How do we handle a change in variable for the method of characteristics?

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In summary, the conversation discusses the equation $u_t + xu_x = t$ with $u(x,t) = g(x)$ and its solution $u(x,t) = \frac{1}{2}t^2 + g(xe^{-t})$. The method of characteristics is used to solve the equation, with equations $\frac{dt}{ds} = 1$, $\frac{dx}{ds} = x$, and $\frac{du}{ds} = t$ and their solutions being $t=s$, $x=x_0e^t$, and $u = \frac12t^2 + g(xe^{-t})$. If instead of $x$, the equation was $2u$, the
  • #1
Dustinsfl
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$u_t + xu_x = t$ 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?
 
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  • #2
dwsmith said:
$u_t + xu_x = t$ 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 $u = \frac12s^2 + g(x_0) = \frac12t^2 + g(xe^{-t}).$
 
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  • #3
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 $\frac{du}{ds} = 0$
 

FAQ: How do we handle a change in variable for the method of characteristics?

1. What is the Method of Characteristics?

The Method of Characteristics is a mathematical technique used to solve partial differential equations, particularly those that describe wave-like phenomena.

2. How does the Method of Characteristics work?

The Method of Characteristics involves transforming the original partial differential equation into a system of ordinary differential equations along specific characteristic curves. These curves represent the paths along which the solution of the equation is constant.

3. What types of problems can be solved using the Method of Characteristics?

The Method of Characteristics is commonly used to solve problems related to heat transfer, fluid flow, and wave propagation.

4. What are the advantages of using the Method of Characteristics?

The Method of Characteristics is advantageous because it can provide a more accurate and precise solution compared to other numerical methods. It also allows for the visualization of characteristic curves, making it easier to understand the behavior of the solution.

5. Are there any limitations to the Method of Characteristics?

One limitation of the Method of Characteristics is that it can only be applied to linear partial differential equations. It can also be computationally expensive and time-consuming for complex problems.

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