Below is a problem from a recent exam and I have some questions about it.
Given the boundary value problem:$$\dfrac{\partial^4 u}{\partial x^4} = f,\quad f\in L^2(0,1)$$ $$u(0) = u''(0) = u'(1) = u'''(1) = 0,$$ derive a variational formulation: $$a(u,v) = l(v)$$ by defining the appropriate Hilbert space $V$ and symmetric, continuous, coercive bilinear form $a(\cdot, \cdot),$ along with a continuous linear form $L(\cdot)$ on $V.$
So I let $$V = H_0^4(0,1) = \{v\in L^2(0,1)\vert\, v',v'',v''',v^{(4)}\in L^2(0,1)\text{ and } v(0) = v''(0) = v'(1) = v'''(1) = 0\}$$ and equip it with the norm $||\cdot||_V$ which I will get to later. Given the boundary conditions and starting with $u^{(4)} = f,$ it's fairly straightforward to derive: $$(f,v)=(u^{(4)},v) = -(u''',v) = (u'',v'') = -(u',v'''),\,\, \forall v\in V.$$ This clearly suggests that we take: $$l(v) = (f,v)\quad\text{ and }\quad a(u,v) = (u'',v'') = \int_{0}^1u''(x)v''(x)dx.$$ With this choice, the continuity of $a(\cdot, \cdot)$ and $l(\cdot)$ are just simple Cauchy-Schwartz. However, the coerciveness condition states: $$\exists \alpha >0 :a(v,v)\geq\alpha||v||_V^2,\,\,\forall v\in V.\quad (1)$$
This is where my confusions are. For instance, if I pick the standard norm on the Sobolev space, that is: $$||v||_V = \left(\int_0^1v^2+(v')^2+(v'')^2+(v''')^2+(v^{(4)})^2dx\right)^{\frac 12},$$ then I am not sure how to establish $(1).$On the other hand, I can improvise and take the norm: $$||v||_2 =\left(\int_0^1v^2+(v')^2+(v'')^2dx\right)^{\frac 12},$$ then I can easily establish $(1)$ with the choice $\alpha = \dfrac 13.$ This follows from proving: $$||v||_{L^2(0,1)}\leq ||v'||_{L^2(0,1)}\leq ||v''||_{L^2(0,1)}.$$
So to sum up, my questions are:
$1)$ Is there a problem with picking a norm of my choice to make the inequality easier like I did?
$2)$ If I have to go with the standard norm $||\cdot||_V,$ then is there an easy way to establish the coerciviness?
$3)$ Are my choices of $a(u,v)$ and $V$ correct?
Thanks in advance.
So, your choice of $V$ is wrong. The space $V=H_0^2(0,1)$. The idea of FEM is to reduce the differentiability condition of the strong solution. Here, as the problem is $4^{th}$ order after using integration by parts twice, we get a $2^{nd}$ order weak formulation.
The original problem is reduced to $$a(u,v)=(u'',v'')$$ where $u$ needs to be only second order differentiable and not fourth. Hence, $V=H_0^2(0,1)$.