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Solve hyperbolic PDE problem
u1 = hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d) u1 = hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d,rtol) u1 = hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d,rtol,atol) u1 = hyperbolic(u0,ut0,tlist,K,F,B,ud,M) u1 = hyperbolic(u0,ut0,tlist,K,F,B,ud,M,rtol) u1 = hyperbolic(u0,ut0,tlist,K,F,B,ud,M,rtol,atol) u1 = hyperbolic(u0,ut0,tlist,K,F,B,ud,M,___,'DampingMatrix',D) u1 = hyperbolic(___,'Stats','off')
u1 = hyperbolic(u0,ut0,tlist,b,p,e,t,c,a,f,d) produces the solution to the FEM formulation of the scalar PDE problem
$$d\frac{{\partial}^{2}u}{\partial {t}^{2}}-\nabla \cdot \left(c\nabla u\right)+au=f,$$
for (x,y) ∊ Ω, or the system PDE problem
$$d\frac{{\partial}^{2}u}{\partial {t}^{2}}-\nabla \cdot \left(c\otimes \nabla u\right)+au=f,$$
on a mesh described by p, e, and t, with boundary conditions given by b, and with initial value u0 and initial derivative ut0.
In the scalar case, each row in the solution matrix u1 is the solution at the coordinates given by the corresponding column in p. Each column in u1 is the solution at the time given by the corresponding item in tlist. For a system of dimension N with n_{p} node points, the first n_{p} rows of u1 describe the first component of u, the following n_{p} rows of u1 describe the second component of u, and so on. Thus, the components of u are placed in blocks u as N blocks of node point rows.
b describes the boundary conditions of the PDE problem. For the recommended way of specifying boundary conditions, see Steps to Specify a Boundary Conditions Object. For all methods of specifying boundary conditions, see Boundary Condition Specification.
The geometry of the PDE problem is given by the mesh data p, e, and t. For details on the mesh data representation, see initmesh.
The coefficients c, a, d, and f of the PDE problem can be given in a variety of ways. The coefficients can depend on t, the time. They can also depend on u, the solution, and on the components of the gradient of u, namely ux and uy. For a complete listing of all options, see Scalar PDE Coefficients and Coefficients for Systems of PDEs.
atol and rtol are absolute and relative tolerances that are passed to the ODE solver.
u1 = hyperbolic(u0,ut0,tlist,K,F,B,ud,M) produces the solution to the ODE problem
$$\begin{array}{c}{B}^{\prime}MB\frac{{d}^{2}{u}_{i}}{d{t}^{2}}+K\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}{u}_{i}=F\\ u=B{u}_{i}+{u}_{d}\end{array}$$
with initial values for u being u0 and ut0. The K, F, B, ud, and M matrices have the same meaning as in the assempde syntax using those matrices.
Add the DampingMatrix name-value pair at the end of any matrix-form syntax to solve the damped equation
$$\begin{array}{c}{B}^{\prime}MB\frac{{d}^{2}{u}_{i}}{d{t}^{2}}+{B}^{\prime}DB\frac{d{u}_{i}}{dt}+K\text{\hspace{0.17em}}\xb7\text{\hspace{0.17em}}{u}_{i}=F\\ u=B{u}_{i}+{u}_{d}.\end{array}$$
For a worked example using this syntax, see Dynamics of a Damped Cantilever Beam.
Add the Stats name-value pair at the end of any syntax to control the display of internal ODE solver statistics. Valid values for Stats are 'off' and 'on' (default).
Solve the wave equation
$$\frac{{\partial}^{2}u}{\partial {t}^{2}}=\Delta u$$
on a square geometry -1 ≤ x,y ≤ 1 (squareg), with Dirichlet boundary conditions u = 0 for x = ±1, and Neumann boundary conditions
$$\frac{\partial u}{\partial n}=0$$
for y = ±1 (squareb3). Choose
u(0) = atan(cos(πx))
and
$$\frac{du}{dt}(0)=3\mathrm{sin}(\pi x)\mathrm{exp}\left(\mathrm{cos}(\pi y)\right).$$
Compute the solution at times 0, 1/6, 1/3, ... , 29/6, 5.
[p,e,t] = initmesh('squareg'); x = p(1,:)'; y = p(2,:)'; u0 = atan(cos(pi/2*x)); ut0 = 3*sin(pi*x).*exp(cos(pi*y)); tlist = linspace(0,5,31); uu = hyperbolic(u0,ut0,tlist,'squareb3',p,e,t,1,0,0,1);
The file pdedemo6 contains a complete example with animation.
Note In expressions for boundary conditions and PDE coefficients, the symbol t is used to denote time. The variable t is often used to store the triangle matrix of the mesh. You can use any variable to store the triangle matrix, but in the Partial Differential Equation Toolbox™ expressions, t always denotes time. |