reduces to an ordinary differential equation are called

Solution: Characteristic lines Consider ut+uxx=0u_{t}+u_{x x}=0 what are the characteristic lines for the above equation? dtdx=0\frac{dt}{dx}=0

Consider utt−uxx=0u_{tt}-u_{x x}=0 a=−1,b=0,c=1a=-1,b=0,c=1 b2−4ac=4>0b^{2}-4ac=4>0 dtdx=±1\frac{dt}{dx}=\pm{1}

Question 2

Which of these apply to parabolic equations?

Solution: They have one real characteristic line

Question 3

Which of these characteristics apply to hyperbolic equation?

Solution: finite domain of dependence and finite domain of influence

Question 4

The nature of the second order partial differential equation ∂ϕ∂t+α∂ϕ∂x+β∂2ϕ∂x2=0\frac{ \partial \phi }{ \partial t }+\alpha \frac{ \partial \phi }{ \partial x }+\beta \frac{ \partial^{2} \phi }{ \partial x^{2} }=0

Solution: parabolic

a=β,b=0,c=0a=\beta,b=0,c=0 D=b2−4ac=0−β×0×4=0D=b^{2}-4ac=0-\beta \times 0 \times 4=0

Question 5

The value of α\alpha for which the equation α∂2ϕ∂x2+4∂2ϕ∂x∂y+2∂2ϕ∂y2=0\alpha \frac{ \partial^{2}\phi }{ \partial x^{2} }+4 \frac{ \partial^{2}\phi }{ \partial x \partial y} + 2 \frac{ \partial^{2} \phi }{ \partial y^{2} }=0 becomes parabolic is

Solution: 2

42−4×2×α=0⟹α=?4^{2}-4\times 2 \times \alpha=0 \implies \alpha=?

Question 6

The steady compressible flow defined by (1−M∞2)∂2ϕ∂x2+∂2ϕ∂y2=0(1-M_{\infty}^{2})\frac{ \partial^{2}\phi }{ \partial x^{2} }+\frac{ \partial^{2}\phi }{ \partial y^{2} }=0 becomes hyperbolic under the condition

Solution: M∞>1M_{\infty}>1

0−4(1−Minf2)>00-4(1-M_{inf}^{2})>0 ⟹\implies Minf>1M_{inf}>1

Question 7

The Tricomi equation y∂2u∂x2+∂2u∂y2=0y \frac{ \partial^{2} u }{ \partial x^{2} }+\frac{ \partial^{2}u }{ \partial y^{2} }=0 shows elliptic nature under the condition

Solution: y>0y>0

0−4y<00-4y < 0 ⟹\implies y>0y>0

Question 8

The following system of equations is classified as ∂u∂t+8∂v∂x=0∂v∂t+2∂u∂x=0 \begin{align} \frac{ \partial u }{ \partial t } +8 \frac{ \partial v }{ \partial x } & =0 \\ \frac{ \partial v }{ \partial t }+2 \frac{ \partial u }{ \partial x } & = 0 \end{align}

Solution: hyperbolic

2D NS : u,v,pu,v,p continuity x & y momentum ux+vy=0u_{x}+v_{y}=0

U=(uv)U = \begin{pmatrix}u \\ v\end{pmatrix}, Ux=(uxvx)U_{x}=\begin{pmatrix}u_{x} \\ v_{x}\end{pmatrix} Ut+AUx=0U_{t}+AU_{x}=0 A=(0820)A=\begin{pmatrix}0 & 8 \\ 2 & 0\end{pmatrix}

First eqn wrt t and then second eqn wrt x ut+8vx=0u_{t}+8v_{x}=0 utt+8vxt=0u_{tt}+8v_{xt}=0

vt+2ux=0v_{t}+2u_{x}=0 vtx+2uxx=0v_{tx}+2u_{x x}=0

vtx=vxtv_{tx}=v_{xt} utt−16uxx=0u_{tt}-16u_{x x}=0 0−4×−16>00-4 \times -16>0 ⟹\implies hyperbolic

Calculate eigen values for matrix A λ2−8×2=0\lambda^{2}-8\times{2}=0

Question 9

The nature of the following Cauchy-Reimann equations is ∂u∂x+∂v∂y=0∂v∂x−∂u∂y=0 \begin{align} \frac{ \partial u }{ \partial x } + \frac{ \partial v }{ \partial y } & =0 \\ \frac{ \partial v }{ \partial x } -\frac{ \partial u }{ \partial y } & =0 \end{align}

Solution: elliptic

U=(uv)U=\begin{pmatrix}u \\ v\end{pmatrix}

Ux+AUy=0U_{x}+ A U_{y}=0 where Ux=(uxvx)U_{x}=\begin{pmatrix}u_{x} \\ v_{x}\end{pmatrix} and Uy=(uyvy)U_{y}=\begin{pmatrix}u_{y} \\ v_{y}\end{pmatrix}

A=(01−10)A=\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}

characteristic equation for matrix AA: λ2+1=0\lambda^{2}+1=0 eigen values for AA are ±i\pm i

Question 10

The following system of equations is classified as (β\beta has real value) β2∂u∂x−∂v∂y=0∂v∂x−∂u∂y=0 \begin{align} \beta^{2} \frac{ \partial u }{ \partial x } -\frac{ \partial v }{ \partial y } & =0 \\ \frac{ \partial v }{ \partial x } - \frac{ \partial u }{ \partial y } & =0 \end{align}

Solution: hyperbolic

U=(uv)U=\begin{pmatrix}u \\ v\end{pmatrix}

Ux+AUy=0U_{x}+ A U_{y}=0 where Ux=(uxvx)U_{x}=\begin{pmatrix}u_{x} \\ v_{x}\end{pmatrix} and Uy=(uyvy)U_{y}=\begin{pmatrix}u_{y} \\ v_{y}\end{pmatrix}

(uxvx)+(abcd)(uyvy)=0 \begin{pmatrix} u_{x} \\ v_{x} \\ \end{pmatrix} + \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} u_{y} \\ v_{y} \end{pmatrix}=0

A=(0−1β2−10)A=\begin{pmatrix}0 & -\frac{1}{\beta^{2}} \\ -1 & 0\end{pmatrix}

λ2−1β2=0\lambda^{2}-\frac{1}{\beta^{2}}=0

β∈ℝ\beta \in \mathbb{R} ⟹\implies λ\lambda ∈ℝ\in \mathbb{R} so hyperbolic

Thank You

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