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I am trying to solve the following PDE: $u_{xt} + uu_{xx} = -\frac{1}{2}u_x^2$, with initial condition: $u(x,0) = u_0(x) \in C^{\infty}$ using the method of characteristics.

I am a beginner with the method of characteristics and PDE in general. Here is what I have so far.

Define $\gamma(x,t)$ as the characteristic curves.

$\frac{\partial}{\partial t} u_x(\gamma(x,t),t) = u_{xt} + u_{xx}\gamma_t(x,t) = - \frac{1}{2}u_x^2$

Set $u_t = u_x$

$\Rightarrow \frac{\partial}{\partial t} u_x(\gamma(x,t),t)= (u_t)_x + u_{xx}\gamma_t(x,t)$

$ = u_{xx} + u_{xx}\gamma_t = - \frac{1}{2}u_x^2$

From this I get $\gamma_t = -\frac{1}{2}\frac{u_x^2}{u_{xx}} - 1$

However, I am not sure this is the right approach and do not fully understand how to use the method of characteristics when the solution $u(x,t)$ is constant on the characteristic curves.

Any help is much appreciated.

Edit: I made some progress by using $v=u_x$ and getting $\frac{dv}{dt} = \frac{-1}{2} v^2$ and $\frac{\partial x}{\partial{t}} = 1$. Then separating the first ODE, I get $\frac{2}{v} = t + c$. However, I am not sure if my solution after integrating with respect to $x$ and using the initial condition is correct. I end up with $u(x,t) = \frac{2}{t+c}x + c_1$, $u(x,0) = \frac{2}{c}x + c_1$.

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    $\begingroup$ Have you tried $v=u_x$? $\endgroup$ – Rumplestillskin Jan 23 '17 at 5:01
  • $\begingroup$ Could you explain more about how I would go about using this? Sorry, I am very new to PDE and characteristics. I am not sure what $v$ is in your comment. Is it the characteristic curves? I just edited for clarity, but $\gamma$ is the characteristic curve in my question. $\endgroup$ – Don Jan 23 '17 at 15:08
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    $\begingroup$ Substituting $v(x,t)=u_x(x,t)$ is quite problematic since the coefficient $u$ gets quite nasty. If the coefficients are independent of $u$ then this is a quite standard approach. But here you'd get 'something' like $$v_t+ \int v \ dx \ v_x+0.5v^2=0$$ and good luck with this. Just to mention this (since it seems you have not seen this before): Say we have the PDE $u_{xt}+u_{xx}+u_x=0$ then we first solve $v_t+v_x+v=0$ which is quite nice and then plug this into $u_x=v$ to get the solution $u$ by integrating. $\endgroup$ – Fritz Jan 23 '17 at 17:26
  • $\begingroup$ @Don apologies see the comment above mine by MarvinF. $\endgroup$ – Rumplestillskin Jan 23 '17 at 23:06
  • $\begingroup$ @Rumplestillskin Thanks for letting me know. I did end up getting somewhere with this approach. I got $\frac{d v}{d t} = \frac{-1}{2}v^2$, $\frac{\partial x}{\partial t} = 1$. Then I separated the first ODE and got to $\frac{2}{u_x} = t + c$. But I'm not sure if the solution I get when integrating with respect to $x$ is correct. I get $u(x,t) = \frac{2}{t+c} x + c_1$ and am not sure what to do with that besides plug in the initial condition and get $u(x,0) = \frac{2}{c}x + c_1$. Is this correct? $\endgroup$ – Don Jan 24 '17 at 4:39
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Modifying the problem. Rather than consider the PDE $u_{xt}+u\ u_{xx}+\frac{1}{2}u_{x}^2=0$ with initial condition $u(x,0)=u_0(x)$ as asked above, I will consider the following variant. $$ \text{Solve }u_{xy}+u\ u_{xx} + u_x^2=0\text{ subject to }u(x,0)=f(x).\qquad(\star) $$ There are three differences between this question and that which was asked originally.

  1. The coefficient of $u_x^2$ has changed from $\frac{1}{2}$ to $1$.
  2. The variable $t$ has been renamed to $y$.
  3. The initial function $u_0(x)$ has been renamed to $f(x)$.

Only (1) represents a significant modification of the problem. It makes the solution more tractable and enables it to be found using an elementary application of the method of characteristics. For these reasons, it is conceivable that this was the intended question.

Note: I will not delve into regularity of the solutions in this answer.

Reduction to a first order quasilinear PDE. Write the equation as $$ \frac{\partial}{\partial x}\left(u_y+u\ u_x\right)=0. $$ Thus $(\star)$ is equivalent to $$ u_y+u\ u_x=g(y),\qquad u(x,0)=f(x),\qquad (\star\star) $$ where $g(y)$ is an arbitrary function of $y$ (with sufficient regularity).

Method of characteristics. Perhaps the simplest formulation of the method of characteristics is for quasilinear first order PDEs. These are PDEs of the form $$a(x,y,u)u_x+b(x,y,u)u_y=c(x,y,u).$$ To solve this equation, one regards the solution as a surface $z=u(x,y)$ in $xyz$-space. Let $s$ parametrize the initial curve $\bigl(s,0,f(s)\bigr)$ and let $t$ be a second parameter, which can be thought of as the distance flowed along a characteristic curve emanating from $\bigl(s,0,f(s)\bigr)$.

The characteristic equations are then $$ \frac{dx}{dt}=a(x,y,z),\quad \frac{dy}{dt}=b(x,y,z),\quad \frac{dz}{dt}=c(x,y,z). $$ Returning to our equation $(\star\star)$, this reduces to $a(x,y,u)=u$ and $b(x,y,u)=1$ and $c(x,y,u)=g(y)$. Thus $$ \frac{dx}{dt}=z,\quad \frac{dy}{dt}=1,\quad \frac{dz}{dt}=g(y) $$ with initial conditions $x(0)=s$ and $y(0)=0$ and $z(0)=f(s)$.

The solution to this system is $$ x=s+zt,\quad y=t,\quad z=f(s)+h(t), $$ where $h(t)$ is the antiderivative of $g(t)$ satisfying $h(0)=0$. Since $g$ was arbitrary, so is $h$ given $h(0)=0$.

The solution. Now we eliminate all occurrences of $t$ by replacing them with $y$, then eliminate $s$ by writing $s=x-zy$. Finally, replace $z$ with $u$ to obtain the implicit equation $$ \boxed{u=f(x-uy)+h(y)}, $$ where $h(y)$ is any sufficiently regular function satisfying $h(0)=0$. This is an implicit equation for the general solution of $(\star)$.

TL;DR. Change the $\frac{1}{2}$ in the original question to $1$ to obtain a PDE solvable by the method of characteristics.

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  • $\begingroup$ Thank you for the answer, but I am interesting in the original equation per se $\endgroup$ – Vlad Apr 25 '17 at 1:59
  • $\begingroup$ Why? (Perhaps if you provide some context into why you care about a certain equation, it will lead to better answers.) $\endgroup$ – pre-kidney Apr 25 '17 at 3:09
  • $\begingroup$ Because I know how to solve simpler cases with different coefficients. I am interested in this particular problem because I am preparing for the preliminary exam on PDEs, and once I saw this question on SE, I thought it was interesting and asked my study group to help. Once professor saw this problem on the board, she confirmed it would be relevant and very useful as a practice before exam. However, we still do not have clean solution by method of characteristics. $\endgroup$ – Vlad Apr 25 '17 at 7:17
  • $\begingroup$ Is it possible that when your professor looked at the board she THOUGHT that she recognized the simpler problem. As posed, I think this is too hard for a prelims question. $\endgroup$ – Philip Roe Apr 25 '17 at 16:18
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Not so sure:

$u_{xt} + uu_{xx} + \frac{1}{2}u_x^2 = 0$

$2u_{xt} + 2uu_{xx} + u_x^2 = 0$

Let's define

$u = x t$

Then:

$u_{x} = x_{x} t$

$u_{t} = x t_{t}$

$u_{xt} = x_{x} t_{t}$

$u_{xx} = x_{xx} t$

Now:

$2 u_{xt} + 2u u_{xx} + u_x^2 = 0$

$2 x_{x} t_{t} + 2 x t x_{xx} t + x_{x}^2 t^2 = 0$

We divide by $x_{x} t^2$:

$2 t_{t} / t^2 + 2 x x_{xx} / x_{x} + x_{x} = 0$

Please, verify.

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  • $\begingroup$ I do not see how you are able to solve for $x$ or $u$ using your result at the end. Could you explain? $\endgroup$ – Don Jan 23 '17 at 16:23
  • $\begingroup$ $2 t_{t} / t^2 = - 2 x x_{xx} / x_{x} - x_{x} = \lambda$ $\endgroup$ – cgiovanardi Jan 23 '17 at 16:27
  • $\begingroup$ So are you defining $\lambda$ as the characteristic curve? $\endgroup$ – Don Jan 23 '17 at 17:14
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    $\begingroup$ I don't think that works. At the beginning setting $u=xt$ means that you say that $xt$ is a solution to the PDE. But obviously (since $u_x=t, u_t=x$ and so on) $$u_{xt}+uu_{xx}+\frac12 u_x^2=0+0+\frac12 t^2$$ and this is not zero. $\endgroup$ – Fritz Jan 23 '17 at 17:18

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