Differential

Transform

Method

Meet ShahU12ME194

MED,SVNITu12me194@med.svnit.ac.in

Guided by:Dr. H.B.MehtaMED,SVNIT

Head of the Department

AcknowledgementThis internship provided a learning experience in the field of research.I acquired a fewimportant skills during the intern.I would like to thank my guide Dr. Hemant B. Mehta, MED, SVNIT for providinginsights as and when needed during the internship.I would also like to thankMechanical Engineering Department, SVNIT for providing this internship opportunity.

1 Introduction to DTM

The differential transformation method (DTM) is one of the numerical methods inordinary differential equations, partial differential equations and integral equations.Since proposed by Zhou(1986), there are tremendous interests on the applications ofthe DTM to solve various scientific problems.The method provides the solution interms of convergent series with easily computable components. The DTM,however,doeshave some drawback.By using the DTM, we obtain a truncated series solution. Thistruncated solution does not exhibit the real behaviours of the problem but, in mostcases, provides an excellent approximation of the true solution in a very small region.

1.1 Description of DTM

It is well known that if a function u is innitely continuously differentiable, then u canbe expressed in Taylor series as:

u(x) =∞∑k=0

1

k!

dku(x0)

dxk(x− x0)k (1)

We dene the differential transform (DT) of u of order k, denoted by U(k), by

U(k) =1

k!

dku(x0)

dxk(2)

In order to solve a given ODE by differential transform, we make use of the differentialtransform of order k in Equation (2).The differential inverse transform of U(k) is denedas follows:

u(x) =∞∑k=0

U(k)(x− x0)k (3)

In real applications, the function u(x) is expressed by a nite series and Equation (3)can be truncated, and will be denoted by as:

uK(x) =K∑k=0

U(k)(x− x0)k (4)

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1.2 Basic Operations of DTM 1-D

The following basic operations of differential transformation can be deduced fromequations (3) and (4):

1. If t(x) =r(x)±p(x) then, T (k) =R(k)±P(k).

2. If t(x) =αr(x) then, T (k) =αR(k).

3. If t(x) =dr(x)dx then, T (k) = (k+1)R(k+1).

4. If t(x) =d2r(x)dx2

then,T(k) = (k+1)(k+2)R(k+2)

5. If t(x) =dbr(x)dxb

then,T(k) = (k+1)(k+2)..(k+b)R(k+b)

6. If t(x) =r(x)p(x)then,T(k) =∑kl=0 P(l)R(k-l)

7. If t(x) = xb then, T (k) = δ(k-b)

8. If t(x) = exp(λx) then, T (x) = λk

k!

9. If t(x) = (1 + x)b then, T (x) = b(b−1)...(b−k+1)k!

10. If t(x) = sin(jx+ α) then, T (k) = jk

k! sin(πk2 + α)

11. If t(x) = cos(jx+ α) then, T (k) = jk

k! cos(πk2 + α)

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2 Examples

1.

dx

dt− 2x+ 3y = 0

dy

dt+ 2x− 3y = 0

x(0) = 8, y(0) = 3

DTM:(k + 1)X(k + 1) = 2X(k)− 3Y (k)

(k + 1)Y (k + 1) = Y (k)− 2X(k)

Transform of initial conditions:

X(0) = 8, Y (0) = 3

By substituting the above values we get,

k = 0→ X(1) = 7, Y (1) = −13

k = 1→ X(2) = 53/2, Y (2) = −27/2

k = 1→ X(3) = 187/6, Y (3) = −135/6

Therefore,

x(t) =3∑

k=0

X(k)tk

y(t) =3∑

k=0

Y (k)tk

2.

d2T

dx2+ 100 = 0

x = 0→ T = 100, x = 1→ T = 50

DTM:(k + 1)(k + 2)U(k + 2) + δ(k)100 = 0

U(0) =d0T (0)

0!dx0= 100

By substituting k=0 in the transformed equation,

U(2) = −50

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U(k) = 0,∀k ≥ 3

Solution:

T (x) =∞∑k=0

U(k)xk

T (x) = U(0)x0 + U(1)x1 + U(2)x2

T (x) = 100 + U(1)x1 − 50x2

T (1) = 100 + U(1)− 50 = 50

U(1) = 0

T (x) = 100− 50x2

3.

d2T

dx2+ 100 = 0

x = 0→ T = 100, x = 1→ dT

dx= 0

DTM:(k + 1)(k + 2)U(k + 2) + δ(k)100 = 0

U(0) =d0T (1)

0!dx0= T (1), x0 = 1

U(1) =dT (1)

dx= 0

By substituting k=0 in the transformed equation,

U(2) = −50

U(k) = 0,∀k ≥ 3

Solution:

T (x) =∞∑k=0

U(k)(x− x0)k

T (x) = U(0) + U(1)(x− x0) + U(2)(x− x0)2

T (x) = T (1)− 50x2

T (0) = T (1)− 50 = 100

T (1) = 150

T (x) = 150− 50(x− 1)2

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4.

d2T

dx2+ 100 = 0

x = 0→ T = 100, x = 1→ dt

dx= −25

DTM:(k + 1)(k + 2)U(k + 2) + δ(k)100 = 0

U(0) =d0T (1)

0!dx0= T (1), x0 = 1

U(1) =dT (1)

dx= −25

By substituting k=0 in the transformed equation,

U(2) = −50

U(k) = 0,∀k ≥ 3

Solution:

T (x) =∞∑k=0

U(k)(x− x0)k

T (x) = U(0) + U(1)(x− x0) + U(2)(x− x0)2

T (x) = T (1)− 25(x− 1)− 50x2

T (0) = T (1) + 25− 50 = 100

T (1) = 125

T (x) = 125− 25(x− 1)− 50(x− 1)2

5.

d

dx(KdT

dx) + 100 = 0

K = K0(1 + βT ),K0 = 0.85, β = 0.0007

T (0) = 100, T (1) = 50

We can write the above equation as:

Kd2T

dx2+dT

dx

dK

dx+ 100 = 0

→ K0(1 + βT )d2T

dx2+dT

dx

d(K0 +K0βT )

dx+ 100 = 0

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K0d2T

dx2+K0β

Td2T

dx2+K0β(

dT

dx)2 + 100 = 0

DTM:

K0(m+ 1)(m+ 2)U(m+ 2) +K0β[m∑r=0

(m− r + 1)(m− r + 2)U(r)U(m− r + 2)]

+K0β[m∑r=0

(r + 1)(m− r + 1)U(r + 1)U(m− r + 1)] + δ(m)100 = 0

for m=0 we get,

2K0U(2) +K0β[2U(0)U(2)] +K0β[U(1)U(1)] + δ(0)100 = 0

Thus,2K0U(2) + 200K0βU(2) +K0βU(1)2 + 100 = 0

T (x) =∞∑k=0

U(k)xk

Neglecting the higher terms,

T (x) = U(0) + U(1) + U(2)

T (x) = 100 + U(1) + U(2)

U(2) =−100−K0βU(1)2

2K0 + 200K0β

Therefore,

−50 = U(1) +−100−K0βU(1)2

2K0 + 200K0β

−50 = −3.27× 10−4U(1)2 + U(1)− 54.975

3.27× 10−4U(1)2 − U(1) + 4.975 = 0

By solving the above equation: U(1) = 4.983;U(2) = 54.983Therefore,

T (x) = 100 + 4.983x− 54.983x2

The above results have been compared with actual solutions in Table 1.6.

d

dx(KdT

dx) + 100 = 0

K = K0(1 + βT 2),K0 = 60, β = 0.000025

T (0) = 100,dT

dx |x=1= 0

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Table 1: Comparison with actual solution

x DTM Value Actual value Percentage error

0 100 100 -1.42109E-140.05 100.1116925 100.1522052 0.0404511080.1 99.94847 100.0295489 0.0810549660.2 98.79728 98.95923887 0.1636622030.3 96.54643 96.78675462 0.2483032130.4 93.19592 93.50737673 0.3330825260.5 88.74575 89.11392997 0.4131564760.6 83.19592 83.59670394 0.4794255290.7 76.54643 76.94334382 0.5158520560.8 68.79728 69.13870836 0.4938309760.9 59.94847 60.16469113 0.3593820941 50 50 1.56319E-13

We can write the above equation as,

Kd2T

dx2+dT

dx

dK

dx+ 100 = 0

→ K0(1 + βT 2)d2T

dx2+dT

dx

d(K0 +K0βT2)

dx+ 100 = 0

K0d2T

dx2+K0βT

2d2T

dx2+ 2K0βT (

dT

dx)2 + 100 = 0

K0(m+ 1)(m+ 2)U(m+ 2) +K0β[m∑r=0

U(r)(m−r∑i=0

U(i)W (m− r − i)]

+2K0β[m∑r=0

U(r)(m−r∑i=0

V (i)V (m− r − i))] + δ(m)100 = 0

K0(m+ 1)(m+ 2)U(m+ 2)

+K0β[m∑r=0

m−r∑i=0

(m− r − i+ 1)(m− r − i+ 2)U(r)U(i)U(m− r − i+ 2)]

+2K0β[m∑r=0

m−r∑i=0

(i+ 1)(m− r − i+ 1)U(r)U(i+ 1)U(m− r − i+ 1) + δ(m)100 = 0

for m=0 we get,

2K0U(2) +K0β[2U(0)2U(2)] + 2K0β[U(0)U(1)2] = 0

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U(1) =dT (1)

dx |x=1= 0

Therefore,

2× 60U(2) + 0 + 60× 0.000025[2U(0)2U(2)] + 100 = 0

120U(2) + 3× 10−3U(0)2U(2) + 100 = 0

U(2) =−100

120 + 3× 10−3U(0)2

T (x) =∞∑k=0

U(k)(x− x0)k

Neglecting higher terms,

T (x) = U(0) + U(1) + U(2)

T (x) = U(0) + U(2)

for x = 0100 = U(0) + U(2)

100 = U(0)− 100

120 + 3× 10−3U(0)2

3× 10−3U(0)3 − 0.3U(0)2 + 120U(0)− 12100 = 0

By solving the equation, U(0) = 100.664, U(2) = −0.66489Therefore, T (x) = U(0) + U(2)(x− 1)2 = 100.664− 0.664(x− 1)2

The above results have been compared with actual solutions in Table 2.

Table 2: Comparison with actual solution

x DTM Value Actual value Percentage error

0 100 100 3.96639E-070.1 100.12616 100.126634 0.0004733780.2 100.23904 100.2398816 0.0008396040.3 100.33864 100.3397682 0.0011243320.4 100.42496 100.4263027 0.0013369870.5 100.498 100.4994882 0.0014808010.6 100.55776 100.5593731 0.0016041660.7 100.60424 100.6059309 0.0016807160.8 100.63744 100.6391812 0.0017300970.9 100.65736 100.6591292 0.0017576581 100.664 100.6657783 0.001766565

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3 DTM 2-D

2-D Differential Transform:

W (k, h) =1

k!h!

[∂k+hw(x, y)

∂xk∂yh

](0,0)

Inverse Differential transform:

w(x, y) =∞∑k=0

∞∑h=0

W (k, h)xkyh

3.1 Basic operations of 2-D DTM

The following basic operations of 2-d differential transformation can be deduced fromabove equations:

1. If w(x, y) = u(x, y)± v(x, y), then W (k, h) = U(k, h)± V (k, h).

2. If w(x, y) = λu(x, y), then W (k, h) = λU(k, h).Here,λ is a constant.

3. If w(x, y) = ∂u(x,y)∂x ,then W (k, h) = (k + 1)U(k + 1, h).

4. If w(x, y) = ∂u(x,y)∂y ,then W (k, h) = (h+ 1)U(k, h+ 1).

5. If w(x, y) = ∂r+su(x,y)∂xr∂ys ,then

W (k, h) = (k + 1)(k + 2)....(k + r)(h+ 1)(h+ 2)....(h+ s)U(k + r, h+ s)

6. If w(x, y) = u(x, y)v(x, y),thenW (k, h) =

∑kr=0

∑ks=0 U(r, h− s)V (k − r, s).

7. If w(x, y) = xmyn,then W (k, h) = δ(k −m,h− n) = δ(k −m)δ(h− n)

8. If w(x, y) = ∂u(x,y)∂x

∂v(x,y)∂x , then

W (k, h) =∑kr=0

∑hs=0(r + 1)(k − r + 1)U(r + 1, h− s)V (k − r + 1, s).

9. If w(x, y) = ∂u(x,y)∂x

v(x,y)∂y ,then

W (k, h) =∑kr=0

∑hs=0(k − r + 1)(h− s+ 1)U(k − r + 1, s)V (r, h− s+ 1).

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3.2 2-D DTM Examples:

1.

∂

∂x

(K∂T

∂x

)+ q = ρCp

∂T

∂t

T = f(x, t)K = K0(1 + βT )t = 0→ T = 75, x = 0→ T = 100, x = 1→ T = 50Failed to find a solution to the above transient problem.

2.

ρ∂uz∂t

= −∂p∂z

+ µ∂2uz∂y2

u(0, t) = 0u(h, t) = uu(y, 0) = 0Failed to find a solution to the above transient problem.

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References

1. J.M.W. Munganga, J.N. Mwambakana, R. Maritz, T.A. Batubenge and G.M.Moremedi (2014) Introduction of the differential transform method to solvedifferential equations at undergraduate level, International Journal ofMathematical Education in Science and Technology, 45:5, 781 - 794

2. Fatma Ayaz, On the two-dimensional dierential transform method, Appl. Math.Comput. 143(2-3) (2003) 361 - 374.

3. New analytical method for solving Burgers and nonlinear heat transfer equationsand comparison with HAM, M.M. Rashidi, E. Erfani / Computer PhysicsCommunications 180 (2009) 1539 - 1544

4. J.K. Zhou, Differential Transformation and Its Application for Electrical Circuits,Huazhong University Press, Wuhan, China, 1986.

5. Two Dimensional differential transform method for solving Non-linear PartialDifferential Method for solving Non-linear Partial Differntial equations, HosseinJafari,Maryam Alipour and Hale Tajadodi,International Journal of Research andReviews in Applied Sciences ISSN:2076-734X,EISSN:2076-7366Volume2,Issue1(January - 2010)

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