Solve $\dfrac{\partial u}{\partial t}+u\dfrac{\partial u}{\partial x}=x$ subject to the initial condition $u(x,0)=f(x)$.

I let $\dfrac{dt}{ds}=1$ , $\dfrac{dx}{ds}=u$ , $\dfrac{du}{ds}=x$ and the initial conditions become: $t=0$ , $x=\xi$ and $u=f(\xi)$ when $s=0$ .

I believe this leads to $t=s$ , but I am unsure how to deal with $\dfrac{dx}{ds}=u$ and $\dfrac{du}{ds}=x$ .

  • $\begingroup$ sorry i cant see how to make the math text larger $\endgroup$ – Permian Feb 8 '14 at 18:09
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    $\begingroup$ I believe you should have a look at the Burger's equation. The way it is being solved will help you understand this one. $\endgroup$ – user88595 Feb 8 '14 at 18:27
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    $\begingroup$ I could not see how to do this as burger's equation is = 0 and my equation is equal to x. This means that u is not constant and therefore I cannot see how to use the solution to burgers equation to help solve my equation. $\endgroup$ – Permian Feb 8 '14 at 20:34
  • $\begingroup$ Studying the Burger's equation, as @user88595 suggest, will not only help you to derive the solution, but also to understand how it behaves. $\endgroup$ – Pragabhava Apr 10 '14 at 2:38

Your PDE is quasilinear, meaning it migh not have a classic solution for all time $t$. That said, we know that the quasilinear equation $$ a\big(x,t,u(x,t)\big) u_x(x,t) + b\big(x,t,u(x,t)\big)u_t(x,t) = c\big(x,t,u(x,t)\big) $$ where $a,\,b,\,c \in C^1$ with data $\mathcal{C}(\xi) = \big(x(\xi), t(\xi), u(\xi)\big) \in C^1$ and with $$ \begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dt}{d\xi} & b\end{vmatrix} \neq 0 $$ has a unique solution near $\mathcal{C}$ given by \begin{align} \frac{d x}{d \eta} &= a & x\big|_{\eta = 0}&= x(\xi)\\ \frac{d t}{d \eta} &= b & t\big|_{\eta = 0}&= t(\xi)\\ \frac{d u}{d \eta} &= c & u\big|_{\eta = 0}&= u(\xi)\\ \end{align}

In your case, $\mathcal{C}(\xi) = \big(\xi,0,f(\xi)\big)$. Near $\eta \sim 0$ $$ \begin{vmatrix} \frac{dx}{d\xi} & a \\ \frac{dt}{d\xi} & b\end{vmatrix} = \begin{vmatrix} 1 & u \\ 0 & 1\end{vmatrix} = 1 $$ and the solution is unique near the initial condition. Now,

\begin{align} \frac{d x}{d \eta} &= u & x\big|_{\eta = 0}&= \xi\\ \frac{d t}{d \eta} &= 1 & t\big|_{\eta = 0}&= 0\\ \frac{d u}{d \eta} &= x & u\big|_{\eta = 0}&= f(\xi)\\ \end{align} has as solution $$ t = \eta, \quad u = f(\xi)\cosh \eta + \xi \sinh \eta, \quad x =f(\xi) \sinh \eta + \xi \cosh \eta. $$ and the characteristics are given by the equation $$ t= \text{arctanh}\left(\frac{x - \xi}{f(\xi)}\right) $$ (you have to be carefull while inverting $\tanh t$).

Depending on the behavior of $f$, if the characteristics can meet. If the do, there will be a shock and the classical solution will cease to exist. We can see this by studying the transformation $$ (x,t) \longrightarrow (\xi,\eta). $$ The change of variables will be invertible iif $$ \begin{vmatrix} \partial_\xi x & \partial_\eta x \\ \partial_\xi t & \partial_\eta t \end{vmatrix} = f'(\xi) \sinh\eta + \cosh \eta \neq 0 $$ meaning there is no solution when $$ f'(\xi) \sinh\eta + \cosh \eta = 0 $$ or, inverting the transformation, the solution will develop a shock at time $$ t = -\text{arctanh}\left(f'(\xi)\right), $$ and the profile of $f(x)$ is crucial in the shock development. This is why, if you want to understand how this works, you have to study the Burger's equation, as suggested by user88595.

In the region where no shock has developed (might be all the domain), the solution is given by $$ u(x,t) = f\big(\xi(x,t)\big)\cosh t + \xi(x,t) \sinh t $$ where $\xi(x,t)$ is determined by inverting $$ x = f(\xi) \sinh t + \xi \cosh t, $$ which can be written in implicit form as $$ f\big(x \cosh t - u(x,t) \sinh t\big) = u(x,t) \cosh t - x \sinh t $$


Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$



$x=C_1\sinh s+C_2\cosh s$

$\therefore u=C_1\cosh s+C_2\sinh s$

Hence $\begin{cases}x=C_1\sinh s+C_2\cosh s\\u=C_1\cosh s+C_2\sinh s\end{cases}$

$x(0)=x_0$ , $u(0)=F(x_0)$ :


$\therefore\begin{cases}x=F(x_0)\sinh s+x_0\cosh s\\u=F(x_0)\cosh s+x_0\sinh s\end{cases}$

$\therefore\begin{cases}x_0=x\cosh s-u\sinh s=x\cosh t-u\sinh t\\F(x_0)=u\cosh s-x\sinh s=u\cosh t-x\sinh t\end{cases}$

Hence $u\cosh t-x\sinh t=F(x\cosh t-u\sinh t)$

$u(x,0)=f(x)$ :


$\therefore u\cosh t-x\sinh t=f(x\cosh t-u\sinh t)$


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