I found a statement in the book "Random Perturbations of Dynamical Systems" written by M.I.Freidlin and A.D. Wentzell that is not justified. Specifically, in a smooth domain $D \subset \mathbb{R}^n$ let us consider the Dirichlet's problem $$\mathcal{L}u(x) + c(x)u(x) = f(x),~~x\in D;\quad u(x)|_{x\in \partial D} = g(x),$$ in which $\mathcal{L}u(x) = \frac 12\sum_{i,j} a^{ij}(x)\frac{\partial^2 u}{\partial x^i\partial x^j}(x) + \sum_i b^i(x)\frac{\partial u}{\partial x^i}(x) $ with $a(x) = (a^{ij}(x))=\sigma(x)\sigma^T(x)$. Here $c(x)$, $f(x)$, for $x \in \mathbb R^n$, and $g(x)$, for $x\in \partial D$, are bounded continuous functions. If we suppose that $\mathcal{L}$ is uniformly elliptic, then under the additional assumption that $c(x) \leq 0$, this problem admits a solution provided by the Feynman-Kac formula, i.e., $$u(x) = -\mathbb{E}_x \int_0^\tau f(X_t)\exp\left\{\int_0^t c(X_s) ds\right\} + \mathbb{E}_x~ g(X_\tau)\exp\left\{\int_0^t c(X_s) ds\right\},$$ where $\tau = \inf\{t\colon X_t\notin D\}$ is the first exit time of the process defined via $dX_t = b(X_t)dt + \sigma(X_t)dW_t$ from the domain $D$. The authors claimed that "If $c(x) > 0$, then as is well known, Dirichlet's problem can "go out to the spectrum"; the solution of the equation $\mathcal{L}u + c(x)u = 0$ with vanishing boundary values may not be unique in this case." May I know why the last statement holds? Can you provide a baby example showing such non-uniqueness of solutions?
Non-uniqueness of a Dirichlet’s problem
elliptic-equationspartial differential equations
Related Solutions
First of all there is a typo in the partial differential equation(PDE) that you state. Let us firstly recall that PDE describes the time evolution of the one-point probability density function of the stochastic process in question. That PDE is termed the backward Chapman-Kolmogorov equation; you can look up the Wikipedia page to learn how to get from the Langevin equation (the stochastic differential equation) to the Chapman-Kolmogorov equation. Now the correct version is : \begin{equation} \frac{\partial f}{\partial t} + \mu x \frac{\partial f}{\partial x} + \frac{1}{2} \sigma^2x^2 \frac{\partial^2 f}{\partial x^2} = 0 \end{equation} Now solving PDEs is in general a hard task but in this particular case the substitution $y=\log(x)$ turns the PDE into one with constant coefficients. Indeed we have: \begin{eqnarray} \frac{\partial f}{\partial y} &=& x \frac{\partial f}{\partial x}\\ \frac{\partial^2 f}{\partial y^2} &=& x \frac{\partial f}{\partial x} + x^2 \frac{\partial ^2 f}{\partial x^2} \end{eqnarray} and therefore our PDE takes the form: \begin{equation} \frac{\partial f}{\partial t} + \left(\mu-\frac{1}{2} \sigma^2\right) \frac{\partial f}{\partial y} + \frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial y^2} = 0 \end{equation} which is basically saying that the logarithm of a geometrical Brownian motion with parameters $(\mu,\sigma^2)$ is a Gaussian stochastic process with drift $\mu - 1/2 \sigma^2$ and variance $\sigma^2$. Interestingly enough we would have arrived at the same conclusion by using the original Langevin equation and Ito's lemma only. Now going back to the PDE in question the standard way of solving it is to take Fourier transforms which is actually something you seek to obtain. We define: \begin{equation} \tilde{f}(k,t) := \frac{1}{2\pi} \int\limits_{\mathbb R} e^{\imath k y} f(y,t) d y \end{equation} and we transform the PDE above into a ODE: \begin{equation} \frac{d \tilde{f}}{d t} + \left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right] \cdot \tilde{f} = 0 \end{equation} which, given an initial condition at time $t=T$, is solved by: \begin{equation} \tilde{f}(k,t) = \tilde{f}(k,T) e^{\left[(\mu-\sigma^2) (-\imath k) + \frac{1}{2} \sigma^2 (\imath k)^2 \right](T-t)} \end{equation} This is the characteristic function that you were seeking. In applications like option pricing the initial condition $\tilde{f}(k,T)$ is equal to a Fourier transform of the option's payoff at maturity and the fair price of an option at some time before maturity is obtained by taking an inverse Fourier transform of the characteristic function.
Now, the really interesting question is to generalize that approach to more generic stochastic processes for example to L\'{e}vy stable processes. This leads to a fractional PDE. Those are much more difficult to handle yet it is worthwhile to pursue this approach because in here the Ito's lemma does not work.
Almost. It should be $u(t,x)=\Bbb E^x\left(\int_0^t f(t-s,X_s)\,ds\right)$.
Best Answer
The simplest example is to just take $c$ to be a (constant) eigenvalue of the operator. For example, if we take $\mathcal{L} = \Delta$, then we can find an infinite sequence of eigenfunction/eigenvalue solution to $$ \begin{cases} -\Delta u_k = \lambda_k u_k &\text{in } D \\ u_k = 0 & \text{on } \partial D. \end{cases} $$ These can be found in a number of ways. For example, we can minimize the functional $E(u) = \int_D |\nabla u|^2$ over $H^1_0(D)$ subject to the constraint that $J(u) = \int_D |u|^2 =1$ in order to find the principal eigenvalue, $\lambda_0>0$.
Once we have these, we immediately see that we cannot have unique solutions to $$ \begin{cases} \Delta u + \lambda_k u = f &\text{in } D \\ u = 0 & \text{on } \partial D \end{cases} $$ for any $k$.