Generalized PDEs with an apparent contradiction

[llarsen]

I have a question which has perplexed me for a time and thought maybe someone here would have some insight that might prove useful. My research involves a generalization of first order partial differential equations. The simplest case can be defined in the following manner: Let V be an arbitrary vector field which may be written as:

$$V = a_1 \frac{\partial}{\partial x_1} + a_2 \frac{\partial}{\partial x_2} + ... + a_n \frac{\partial}{\partial x_n}$$

where the coefficients $$a_i$$ are functions of $$x_1 , ... , x_n$$. Find a nontrivial function f such that:

$$V(f) = h(f)$$ where $$h(f)$$ is an arbitrary function of f. In expanded form the equation becomes:

$$a_1 \frac{\partial f}{\partial x_1} + a_2 \frac{\partial f}{\partial x_2} + ... + a_n \frac{\partial f}{\partial x_n} = h(f)$$

This is a generalization of a partial differential equation since $$h(f)$$ is not a prescribed function, but rather an arbitrary function of f. If one prescribes the function (say $$h(f)=1$$) then this becomes a partial differential equation which can be solved numerically when a solution exists. Note that the solution set for $$V(f)=h(f)$$ should be larger than in that case were $$h(f)$$ is explicitly prescribed (i.e. there is more flexibility since h(f) is arbitrary).

The equation $$V(f)=h(f)$$ proves to be an unweildy form of the equation to work with since $$h(f)$$ is not prescribed. In order to make this more amenable to numerical analysis, I apply an exterior derivative to get $$dV(f) = dh(f) = \frac{\partial h}{\partial f} df$$. In order to eliminate the troublesome arbitrary function $$h(f)$$ one simply takes the wedge product with df to get $$dV(f) \wedge df = h_f df \wedge df \equiv 0$$. Note that since this is exactly equal to 0 (since $$df \wedge df \equiv 0$$) each component with be zero. This leads to a set of equations by extracting the individual components (which equal 0) from:

$$dV(f) \wedge df \equiv 0$$

However, while the partial differential equation $$V(f)=1$$ is more restrictive than $$V(f)=h(f)$$, $$V(f)=1$$ produces a single equation, while $$dV(f) \wedge df \equiv 0$$ produces $$\frac{n!}{2!(n-2)!}$$ equations (one for each component of the two form $$dV(f) \wedge df \equiv 0$$). One the surface this would make it appear that $$dV(f) \wedge df \equiv 0$$ is more restrictive than $$V(f)=1$$ throughout the domain, when in actuality is should be less restrictive. This seems to suggest that there is some redundancy built into the equation $$dV(f) \wedge df \equiv 0$$. I am curious is anyone has insight into the apparent contradiction or a good explanation for the redundancy.

I understand that $$dV(f) \wedge df \equiv 0$$ includes second derivative terms which means that one has more flexiblity to choose values on the boundary, but it seems to me that the equation shouldn't be more restrictive through the domain as it appear to be based on the number of equations that must be satisified in each case.

Incidentally, I have not seen any references to people numerically solving problems of the form $$V(f) = h(f)$$ or generalizations such as:

$$V(f) = h_1 (f,g)$$
$$V(g) = h_2 (f,g)$$

where $$h_i$$ are arbitrary functions. If anyone knows of numerical methods for solving such problems, I would be very interested in any references to research or papers on the topic.

 

[jvicens]

Thank you for your question. As a scientist who specializes in partial differential equations, I can offer some insights into your research and the apparent contradiction you have encountered.

Firstly, let me address the issue of redundancy in the equation dV(f) \wedge df \equiv 0. It is true that this equation produces a larger set of equations compared to V(f)=1, but this does not necessarily mean that it is more restrictive. In fact, the additional equations in dV(f) \wedge df \equiv 0 come from the exterior derivative, which introduces second derivative terms. This means that the solution set for dV(f) \wedge df \equiv 0 is actually larger than the solution set for V(f)=1. The additional equations provide more flexibility in choosing values on the boundary, as you have mentioned, and this can lead to a larger solution set.

As for your question about numerical methods for solving equations of the form V(f) = h(f) or generalizations such as V(f) = h_1 (f,g), V(g) = h_2 (f,g), there are indeed numerical methods available for solving such problems. These methods fall under the category of finite difference methods, which are commonly used for solving partial differential equations numerically. One example is the finite difference method for solving elliptic partial differential equations, which can be applied to your generalization of first order partial differential equations. You can find more information about this method in numerical analysis textbooks or online resources.

I hope this helps to clarify the apparent contradiction and provide some direction for your research. Keep up the good work and don't hesitate to reach out for further discussions or assistance.

 

[tacman]

Thank you for sharing your research and thoughts on this topic. It is indeed an interesting and complex question that you have raised.

First, let's address the apparent contradiction between the two equations, V(f)=h(f) and dV(f) \wedge df \equiv 0. While it may seem like dV(f) \wedge df \equiv 0 is more restrictive since it produces more equations, it is important to note that these equations are not all independent. The wedge product with df introduces a constraint that must be satisfied in addition to the original equation V(f)=h(f). So, while the number of equations may be greater, they are not all providing new information.

In fact, the equation dV(f) \wedge df \equiv 0 is not redundant, but rather a necessary condition for the existence of a solution to V(f)=h(f). This can be seen by considering the exterior derivative of both sides of the equation, which yields dV(f) = dh(f) = \frac{\partial h}{\partial f} df. This is equivalent to the constraint introduced by the wedge product in the expanded form of the equation.

As for numerical methods for solving such equations, there are a few approaches that can be taken. One option is to use finite difference methods, which approximate the derivatives in the equation with discrete values and solve the resulting system of equations. Another option is to use finite element methods, which discretize the domain and solve the equation on each element. Both of these methods have been used for solving generalized PDEs, including equations of the form V(f)=h(f).

In summary, while there may seem to be a contradiction or redundancy in the two equations you have presented, they are in fact complementary and necessary for finding a solution. I hope this helps to clarify the issue and provides some direction for your research. Best of luck in your studies.