Numerical Methods for Partial Differential Equations (1st Edition, 2015) – Solutions – Mazumder

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,© Sandip Mazumder, 2015


Solution to Exercise Problems in Chapter 1
Exercise 1.1

Classify the following partial differential equations into the three canonical types or a
combination thereof. In each case, φ is the dependent variable.

∂φ
a) + U∇φ =0 , where U is a vector (such as velocity).
∂t
∂φ
b) + U∇φ = ∇(Γ∇φ ) , where Γ is a positive real number.
∂t
∂  ∂φ ∂φ  ∂  ∂φ ∂φ 
c)  Γ1 − Γ2  + Γ1 − Γ2 = 0 , where Γ1 and Γ 2 are real numbers,
∂x  ∂x ∂y  ∂y  ∂y ∂x 
either positive or negative.
∂φ 1 ∂  ∂φ 
d) − r = 0
∂t r ∂r  ∂r 
∂ 2φ ∂φ
e) + − ∇(Γ∇φ ) = φ , where Γ is a positive real number.
∂t 2 ∂t


Solution

Part 1(a)

Since U is a vector, in Cartesian coordinates, it may be written as
U = u ˆi + v ˆj + w kˆ .
Using Eq. (B.2a) and the above equation, we obtain
 ∂φ ∂φ ˆ ∂φ ˆ  ∂φ ∂φ ∂φ
( )
U∇φ= u ˆi + v ˆj + w kˆ  ˆi +
 ∂x ∂y
j + k = u
∂z  ∂x
+v
∂y
+w
∂z
.

Therefore, the governing PDE may be written as
∂φ ∂φ ∂φ ∂φ
+u +v +w = 0.
∂t ∂x ∂y ∂z
The above equation is the three-dimensional Euler equation [cf. Eq. (1.7)]. Since it has
no second derivatives, it cannot be compared to the standard PDE form shown in Eq.
(1.1). Therefore, we differentiate it with respect to time to yield
∂ 2φ ∂ 2φ ∂ 2φ ∂ 2φ ∂u ∂φ ∂v ∂φ ∂w ∂φ
+u +v +w + + + =0.
∂t 2 ∂t ∂x ∂t ∂y ∂t ∂z ∂t ∂x ∂t ∂y ∂t ∂z
Selecting t and one of x , y or z as the other independent variable, and comparing the
above equation with Eq. (1.1), we obtain A = 1 , B = u or v or w , and C = 0 , which yields
B 2 − 4 AC =
u 2 or v 2 or w2 . Thus, the PDE is hyperbolic.

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,© Sandip Mazumder, 2015




Part 1(b)

Following the same procedure as Part 1(a), the PDE may be written in Cartesian
coordinates as
∂φ ∂φ ∂φ ∂φ ∂  ∂φ  ∂  ∂φ  ∂  ∂φ 
+u +v +w = Γ  + Γ  + Γ 
∂t ∂x ∂y ∂z ∂x  ∂x  ∂y  ∂y  ∂z  ∂z 
.
 ∂ 2φ ∂ 2φ ∂ 2φ  ∂Γ ∂φ ∂Γ ∂φ ∂Γ ∂φ
=
Γ 2 + 2 + 2  + + +
 ∂x ∂y ∂z  ∂x ∂x ∂y ∂y ∂z ∂z
Comparison with Eq. (1.1) yields A = Γ , B = 0 , and C = Γ . Thus, B 2 − 4 AC =−4Γ 2 , and
the PDE is elliptic.

Part 1(c)

Differentiating through, the governing PDE may be written as

∂ 2φ ∂ 2φ ∂ 2φ ∂φ ∂Γ1 ∂φ ∂Γ 2 ∂φ ∂Γ1 ∂φ ∂Γ 2
Γ1 − 2Γ + Γ + − + − =
0.
∂x 2 ∂x∂y ∂y 2 ∂x ∂x ∂y ∂x ∂y ∂y ∂x ∂y
2 1




Comparison with Eq. (1.1) yields A = Γ1 , B =−2Γ 2 , and C = Γ1 . Therefore,
B 2 − 4 AC = 4Γ 22 − 4Γ12 = 4(Γ 22 − Γ12 ) . Hence, the type of PDE is dictated by the relative
magnitudes of the coefficients Γ1 and Γ 2 .

If Γ1 =Γ 2 , the PDE is parabolic.

If Γ1 > Γ 2 , the PDE is hyperbolic.

If Γ1 < Γ 2 , the PDE is elliptic.


Part 1(d)
We first rewrite the governing equation as

∂φ 1 ∂  ∂φ  ∂φ ∂ 2φ 1 ∂φ
− r = − − =0
∂t r ∂r  ∂r  ∂t ∂r 2 r ∂r

Comparison with Eq. (1.1) yields A = −1 , B = 0 , and C = 0 . Therefore, B 2 − 4 AC =
0 , and
the PDE is parabolic.



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, © Sandip Mazumder, 2015


Part 1(e)
Expanding in Cartesian coordinates, the governing PDE may be written as

∂ 2φ ∂φ ∂  ∂φ  ∂  ∂φ  ∂  ∂φ 
+ − Γ  − Γ  − Γ  = φ.
∂t 2 ∂t ∂x  ∂x  ∂y  ∂y  ∂z  ∂z 

Differentiating through, we obtain

∂ 2φ ∂φ ∂ 2φ ∂ 2φ ∂ 2φ ∂Γ ∂φ ∂Γ ∂φ ∂Γ ∂φ
+ − Γ − Γ − Γ − − − =φ .
∂t 2 ∂t ∂x 2 ∂y 2 ∂z 2 ∂x ∂x ∂y ∂y ∂z ∂z
In this case, we have 4 independent variables. Therefore, the type of PDE will depend on
which 2 of the 4 independent variables are considered for classification purposes.
If t and one of x , y or z is considered the other independent variable, then A = 1 ,
B = 0 , and C = −Γ . Hence, B 2 − 4 AC =
4Γ . Since Γ is a positive real number, the PDE is
hyperbolic.
If t is not considered as one of the independent variables, then A = −Γ , B = 0 , and
C = −Γ . Hence, B 2 − 4 AC =−4Γ 2 , which makes the PDE is elliptic.

To summarize, the PDE is hyperbolic in time, and elliptic in space.




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