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[SOLVED] Method of Characteristics

dwsmith

Well-known member
Feb 1, 2012
1,673
$u_t + xu_x = t$ with $u(x,t) = g(x)$. The solution is $u(x,t) = \frac{1}{2}t^2 + g(xe^{-t})$.

Using the method of characteristics (I am trying to understand).

So we would have then
$\frac{dt}{ds} = 1$, $\frac{dx}{ds} = x$, and $\frac{dz}{ds} = z$.
$$
\begin{align*}
t(s) & = & s + c_1\\
x(s) & = & xs + c_2\\
z(s) & = & zs + c_3
\end{align*}
$$
What next?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,706
$u_t + xu_x = t$ with $u(x,t) = g(x)$ (should be $\color{red}{u(x,0) = g(x)}$). The solution is $u(x,t) = \frac{1}{2}t^2 + g(xe^{-t})$.

Using the method of characteristics (I am trying to understand).

So we would have then
$\frac{dt}{ds} = 1$, $\frac{dx}{ds} = x$, and $\frac{dz}{ds} = z$.
$$
\begin{align*}
t(s) & = & s + c_1\\
x(s) & = & xs + c_2\\
z(s) & = & zs + c_3
\end{align*}
$$
What next?
You may find it helpful to follow the method used in the Example in the Wikipedia page on this topic.

For a start, your equations should be $\dfrac{dt}{ds} = 1$, $\dfrac{dx}{ds} = x$ and $\dfrac{du}{ds} = t$. Their solutions are:

• For $t$, $\dfrac{dt}{ds} = 1$ gives $t = s+{}$const. Take $t(0)=0$ to get $t=s$.

• For $x$, $\dfrac{dx}{ds} = x$ gives $x= {}$(const.)$e^s$. Let $x(0) = x_0$ to get $x=x_0e^s = x_0e^t$, from which $x_0 = xe^{-t}.$

• For $u$, the equation is $\dfrac{du}{ds} = t$. But we are taking $t=s$. So it becomes $\dfrac{du}{ds} = s$, giving $u = \frac12s^2 + {}$const. To evaluate the constant, notice that when $s=0$, $u$ will be equal to $u(x_0,0) = g(x_0)$. Thus $u = \frac12s^2 + g(x_0) = \frac12t^2 + g(xe^{-t}).$
 
Last edited:

dwsmith

Well-known member
Feb 1, 2012
1,673
How would it be handled if instead of x, we had 2u.

So
$$
\frac{dx}{ds} = 2u
$$
$t(s) = s + c_1$ and $s= t$ but $\frac{du}{ds} = 0$