Biharmonic equation

In mathematics, the biharmonic equation is a fourth-order partial differential equation which arises in areas of continuum mechanics, including linear elasticity theory and the solution of Stokes flows. Specifically, it is used in the modeling of thin structures that react elastically to external forces.

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. . In this segment we're going to take an example of Finite Difference Method of solving boundary value ordinary differential equations. So let's go and see that how we can use the theory behind the finite difference methods to solve ordinary differential equation boundary value problems. And we are going to look at an example of doing that. So let's go and take a simple, so this is the problem statement which is given to us for the example. You have a second order differential equation which is of this particular form. . So you are given this and you are given the boundary conditions here that the value of u at 2 is 0.008 and the value of u at 6.5 is 0.003. So what you want to be able to do is, you want to be able to find out what u is as a function of r because u is the dependent variable and r is the independent variable and you want to be able to find what u is at the different points. In order to keep the problem simple, so somebody's asking so basically the problem statement is find u as a function of, function of r. But since we are given the numerical methods we cannot find u as a function of r at every data point, so somebody might say hey, use 4 nodes. Use 4 nodes to do the problem. Let me say 4, equidistant. Equidistant meaning that the nodes have the same distance between each other. So what that implies is that we need to break up our interval which is going from 2 to 6.5 into 4 nodes to be able to solve the problem. So I am here, so let's suppose if I am here, at r equal to 2 and then here I am at r equal to 6.5. So if I am going to break it up into 4 nodes I'll have another node here and another node right here so that means that three segments here. So the segment width will be divided by 3, the difference between the two. So if I want to calculate delta r it's just (6.5-2)/3 which is 1.5. And I have four nodes here so this will become the first, second node will be at r + delta r which will be at 3.5 and the next node will be at 3.5 plus delta r which will be at 5.0 so that's where the next node is going to be at 5.0. And what I am going to do is I'm going to number these nodes. I am going to number this node as 1, number this node as 2, and number this node as 3, and number this node as 4 and that's how I'll be designating that. So I already know what the value of u at 1 is because it is already given to me. The value of u is already given to me as 0.008 and I already know what the value at 4 is which is given to me as the value at r=6.5 is 0.003. So what that implies that I have to find out what the value of the u is, the dependent variables at the second node and the third node and if I am able to do that, that means that I have been able to solve the problem. That I can draw the profile of u, then I can do things like interpolation or spline interpolation to be able to find the value of u at some other point approximately. Even the values which I get at 2 and 3 are known approximately because I will be using approximations for my derivatives. So let's go and see that how do we go about doing this problem here by using Finite Difference Methods is by, what I am going to do is I am going to say hey, d2u/dr2 which is one of the terms in the ordinary differential equation which was given to me, that I can approximate it at any node i. So if I have a node i because I have to approximate this second derivative at each node, what is this approximation equal to? So this approximation will be equal to u_1+1 - 2(u_i) + u_i-1, so you're basically taking the information of the dependent variable at the node ahead and the node where you are and the node behind, divided by delta r squared. And that gives you the approximation of the second derivative. So, again, there are different approximations that you can choose, this is the central divided difference approximation of the second derivative of the function and then also what I am going to do is, another derivative; du/dr which I need to approximate at node i. And let's suppose I am going to use a forward divided difference scheme for that, so I am going to say u_i+1 - u_i divided by delta r. I am going to use that as my approximation to do that. So what that implies is that what I am trying to do is to change these derivative into unknown still, I still have unknowns of u_i, u_i-1 and things like that but now there are unknowns at specific points and that is what numerical methods is all about, being able to convert a problem into, into basic operations of multiplication, addition, division, and subtraction, that's what you are doing, that you are setting these up so that you can get simultaneous linear equations. So let's go and see how we end up doing that. So the ordinary differential equation woudl turn out to be [u_i+1 - 2(u_i) + u_i-1]/(delta r)^2 [u_i+1 - 2(u_i) + u_i-1]/(delta r)^2, because that is the approximation of the first derivative, plus (1/r), so since you are writing the equation at a particular node it will be r_i, that's what I will get there, times (u_i+1 - u_i)/(delta r) and then you have minus u/r^2, so since we are writing the equation at node i would be u_i and since we are writing equation node i, it will be the value of r which you have at that point equal to 0. So that's what you have reduced your ordinary differential equation into but you can see that it's not just solving one equation, one unknown at a time because you have, let's suppose, in this case at node i you have 3 unknowns. You have this, this, and this so you cannot solve three equations, three unknowns at this, three equations, so three unknowns in one equation in a single instance. You will need multiple equations to be able to, to be able to do that. So that's why what we are going to do is we are going to write down this equation at node 2 and 3 because we already know what the value at node 1 and node 4 is because those are the boundary conditions which are given to us. So if we write down this equation for node 2 and 3 we will be able to set up two equations and 4 unknowns but since we know the value of the function at the first node and the, and the last node, that will set us up a four equation, four unknowns. So let's go and see how we go about doing that. So I am going to write down this equation at, what I am going to do is I am going to write it one at a time, so node i, so which is node 1, so let me look at node 1. Node 1, I have u_1 is same as u at, at 2, is the value of the displacement of u at 2, it is 0.008. So that's my first equation. Now if I write down the equation at node 2, let me see what do I get from there, I get, I'll put i equal to 2 in there. I'll get u_3 minus 2(u_2) + u_1, I am substituting i=2 now, and that's what I will get when I subtitute it there, divided by delta r. Delta r is nothing but 1.5, squared, plus 1 divided by r_i which is r_2, (u_3 - u_2)/(delta r), so maybe I should just keep it as delta r for the time being here, divided by delta r, minus u_i which is u_2 divided by (r_i)^2 which is (r_2)^2 equal to, equal to 0. So since I am getting that, what that means is that I got to substitute the values of delta r and r_2 which I know, so I will get (u_3 - 2(u_2) + u_1)/(delta r)^2 which is what, 1.5 squared. plus (1/r_2) which is what, r_2 is 3.5, times (u_3 - u_2) divided by delta r, which is 1.5 squared, minus u_2 divided by r_2 squared which is (3.5)^2 equal to 0. If I expand this I'll be getting 0.444 for u_1, so combining all the u_1 terms, combining all the u_2 terms, and then combining all the u_3 terms and that's what I'll get as my equation number 2. And that's the end of this segment. . . .

Notation

It is written as

\nabla ^{4}\varphi =0

\nabla ^{2}\nabla ^{2}\varphi =0

\Delta ^{2}\varphi =0

where $\nabla ^{4}$ , which is the fourth power of the del operator and the square of the Laplacian operator $\nabla ^{2}$ (or $\Delta$ ), is known as the biharmonic operator or the bilaplacian operator. In Cartesian coordinates, it can be written in $n$ dimensions as:

\nabla ^{4}\varphi =\sum _{i=1}^{n}\sum _{j=1}^{n}\partial _{i}\partial _{i}\partial _{j}\partial _{j}\varphi =\left(\sum _{i=1}^{n}\partial _{i}\partial _{i}\right)\left(\sum _{j=1}^{n}\partial _{j}\partial _{j}\right)\varphi .

Because the formula here contains a summation of indices, many mathematicians prefer the notation $\Delta ^{2}$ over $\nabla ^{4}$ because the former makes clear which of the indices of the four nabla operators are contracted over.

For example, in three dimensional Cartesian coordinates the biharmonic equation has the form

{\partial ^{4}\varphi  \over \partial x^{4}}+{\partial ^{4}\varphi  \over \partial y^{4}}+{\partial ^{4}\varphi  \over \partial z^{4}}+2{\partial ^{4}\varphi  \over \partial x^{2}\partial y^{2}}+2{\partial ^{4}\varphi  \over \partial y^{2}\partial z^{2}}+2{\partial ^{4}\varphi  \over \partial x^{2}\partial z^{2}}=0.

As another example, in n-dimensional Real coordinate space without the origin $\left(\mathbb {R} ^{n}\setminus \mathbf {0} \right)$ ,

\nabla ^{4}\left({1 \over r}\right)={3(15-8n+n^{2}) \over r^{5}}

where

r={\sqrt {x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}}.

which shows, for n=3 and n=5 only, ${\frac {1}{r}}$ is a solution to the biharmonic equation.

A solution to the biharmonic equation is called a biharmonic function. Any harmonic function is biharmonic, but the converse is not always true.

In two-dimensional polar coordinates, the biharmonic equation is

{\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial }{\partial r}}\left({\frac {1}{r}}{\frac {\partial }{\partial r}}\left(r{\frac {\partial \varphi }{\partial r}}\right)\right)\right)+{\frac {2}{r^{2}}}{\frac {\partial ^{4}\varphi }{\partial \theta ^{2}\partial r^{2}}}+{\frac {1}{r^{4}}}{\frac {\partial ^{4}\varphi }{\partial \theta ^{4}}}-{\frac {2}{r^{3}}}{\frac {\partial ^{3}\varphi }{\partial \theta ^{2}\partial r}}+{\frac {4}{r^{4}}}{\frac {\partial ^{2}\varphi }{\partial \theta ^{2}}}=0

which can be solved by separation of variables. The result is the Michell solution.

2-dimensional space

The general solution to the 2-dimensional case is

xv(x,y)-yu(x,y)+w(x,y)

where $u(x,y)$ , $v(x,y)$ and $w(x,y)$ are harmonic functions and $v(x,y)$ is a harmonic conjugate of $u(x,y)$ .

Just as harmonic functions in 2 variables are closely related to complex analytic functions, so are biharmonic functions in 2 variables. The general form of a biharmonic function in 2 variables can also be written as

\operatorname {Im} ({\bar {z}}f(z)+g(z))

where $f(z)$ and $g(z)$ are analytic functions.

References

Eric W Weisstein, CRC Concise Encyclopedia of Mathematics, CRC Press, 2002. ISBN 1-58488-347-2.
S I Hayek, Advanced Mathematical Methods in Science and Engineering, Marcel Dekker, 2000. ISBN 0-8247-0466-5.
J P Den Hartog (Jul 1, 1987). Advanced Strength of Materials. Courier Dover Publications. ISBN 0-486-65407-9.

External links

This page was last edited on 23 October 2022, at 09:43

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