Consider the differential equation $$ \frac{\text d^2 u}{\text d x^2} + \lambda_1 \frac{\text d u}{\text d x} + \lambda_2 u = -f(x),\quad\text{for}\quad x\in [a,b], $$ with boundary conditions $$ u(a) = u(b) = 0, $$ where $\lambda_1$ and $\lambda_2$ are both constant. The function $f$ is a continuous function of $x$.
(a) Write down a weak formulation of this differential equation, including definitions of the inner product and the function space $V$ used.
I need help with formulating the weak form of this PDE. i have done it but not sure if it is correct,
my working:
$ u_{xx} +\lambda_{1} u_x + \lambda_2 u = -f(x) $
inner product is defined as $ \left< g,h\right> = \int_{a}^{b} g(x)h(x) dx$
choosing $v$ such that $v(a) = v(b) = 0$
$v \in V$ and $V = \lbrace v:v \text{ is continuous on } [a,b], \text{ piecewise continuous and bounded on } [a,b], v(a) = v(b) =0\rbrace$
multiply by $v$ and integrate over $[a,b]$
$ \int_{a}^{b} v u_{xx} + \lambda_{1} \int_{a}^{b}\ v u_x + \lambda_{2}\int_{a}^{b} uv = \int_{a}^{b}-vf(x) dx$
using integration by parts i get
$ [vu_x]_{a}^{b}- \int_{a}^{b}v_x u_x + \lambda_{1}[vu]_{a}^{b} - \lambda_{1}\int_{a}^{b}v_xu+ \int_{a}^{b}uv = \int _{a}^{b}-vf(x) dx $
write this in inner product form, using the above definition, and the fact that $v(a)=v(b)=0$
$ \left< v_x,u_x\right> -\lambda_{1}\left< v_x, u\right> + \lambda_{2}\left<v,u\right> = -\left<v,f\right> $ is this correct?
also am not sure weather the second term in my expression is zero, or it should be included in this expression?
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– Han de Bruijn May 20 '18 at 19:48