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Chapter 5

Boundary and initial conditions

5.1 Solved Problems

Exercise 103. Solve the boundary problem

2
d u du
t + = 0
dt 2 dt

u(t ) = T , u(t ) = T , 0 < t < t ,
1
2
1
2
2
1
Solution: Let v(t) = du . Then
dt
Z Z
dv v dv dt
= − , = − , ln v = − ln t + ln C ,
1
dt t v t
C 1
v(t) = , u(t) = C ln t + C .
2
1
t
Taking into account the initial condition, we obtain
(
C ln t + C = T ,
1
1
1
2
C ln t + C = T .
2
1
1
1
Solving this system of equations, one deduces
T − T 1 T − T 1
2
2
C = , C = T − ln t 1
1
1
2
ln t 2 ln t 2
t 1 t 1
Hence, the problem has the following solution
T − T 1 T − T 1
2
2
u(t) = ln t + T − ln t 1 .
1
ln t 2 ln t 2
t 1 t 1

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