Computational Fluid Dynamics Interview Questions and Answers on “Classification of PDE – 2”.
1. What are the two methods used to find the type of PDEs?
a) Lagrangian Method and Eulerian method
b) Cramer’s method and Eulerian method
c) Cramer’s method and Lagrangian Method
d) Cramer’s method and Eigenvalue method
Answer: d
Clarification: Partial differential equations can be classified using their characteristic lines. These are located using either the Cramer’s method or the Eigenvalue method.
2. Let u be a variable dependent on x and y. In the diagram, (frac{du}{dy}, represents, frac{partial u}{partial y}).
What does this line in the diagram represent?
a) Characteristic line
b) Eigenvalue line
c) Lagrange line
d) Cramer line
Answer: a
Clarification: Characteristic lines are those where the derivatives of the dependent variable are indeterminate. In the diagram, (du/dy) is indeterminate and hence the line represents a characteristic line.
3. How the type of PDE is identified using Cramer’s rule?
a) By equating the Cramer’s denominator to 1
b) By equating the Cramer’s numerator to 1
c) By equating the Cramer’s denominator to 0
d) By equating the Cramer’s numerator to 0
Answer: c
Clarification: The denominator of Cramer’s solution is equated to zero to find the type of PDE. The denominator is equated to zero to make the solution indeterminate.
4. What are the Cramer’s solutions equated to while using Cramer’s method of classifying a PDE?
a) The dependent variables
b) The derivatives of dependent variables
c) The second derivatives of dependent variables
d) The highest derivatives of dependent variables
Answer: b
Clarification: For characteristic lines, the derivatives of dependent variables are zero. Cramer’s rule is used to find these derivatives and then it is made indeterminate to find the type of PDE.
5. _________ of the characteristic curves is used to find the type of PDE.
a) Starting point
b) Centre
c) Length
d) Slope
Answer: d
Clarification: The nature of the slope of the characteristic curves gives the nature of the characteristic curves. This directly matches with the type of the PDE too.
6. What is the Cramer’s numerator when the solution is the derivative of dependent variables?
a) any negative value
b) any positive value
c) 1
d) 0
Answer: d
Clarification: The solution should be indeterminate. Therefore, if the denominator is 0, the numerator must also be zero. (0/0=indeterminate).
7. Consider the following system of PDEs.
(a_1frac{partial u}{partial x}+b_1frac{partial u}{partial y}+c_1frac{partial v}{partial x}+d_1frac{partial v}{partial y}=0)
(a_2frac{partial u}{partial x}+b_2frac{partial u}{partial y}+c_2frac{partial v}{partial x}+d_2frac{partial v}{partial y}=0)
The Eigenvalues of which of these matrices can be used to classify this system of PDEs?
a) (begin{bmatrix}
a_1 & b_1 \
a_2 & b_2
end{bmatrix}^{-1})
(begin{bmatrix}
c_1 & d_1 \
c_2 & d_2
end{bmatrix})
b) (begin{bmatrix}
a_1 & c_1 \
a_2 & c_2
end{bmatrix}^{-1})
(begin{bmatrix}
b_1 & d_1 \
b_2 & d_2
end{bmatrix})
c) (begin{bmatrix}
a_1 & c_1 \
a_2 & c_2
end{bmatrix})
(begin{bmatrix}
b_1 & d_1 \
b_2 & d_2
end{bmatrix})
d) (begin{bmatrix}
a_1 & b_1 \
a_2 & b_2
end{bmatrix})
(begin{bmatrix}
c_1 & d_1 \
c_2 & d_2
end{bmatrix})
Answer: b
Clarification: From the given system of PDEs,
(a_1frac{partial u}{partial x}+b_1frac{partial u}{partial y}+c_1frac{partial v}{partial x}+d_1frac{partial v}{partial y}=0)
(a_2frac{partial u}{partial x}+b_2frac{partial u}{partial y}+c_2frac{partial v}{partial x}+d_2frac{partial v}{partial y}=0)
Let, (W=begin{bmatrix}
u \
v
end{bmatrix}) then
(begin{bmatrix}
a_1 & c_1 \
a_2 & c_2
end{bmatrix}
frac{partial W}{partial x}+
begin{bmatrix}
b_1 & d_1 \
b_2 & d_2
end{bmatrix}
frac{partial W}{partial y}=0 )
(frac{partial W}{partial x}+
begin{bmatrix}
a_1 & c_1 \
a_2 & c_2
end{bmatrix}^{-1}
begin{bmatrix}
b_1 & d_1 \
b_2 & d_2
end{bmatrix} frac{partial W}{partial y}=0)
The Eigenvalues of (begin{bmatrix}
a_1 & c_1 \
a_2 & c_2
end{bmatrix}^{-1})
(begin{bmatrix}
b_1 & d_1 \
b_2 & d_2
end{bmatrix}) determines the class of PDE.
8. The Eigenvalues in the Eigenvalue method are ____________
a) the type of the characteristic lines
b) the type of PDE
c) the slope of the characteristic lines
d) the slope of PDE
Answer: c
Clarification: The Eigenvalues will give you the slope of the characteristic lines. Using this slope, we determine the type of the characteristic line and the type of PDE.
9. When the Eigenvalues are a mixture of real and imaginary values, the PDE is ___________
a) elliptic-hyperbolic
b) parabolic
c) elliptic
d) hyperbolic
Answer: a
Clarification: When we get a mixed type of Eigenvalues, the type of PDE is also mixed. Many practical equations have mixed behaviour also.
10. Solutions of a system of PDEs can be obtained by equating the numerator of Cramer’s solution while using Cramer’s rule. This method is used by __________
a) Integral transform
b) Change of variables
c) Separation of variables
d) Method of characteristics
Answer: d
Clarification: The Cramer’s numerator is equated to zero and compatibility equations are obtained. These are then solved to get the solution of PDEs.
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