Dirichlet boundary condition

In connection with differential equations (more precisely: boundary value problems ), the Dirichlet boundary condition (according to Peter Gustav Lejeune Dirichlet ) refers to values ​​that the function should assume on the respective boundary of the domain of definition.

Further boundary conditions are, for example, Neumann boundary conditions or inclined boundary conditions .

Ordinary differential equation

The Dirichlet problem

In the case of an ordinary differential equation , the domain of the function is a closed interval. As a result, the edge of the definition area consists only of the right and left end of the interval. Due to the freedom in ordinary differential equations, Dirichlet boundary conditions only make sense for equations of the second or higher order. In this case a Dirichlet problem, i.e. H. a differential equation with Dirichlet boundary condition as follows:

${\displaystyle {\begin{cases}f(x,y(x),y'(x),y''(x))=0,\quad x\in (a,b)\\y(a)=\alpha ,\quad y(b)=\beta \end{cases}}}$

Here is a prescribed function, and are prescribed real numbers for the function values ​​of a solution at the interval ends. Finally we are looking for a (classical) solution from the given regularity class . ${\displaystyle f}$ ${\displaystyle \alpha }$ ${\displaystyle \beta }$ ${\displaystyle y\in C^{2}(a,b)\cap C^{0}[a,b]}$

example

We choose as our interval and consider the following Dirichlet problem: ${\displaystyle [0,\pi ]}$

${\displaystyle {\begin{cases}y''=-y\\y(0)=0,\quad y(\pi )=0\end{cases}}}$

With the theory of linear ordinary differential equations with constant coefficients, we first get as a general (classical) solution of the differential equation:

${\displaystyle y(x)=C\cos x+D\sin x}$

with two freely selectable real constants and . We use the boundary conditions to fix these constants. We get a linear system of equations in the unknowns and : ${\displaystyle C}$ ${\displaystyle D}$ ${\displaystyle C}$ ${\displaystyle D}$

${\displaystyle C=0,}$

${\displaystyle -C=0.}$

Remarkably, this system is not uniquely solvable, but for any real thing a solution is given by ${\displaystyle D}$

${\displaystyle y(x)=D\sin x.}$

Existence and uniqueness

The following sentence is formulated for homogeneous ( ) data. However, this is not a restriction, because a transformation with ${\displaystyle \alpha =\beta =0}$ ${\displaystyle {\tilde {u}}(x)=u(x)-r(x)}$

${\displaystyle r(x)={\frac {(b-x)\alpha +(x-a)\beta }{b-a}}}$

an inhomogeneous problem can always be transformed into a homogeneous problem.

The task is given

${\displaystyle {\begin{cases}-u''(x)=f(x,u(x),u'(x)),\quad x\in (a,b)\\u(a)=u(b)=0\end{cases}}}$

Let it be a continuous function. In addition, it fulfills a Lipschitz condition , that is, there are numbers such that for all and for all the inequality ${\displaystyle f\colon [a,b]\times \mathbb {R} ^{2}\to \mathbb {R} }$ ${\displaystyle L,K>0}$ ${\displaystyle x\in [a,b]}$ ${\displaystyle s,t,s',t'\in \mathbb {R} }$

${\displaystyle |f(x,s,s')-f(x,t,t')|\leq L|s-t|+K|s'-t'|}$

is fulfilled. Continue to apply

${\displaystyle L{\frac {(b-a)^{2}}{8}}+K{\frac {b-a}{2}}<1.}$

Be a solution of ${\displaystyle w}$

${\displaystyle w''(x)+Lw'(x)+Kw(x)=0,\quad x\in (a,b).}$

${\displaystyle w}$vanish for and be the first unique number such that for . Then the underlying task has exactly one solution, if ${\displaystyle x=a}$ ${\displaystyle \alpha (L,K)}$ ${\displaystyle w'(x)=0}$ ${\displaystyle x=a+\alpha (L,K)}$

${\displaystyle b-a<2\alpha (L,K).}$

If, on the other hand , applies , no solution need exist or it does not need to be unique. Furthermore applies ${\displaystyle b-a\geq 2\alpha (L,K)}$

${\displaystyle \alpha (L,K)={\begin{cases}{\frac {2}{\sqrt {4K-L^{2}}}}\arccos {\frac {L}{2{\sqrt {K}}}},&4K-L^{2}>0\\{\frac {2}{L^{2}-4K}}\operatorname {arcosh} {\frac {L}{2{\sqrt {K}}}},&4K-L^{2}<0;L,K>0\\{\frac {2}{L}},&4K-L^{2}=0,L>0\\+\infty ,&{\text{sonst}}\end{cases}}}$

A proof of this theorem can be found in Bailey, Shampine, Waltman. Nonlinear two-point boundary value problems . Academic Press, 1968.

If the right side of the differential equation is only continuous and bounded, then Scorza Dragoni's theorem guarantees the existence of a solution. ${\displaystyle f}$

Partial differential equations

The Dirichlet problem

In the case of a partial differential equation , the sole specification of Dirichlet boundary conditions only makes sense for elliptical equations in a restricted area , since the other types also require specifications for the initial values. Dirichlet boundary conditions are prescribed on the edge of the area . Here we define the Dirichlet problem for quasi-linear partial differential equations ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ ${\displaystyle \partial \Omega }$

${\displaystyle u\in C^{2}(\Omega )\cap C^{0}({\overline {\Omega }})}$

${\displaystyle \sum _{i,j=1}^{n}a_{ij}(x,u,\nabla u){\frac {\partial ^{2}}{\partial x_{i}\partial x_{j}}}u=f(x,u,\nabla u)}$

${\displaystyle u(x)=g(x),\qquad x\in \partial \Omega }$

Here the function represents the prescribed function values ​​of the solution on the edge. The question of the solvability of such a problem is already very demanding and is the focus of current research. It is also very difficult to give a general solution method. ${\displaystyle g\colon \partial \Omega \rightarrow \mathbb {R} }$

example

In this example we consider the following boundary value problem in the field : ${\displaystyle \Omega =(0,\pi )^{n}=\{x=(x_{1},\dots ,x_{n})\in \mathbb {R} ^{n}\,:\,0<x_{i}<\pi ,\quad i=1,\dots ,n\},}$

${\displaystyle u\in C^{2}(\Omega )\cap C^{0}({\overline {\Omega }})}$

${\displaystyle \Delta u(x)=-nu(x),\qquad x\in \Omega }$

${\displaystyle u(x)=0,\qquad \qquad \quad \;x\in \partial \Omega .}$

Here denotes the Laplace operator . First, we note that there is a solution to the problem. We want to find more solutions. We now assume for and make the following product approach ${\displaystyle \Delta }$ ${\displaystyle u\equiv 0}$ ${\displaystyle u(x)\neq 0}$ ${\displaystyle x\in \Omega }$

${\displaystyle u(x)=v_{1}(x_{1})\cdot \dots \cdot v_{n}(x_{n})=\prod _{k=1}^{n}v_{k}(x_{k}).}$

For the functions we derive ordinary differential equations with corresponding Dirichlet boundary conditions. It follows ${\displaystyle v_{k}}$

${\displaystyle {\begin{aligned}\Delta u&=\left({\frac {\partial ^{2}}{\partial x_{1}^{2}}}+\dots +{\frac {\partial ^{2}}{\partial x_{n}^{2}}}\right)u(x)\\&=v_{1}''(x_{1})v_{2}(x_{2})\cdot \dots \cdot v_{n}(x_{n})+\dots +v_{1}(x_{1})\cdot \dots \cdot v_{n-1}(x_{n-1})v_{n}''(x_{n})\\&=u(x)\sum _{k=1}^{n}{\frac {v_{k}''(x_{k})}{v_{k}(x_{k})}}.\end{aligned}}}$

If the boundary value problem ${\displaystyle v_{k}}$

${\displaystyle v_{k}\in C^{2}(0,\pi )\cap C^{0}[0,\pi ]}$

${\displaystyle v_{k}''\,=\,-v_{k}}$

${\displaystyle v_{k}(0)=0,\quad v_{k}(\pi )=0}$

suffice, then the function defined above is a solution of the Dirichlet boundary value problem for the partial differential equation. Using the example of ordinary differential equations, we get ${\displaystyle u}$

${\displaystyle v_{k}(x)\,=\,D_{k}\sin(x_{k})}$

and thus

${\displaystyle u(x)=D\prod _{k=1}^{n}\sin(x_{k})}$

as a solution to our problem of partial differential equations to Dirichlet boundary conditions. The question remains whether there are any other solutions.

literature

• D. Gilbarg, NS Trudinger: Partial Differential Equations of Second Order , Springer-Verlag, Berlin 1998, ISBN 3-540-41160-7 .