1) Find a partial differential equation whose solution is:
z = (a + 2) x + ( a2 + 1) y + b
Soln.
Differentiate the function partially with respect to x and then y to obtain:
¶ z/ ¶ x = p = a + 2
¶ z/ ¶ y = q = a2 + 1
Eliminate a from these eqns. to obtain:
q = (p - 2) 2 + 1
Then:
¶ z/ ¶ y = (¶ z/¶ x - 2) 2 + 1
2)Use separation of variables to solve the partial differential equation:
¶2 y/ ¶ x2 + ¶2 y/ ¶ y2 + ¶2 y/ ¶ z2 + 2mE/ ħ2 y = 0
For a particle in a cubic box. Include the quantized energy for the particle.
Soln.:
For a cubic box: a = b = c
By the separation of variables:
y = X(x) Y(y) Z(z)
Then:
X’ = ¶X/ ¶ x Y’ = ¶Y/ ¶ y and Z’ = ¶Z/ ¶ z
This leads to the equation:
X”YZ + XY” Z + XYZ” + 2mE/ ħ2 XYZ = 0
Dividing the equation by XYZ:
X”/ X + Y”/ Y + Z”/Z + 2mE / ħ2 = 0
We let:
X”/ X = - a2, Y”/ Y = -b2, Z”/Z = -g2
The independent solutions will be:
x= Ö(2/a) sin ( n xpx/a)]
y= Ö(2/a) sin ( n y px/a)]
z= Ö(2/a) sin ( n z px/a)]
where: n x = 1, 2, 3, 4.. etc.
Then the quantized energy is:
E = p2 ħ2 / 2m a2 (n2x + n2 y + n2 z )
3) The partial differential equation for a deflected beam is given as:
¶ 2 u/ ¶ t 2 + c 2 [¶ 4 u/ ¶ x4 ] = 0
And: c 2 = EI/ r A
Where EI is the flexural rigidity, r is the density of the wood, and A is the area.
a) If u(x,t) is defined such that: t > 0 and
0 < x < ℓ
Use separation of variables to arrive at an initial general (but not optimal) solution and write the two ordinary differential equations arising from the approach.
Soln.
u(x,t) = X(t)T(t)
¶ 2 u/ ¶ t 2 = X(x) (d 2 T / d t 2 )
¶ 4 u/ ¶ x4 = T(t) d 4 X / dx4
XT" - c 2 X4 T = 0
Separate variables to get;
X4 / X = - T / c 2 T = k
Let u(x,t) be defined over 0 < x < ℓ (t>0)
T(t) = c1 exp ( cr t) + exp (- cr t)
b) Apply the characteristic equati0n (i.e. in order 4) to obtain improved general solutions X(x) and T(t) and also suggest values for the constants arising: C1, C2, C3 and C4.
Soln.:
Use characteristic eqn.: m 4 - b4 = 0
Then: (m 2 + b2 ) (m 2 - b2) = 0
Further factoring:
( m + ib ) (m - ib ) (m + b ) (m - b ) = 0
m = + ib m = + b
Then:
X(x) = C1 cos b x + C2 sin b x + C3 exp (b x) + C4 exp (-b x)
C1 = A, C2 = B, C3 = ½ (C + D), C4 = ½ (C - D)
This works since:
X(x) = A cos b x + B sin b x + ½ (C + D) exp (b x)
+ ½ (C - D) exp (-b x)
= A cos b x + B sin b x + C (e b x + e -b x /2) +
D (e b x - e -b x /2)
= A cos b x + B sin b x + C cosh b x + D sinh b x
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