Considering the linear 1 st order PDE
u
t
+
u
x
x, x>0, t>0
u(x,0)sinx, x≥0
u(0,t)t,t≥0

Mathematical solution

The method of characteristics gives
dx
1

dt
1

du
x
.
From the first equality we obtain
dt
1

dx
1
⟹xt+
c
1
⟹
c
1
x-t,
and from the second one
dt
1

du
x

du
t+
c
1
⟹u
c
2
+
2
(t+
c
1
)
2
⟹
c
2
u-
2
(x-t)
2
.
The general solution of the PDE can be expressed as
c
2
F(
c
1
).
So, we have
uF(x-t)+
2
(x-t)
2
.
Now, applying the initial conditions, we have
​
u(x,0)sinxF(x),x≥0​​u(0,t)tF(-t),t≥0
Hence, it suffices to say that a solution
u(x,t)
of the given Boundary Value Problem is the function defined as follows :
u(x,t)=
F(x-t)+
2
(x-t)
2
where
F
(
x
)
s
i
n
x
.
F
(-
t
)
t
.
​​
​
Therefore, the solution of the system is
​​
u(x,t)=
2
x
2
+(
t
-
x
)
,
0<
x
<
t
x
t
-
2
t
2
+
s
i
n
(
x
-
t
),
x
>
t
>0
The solution needs to be checked for consistency with the original differential equation . Also, continuity conditions need to be checked to determine if the solution is classical or weak . The continuity is assured if
sothesolutionisclassical.

Solution on Mathematica

(*Definethepartialdifferentialequation:*)
In[]:=
pde=D[u[x,t],t]+D[u[x,t],x]-x==0;
(*Definetheinitialandboundaryconditions:*)
In[]:=
bc1=u[x,0]==Sin[x];​​bc2=u[0,t]==t;​​
(*Solvetheequation:*)
In[]:=
uSol[x_,t_]:=If[x<t,x^2/2+(t-x),x*t-t^2/2+Sin[x-t]];​​
(*Inspectthesolution:*)
In[]:=
Plot3D[uSol[x,t],{x,-10,10},{t,0,10}]
Out[]=
After you plot the solution
using the Plot3D function in Mathematica, you will get a three - dimensional plot . This plot represents the solution to the PDE for different values of
x
and
. The
- axis and
- axis (horizontal axes) in the plot represent the variables
and
respectively . The z - axis (vertical axis) represents the value of the solution
. The form of
changes based on whether
or
, hence, you will observe a discontinuity or "break" along the line
. 1. For the region where
, the solution
is
. This is a linear function with respect to both
and
, which should appear as a plane in the
-plot . As
x
or
increases, the value of
increases linearly. 2. For the region where
, the solution
is
. This is more complex, due to the presence of the trigonometric term
, and you will see a surface with a wave - like pattern (due to the sine function) in the 3 D plot . The
term creates an oscillation in the
values, which will appear as regular peaks and valleys in the 3 D plot . ​It' s important to input the code into Mathematica and study the plot to gain a deeper understanding of the solution' s behavior . As you rotate and zoom the
- plot, you' ll be able to explore how the solution
changes with different
and
values .