# Fourier Analysis Questions and Answers – Solution of 1D Heat Equation

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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) ut = c2uxx
b) ut = puxx
c) utt = c2uxx
d) ut = – c2uxx

Explanation: The one-dimensional heat equation is given by ut = c2uxx where c is the constant and ut represents the one time partial differentiation of u and uxx 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

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

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 = – p2.

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) ux(0,t) = 0 and ux(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) ux(0,t) = ux(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

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) ux(0,t) = 0 and ux(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) =$$\frac{l^2}{3}/2 + ∑cos⁡(\frac{nπx}{l}) e^{\frac{-c^2 n^2 π^2 t}{l^2}} \frac{4l^2}{(2m)^2+π^2}$$

Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e-p2 c2t)
ux = (-cp sinpx + c’p cospx) (c’’ e-p2 c2t)
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) ux(0,t) = 0 = ux(30,t) and u(x,0) = 20 + 60/30 x
c) ut(0,t) = 0 = ut(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

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

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 ut = uxx 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)

Explanation: u(x,t) = (c cospx + c’ sinpx) (c’’ e-p2t)
When x=0, c=0 and when x=1, p=nπ.
When t=0, 3 sin (nπx) = $$∑_{n=1}^∞$$ bn e-n2 π2 t sin⁡(nπx)
Therefore bn =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) ux(0,t) = 0 = ux(l,t)
b) ut(0,t) = 0 = ut(l,t)
c) u(0,t) = 0 = u(l,t)
d) uxx(0,t) = 0 = uxx(l,t)

Explanation: Since the ends are insulated no heat can flow through the ends of the bar. Therefore ux(0,t) = 0 = ux(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 + 52 x
b) 30 + 52 x
c) 30 + 2x
d) 20 + 2x

a = 30 and b = $$\frac{(80-30)}{20} = \frac{5}{2}$$
Therefore, u= 30 + $$\frac{5}{2} x$$.