How can I solve $u_{xt} + uu_{xx} + \frac{1}{2}u_x^2 = 0$ with the method of characteristics.

Question

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 $u(x,t) = \frac{2}{t+c}x + c_1$, $u(x,0) = \frac{2}{c}x + c_1$.

Answer 1

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.

Answer 2

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.