BKS-MidsemPPT.pptx

SIMILARITY SOLUTIONS OF FPDEs
07-11-2022
Varun J Kaushik 2019B4A10681P
Varun J Kaushik
2019B4A10681P
Faculty in-charge: BK Sharma
1

Fractional Derivatives
• You might be familiar with the notation, 𝐷𝑛
𝑓 or 𝐼𝑛
𝑓. Where D is the differential operator and I is the integral operator. N is the
order of operation. You must be used to seeing N ∈ ℤ.
• But sometimes n can be genialized to any real value. This gave birth to the field of fractional Calculus. It was the brainchild of the
German magician (well, Mathematician he was literally a magician so technically I am right). The field was further explored and
improved upon in the last century.
• Mathematicians like Riemann, Reisz, Caputo etc. have all given their definitions of fractional derivatives and integrals.
• The ordinary or integral order derivative is localized in nature i.e the value of the operator only depends on the point where it is
being operated. Whereas Fractional Derivatives have a property called memory effect. This makes it useful for modelling many
natural phenomena
• Today, many papers have been published where the standard laws have been modifed by fractional order derivative and integral
terms. We are doing a case study of one such modification where the Fick’s law of diffusion has been modified by fractional order
derivatives.
2
Varun J Kaushik 2019B4A10681P 07-11-2022

Similarity Transformations- the Scaling
Method
• Many functions (or rather almost all) are multivariate in nature. So the governing equations are also PDEs rather than ODEs.
• Many techniques have been developed to reduce a PDE into ODE, or rather a PDE in n+1 independent variables into PDE of n
variables. These techniques come under the umbrella of similarity transformations.
• In our study our focus is mainly on the scaling method.
• For example, the BVP
,
𝑢𝑥 + 𝑢𝑦 = 0
𝑣𝑢𝑥 + 𝑣𝑢𝑦 = 𝑈𝑈𝑥 + 𝑢𝑦𝑦
⟶ (1)
With boundary conditions…
𝑢(𝑥, 0) = 0; 𝑣 𝑥, 0 = 0; 𝑢(𝑥, ∞) = 𝑈(𝑥)
3
07-11-2022

Method
Can be reduced to an ODE…
Step 1: Transform the variables as follows:
𝑥∗ = 𝑒∈𝑎𝑥; 𝑥∗ = 𝑦∈𝑏𝑦; 𝑢∗ = 𝑒∈𝑐𝑢; 𝑣∗ = 𝑒∈𝑑𝑣; 𝑈∗ = 𝑒∈𝑒 𝑈
 Step 2: Substituting the transformed variables in (1), we get
𝑢𝑥
∗ + 𝑒(𝑏+𝑐−𝑎−𝑑)𝑣𝑥
∗ = 0
𝑢∗𝑢𝑥
∗ + 𝑒(𝑏+𝑐−𝑎−𝑑)𝑣𝑦
∗ = 𝑒(2𝑏+𝑐−𝑎)𝑢𝑦𝑦
∗ + 𝑒(𝑐−𝑒)𝑈𝑥
∗
⟶ (2)
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07-11-2022

Method 5
Varun J Kaushik 2019B4A10681P 07-11-2022
Step 3: In (2) we omit ∈ and * from some of the subscripts for simplicity. For (2) to satisfy the invariance conditions, i.e same form as equation (1),
the following conditions must hold:
b= (a-c)/2 , d= (c-a)/2, e=c.
These conditions are called the invariance conditions. Required for the PDE (2) to be of the same form as PDE (1)
Now, if we observe parameters a and c are arbitrary, if we establish some kind of relationship between them (say c=ma)

Method 6
07-10-2022
 Step 4: Substitute the invariance realtions and try forming a relationship between variables,
𝑑𝑥
𝑥
=
2𝑑𝑦
(1 − 𝑚)𝑦
=
𝑑𝑢
𝑚𝑢
=
𝑑𝑣
(𝑚 − 1)𝑣
=
𝑑𝑈
𝑚𝑈
 Step 5: Solving the above relations by a method of characterestics, we get the following set of equations:
𝑚𝑓 +
𝑚 − 1
2
𝜉𝑓′
+ 𝑔′
= 0
𝑚𝑓2
+
𝑚 − 1
2
𝜉𝑓𝑓′
+ 𝑔𝑓′
= 𝑓" + 𝑚𝑘′
With Boundary Conditions:
𝑓 0 = 0; 𝑔 0 = 0; 𝑓 ∞ = 𝑘

Case Study- Fractional Fick’s Law 7
07-10-2022
 The traditional Fick’s law of diffusion was modified using fractional Derivatives and a sacling similarity approach was used to find
the analytical solution of the problem.
 The modified equation was given by:
Ω∗𝐷𝑡
𝛽
𝑢 𝑥, 𝑡 =
𝜕[𝑢𝑚 𝑥, 𝑡 ]𝑥 𝑅𝜃1
𝛼
𝑢 𝑥, 𝑡 + 𝑢𝑚 𝑥, 𝑡 𝑥𝑅𝜃2
𝛾
𝑢 𝑥, 𝑡
𝜕𝑥
Here, ∗𝐷𝑡
𝛽
is the 𝛽 order Caputo derivative with respect to time 𝑥 𝑅1
𝛼
is the Reisz Feller derivative with 𝜃=1 of order 𝛼 and 𝑥 𝑅0
𝛼
is the Reiss Feller
derivative with 𝜃=0 or the simple Reisz derivative. If Ω∗=1, we the above equation becomes simplified as..
∗𝐷𝑡
𝛽
𝑢 𝑥, 𝑡 =
𝜕[𝑢𝑚 𝑥,𝑡 ]𝑥 𝑅1
𝛼𝑢 𝑥,𝑡 +𝑢𝑚 𝑥,𝑡 𝑥𝑅0
𝛾
𝑢 𝑥,𝑡
𝜕𝑥
⟶ (3)

07-10-2022
We take,
0 < 𝛼 ≤ 1 and 1 ≤ 𝛾 ≤ 2
With respect to the value of 𝛽 we have two cases.
If 0 < 𝛽 ≤ 1 then it is called the fractional diffusion equation and if 1 < 𝛽 ≤ 2 it is called the fractional wave equation.

07-10-2022
Now, in order to make this PDE into an ODE, we introduce the transformations:
𝑢 𝑥, 𝑡 = 𝑢 𝑥, 𝑡 = 𝑡𝑎
𝑈(𝜂) and 𝜂 = 𝑥𝑡−𝑏
.
a and b are arbitrary and must be obtained from the invariance conditions.
Now to solve each fractional derivative separately

07-10-2022
Lets start with the Caputo Derivative,
By substituting the Caputo operator in the LHS of equation (6) and by applying the similarity transforms and performing some manipulations and
chain rules we obtain the following expression for the LHS:
𝑡𝑎−𝑏
1 − 𝛽 + 𝑎 − 𝑏𝜂
𝑑
𝑑𝜂
𝐹𝑎,𝑏
𝑈 𝜂
Where, 𝐹𝑎,𝑏
𝑈 𝜂 is given by:
𝐹𝑎,𝑏𝑈 𝜂 =
1
Γ(1 − 𝛽) 0
1
1 − 𝜏 𝛽𝜏𝛼𝑈 𝜂𝜏−𝑏 𝑑𝜏

07-10-2022
For the Reiss Feller Derivative with 𝜃 = 1, By feeding the value of 𝑢 𝑥, 𝑡 and by applying the similarity transform, we obtain the
following expression for the Reiss Feller Derivative:
𝑥 𝑅1
𝛼
𝑢 𝑥, 𝑡 =
𝑡𝑎−𝑏𝛼
2 sin
𝜋𝛼
2
𝜂−𝛼 𝑅𝛼+1𝑈 𝜂 − 𝐺𝛼+1𝑈 𝜂

07-10-2022
The notations 𝑅𝛼 and 𝐺𝛼 are given by:
𝑅𝛼𝑈 𝜂 =
1
Γ(1 − 𝛼) −∞
1
1 − 𝑠 −𝛼
𝑈 𝜂𝑠 𝑑𝑠
𝐺𝛼 =
1
Γ(1 − 𝛼) 1
∞
𝑠 − 1 −𝛼𝑈 𝜂𝑠 𝑑𝑠

07-10-2022
To evaluate, 𝑥𝑅0
𝛾
𝑢(𝑥, 𝑡) = 𝑅𝑥
𝛾
𝑢(𝑥, 𝑡) (if 𝜃 = 0, the Reisz Fellar derivative is the same as the Reisz derivative:
𝑅𝑥
𝛾
𝑢 𝑥, 𝑡 = −
𝑡𝑎−𝑏𝛾
2 cos
𝜋𝛾
2
𝑑2
𝑑𝜂2
[𝜂2−𝛾(𝐿𝛾𝑈 𝜂 + 𝑁𝛾𝑈 𝜂 )]

07-10-2022
Where 1 < 𝛾 ≤ 2
Where,
𝐿𝛾𝑈 𝜂 =
1
Γ(2 − 𝛼) −∞
1
1 − 𝑠 1−𝛾
𝑁𝛾𝑈 𝜂 =
1
Γ(2 − 𝛼) 1
∞
𝑠 − 1 1−𝛾

07-10-2022
Substituting the different fractional derivative and similarity terms in equation (3), we get
𝑡𝑎−𝛽 1 − 𝛽 + 𝑎 − 𝑏𝜂
𝑑
𝑑𝜂
𝐹𝛽
𝑎,𝑏
𝑈 𝜂
=
𝑡𝑎 1+𝑚 −𝑏 1−𝛼
2 sin
𝛼𝜋
2
𝑑
𝑑𝜂
[𝑈𝑚
𝜂 ]𝜂−𝛼
𝑅𝛼+1𝑈 𝜂 − 𝐺𝛼+1𝑈 𝜂
−
𝑡𝑎 1+𝑚 −𝑏𝛾
2 cos
𝜋𝛾
2
[𝑈𝑚(𝜂)]
𝑑2
𝑑𝜂2
[𝜂2−𝛾(𝐿𝛾𝑈 𝜂 + 𝑁𝛾𝑈 𝜂 )] ⟶ (4)

07-10-2022
Now, in order to find the analytical solution to (11), we use the following approach:
We put:
𝑈 𝜂 = 𝐴𝜂𝜅
Where the parameters A and 𝜅 are real constants.
If m=0, the equation loses its fractional nature so it is considered a special case, therefore:
If m ≠ 0, we obtain the following solution

07-10-2022
If m ≠ 0, we obtain the following solution
A=
2 sin
𝛼𝜋
2
Γ −1−𝑘 Γ 1+𝑎−𝑏𝑘
(1− −1 𝛼Γ 𝛼−𝑘 Γ(1+𝑎−𝛽−𝑏𝑘)
1
𝑚
Hence,
𝑢(𝑥, 𝑡) =
2 sin
𝛼𝜋
2
Γ −1−𝑘 Γ 1+𝑎−𝑏𝑘
(1− −1 𝛼Γ 𝛼−𝑘 Γ(1+𝑎−𝛽−𝑏𝑘)
1
𝑚
𝑥
𝑡
𝛽
1+𝛼
1+𝛼
𝑚

07-10-2022
Or, after further manipulations, the solution can be written as:
𝑢(𝑥, 𝑡) =
exp(
𝑖𝜋𝛼
2
)Γ −1 − 𝑘 Γ 1 − 𝛽/𝑚
(1 − −1 𝛼Γ 𝛼 − 𝑘 Γ(1 − 𝛽 − 𝛽/𝑚)
1
𝑚
The solution is only defined for t > 0 and x ∈ 0, ∞ .We thus define its value at the origin as u(0,0) = 0.

07-10-2022
• Now different cases of the fractional wave and diffusion equation depending on the 𝛽 values are plotted. The values of u were plotted with
varying x and t for different fractional parameters, 𝛽. 𝛾 and m were kept constant at 1.5 and 2 respectively.
• Then there were special cases where 𝛼 and/or 𝛽 were kept integer. The plots with varying x and t were obtained.
• The results obtained in the paper were verified by plotting the equations on MATLAB. There were variations in the orders of the plot (greater
than/lesser than nature) with respect to varying 𝛽 but the trajectory of the plot looked the same for both varying x and t.

Further Directions…. 20
07-10-2022
• Study of different methodologies to compute the solutions of fractional differential equations
• Application of similarity transform to systems of equations
• Exploring numerical methods to solve those systems of equations.
• Application of fractional calculus in Heat and Mass Transfer problems.

BKS-MidsemPPT.pptx

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