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Mathematics (maths) - Applications of Partial Differential Equations - Important Short Objective Questions and Answers: Applications of Partial Differential Equations

**APPLICATIONS
OF PARTIAL DIFFERENTIAL EQUATIONS**

**PART
–A**

**1. What conditions are
assumed in deriving the one dimensional wave equation?**

The wave equation is

In
deriving this equation we make the following assumptions.

(i) The
motion takes place entirely in one plane i.e., XY plane.

(ii) We consider
only transverse vibrations
the horizontal displacement
of the

particles
of the string is negligible.

(iii)The
tension T is constant at all times and at all points of the deflected string.

(iv)
T is considered to be so large compared
with the weight of the string and hence the force of gravity is negligible.

(v) The
effect of friction is negligible.

(vi)
The string is perfectly flexible.

**2.State the wave
equation and give the various solutions of it?**

The
various possible solutions of this equation are

(i) *y*(*x*,*t*)*
*=(* A*_{1}*e ^{px}
*+

(ii) *y*(*x*,*t*)*
*=(* A*_{5}* *cos*
px *+*A*_{6}*
*sin*
px*)(* A*_{7}* *cos* pat *+*A*_{8}*
*sin*
pat*)* *

(iii) *y*(*x*,*
t*)* *=(* A*_{9}*
x *+*A*_{10}*
*)(*
A*_{11}*t *+*A*_{12}*
*)*
*.

**3. Find the nature of
PDE **

Here
*A* =4, *B* =4,*C*=.1

B^{2}-4AC=16-4(4)(1)=0.

Therefore
the equation is Parabolic.

**Classify the equation u _{xx}-y^{4}u_{yy}=2y^{3}u_{y}.
Solution:**

This
is of the form Au_{xx}+Buxy+Cu_{yy}+f(x,y,u,u_{x},u_{y})=0.

Here
A=1, B=0, C=-1.

B^{2}-4AC=0-4(1)(-1)=4>0.

Therefore
the equation is Hyperbolic.

2. **Classify:
x ^{2}uxx+2xyu_{xy}+(1+y^{2}) u_{yy}-2u_{x}=0.
Solution: **

This
is of the form Au_{xx}+Buxy+Cu_{yy}+f(x,y,u,u_{x},u_{y})=0.

Here
A=x^{2}, B=2xy, C=1+y^{2}.

B^{2}-4AC=4x^{2}y^{2}-4(x^{2})(1+y^{2})

=
4x^{2}y^{2}-4 x^{2}-4(x^{2} y^{2})

=-4x^{2}<0.

Therefore
the equation is Elliptic.

3. **A string
is stretched and fastened
to two point l
apart. Motion is started by**

**displacing the string
into the form ***y***
**=*y*^{0}**
**sin
__p__x/* l ***from which it is released at time t=0. Formulate
this problem as the boundary value problem. Solution:**

The displacement y(x,t)
is the solution of the wave equation.

The boundary conditions
are:

4. **What is the constant a ^{2}**

**Solution:**

*u*(*x*,*
t*)* *=(* A*cos* px *+*B *sin* px*)*e*^{-}^{c}^{2}*
^{p}*

6. **State
the governing equation for one dimensional heat equation and necessary
conditions to solve the problem. **

**Solution:**

temperature at time t at a point
distance x from the left end of the rod.

The
boundary conditions are

i)
*u*(0, *t*) =*k* ^{0}_{1}*C
for all t *³0

ii)
*u*(*l*, *t*)
=*k*
^{0}_{2}*C for all t *³0

iii)
the initial condition is *u*(*x*,0)
=*f*
(*x*), 0 <*x*
<*l*
.

**7.Write all variable separable solutions
of the one dimensional heat equation**

**Solution:**

i) *u*(*x*,*
t*)* *=(* A*_{1}*e*^{l}^{x}* *+*B*_{2}*
e*^{-l}^{x}*
*)*C*_{1}*e*^{a}^{2}^{l}^{2t}*
*

ii)*u*(*x*,*t*)*
*=(* A*_{2}* *cos*
**l**x *+*B*_{2}*
*sin*
**l**x*)*
C*_{2}* e*^{-a}^{2}^{l}^{2t}

iii) *u*(*x*,*
t*)* *=(* A*_{3}*
x *+*B*_{3}*
*)*C*_{3}*
*.* *

8. **Write
down the diffusion problem in one dimension as a boundary value problem in two
different forms. **

**Solution: **

9. **State
any two laws which are assumed to derive one dimensional heat equation.
Solution: **

i)
Heat flows from higher to lower
temperature

ii)
The rate at which the heat flows across
any area is proportional to the area and to the temperature gradient normal to
the curve. This constant is proportionality is known as the thermal
conductivity (k) of the material. It is known as Fourier law of heat
conduction.

10.
**Write any two solutions of the
Laplace equation U _{xx}+U_{yy}=0 involving exponential terms in
x or y. **

**Solution: **

i) *u*(*x*,*
y*)* *=(* A*_{1}*e
^{px} *+

ii) *u*(*x*,*
y*)* *=(* A*_{1}*
*cos* px *+*A*_{2}*
*sin*
px*)(* A*_{3}*e ^{py}
*+

11.
**In steady state conditions derive
the solution of one dimensional heat flow equation. Solution: **

The PDE of unsteady one dimensional heat flow is

In steady state condition, the
temperature u depends only on x and not on t.

The
general solution is u=ax+b, where a, b are arbitary.

**12.
****Write the boundary condition and
initial conditions for solving the vibration of string equation, if the string
is subjected to initial displacement f(x) and initial velocity g(x). **

**Solution:**

The
initial and boundary conditions are

i) *y*(0,*
t*)* *=0* *.* *

ii) *y*(*l*,*
t*)* *=0* *.* *

iii) ^{¶}^{y}*
*(*x*,0)=*g*(*x*)* *.* *

¶*t *

iv) *y*(*x*,0)*
*=*f *(*x*)*
*

**Write down the governing equation of two
dimensional steady state heat equation. Solution:**

**14.
****The ends A and B of a rod of length
10cm long have their temperature distribution kept at 20 ^{o}C and 70^{o}C.
Find the steady state temperature distribution of the rod. Solution: **

The
steady state equation of one dimensional heat flow is

The boundary conditions are u(0)=20, u(*l*)=70.

Put x=0 in (2) we get u(0)=a(0)+b

Þb=20

Put x=*l*
in (2) we get u(*l*)=a*l*+b

70= a*l*+20 a*l=*50

a=
50/*l*

Therefore
equation (2) Þ u(x)= 50x/*l*+20

Here
*l=*10 cm

Therefore u(x)= 50x/10+20 u(x)=5x+20.

**15.
****Write down the different solutions
of Laplace equation in polar coordinates.**

i) *u*(*r*,*q*) =(*C*_{1}*r*
* ^{p}* +

ii) *u*(*r*,*q*)* *=(*C*_{5}*
*cos(* p *log* r*)* *+*C*_{6}*
*sin(*
p *log* r*)(*C*_{7}* e ^{p}*

iii) *u*(*r*,*q*)* *=(*C*_{9}*
*log* r *+*C*_{10}*
*)(*C*_{11}*q*+*C*_{12}*
*)*
*.* *

**16.
****What is the general solution of a
string of length l whose end points are fixed and which starts from
rest? **

**Solution:**

**17.
****How many boundary conditions and
initial conditions are required to solve the one dimensional wave equation? **

**Solution: **

Two boundary conditions and two initial conditions
are required.

**PART
B**

**1.A string is stretched and fastened to
two points x = 0 and x= l apart. Motion is started by displacing the string
into the form y = k (l x –x ^{2} ) from which it is released at time
t=0. Find the displacement of any point on the sting at a distance of x from
one end at time t.**

Solution: The
ODWE y_{tt}=c 2 y_{cc}

Solution : y(x,t)=
(Acos px +Bsin px)(Ccos pct +Dsin pct)

Boundary
and initial conditions are (i) y(0,t) = 0 (ii) y(*l*,t) = 0

(iii)
y * _{t}* (x,0)=0 (iv) y(x,0)=f(x), 0< x <

Using
Boundary and initial conditions:

i)
y(0,t) = 0, put x=0

A(Ccos pct +Dsin pct)=0 \
A=0

\ Suitable solution y(x,t)= Bsin px (Ccos pct +Dsin pct)

ii) y(*l*,t) = 0 , put x=*l*

Bsin
p*l* (Ccos pct +Dsin pct)=0 ÞB
¹0 Bsin p*l* =0

**2.A taut string of
length 2 l is fastened at both ends . The midpoint of the string is taken
to a height b and then released from rest in that position. Find the
displacement of the string at any time.**

**Solution**:
let L=2*l*

Suitable solution y(x,t)=
(Acos px +Bsin px)(Ccos pct +Dsin pct)

Boundary and initial conditions are

(i) y(0,t) = 0 (ii) y(*l*,t) = 0 (iii) y * _{t}*
(x,0)=0 (iv) y(x,0)=f(x), 0< x <

Using Boundary and initial conditions:

i) y(0,t) = 0, put x=0

A(Ccos
pct +Dsin pct)=0 \ A=0

Suitable solution y(x,t)= Bsin px (Ccos pct +Dsin
pct)

ii) y(*l*,t)
= 0 , put x=*l*

Bsin p*l* (Ccos pct +Dsin pct)=0 ÞB ¹0
Bsin p*l* =0

**3.If a string of length **'** ***l*** **'**
is initially at rest in its equilibrium position and each of its points **

**Suitable solution **y(x,t)=
(Acos px +Bsin px)(Ccos pct +Dsin pct)

Boundary and initial conditions are (i) y(0,t) = 0 (ii) y(*l*,t) = 0

(iii) y (x,0)=0 (iv) y *t* (x,0)=f(x) =*v*0 sin^{3} ^{p}* ^{x}* /

Using Boundary and initial conditions:

i) y(0,t) = 0, put x=0

A(Ccos
pct +Dsin pct)=0 \ A=0

\ Suitable solution y(x,t)= Bsin px (Ccos pct +Dsin pct)

ii) y(*l*,t)
= 0 , put x=*l*

Bsin p*l* (Ccos pct +Dsin pct)=0 ÞB ¹0
Bsin p*l* =0

**4.A rod 30 cm long has its ends A and B
kept at 20 ^{0}C and 80^{0}C respectively until steady state
conditions prevail the temperature at each end is then suddenly reduced to 0^{0}
c and kept so. Find the resulting temperature function u(x,t) taking x=0 at A.**

**Solution**:

After change
ODHE: *u _{t}* =

**Suitable Solutions:**

u(x,t)=
(Acospx + Bsin px) *e*^{-a}^{2}
^{p}^{2t}

**Boundary and Initial Conditions:**

Using Boundary and initial conditions:

i) u(0,t)=0

Here x=0 Þ
u(0,t)= *Ae* ^{-a}^{2}
^{p}^{2t}

\A=0

**5. An infinitely long
rectangular plate with insulated surface 10 cm wide. The two long edges and one
short edge are kept at 0 ^{0} temperature, while the other short edge
x=0 is kept at temperature given by u=20y, 0 **£

**Solution **:
Steady state two dimensional heat equation:

*u _{xx}
*+

Infinite plate extended in x-direction : Let *l*=10

**I Boundary Conditions**

i) u(x,0) = 0
ii) u(x,*l*) = 0

**II Suitable Solution:**

u(x,y)=
(*Ae*
^{-}* ^{px}*
+

Using boundary conditions:

u(x,0) = (*Ae*^{-}* ^{px}* +

C=0

Suitable Solution:

u(x,y)= (*Ae*
^{-}* ^{px}*
+

ii) u(x,*l*)
= 0

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