This set of Fourier Analysis Interview Questions and Answers for Experienced people focuses on “Solution of 1D Heat Equation”.

1. The partial differential equation of 1-Dimensional heat equation is ___________

a) u_{t} = c^{2}u_{xx}

b) u_{t} = pu_{xx}

c) u_{tt} = c^{2}u_{xx}

d) u_{t} = – c^{2}u_{xx}

View Answer

Explanation: The one-dimensional heat equation is given by u

_{t}= c

^{2}u

_{xx}where c is the constant and u

_{t}represents the one time partial differentiation of u and u

_{xx}represents the double time partial differentiation of u.

2. When using the variable separable method to solve a partial differential equation, then the function can be written as the product of functions depending only on one variable. For example, U(x,t) = X(x)T(t).

a) True

b) False

View Answer

Explanation: When solving a partial differential equation using a variable separable method, then the function can be written as the product of functions depending on one variable only.

3. The one dimensional heat equation can be solved using a variable separable method. The constant which appears in the solution should be __________

a) Positive

b) Negative

c) Zero

d) Can be anything

View Answer

Explanation: Since the problems are dealing on heat conduction, the solution must be a transient solution. Therefore the constant should be negative, i.e., k = – p

^{2}.

4. When solving the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:

i) u(x,t) is finite for t tends to infinite

ii) u_{x}(0,t) = 0 and u_{x}(l,t) = 0

iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l, which condition is the best to use in the first place?

a) u_{x}(0,t) = u_{x}(l,t) = 0

b) u(x,t) = x(l-x) for t=0 between x=0 and x=l.

c) u(x,t) = x(l-x) for x=0 between t=0 and t=l.

d) u(0,t) = u(l,t) = 0

View Answer

Explanation: Boundary conditions are always used first to solve the partial differential equations. Using these boundary conditions, we can remove one constant thus making only one constant remaining to remove. The last constant is removed using the initial conditions.

5. Solve the 1-Dimensional heat equation for the conduction of heat along the rod without radiation with conditions:

i) u(x,t) is finite for t tends to infinite

ii) u_{x}(0,t) = 0 and u_{x}(l,t) = 0

iii) u(x,t) = x(l-x) for t=0 between x=0 and x=l.

a) U(x,t) =\(\frac{l^2}{3}/2 + ∑cos(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{-4l^2}{(2m)^2+π^2} \)

b) U(x,t) =\(\frac{l^2}{3} + ∑cos(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{-4l^2}{(2m)^2+π^2} \)

c) U(x,t) =\(\frac{l^2}{3} + ∑cos(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{4l^2}{(2m)^2+π^2} \)

d) U(x,t) =\(\frac{l^2}{3}/2 + ∑cos(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{4l^2}{(2m)^2+π^2} \)

View Answer

Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e

^{-p2}c

^{2}t)

u

_{x}= (-cp sinpx + c’p cospx) (c’’ e

^{-p2}c

^{2}t)

Applying the first condition of second condition,

C’ = 0

Now applying the second condition of the second condition,

\(p= \frac{nπ}{l} \)

Now we have only one constant left. This can be solved using the third condition.

U(x,t) =\( \frac{a_0}{2} + ∑cos(\frac{nπx}{l}) a_n e^{\frac{-c^2 n^2 π^2 t}{l^2}} \)

\(a_0 = \frac{2}{l} ∫_0^l x(l-x)dx = \frac{l^2}{3} \)

\(a_n = \frac{2}{l} ∫_0^l x(l-x) cosnx dx = \frac{-4l^2}{(2m)^2+π^2}. \) \(U(x,t) = \frac{l^2}{3}/2 + ∑cos(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{-4l^2}{(2m)^2+π^2}.\)

6. A rod of 30cm length has its ends P and Q kept 20°C and 80°C respectively until steady state condition prevail. The temperature at each point end is suddenly reduced to 0°C and kept so. Find the conditions for solving the equation.

a) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/10 x

b) u_{x}(0,t) = 0 = u_{x}(30,t) and u(x,0) = 20 + 60/30 x

c) u_{t}(0,t) = 0 = u_{t}(30,t) and u(x,0) = 20 + 60/10 x

d) u(0,t) = 0 = u(30,t) and u(x,0) = 20 + 60/30 x

View Answer

Explanation: 0 and 30 are the end points. So, at these points the function is zero. Hence u(0,t) = 0 = u(30,t). Next due to steady state conditions, at the beginning that is initial conditions, we have u(x,0) = 20 + 60/30 x.

7. Is it possible to have a solution for 1-Dimensional heat equation which does not converge as time approaches infinity?

a) Yes

b) No

View Answer

Explanation: It is not possible to have a solution which does not converge as time approaches infinity because the solution to a heat equation must be transient.

8. Solve the equation u_{t} = u_{xx} with the boundary conditions u(x,0) = 3 sin (nπx) and u(0,t)=0=u(1,t) where 0<x<1 and t>0.

a) \(3∑_{n=1}^∞ \) e^{-n2 π2 t} cos(nπx)

b) \(∑_{n=1}^∞ \) e^{-n2 π2 t} sin(nπx)

c) \(3∑_{n=1}^∞ \) e^{-n2 π2 t} sin(nπx)

d) \(∑_{n=1}^∞ \) e^{-n2 π2 t} cos(nπx)

View Answer

Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e

^{-p2}t)

When x=0, c=0 and when x=1, p=nπ.

When t=0, 3 sin (nπx) = \(∑_{n=1}^∞ \) b

_{n}e

^{-n2 π2 t}sin(nπx)

Therefore b

_{n}=3 for all n

Hence the solution is \(3∑_{n=1}^∞ \) e

^{-n2 π2 t}sin(nπx).

9. If two ends of a bar of length l is insulated then what are the conditions to solve the heat flow equation?

a) u_{x}(0,t) = 0 = u_{x}(l,t)

b) u_{t}(0,t) = 0 = u_{t}(l,t)

c) u(0,t) = 0 = u(l,t)

d) u_{xx}(0,t) = 0 = u_{xx}(l,t)

View Answer

Explanation: Since the ends are insulated no heat can flow through the ends of the bar. Therefore u

_{x}(0,t) = 0 = u

_{x}(l,t).

10. The ends A and B of a rod of 20cm length are kept at 30°C and 80°C until steady state prevails. What is the condition u(x,0)?

a) 20 + ^{5}⁄_{2} x

b) 30 + ^{5}⁄_{2} x

c) 30 + 2x

d) 20 + 2x

View Answer

Explanation: u

_{xx}=0.

The solution to this equation is u=a+bx. Since at x=0, u=30 and at x=20, u=80,

a = 30 and b = \(\frac{(80-30)}{20} = \frac{5}{2} \)

Therefore, u= 30 + \(\frac{5}{2} x\).

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