Transformation from the PDE problem with a source to the PDE problem without it and viceversa

Question

In the study of nonlinear conservation laws a lot of time I work on the two problems given bellow:

$$(1) \hspace{1cm} \begin{cases} u_t+(f_{1}(u))_x=\lambda \cdot g(u) \\[2ex] u(x,0)=h_{1}(x) \end{cases} $$

$$(2) \hspace{1cm} \begin{cases} u_t+(f_{2}(u))_x=0 \\[2ex] u(x,0)=h_{2}(x) \end{cases} $$

Here u $\in \mathbb{R}^n$ and $\lambda$ is a constant. Besides $u$, the functions $g,f_1, f_2$ could also depend of $x$ and/or $t$.

A long time ago a professor of mine told me that he read somewhere that there is a way of transforming problem $(1)$ to the problem $(2)$ for some special cases of functions $g,f_1, f_2$. And vice versa. But he can't remember where he has read that.

Obviously it would be very useful if we can switch between problems $(1)$ and $(2)$ from time to time. In one moment you work on a problem that has a source term $\lambda \cdot g(u)$, i.e. problem $(1)$. In the other moment you use change of coordinates (or something similar) and now you work with a problem with no source term and with different initial condition (and possibly different flux too), i.e. problem $(2)$.

I am pretty sure that some general transformation doesn't exist. In the problem $(1)$ source is time dependent and in the problem $(2)$ new initial data are not time dependent. But

• maybe source could go partially into the flux and partially into the initial data or something similar.
• Or maybe this kind of transformation exists for some functions $g,f_1, f_2, h_1, h_2$ with the special properties.
• Or maybe it exists for some other kind of PDE problems - that don't deal with conservation laws.

The only example of transformation of this kind I know is: if $g(u)=(w(u))_x$ (assuming that function $w$ could be found), then the source term would be absorbed in the flux term. So we would get the problem $(2)$ form where $f_2 (u) = f_1 (u) - \lambda \cdot w(u)$ and $h_2(x)=h_1 (x).$

My question is: is anyone encountered some similar transformation and where (reference in the literature would be nice)? Of course you could share your ideas or answers too. This question intrigues me from the moment the professor told me that.

Answer 1

When $n=1$, you can always do this, at least near $t=0$, by solving a single inhomogeneous, linear first-order PDE; you can even arrange that $h_2 = h_1$. When $n>1$, there is a geometrical obstruction that can be computed in terms of $f_1$ and $\lambda g$. This is a classical fact in the geometry of PDE and characteristic systems. Here is a summary of the argument:

First, let's suppose that $n=1$ and that $f_1$ and $g$ are differentiable functions of $x,t,u$. Then the condition that $u(x,t)$ satisfies the equation and initial condition is that the surface graph $\bigl(u(x,t),x,t\bigr)$ in $\mathbb{R}^3$ should contain the curve $\bigl(h_1(x),x,0\bigr)$ and that the $2$-form $$ \Omega = \mathrm{d}u\wedge\mathrm{d}x -\mathrm{d}(f_1)\wedge\mathrm{d}t - \lambda\,g\ \mathrm{d}t\wedge\mathrm{d}x $$ should vanish on the surface graph. This is equivalent to saying that the surface graph is tangent to the characteristic vector field $$ X = \frac{\partial}{\partial t} + \frac{\partial f_1}{\partial u}\,\frac{\partial}{\partial x} + \left(\lambda g - \frac{\partial f_1}{\partial x}\right)\,\frac{\partial}{\partial u}. $$ Now let $\mu$ be the solution to the first-order, inhomogeneous, linear, scalar PDE $\mathrm{d}\mu(X) = \lambda g_u$, i.e., $$ \frac{\partial\mu}{\partial t} + \frac{\partial f_1}{\partial u}\,\frac{\partial\mu}{\partial x} + \left(\lambda g - \frac{\partial f_1}{\partial x}\right)\,\frac{\partial\mu}{\partial u} = \lambda\,\frac{\partial g}{\partial u} $$ such that $\mu(u,x,0) = 1$. (Such a solution always exists on a neighborhood of the plane $t=0$. With the right conditions on $f_1$ and $\lambda g$, the solution $\mu$ could exist globally on $\mathbb{R}^3$.)

Then $\bar\Omega = \mathrm{e}^{-\mu}\Omega$ will be closed (i.e., $\mathrm{d}\bar\Omega = 0$) and then it is easy to see that it can be written in the form $$ \bar\Omega = \mathrm{d}v\wedge\mathrm{d}x -\mathrm{d}f_2\wedge\mathrm{d}t $$ for some functions $v$ and $f_2$ on $\mathbb{R}^3$ with $v(u,x,0) = u$. Since $\mathrm{d}v\wedge\mathrm{d}x\wedge\mathrm{d}t = \bar\Omega\wedge\mathrm{d}t = \mathrm{e}^{-\mu}\mathrm{d}u\wedge\mathrm{d}x\wedge\mathrm{d}t\not=0$, it follows that $v,x,t$ are (local) coordinates on $\mathbb{R}^3$ (and, given reasonable assumptions on $f_1$ and $g$, will actually be global coordinates on $\mathbb{R}^3$). Writing $f_2$ as a function of $v,x,t$ we now have $$ \bar\Omega = \mathrm{d}v\wedge\mathrm{d}x -\mathrm{d}\bigl(f_2(v,x,t)\bigr)\wedge\mathrm{d}t $$ It follows that the coordinate transformation $(u,x,t)\mapsto\bigl(v(u,x,t),x,t\bigr)$ takes the first sytem to the second system (with $h_2 = h_1$), as desired.

What happens when $n>1$ is that the $2$-form $\Omega$ on $\mathbb{R}^3$ is replaced by $n$ $2$-forms on $\mathbb{R}^{n+2}$. In order to get to a system of the second kind, you would need to find $n$ linearly independent closed $2$-forms that have the same linear span as the given $n$ $2$-forms on $\mathbb{R}^{n+2}$. However, when $n>1$, there are `integrability conditions' on the $2$-forms in order to be able to do this.

For example, when $n=2$, you are asking when a pair of $2$-forms on $\mathbb{R}^4$ has the same linear span as a pair of closed $2$-forms, and there are well-known 'curvature' obstructions to this. Most of the time, the space of local, closed $2$-forms in the linear span of the two given $2$-forms is a finite dimensional space; in fact, nearly always, it's the zero dimensional space.

Answer 2

Here's one possible solution, but this may or may not be what your professor had in mind.

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Since $\lambda$ is a constant, we can ignore it by absorbing it into $g$.

Assume $u$ is scalar (takes value in $\mathbb{R}^1$).

Assume that $g$ is a function of $u$ only.

Let $G(s; u_0)$ be the solution to $\partial_s G = g(G)$ with initial data $G(0;u_0) = u_0$.

Suppose for every fixed $s$, the two functions $f(s;j)$ and $F(s;k)$ are related by $$(\partial_k F)(s;G(s;j)) = \partial_jf(s;j). \tag{*}$$

Then if $v(t,x)$ is a solution to the homogeneous equation

$$ \partial_t v + \partial_x [f(t;v)] = 0 $$

then the function $u(t,x) = G(t; v(t,x))$ satisfies

$$ \partial_t u + \partial_x [F(t; u)] = g(u) $$

You even have that $u(0,x) = v(0,x)$.

In particular, you can definitely convert from (1) to (2) in this setting, just by solving two ODEs (for $G$ and for $f$).

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When $u$ is not a scalar (has multiple components), the replacement for (*) may not be integrable, giving an obstruction to the sort of transformation you seek using this naive method. (Roughly, the equation is $ \partial f = (\partial G)^{-1} \cdot [(\partial F)\circ G] \cdot (\partial G)$.)