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Let $$a(v,w) = \int_0^1 a_o(x)v(x)w(x)dx + \lambda \int_0^1 v(y)w(x)dxdy, \; \; \forall v,w \in L^2(0,1)$$ be a bilinear, continuous, elliptic functional with $a_0(x) \in C^0[0,1], \; \;a_0(x) > 0$ on $[0,1]$ and $\lambda \in \mathbb{R}.$ Suppose that $f \in L^2(0,1)$ and consider the variational problem:

$$u \in L^2(0,1) \; \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$ a(u,v) = \int_0^1f(x)v(x)dx, \; \; \forall v \in L^2(0,1) $$

(i) Suppose that $a_0 = 1$. Show that if $\lambda \ne -1$ then this problem has a closed form solution and find it.

(ii) if $\lambda = -1$, the problem has no solutions unless $\int_0^1f(y)dy = 0$ and if this holds, then it actually has infinitely many solutions of the form $u = f + C$ where $C$ is a constant.

What I have so far:

I have manipulated the formulation of the problem to arrive at:

$$ a_0(x)u(x) + \lambda \int_0^1u(y)dy = f(x), x \in (0,1) $$

Substituting the given conditions, this becomes:

$$u(x) + \lambda\int_0^1u(y)dy = f(x), x \in (0,1) $$

I'm kind of stuck here. I was thinking that this could be differentiated somehow to turn it into an ODE that can be shown to be solvable/not solvable based on what lambda is, but I'm not sure how to make that work... or if it's even a valid idea.

Thanks for your input!

1 Answers 1

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i) integrate $u(x) + \lambda \int_{0}^{1} u(y) dy = f(x)$ from $0$ to $1$ to get \begin{equation} \int_{0}^{1} u(x) dx + \lambda \int_{0}^{1} \int_{0}^{1} u(y) dy dx = \int_{0}^{1} f(x) dx \end{equation} now giving $\int_{0}^{1} u(y) dy$ a name, say, $\int_{0}^{1} u(y) dy = q$, we have \begin{equation} q + \lambda \int_{0}^{1} q dx = q (1 + \lambda) = \int_{0}^{1} f(x) dx \end{equation} so, provided, $\lambda \neq -1$, we have \begin{equation} \int_{0}^{1} u(y) dy = q = \frac{1}{(1 + \lambda)}\int_{0}^{1} f(x) dx \end{equation} plugging this back into the original equation gives \begin{equation} u(x) = f(x) - \lambda q \end{equation} or simply \begin{equation} u(x) = f(x) - \frac{\lambda}{(1 + \lambda)}\int_{0}^{1} f(x) dx \end{equation}

ii) falls out directly from the Fredholm alternative: https://en.wikipedia.org/wiki/Fredholm_alternative