Chapter 3 Strong and weak forms of one-dimensional problems

Strong form: Governing equations along with the boundary conditions

Weak form: Integral form of the equations in the strong form ----> Weak continuity requirements

Finite difference method (FDM):

Strong form -----> A set of discrete equations

Finite element method (FEM):

Strong form -----> Weak form -----> (Combined with the approximation of functions) -----> Discrete equations

The strong form in one-dimensional problems

An axial loaded elastic bar

1DBar

Internal force $p(x)$; body force $b(x)$ (force per unit length); traction $\overline{t}$ (force per unit area)

$-p(x)+b(x+\frac{\Delta x}{2})\Delta x+p(x+\Delta x)=0$

$\frac{p(x+\Delta x)-p(x)}{\Delta x}+b(x+\frac{\Delta x}{2})=0$

take the limit $\Delta x \rightarrow 0$

$\frac{d p(x)}{dx}+b(x)=0$

Then, linear assumption (def of strain & stress-strain law) is adopted $\Downarrow$

strain is given by:

$\varepsilon(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{u(x+\Delta x)-u(x)}{\Delta x}=\frac{du}{dx}$

Internal force is given by

$p(x)=A(x)\sigma (x)=A(x)E\varepsilon (x)=AE\frac{du}{dx}$

So, governing equation (2-order ODE) is given by

$\frac{d}{dx}(AE\frac{du}{dx})+b=0$, ($0<x<l$)

With the boundary conditions at the two ends, the strong form becomes

$\frac{d}{dx}(AE\frac{du}{dx})+b=0$ on $0<x<l$

$\sigma (x=0)=(E\frac{du}{dx})_{x=0}=-\overline{t}$

$u(x=l)=\overline{u}$

Note that the sign convention for $\sigma$ and $\overline{t}$, former is tension + and compression -, latter is positive x-axis +,negative x-axis -

Steady-state heat conduction in one dimension

The form of heat transfer: conduction, convection, thermal radiation

1DCondcution

Conservation of energy:

$s(x+\frac{\Delta x}{2})\Delta x+q(x)A(x)-q(x+\Delta x)A(x+\Delta x)=0$

$\frac{q(x+\Delta x)A(x+\Delta x)-\Delta x+q(x)A(x)}{\Delta x}-s(x+\frac{\Delta x}{2})=0$

Take the limit $\Delta x \rightarrow 0$

$\frac{dqA}{dx}-s=0$

The constitutive equation for heat flow is given by

$q=-k\frac{dT}{dx}$

Then, the governing equation is given by

$\frac{d}{dx}(Ak\frac{dT}{dx})+s=0$

With the boudany conditions at two ends, the strong form becomes

$\frac{d}{dx}(Ak\frac{dT}{dx})+s=0$ on $0<x<l$

$q=-k\frac{dT}{dx}=-\overline{q}$ on $x=0$

$T=\overline{T}$ on $x=l$

The negative sign in $q=-\overline{q}$ is beacuse the prescribed flux $\overline{q}$ is positive when heat flows out of the bar

Diffusion in one dimension

P46 Similar to 1D heat conduction...

The weak form in one dimension

For the problem of axial loaded elastic bar, ...

use the weight function (test function) $w(x)$ to multiply the gonverning equation and boudary conditions ...

and integrate over the domian they hold

$\int_0^l w[\frac{d}{dx}(AE\frac{du}{dx})+b]dx=0$, $\forall w$ with $w(l)=0$

$(wA(E\frac{du}{dx}+\overline{t}))_{x=0}=0$, $\forall w$ with $w(l)=0$

The arbitariness of the weight function $w(x)$ is crucial

Use the integration by parts, the weak form is transformed into

$\int_0^l w[\frac{d}{dx}(AE\frac{du}{dx})+b]dx=(wAE\frac{du}{dx})|_0^l-\int_0^l \frac{dw}{dx}AE\frac{du}{dx}dx+\int_0^lwbdx=0$

For $w(l)=0$ and $(wA(E\frac{du}{dx}+\overline{t}))_{x=0}=0$

$(wA\sigma)_{x=l}-(wA\sigma)_{x=0}-\int_0^l \frac{dw}{dx}AE\frac{du}{dx}dx+\int_0^lbdx=(wA\overline{t})_{x=0}-\int_0^l \frac{dw}{dx}AE\frac{du}{dx}dx+\int_0^lwbdx=0$

So the weak form of the above problem becomes:

$--------------------------$

Find $u(x)$ among the smooth functions that satisfy $u(l)=\overline{u}$ such that

$\int_0^l \frac{dw}{dx}AE\frac{du}{dx}dx=(wA\overline{t})_{x=0}+\int_0^lwbdx$, $\forall w$ with $w(l)=0$

$--------------------------$

Trial solutions/ candidate solutions: a set of admissible solutions $u(x)$ that satisfy certain conditions

Boundary conditions

Essential boundary conditions: BCs that trial solutions $u(x)$ must saitisfy (Dispalcement)

Natural boundary conditions: BCs that emnate naturally from the weak form and trial solution $u(x)$ needed to satisfy (Traction)

Requirements for using weak form: Admissible

Admissible of trial solution: smooth and satisifies the essential boundary conditions

Admissible of weight function: smooth and vanishes the essential boundaries

Continuity

$C^n$ fucntion: the derivatives of the function of order $j$ exist (0$\leq j \leq n$) and are continuous functions in the entire domain

Continuity

Jumps: Strong discontinuities; Kinks: Weak discontinuities

In general, the derivative of $C^n$ fucntion is $C^{n-1}$ fucntion

The equivalence between the weak and strong forms

Key to the proof: The arbitrariness of the weight function $w(x)$

The trial solution $u(x)$ that satisfies the weak form satisfies the strong form

One-dimensional stress analysis with arbitary boundary conditions

$\Gamma_u \cup \Gamma_t=\Gamma$

Any boudanry is either an essential boundary or a natural boundary and their union is the enire boundary

$\Gamma_u \cap \Gamma_t=0$

Natural boundary conditions and essential boundary conditions cannot be applied at the same boundary points

P58 The two boundaries are said to be complementary

Strong form for 1D stress analysis

$\frac{d}{dx}(AE\frac{du}{dx})+b=0$ on $0<x<l$,

$\sigma n=En\frac{du}{dx}=\overline{t}$ on $\Gamma_t$,

$u=\overline{u}$ on $\Gamma_u$

$n$ -----> the normal vector pointing outwards

Weak form for 1D stress analysis

$H^1$ is a space of functions with square integrable derivatives and $H^1 \subset C^0$

P60 $H^1$ contains an infinite number of functions and is called an infinite set

Define two function spaces $U$ and $U_0$ :

$U=\{u(x)|u(x) \in H^1, u=\overline{u}$ on $\Gamma_u\}$

$U_0=\{w(x)|w(x) \in H^1, w=0$ on $\Gamma_u\}$

Use the weight function $w(x)$, the strong form becomes

$\int_{\Omega}w(\frac{d}{dx}(AE\frac{du}{dx})+b)dx=0$, $\forall w$ ,

$(wA(\overline{t}-\sigma n))|_{\Gamma_t}=0$, $\forall w$

For the boundary parts $\Gamma$ can be divided into $\Gamma_t$ and $\Gamma_u$ and $w$ vanishes on $\Gamma_u$

$\int_{\Omega} w[\frac{d}{dx}(AE\frac{du}{dx})+b]dx=(wAE\frac{du}{dx}n)|_{\Gamma_u}+(wAE\frac{du}{dx}n)|_{\Gamma_t}-\int_{\Omega} \frac{dw}{dx}AE\frac{du}{dx}dx+\int_{\Omega}wbdx=0$

So, $\int_{\Omega} \frac{dw}{dx}AE\frac{du}{dx}dx=(wA\overline{t})_{\Gamma_t}+\int_{\Omega}wbdx$

The weak form becomes

$--------------------------$

Find $u(x) \in U$ such that

$\int_{\Omega} \frac{dw}{dx}AE\frac{du}{dx}dx=(wA\overline{t})_{\Gamma_t}+\int_{\Omega}wbdx$, $\forall w \in U_0$

$--------------------------$

One-dimensional heat conduction with arbitary boundary conditions

Similar with 1D stress analysis...

Two-point boundary value problems wit generalized boundary conditions

TWo-point boundary value problem:

$\frac{\partial}{\partial x}(A\kappa \frac{\partial \theta}{\partial x})+f=0$ on $\Omega$

Diffusion, heat conduction, elastic problems in this Chapter are all of the above form

Generalized boundary conditions:

$(\kappa n \frac{\partial \theta}{\partial x}-\overline{\Phi})+\beta(\theta-\overline{\theta})=0$ on $\Gamma_{\Phi}$

When $\beta$ is a penalty paramerter (large number), the essential boundary conditions become a limiting case of the equation above

then $\Gamma \equiv \Gamma_{\Phi}$

Two approches to deal with the boundary condition: Penalty method and Partition method

Strong form for two-point boundary value problems with generalized boundary conditions

General strong form for 1D problems-penalty method

$\frac{\partial}{\partial x}(A\kappa \frac{\partial \theta}{\partial x})+f=0$ on $\Omega$

$(\kappa n \frac{\partial \theta}{\partial x}-\overline{\Phi})+\beta(\theta-\overline{\theta})=0$ on $\Gamma$

General strong form for 1D problems-partition method

$\frac{\partial}{\partial x}(A\kappa \frac{\partial \theta}{\partial x})+f=0$ on $\Omega$

$(\kappa n \frac{\partial \theta}{\partial x}-\overline{\Phi})+\beta(\theta-\overline{\theta})=0$ on $\Gamma_{\Phi}$

$\theta=\overline{\theta}$ on $\Gamma_{\theta}$

Weak form for two-point boundary value problems with generalized boundary conditions

General weak form for 1D problems-penalty method

Find $\theta(x) \in H^1$ such that

$\int_{\Omega}\frac{dw}{dx}A\kappa\frac{d\theta}{dx}dx-\int_{\Omega}wfdx-wA(\overline{\Phi}-\beta(\theta-\overline{\theta}))|_{\Gamma}=0$, $\forall w \in H^1$

General weak form for 1D problems-partition method

Find $\theta(x) \in U$ such that

$\int_{\Omega}\frac{dw}{dx}A\kappa\frac{d\theta}{dx}dx-\int_{\Omega}wfdx-wA(\overline{\Phi}-\beta(\theta-\overline{\theta}))|_{\Gamma_{\overline{\Phi}}}=0$, $\forall w \in U_0$

Advection-diffusion (1D)

Conservation principle: the species (material ,energy or state) is conserved in each volume $\Delta x$

i.e. the amount of species entering minus the amount of leaving equals the amount produced

Notation:

Concentration of species: $\theta(x)$

Cross section area: $A(x)$

Velocity of the fluid: $v(x)$

Source: $s(x)$

control volume: $\Delta x$

Strong form of advection-diffusion equation

From the conervation principle

$(Av\theta)_x+(Aq)_x-(Av\theta)_{x+\Delta x}-(Aq)_{x+\Delta x}+s(x+\frac{\Delta x}{2})\Delta x=0$

Take the limit $\Delta x \rightarrow 0$

$\frac{d (Av\theta)}{d x}+\frac{d (Aq)}{d x}-s=0$

For an incompressible flow, the volume of material entering the control volume equals the volume leaving.

$(Av)_x=(Av)_{x+\Delta x}$

Take the limit $\Delta x \rightarrow 0$

$\frac{d (Av)}{d x}=0$

So the governing equation becomes

$\frac{d (Av\theta)}{d x}+\frac{d (Aq)}{d x}-s=\theta \frac{d (Av)}{d x}+Av\frac{d \theta}{dx}+\frac{d (Aq)}{d x}-s=0$

$Av\frac{d \theta}{dx}+\frac{d (Aq)}{d x}-s=0$

Assume the diffusion is linear and use the Fick's first law

$q=-k\frac{d\theta}{dx}$

Then the advection-diffusion equation is as follows

$\underbrace{Av\frac{d \theta}{dx}}_{advection}-\underbrace{\frac{d}{dx}(Ak\frac{d\theta}{dx})}_{Diffusion}-\underbrace{s}_{Source}=0$

With the boundary conditions

$\theta=\overline{\theta}$ on $\Gamma_{\theta}$

$qn=-k\frac{d\theta}{dx}n=\overline{q}$ on $\Gamma_q$

Weak form of advection-diffusion equation

Multipy the governing eqaution and intergrate over the domain

Intergrate the diffusion term by parts, we can get the weak form:

Find $\theta(x)\in U$ such that

$\int_{\Omega}wAv(\frac{d\theta}{dx})dx+\int_{\Omega}\frac{dw}{dx}Ak\frac{d\theta}{dx}dx-\int_{\Omega}wsdx+(wA\overline{q})|_{\Gamma_q}=0$ for $\forall w \in U_0$

Instead of using the flux boundary condition (natural) above, follow generalized boundary condition can be used

$(-k\frac{d\theta}{dx}+v\theta)n=\overline{q}_T$

Then integrate the first term by parts in the above weak form, we can get

$-\int_{\Omega}\frac{dw}{dx}Av\theta dx+\int_{\Omega}\frac{dw}{dx}Ak\frac{d\theta}{dx}dx-\int_{\Omega}wsdx+(wA\overline{q}_T)|_{\Gamma_q}=0$

... ...

Minimum potential energy

Theorem of minimum potential energy

$------------------------$

The solution of the strong from is the minimizer of

$W(u(x))=\underbrace{\frac{1}{2}\int_{\Omega}AE(\frac{du}{dx})^2dx}_{W_{int}}-\underbrace{(\int_{\Omega}ubdx+(uA\overline{t})|_{\Gamma_t})}_{W_{ext}}$

$------------------------$

Potential energy of the system : $W(u(x))$ -----> functional (function of function)

The physical meaning:

The solution is minimizer (a stationary point) of the potential energy $W$ among all admissible displacement functions

Proof of equivalence:

A variation of the function $u(x)$ : $\delta u(x) \equiv \zeta w(x)$

where $w(x)$ is an arbitary function and $0<\zeta\ll1$

Variation in the functional : $\delta W=W(u(x)+\zeta w(x))-W(u(x))\equiv W(u(x)+\delta u(x))-W(u(x))$

For $(U(x)+\zeta w(x)) \in U$, $w(x)$ must vanish on the essential boundary, $w(x)\in U_0$

Variation of in the internal and external work

$\delta W_{int}=\frac{1}{2}\int_{\Omega}AE(\frac{du}{dx}+\zeta\frac{dw}{dx})^2dx-\frac{1}{2}\int_{\Omega}AE(\frac{du}{dx})^2dx=\zeta\int_{\Omega}AE(\frac{du}{dx})(\frac{dw}{dx})dx$

$\delta W_{ext}=\delta W_{ext}^{\Omega}+\delta W_{ext}^{\Gamma}=\int_{\Omega}(u+\zeta w)bdx-\int_{\Omega}(u)bdx+(u+\zeta w)A\overline{t}|_{\Gamma_t}-(u\overline{t})A|_{\Gamma_t}=\zeta(\int_{\Omega}wbdx+(wA\overline{t})|_{\Gamma_t})$

At the minimum of $W(u(x))$, there should be $\delta W=\delta W_{int}-\delta W_{ext}=0$

So we can get the follwoing statement

Find $u\in U$ such that

$\frac{\delta W}{\zeta}=\int_{\Omega}AE(\frac{du}{dx})(\frac{dw}{dx})dx-\int_{\Omega}wbdx-(wA\overline{t})|_{\Gamma_t}=0$ , $w\in U_0$

Alternatively, use the $\delta u$ to replace the $\zeta w$, i.e.

Find $u\in U$ such that

$\delta W=\int_{\Omega}AE(\frac{du}{dx})(\frac{d(\delta u)}{dx})dx-\int_{\Omega}\delta u bdx-(\delta u A\overline{t})|_{\Gamma_t}=0$ , $w\in U_0$

The minimizer of the potential energy functional <--------> The weak form <--------> The strong form

Principle of virtual work

Use the strain-displacement equation and the stress-strain law

$\delta W=\underbrace{\int_{\Omega}A\sigma\delta \varepsilon dx}_{\delta W_{int}}-\underbrace{\int_{\Omega}b\delta udx-(\overline{t}A\delta u)|_{\Gamma_t}}_{\delta W_{ext}}=0$

So the principle of virtual work can be stated as follows:

The admissible displacement field ($u\in U$) for which the variational in the internal work $\delta W_{int}$ equals the variation in the external work $\delta W_{ext}$ for $\forall \delta u\in U_0$ satisfies equilibrium and boundary conditions.

$W_{int}=\int_{\Omega}w_{int}Adx=\frac{1}{2}\int_{\Omega}AE\varepsilon^2dx$

where $w_{int}$ is the energy per volume called energy density

Some weak forms can be converted to variational principles

The potential energy theorem holds for any elastic system and the similar variational principle can be formulated for heat conduction

Variational principles can only be developed for systems that are self-adjoint

Integrability

Integrability for the weak form :

The integrals in the weak form can be evaluated, which requires the smoothness of the trial solution and weight function

A derivative of a function $u(x)$ is called square integrable if $W_{int}(\theta)$ is bounded ($W_{int}(\theta)<\infty$)

$W_{int}(\theta)=\frac{1}{2}\int_{\Omega}\kappa A(\frac{d\theta}{dx})^2dx$

The required smoothness in FEM: The weight and trial functions are required to posses square integrable derivatives

In elasticity, $W_{int}(\theta)$ corresponds to the strain energy

The value of $\sqrt{W_{int}(\theta)}$ is often called an energy norm

The notions of required smoothness also have a physical basis (Compatibility of the displacement field, no gaps or overlaps...)

$C^0$ continuity is enough for the weight and trial functions