# Method to find PDE equation coefficient satisfying mean solution?

What is the best approach to go about solving a PDE problem of the type

$$$$k^3\Delta u - k(\mathbf{1}\cdot\nabla u) = 0\, ,\\ u=g\; \text{on}\; \Gamma_D\, ,\\ mean(u) = u_\text{mean}$$$$

where one wants to find for which positive constant scalar coefficient $$k>0$$ the mean of the solution $$u$$ fulfills a prescribed value $$u_\text{mean}$$?

• I assume $k$ is a vector to be dimensionally correct for $k \nabla u$ and $k^3=|k|^3$, is the direction of $k$ known? Otherwise you have to few conditions. If so, you can add your condition $mean(u)-u_\text{mean}=0$ with a lagrange multiplier to your equation and solve the system for $|k|$ as free parameter. – Bort Nov 13 '18 at 17:27
• Ah no, k is just a constant scalar in this case. – JacobP Nov 14 '18 at 3:00
• if $k$ is a constant scalar, then the dimensions don't match up and writing $k^3\Delta u + k\nabla u$ doesn't make much mathematical sense... is there a unit vector field hiding somewhere in there i.e. do you mean to write $\mathbf{1} \cdot \nabla u$? – GoHokies Nov 14 '18 at 14:45
• Yes, that's right a unit vector for the second term, thanks. – JacobP Nov 16 '18 at 11:07

I would recommend an adjoint method also but slightly different. Following is the general idea. Whether this approach is well posed for this problem is left as an excercise :-)

Solve this $$\min_k J(k) = (1/2)[u_{mean} - mean(u)]^2$$ subject to the pde as constraint. If you change $$k$$ to $$k+k'$$, $$u$$ changes to $$u+u'$$ and $$J$$ to $$J+J'$$. Find a linear PDE for $$u'$$ assuming changes are small, which should have following structure $$L(u',k)=F(u,k) k'$$ and it will have $$u'=0$$ on boundary since you have a Dirichlet bc on $$u$$. Then you get $$J' = -[u_{mean} - mean(u)] mean(u') + \int_\Omega v \cdot L(u',k) dx - \int_\Omega F(u,k) k' v dx$$ and we just added a zero term. Integrate by parts so all derivatives are on the adjoint variable $$v$$. Then you get an equation of the form $$J' = - k' \int_\Omega F(u,k) v dx + \int_\Omega L^*(v,k)\cdot u' dx + \textrm{boundary terms}$$ Form adjoint pde $$L^*(v,k)=0$$ and choose bc for $$v$$ so that boundary terms in $$J'$$ vanish. Once you solve adjoint, you can update $$k$$ as $$k' = \int_\Omega F(u,k) v dx$$ which ensures that $$J$$ will decrease.

With $$k^3\Delta u-k(1\cdot\nabla u)=0,$$ you can already remove 1 solution for $$k$$, $$k=0$$ which is not of interest for you. You are really looking at $$k^2\Delta u-(1\cdot\nabla u)=0.$$

Edit: The previous equations enforced for a given $$k$$ and $$u_\text{mean}$$ a solution on the field $$u$$ and its constraint force $$\lambda$$.

If you want to find $$k$$ for a given $$u_\text{mean}$$, it should be rephrased to an optimization under constraints: $$\min_{k>0} \frac{1}{2}||PDE(k,u)||^2+\frac{\lambda}{2}^2||u_\text{mean}-\text{mean}(u)||^2$$

Here $$\lambda$$ is a regularization parameter, basically how much weight you put on the constraint. In this case you end up with a similar equation:

$$k^2\Delta u-(1\cdot\nabla u)+\lambda(u_\text{mean}-\text{mean}(u))=0.$$ In this optimization problem you are no longer solving for $$\lambda$$ (this is input) but for $$k$$.

In case of optimization I suggest to look for the adjoint formulation of PDEs with constraints, e.g. the mathematical description on the dolfin adjoint page and its references.

• Using this technique, how do you find the $k$ which naturally satisfies the condition $\text{mean}(u) = u_\text{mean}$? To me it is quite obvious that such $k$ exists. This technique basically adds an artificial force field $\lambda$ which causes the constraint $\text{mean}(u) = u_\text{mean}$ to hold but then the resulting $u$ does not satisfy the original equation anymore. – knl Nov 20 '18 at 12:35