# na.numerical evaluation – Smoothly connecting PDEs with finite variations Answer

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## A PDE with non-smooth inhomogeneity

Let $$mathcal{L}$$ breathe a second-order, linear, elliptic differential operator appearing on $$mathcal{C}^2([0,2]^2)$$.

I’m numerically fixing the inhomogeneous PDE
commence{align*} mathcal{L}u(x,y)+(x-1)^+=0, aim{align*}
the place $$(cdot)^+$$ denotes the optimistic sever.

Put in a different way, I decipher two PDEs which necessity to breathe related at $$x=1$$:
commence{align*} commence{instances} mathcal{L}u(x,y)=0 & textual content{for }(x,y)in[0,1]occasions[0,2], mathcal{L}u(x,y)+x-1=0& textual content{for }(x,y)in(1,2]occasions[0,2]. aim{instances} aim{align*}

Approximating all partial derivatives by central variations, I get the nine-point stencil
commence{align*} c_1 u_{i-1,j-1} + c_2 u_{i,j-1} + c_3 u_{i+1,j-1} + c_4 u_{i-1,j} + c_5 u_{i,j} & + c_6 u_{i+1,j}+ c_7 u_{i-1,j+1} + c_8 u_{i,j+1} + c_9 u_{i+1,j+1} + (x_i-1)^+ &=0. aim{align*}
Thus, $$u$$ is the answer to a system of linear equations.

## Problem

Plotting the answer $$u$$, all of it seems exquisite and consummate. However, a plot of $$frac{partial u}{partial x}$$ as a duty of $$x$$ reveals that the spinoff just isn’t {smooth} at $$x=1$$. The above FD strategy works exquisite with value-matching (the answer $$u$$ is completely steady) however struggles with smooth-pasting at $$x=1$$ (the spinoff just isn’t {smooth}).

## Question

How do I guarantee smooth-pasting with a finite dissimilarity strategy at $$x=1$$?

Some of my failed makes an attempt comprise

• Impose that route and backward variations at $$x=1$$ equal one another (= 2nd bid central dissimilarity is zero).
• Use increased bid approximations round $$x=1$$ akin to $$u_{xx}approx frac{-u(-2h)+16u(-h)-30u(0)+16u(h)-u(2h)}{12h^2}$$ and $$u_{x}approx frac{u(-2h)-8u(-h)+8u(h)-u(2h)}{12h}$$ and central variations in all places else.
• Approximating partial derivatives utilizing factors solely from one aspect of $$x=1$$ (i.e. solely utilizing both $$u(0), u(h), u(2h)$$ or as a substitute $$u(0),u(-h),u(-2h)$$).
• Imposing that $$u_xapproxfrac{u_{i+1,j}-u_{i-1,j}}{2h}$$ equals an medium of partial derivatives at $$1+h$$ and $$1-h$$.

Note: This drawback arises as sever of a bigger system of free border issues. Thus, it is needful to decipher the PDE numerically. This query is too posted right here.

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