Why is th coerciveness of a bilinear form important for making sure a variational problem is well-posed?

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Hello, I've come across the idea of the coerciveness of a bilinear form in the context of variational problems. Could you help me understand in simpler terms why this property is so important for making sure the problem is well-posed? I appreciate any insights you can share.

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For example, if we consider Poisson equation, it is often written in variational form as follows:

$$ - \Delta u = f~~~~~ ~~in~~ \Omega$$

subject to appropriate boundary conditions, where $u$ is the unknown function, $\Delta$ is the Laplace operator, $f$ is a given function, and $\Omega$ is the domain of the problem.

The corresponding variational problem can be stated as finding $u$ in a certain function space $V$ such that:

$a(u,v)=F(v)~~$ for all $~~v∈V$

Here, $a$ is a coercive bilinear form, and $F$ is a linear functional. The bilinear form $a$ and the linear functional $F$ are defined as:

$a(u,v)=\int_{\Omega}∇u⋅∇vdx$

$F(v)=\int_{\Omega}f⋅v dx$

The coerciveness condition typically involves showing that there exists a constant $C>0$ such that:

$a(u,u)≥C\int_{\Omega} ∣∇u∣^2 dx$

This condition ensures that the bilinear form is coercive and leads to a well-posed variational problem for the Poisson equation.