This is the general solution to the given PDE.

[squenshl]

Homework Statement

Find the general solution of the following equation
u_t + x²u_x = t, u(x,0) = f(x), -inf < x < inf, t > 0

Homework Equations

The Attempt at a Solution

Using method of characteristics I get
du/dt = u_t + dX/dtu_x = u_x(dX/dt - x^2) + t
so along the curve dX/dt = x^2 with x(0) = x₀ we get x = -1/(t+x₀) and x₀ = -1/x - t
So du/dt = t = -1/x - x₀ so
u(x,t) = -t/x - x₀t + f(x₀)
so u(x,t) = t² + f(-1/x - t)
but when I check this I don't get the original PDE.
Someone help.

 

[syj]

hey,
so I am only just starting to understand method of characteristics, but I am gona try to help.

The first step I believe is to get your characteristic equations

\frac{dx}{ds}=x^{2}

\frac{dt}{ds}=1

\frac{du}{ds}=t

solving these equations gives

x=x^{2}s +x_{0}

t=s+t_{0}

u=ts+u_{0}

from the initial condition t_{0}=0

so t=s

x=x^{2}t+x_{0}

u=t^{2}+u_{0}

u(x_{0},0)=f(x_{0})

x_{0}=x-x^{2}t

so u_{0}=f(x-x^{2}t)

 

[Coelum]

The system of ODEs for the characteristic lines is

\frac{dx}{ds}(r,s)=x^2
\frac{dt}{ds}(r,s)=1
\frac{du}{ds}(r,s)=t.

The first ODE can be integrated by separation of variables:

\frac{dx}{ds}(r,s)=x^2
∫\frac{dx}{x^2}=∫ds
-\frac{1}{x(r,s)}=s+g_1(r)
x(r,s)=-\frac{1}{s+g_1(r)}.

The second ODE is straightforward:

\frac{dt}{ds}(r,s)=1
t=s+g_2(r).

The third ODE can be solved using the value of t just computed:

\frac{du}{ds}(r,s)=t=s+g_2(r)
u(r,s)=(s^2)/2+s*g_2(r)+g_3(r).

Note that the constants of integration are functions of r since we are integrating w.r.t. s. The general solution is:

x(r,s)=-\frac{1}{s+g_1(r)}.
t=s+g_2(r)
u(r,s)=(s^2)/2+s*g_2(r)+g_3(r).

The initial conditions are:

x(r,0)=r
t(r,0)=0
u(r,0)=f(r)

which we substitute into the general solution

x(r,0)=-\frac{1}{g_1(r)}=r
t(r,0)=g_2(r)=0
u(r,0)=g_3(r)=f(r)

getting the value for the constants of integration

g_1(r)=-\frac{1}{r}
g_2(r)=0
g_3(r)=f(r).

that we can substitute into u to get it as a function of s,r:

u(r,s)=s^2/2+f(r).

In order to get u as a function of x,t, we need to invert x(r,s),t(r,s) first

t=s
x=-\frac{1}{t-1/r}=\frac{r}{1-rt}
1-rt=\frac{r}{x}
1=r(t+\frac{1}{x})
r=\frac{x}{1+tx}

and then substitute them into the above expression for u:

u(x,t)=t^2/2+f(\frac{x}{1+tx}).