Cauchy Problems of Hyperbolic Heat-Conduction Equations - Heat Conduction: Mathematical Models and Analytical Solutions - page 177

Note that u η η = 0 =

1
a ψ ( ξ )

, u ( P )= ϕ ( x − y ) , u ( Q )= ϕ ( x + y ) .Also, MP and

QM satisfy ξ = x − ( y − η ) , ξ = x +( y − η ) , respectively. The solution of PDS (5.12)

becomes

x + y

1
2 [ ϕ (

1
2 a

u ( x , y )= x − y )+ ϕ ( x + y )]+

y ψ ( ξ )

d ξ

x −

2 a 2 y

x +( y − η )

f

d

1

ξ , a + d η ξ .

0 x − ( y − η )

Step 4. Solution of PDS (5.11).
Since y = at ,wehave η = a τ and d η = a d τ . Thus the solution of PDS (5.11) is

x + at

)= ϕ (

x − at )+ ϕ ( x + at ) 1

2 a

u ( x , y + ψ ( ξ ) d ξ

2

x − at

t x + a ( t − τ )

1
2 a

+ d τ f ( ξ , τ ) d ξ .

0 x − a ( t − τ )

This is the same as what we obtained by using integral transformations, the method
of characteristics or the method of descent.

5.2 Riemann Method and Method of Laplace Transformation
for One-Dimensional Cauchy Problems

In this section we use the Riemann method and the method of Laplace transforma-
tiontosolve

u t + τ 0 u tt =

a 2 u xx +

f ( x , t ) , − ∞ < x < + ∞ , 0 < t ,

(5.15)

u ( x , 0 )= ϕ ( x ) , u t ( x , 0 )= ψ ( x ) .

By the solution structure theorem, we can focus on solving
u t + τ 0 u tt =

a 2 u xx , − ∞ < x < + ∞ , 0 < t ,

(5.16)

u ( x , 0 )= 0 , u t ( x , 0 )= ψ ( x ) .