Differential equations with distributions

[tjackson3]

Homework Statement

Solve [itex]x^2\frac{du}{dx} = 0[/itex] in the sense of distributions.

Homework Equations

<u',f> = -<u,f'> for any test function f.

The Attempt at a Solution

My thinking is that since we want to see the action of the left hand side on a general test function f, we try

<x^2u',f> = -<u,(x^2f)'> = 0

so clearly we can drop the negative

<u,(x^2f)> = 0

But I'm stuck as to where to go from here. On one hand, it would seem like the delta function would satisfy this, since [itex]\int_{-\infty}^{\infty} \delta(x)x^2f(x)\ dx = 0[/itex]. However, besides the fact that this seems too easy, there's an example in the book I'm using (Keener) which shows that the solution for the differential equation [itex]x\frac{du}{dx} = 0[/itex] is [itex]c_1 + c_2H(x)[/itex]. Since the solution for [itex]\frac{du}{dx} = 0[/itex] is just the constant distribution, this would seem to imply to met that the solution to my differential equation involves the constant distribution, the Heaviside distribution, and some other distribution. The book is extremely unclear on how to solve these problems. Any tips?

Thanks!

 

[mighty2000]

Hello,

You are on the right track with using the delta function as a solution to this differential equation. However, as you mentioned, this may seem too easy. In fact, the delta function is a fundamental solution to this type of differential equation, meaning that it forms a basis for the solutions. This means that any linear combination of the delta function will also be a solution. So your solution of c_1 + c_2H(x) is actually correct.

To understand this better, think about what the delta function represents. It is a distribution that has a value of 0 everywhere except at x = 0, where it has a value of infinity. This means that when you integrate the product of the delta function and any test function f, the only term that will contribute to the integral is when x = 0. So in a sense, the delta function acts like a "spike" or "point mass" at x = 0.

Using this understanding, you can see that the solution c_1 + c_2H(x) is actually a combination of two "spikes" - one at x = 0 (represented by c_2H(x)) and one at all other points (represented by c_1). This is why the book mentions the Heaviside distribution as a solution to the equation x\frac{du}{dx} = 0 - because it is a "spike" at all points except x = 0.

So to summarize, the solution to x^2\frac{du}{dx} = 0 in the sense of distributions is c_1 + c_2H(x), where c_1 and c_2 are constants. This can also be written as c_1 + c_2\delta(x) + c_2H(x), showing the two "spikes" at x = 0 and all other points.

I hope this helps clarify things! Keep up the good work in your studies.