# Is a symmetric bilinear form necessary to ensure a weak formulation has a solution?

Problem

I want to convert the general second order linear PDE problem \begin{align} \begin{cases} a(x,y)\frac{\partial^2 u}{\partial x^2}+b(x,y) \frac{\partial^2 u}{\partial y^2} +c(x,y)\frac{\partial^2 u}{\partial x \partial y}\\+d(x,y)\frac{\partial u}{\partial x}+e(x,y)\frac{\partial u}{\partial y}+f(x,y)u=g(x,y) & \text{in } R \text{ PDE} \\ u=u^* & \text{on } S_1 \text{ Dirchlet boundary condition} \\ \dfrac{\partial u}{\partial n}=q^* & \text{on } S_2 \text{ Neumann boundary condition} \\ \dfrac{\partial u}{\partial n}=r^*_1-r^*_2 u & \text{on } S_3 \text{ Robin boundary condition} \\ \end{cases} \end{align} into a weak form suitable for the finite element method. That is into the weak bilinear form $$B(u,v)=L(v)$$ where $$B$$ is bilinear, symmetric and positive definite functional and $$L$$ is a linear functional.

Work thus far

I know to how convert the following
\begin{align} \begin{cases} \dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}+u=g(x,y) & \text{in } R \text{ PDE} \\ u=u^* & \text{on } S_1 \text{ Dirchlet boundary condition} \\ \dfrac{\partial u}{\partial n}=q^* & \text{on } S_2 \text{ Neumann boundary condition} \\ \dfrac{\partial u}{\partial n}=r^*_1-r^*_2 u & \text{on } S_3 \text{ Robin boundary condition} \\ \end{cases} \end{align} into the weak bilinear form $$B(u,v)=L(v)$$ where $$B$$ is bilinear, symmetric and positive definite and $$L$$ is linear. The steps are as follows (note that $$v$$ is our test function) \begin{align} \int \int_{R} \left(\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+u \right) v \ dA &= \int \int_{R} g(x,y) v \ dA \end{align} Using the identity \begin{align} \int \int_{R} v \nabla^2 u\ dA &= \int_{S} v \frac{\partial u}{\partial n}\ ds-\int\int_{R} \nabla u \cdot \nabla v\ dA \end{align} We get \begin{align} \int \int_R -\nabla u \cdot \nabla v +uv \ dA &= \int \int_R g v \ dA - \int_{S} v \frac{\partial u}{\partial n}\ ds \\ \int \int_R -\nabla u \cdot \nabla v +uv \ dA &= \int \int_R g v \ dA - \int_{S_1} v \frac{\partial u}{\partial n}\ ds- \int_{S_2} v \frac{\partial u}{\partial n}\ ds - \int_{S_3} v \frac{\partial u}{\partial n}\ ds \\ \int \int_R -\nabla u \cdot \nabla v +uv \ dA &= \int \int_R g v \ dA - \int_{S_2} v q^* \ ds - \int_{S_3} v (r^*_1-r^*_2 u) \ ds \\ \int \int_R -\nabla u \cdot \nabla v +uv \ dA &= \int \int_R g v \ dA - \int_{S_2} v q^* \ ds - \int_{S_3} v r^*_1\ ds +\int_{S_3} r^*_2 uv \ ds \\ \int \int_R -\nabla u \cdot \nabla v +uv \ dA +\int_{S_3} r^*_2 uv \ ds &= \int \int_R g v \ dA - \int_{S_2} v q^* \ ds - \int_{S_3} v r^*_1\ ds \\ B(u,v)&=L(v) \end{align}

Where I am having trouble

I do not know what to do with the terms $$c(x,y)\frac{\partial^2 u}{\partial x \partial y}+d(x,y)\frac{\partial u}{\partial x}+e(x,y)\frac{\partial u}{\partial y}$$ as using the divergence theorem/integration by parts used in the work thus far section leaves terms that are not symmetric and therefore does not not satisfy the requirements for $$B(u,v)$$.

The other problem are the terms $$a(x,y)\frac{\partial^2 u}{\partial x^2}+b(x,y) \frac{\partial^2 u}{\partial y^2}$$ the identity that I used in the work thus far section does not work (I am probably wrong on this part).

I could really use some guidance on both of these problems.

Notes

• This question is part of a much larger problem in which I have to use the finite element method. Once the problem is in a weak form in which the finite element/galerkin method can be applied I know what to do. From what I know the symmetry of $$B(u,v)$$ is essential. If there is some other weak form that works with the finite element (that is suitable for a numerical solution), that would be an acceptable answer to my problem.
• I have been following "Finite Elements: A Gentle Introduction" I could not find anything in the book that answered the problem. If you have any references that covers my problem that would be great (so far I have found nothing).
• If you have any questions feel free to ask.
• I originally posted this question on math stack exchange. I reposted the question here as it relevant and could bring more interest to my problem.

Notation

• $$n$$ is the vector normal to the boundary surface.
• $$u(x,y)$$ is the solution to the given PDE or ODE. $$v(x,y)$$ is a test function.
• $$\int \int_{R} * \ dA$$ is an integral over region $$R$$. $$\int_{S} * ds$$ is a surface integral over $$S$$.
• $$u^*, q^*, r^*_1, r^*_2, r^*_3$$ are either constants or functions used to define the boundary conditions.
• The surface (S) boundary conditions can be divided into Dirchlet, Neumann, and Robin boundary conditions. That is $$S=S_1\cup S_2 \cup S_3$$.
• Welcome to the Computational Science Stack exchange site. You do realise that i) Not all second order PDEs will give a discretisation in that form (many of the meaningful ones that do will be variations on the diffusion operator $\nabla \cdot \mu(x,y) \nabla u$ ) and ii) symmetry isn't a necessary condition to use the Finite Element method? – origimbo Apr 15 at 16:38
• That depends on whether you have your mathematician or engineer hat on. In the first case you need to show that discretised your bilinear form has solutions with the usual existence, uniqueness and continuity properties. In the second, you're probably happy to show that your method generates a matrix problem with a solution and it exhibits mesh convergence. Céa's lemma helps with the mathematician's version for symmetric elliptic problems, so these tend to be taught early. For examples of the later, consider a hyperbolic problem e.g. advection (which often requires more advanced techniques). – origimbo Apr 15 at 16:56
• Symmetry is not necessary. Ellipticity, boundedness, and coercivity are (lax-milgram theorem). – Paul Apr 15 at 23:25
• Sorry, I was mistaken. I should have said v-continuity. By elliptic, I meant v-elliptic. Each of these three properties are essential to applying lax milgram theorem. – Paul Apr 16 at 4:50
• @AzJ That would depend on what the problem actually is, which might be more easily dealt with by you asking your instructor what they intended. It's possible they just want you to generate a weak form which can be used with (e.g.) first order polynomial (linear) continuous functions to generate a matrix problem, in which case you only absolutely need to integrate the second order terms by parts once, but it's useful to take note that this can give you singular matrices, corresponding to ill-posed weak problems. – origimbo Apr 16 at 9:06

To understand the functional analysis (existence & uniqueness) part of the finite element method, it's helpful to see an analogy in linear algebra.

In the world of linear algebra, when you assemble any system of linear equations, you want to make sure that a solution exists. Theory suggests that you have to make sure that the (square) matrix system is invertible. One way to ensure that is to ensure that all the eigenvalues are non-zero. It is sufficient (but not absolutely necessary) to ensure that all eigenvalues are positive. Symmetric positive definiteness in a linear system of equations (for real systems of equations) ensures that eigenvalues are all positive and thus a solution exists. It's not the only way to guarantee that a matrix system is invertible, but if a linear system is symmetric positive definite, you get to take advantage of a lot of easy-to-verify analysis tools.

Think of functional analysis in a similar way. Weak forms produce linear systems of equations. You want to make sure that the resulting matrix system is invertible. If you have a symmetric bilinear form, you get to take advantage of a lot of readily available theorems (e.g. Riesz Representation, Cea’s Lemma) which enable you to prove whether the resulting system of equations is invertible (like having all-positive eigenvalues). But symmetry is by no means a necessary condition for the system to be invertible.

There are more generalized theorems for bilinear forms, but the one most people are familiar with is Lax-Milgram theorem. It basically says that in order for a solution to exist to the weak form (i.e. invertability of the resulting matrix system), you need to certain conditions on $$a(u,v)=L(v)$$ to be true for the function space $$V$$:

1. Boundedness of $$L(.)$$ on the space $$V$$ : In other words, $$||L(v)||\le M||v||_V$$
2. Coercivity (also known as V-ellipticity) of $$a(.,.)$$ on the space $$V$$: In other words, $$a(v,v)\le C_1||v||^2_V$$
3. Continuity of $$a(.,.)$$ on the space $$V$$: In other words, $$a(u,v)\le C_2||u||_V||v||_V$$

Together, these three conditions guarantee that your weak form (i.e. the system of equations setup by the weak form) is invertible and thus a solution exists. Symmetry is not necessary to use Lax Milgram Theorem. Symmetry is necessary to use Cea's lemma though. The equation $$-\nabla\cdot[k\nabla u]+\nabla\cdot(cu)+au=f$$ is non-symmetric and Lax Milgram theorem applies to it as well.

Of course, if your weak form does not have these conditions, it doesn't guarantee that a solution doesn't exist. But it's harder to prove that a solution exists by other means. There are other choices of finite element spaces that require more sophisticated theorems (e.g. mixed finite elements, petrov-galerkin methods, discontinuous galerkin methods, etc..). I can't speak on those theorems with authority, but there are others here on scicomp who can address questions on those spaces/methods.

• Thanks this clears things up quite a bit. I am going to leave this question open for a bit to encourage other answers but if no one else answers then I will accept your answer. – AzJ Apr 17 at 21:12
• Paul, great answer. I have a minor correction, though, the symmetry of matrices does not imply that the eigenvalues are positive. – nicoguaro Apr 17 at 21:22
• @nicoguaro: Yes, you’re right. Symmertric positive definite guarantees it. – Paul Apr 17 at 22:04