Please can anyone help me solving this problem?

$$ \frac{\partial U}{\partial t} - \left(\frac{\partial U}{\partial x}\right)^2 = 0 $$

where $U=U(x,t)$ with side condition as $U(x,0)=\cos x$.

The problem is given in the following article: A. Thess, D. Spirn, B. Jüttner, "Viscous Flow at Infinite Marangoni Number", Physical Review Letters 75(25), 1995. doi:10.1103/PhysRevLett.75.4614


migrated from physics.stackexchange.com Jun 12 at 8:10

This question came from our site for active researchers, academics and students of physics.

  • $\begingroup$ Shouldn't Burgers' equation have $\partial^2 U/\partial x^2$ rather than $(\partial U/\partial x)^2$ ? $\endgroup$ – John Rennie Jun 12 at 6:37
  • $\begingroup$ it's special case in it $\endgroup$ – Mohan Aditya Sabbineni Jun 12 at 6:38
  • $\begingroup$ the problem is given in the physical review letters volume 75, Number 25 Viscous flow at infinite Marangoni Number. $\endgroup$ – Mohan Aditya Sabbineni Jun 12 at 6:39
  • $\begingroup$ You've asked the same question twice now. $\endgroup$ – Mattos Jun 12 at 9:59
  • $\begingroup$ This question got migrated from physics to mathematics, so it appeared twice in mathematics $\endgroup$ – Mohan Aditya Sabbineni Jun 12 at 10:02

The article cited in OP deals with the convection problem $\partial_t \theta + v \partial_x \theta = 0$, where $v = -H\theta$ is expressed in terms of the Hilbert transform $H\theta$ of $\theta$. While discussing this model, the authors make some statements on the alternative model (p. 4615, top of 2nd column)

$v = -\partial_x\theta$, which leads to the well-known Burgers equation $\partial_t\theta - (\partial_x\theta)^2 = 0$

This is actually an Hamilton-Jacobi equation, which classical resolution relies on the Lax-Hopf formula. A Burgers-like equation is recovered after differentiation of the above PDE with respect to $x$: $$ g_t - 2 g g_x = 0, \qquad\text{with}\qquad g = \theta_x . $$ The solutions deduced from the method of characteristics satisfy the implicit equation $g = -\sin (x +2g t)$ of which no analytical solution is known. This classical solution is valid up to the breaking time $t=1/2$ when it breaks down, as displayed on the plot of the characteristics in $x$-$t$ plane below:


Along the characteristic lines, the variable $g = -\sin(x_0)$ is constant. Moreover, we have $$ \frac{\text d}{\text d t} \theta = \theta_x \frac{\text d}{\text d t} x + \theta_t = -2g\theta_x + \theta_t = -g^2 , $$ so that $\theta = \cos(x_0) - g^2 t$ along the characteristics. Thus, $$ \theta = \cos(x +2g t) - g^2 t, \qquad\text{with}\qquad g = -\sin(x +2g t) . $$ Below is a Matlab script for this computation, along with its output (requires Optimization Toolbox):

nx = 200;
nt = 10;
tf = 0.49;
x = linspace(0,2*pi,nx);
t = linspace(0,tf,nt);
g = -sin(x);
theta = cos(x);

hg = plot(x,g,'k-');
xlim([0 2*pi]);
ylim([-1 1]);
ht1 = title(strcat('t =',num2str(t(1))));
htheta = plot(x(2:nx),theta(1,2:nx),'k-');
xlim([0 2*pi]);
ylim([-1 1]);
ht2 = title(strcat('t =',num2str(t(1))));

for i = 2:nt
    fun = @(g) g + sin(x+2*g*t(i));
    g = fsolve(fun,-((x<pi).*x+(x>pi).*(x-2*pi))/(1+2*t(i)));
    theta = cos(x+2*g*t(i)) - g.^2*t(i);
    set(ht1,'String',strcat('t =',num2str(t(i))));
    set(ht2,'String',strcat('t =',num2str(t(i))));


  • $\begingroup$ I too proceeded in the same way and got the same result g= -sin(x+2gt) but can't we solve further to get thetha? $\endgroup$ – Mohan Aditya Sabbineni Jun 13 at 9:07
  • $\begingroup$ In Matlab, I am facing this problem with the above code "Subscript indices must either be real positive integers or logicals." and thus unable to obtain the second plot. Please help me out $\endgroup$ – Mohan Aditya Sabbineni Jun 14 at 19:07

This equation can be solved by the method of characteristics. Let us start by changing variables, and use the field

$$v(x,t) = -2 \partial_x U(x,t) \, .$$

Indeed, in terms of $v(x,t)$, the problem is

$$ \partial_t v + v \partial_x v= 0 \, , \qquad v(x,0) = f(x) = 2 \sin(x) \, . \qquad (*)$$

Then the solution of this differential equation comes from the observation that

$$ \frac{d}{dt}v(z(t),t) = 0 \, , \qquad \text{if} \qquad \frac{dz}{dt} = v(z(t),t) = v_0 = f(z(0)) \, .$$

The right-hand-side of the second equation is a constant as a consequence of the first equation that tells us that $v(z(t),t)$ is a constant. Then if in order to compute $v(x,t)$, we need to find $x_0$ such that $z(0) = x_0$ and $z(t) = x$. I.e. solve

$$ f(x_0) t + x_0 = x \, , \qquad \rightarrow \qquad x_0 = x_0(x,t) \, . \qquad (**)$$

In the case $f(x) = 2 \sin(x)$, this has to be done numerically for most choices of $x$ and $t$. Finally this solution can be inserted back into

$$v(x,t) = v(z(t),t) = v(z(0),0) = v(x_0(x,t),0) = f(x_0(x,t)) \, .$$

Modulo the numerical solution of ($**$), this answers the question.

Things get difficult (and interesting) when ($**$) can not be solved. With $f(x) = 2 \sin(x)$, this happens when $t\geq 1/2$ and ($**$) has multiple solutions. Then ($*$) can not be solved. It is however possible to circumvent this problem by coming up with a prescription to decide which solution to pick. With such a prescription, it turns out that that $v(x,t)$ jumps from one solution to another as $x$ is changed. $v(x,t)$ develops discontinuities for $t>1/2$ and displays so-called shocks. See e.g. wikipedia.

An intuitive (and physically based) prescription is to introduce a viscosity term

$$ \partial_t v + v \partial_x v= \nu \partial_x^2 v \, , \qquad (***)$$

and define the solution of ($*$) as the solution of ($***$) in the limit $\nu \rightarrow 0$. As long as $\nu >0$, ($***$) has a smooth and well defined solution. This solution becomes however discontinuous in the desired limit $\nu \rightarrow 0$, and shocks emerge naturally. Moreover, in the absence of shocks, the limit $\nu \rightarrow 0$ can be taken straightforwardly and the solutions of limit of the solution of ($*$) coincides with the solution of ($***$).

I conclude with a short comment on terminology: To me (and also to wikipedia), ($*$) is Burgers' equation and $\partial_t U - [\partial_x U]^2=0$ is the KPZ equation without viscosity and noise. These two names are often mixed up because both equations are equivalent as I show in the beginning of my answer.

I quickly wrote a short Mathematica code to solve the above equation up to $t=1/2$:

X0[x_, t_] := x0 /. FindRoot[2 Sin[x0] t + x0 == x, {x0, x}]
v[x_, t_] := 2 Sin[X0[x, t]]
U[x_, t_, Npts_] := -1/2 Sum[v[x (i - 1)/(Npts - 1), t] x/(Npts - 1), {i, 1, Npts}] + 1

The first line solves ($**$), the second line computes $v(x,t)$ and the third line converts it back to $U(x,t)$ with the correct initial conditions. I implemented the integration over $v(x,t)$ as a Rieman sum with $Npts$ the number of discrete elements. With $Npts=50$, I get the following plot: enter image description here

The vertical axis represents $U$, the horizontal axis (from $2\pi$ to $0$) is $x$ and the 'depth' axis (from $0$ to $1/2$) is $t$. The plot is generated in Mathematica with

Plot3D[U[x, t, 50], {x, 0, 2 Pi}, {t, 0, 1/2}]

The formation of a shock at $x=\pi$ and $t=1/2$ is visible here as a kink in $U(x,t)$. Remember that $v = -2 \partial_x U$, so that a kink in $U$ is equivalent to a jump in $v$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.