Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Consider the second order linear elliptic differential equation

$$Lu=(\sum_{i=1}^d{\partial^2\over\partial\theta_i^2}+{\partial b\over\partial \theta_i}{\partial\over\partial\theta_i})u=exp(i\theta_1)$$

This differential equation is defined on the domain $[0,2\pi]^d$ with periodic boundary condition. The function $b$ are some polynomial in terms of sines and cosines in $\theta_k$, $k=1,..,d$. Is there some algorithm to compute the coefficients of the power series or better yet the Fourier series of the solution $u$?

share|improve this question
add comment

1 Answer

In general this will not have a solution. Note that your equation is equivalent to the following:

\[ \Delta_g (u) = \sum_{i} \frac{1}{a} \frac{\partial}{\partial \theta_i}( a \frac{\partial u}{\partial \theta_i} ) = e^{i\theta_1} \]

where by definition $a = e^b$. This is the Laplace equation (see the Wikipedia article "Laplace-Beltrami operator") with respect to the metric

\[ g_{ij} = e^{2b} \delta_{ij} \]

Hence it only has a solution if the average value of $e^{i\theta_1}$ with respect to the volume form of $g$ is zero. The reason for this is the integration by parts formula

\[ \int_{[0,2\pi]^d} (\Delta_g u)\cdot v dvol_g = \int_{[0,2\pi]^d} u \cdot (\Delta_g v) dvol_g \]

which when you plug in $v = 1$ shows that the average value of $\Delta_g u $ must be zero. The formula can be proved in much the same way as for the standard Laplacian. Thus a necessary (and sufficient) condition for a solution to exist is:

\[ \int_{[0,2\pi]^d} e^{b+ i \theta_1} d\theta_1 \cdots d\theta_n = 0\]

In general, even if a solution exists I wouldn't expect a particularly nice description of the Fourier coefficients. You might be able to get estimates of them, particularly if the function $b$ is very small, in which case the equation will closely resemble the standard Laplacian on the torus. If $b$ has very few terms you might try taking the Fourier transform of your equation, in which case the second term will become a convolution and you may be able to solve inductively for the Fourier coefficients if you're lucky.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.