Write an explicit formula for a function
u
{\displaystyle u}
{
u
t
+
b
⋅
D
u
+
c
u
=
0
in
R
n
×
(
0
,
∞
)
u
=
g
on
R
n
×
{
t
=
0
}
.
{\displaystyle \left\{{\begin{array}{r l}u_{t}+b\cdot Du+cu=0&{\text{ in }}\mathbb {R} ^{n}\times (0,\infty )\\u=g&{\text{ on }}\mathbb {R} ^{n}\times \{t=0\}.\end{array}}\right.}
c
∈
R
{\displaystyle c\in \mathbb {R} }
b
∈
R
n
{\displaystyle b\in \mathbb {R} ^{n}}
Consider characteristics
(
x
(
s
)
,
t
(
s
)
)
=
(
x
0
+
b
s
,
s
)
∈
R
n
{\displaystyle (x(s),t(s))=(x_{0}+bs,s)\in \mathbb {R} ^{n}}
x
∈
R
n
,
t
∈
R
{\displaystyle x\in \mathbb {R} ^{n},t\in \mathbb {R} }
z
(
s
)
=
u
(
x
+
s
b
,
t
+
s
)
{\displaystyle z(s)=u(x+sb,t+s)}
z
˙
(
s
)
=
D
u
(
x
+
s
b
,
t
+
s
)
⋅
b
+
u
t
(
x
+
s
b
,
t
+
s
)
=
−
c
z
(
s
)
{\displaystyle {\dot {z}}(s)=Du(x+sb,t+s)\cdot b+u_{t}(x+sb,t+s)=-cz(s)}
where the last inequality is a result of the original PDE.
The above ODE can be solved and we get
z
(
s
)
=
z
(
0
)
e
−
c
s
{\displaystyle z(s)=z(0)e^{-cs}}
Finally, any point
(
x
,
t
)
{\displaystyle (x,t)}
(
x
0
,
0
)
{\displaystyle (x_{0},0)}
x
0
=
x
−
t
b
{\displaystyle x_{0}=x-tb}
u
(
x
,
t
)
=
g
(
x
−
t
b
)
e
−
c
t
{\displaystyle u(x,t)=g(x-tb)e^{-ct}}
