Simplifying Trick for Homework Statement on Differential Equation

[Kreizhn]

Homework Statement

I'm given a differential equation that evolves as
$$\frac{du}{dt} = F(u,t; y(t) ), \qquad u \in \mathbb R^n$$
and told that a vector valued function P(u,t,y) satisfies
$$\frac{\partial P}{\partial t} = - \sum_{i=1}^N \frac{\partial}{\partial u_i} F_i(u,t,y) P$$

If it turns out that
$$\frac{du_i}{dt} = \sum_{j=1}^N A_{ij}(y)u_j$$
this is supposed to simplify $\frac{\partial P}{\partial t}$ greatly.

The Attempt at a Solution

I don't see how this simplifies at all. One possibility is that the question is poorly written and the summation term should actually be

$$- \sum_{i=1}^N \left[\frac{\partial}{\partial u_i} F_i(u,t,y) \right] P$$
In this case this is equivalent to the divergence of a linear vector field, and if we define A such that $(A)_{ij} = A_{ij}$ then indeed this does simplify to
$$\frac{\partial P}{\partial t} = - \text{Tr}[A] P$$
However, I have reason to believe that the equation is not wrongly written, in which case substituting our value for F into the differential equation for P doesn't drammatically simply the problem as suggested.

 

[mighty2000]

Thank you for bringing up this question. Based on the given information, it is possible that there may be a typo in the summation term. However, let's assume that the equation is written correctly and explore the potential simplifications that could arise.

First, let's rewrite the given differential equation as:
\frac{du}{dt} = F(u,t,y(t); y(t))

We can see that the function F depends on the variables u, t, and y(t). Therefore, we can rewrite the summation term as:
- \sum_{i=1}^N \frac{\partial}{\partial u_i} F_i(u,t,y(t); y(t)) P

Now, let's substitute the given equation for du/dt into the above expression:
- \sum_{i=1}^N \frac{\partial}{\partial u_i} \left( \sum_{j=1}^N A_{ij}(y(t))u_j \right) P

Expanding the summation and using the product rule, we get:
- \sum_{i=1}^N \sum_{j=1}^N A_{ij}(y(t)) \frac{\partial}{\partial u_i} (u_j P)

Now, using the chain rule, we can rewrite the partial derivative as:
- \sum_{i=1}^N \sum_{j=1}^N A_{ij}(y(t)) \left( \frac{\partial u_j}{\partial u_i} P + u_j \frac{\partial P}{\partial u_i} \right)

Since u_j does not depend on u_i, the first term in the parentheses is equal to 1. Therefore, we can simplify further to:
- \sum_{i=1}^N \sum_{j=1}^N A_{ij}(y(t)) \left( P + u_j \frac{\partial P}{\partial u_i} \right)

Finally, we can rewrite the summation term as a matrix multiplication:
- \sum_{i=1}^N \left( A(y(t)) P + u \cdot \frac{\partial P}{\partial u_i} \right)

This simplification may not seem significant at first glance, but it allows us to express the partial derivative of P with respect to t as a linear combination of