# Partial Differential Equation $U_t - (U_x)^2 = 0$

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

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

figure(1);
clf;
subplot(1,2,1);
hg = plot(x,g,'k-');
xlim([0 2*pi]);
ylim([-1 1]);
xlabel('x');
ylabel('g');
ht1 = title(strcat('t =',num2str(t(1))));
subplot(1,2,2);
htheta = plot(x(2:nx),theta(1,2:nx),'k-');
xlim([0 2*pi]);
ylim([-1 1]);
xlabel('x');
ylabel('\theta');
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(hg,'YData',g);
set(htheta,'YData',theta(2:nx));
set(ht1,'String',strcat('t =',num2str(t(i))));
set(ht2,'String',strcat('t =',num2str(t(i))));
drawnow;
end


• 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? – Mohan Aditya Sabbineni Jun 13 at 9:07
• 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 – 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:

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 $$v$$.