From: bangerth Date: Fri, 23 Mar 2007 19:33:31 +0000 (+0000) Subject: Explain the transformation to an initial value problem. X-Git-Url: https://gitweb.dealii.org/cgi-bin/gitweb.cgi?a=commitdiff_plain;h=2d96f2895c4ac755a66a523a74c5da70da7aac89;p=dealii-svn.git Explain the transformation to an initial value problem. git-svn-id: https://svn.dealii.org/trunk@14597 0785d39b-7218-0410-832d-ea1e28bc413d --- diff --git a/deal.II/examples/step-24/doc/intro.dox b/deal.II/examples/step-24/doc/intro.dox index 7253135cc6..e785e1d0a9 100644 --- a/deal.II/examples/step-24/doc/intro.dox +++ b/deal.II/examples/step-24/doc/intro.dox @@ -57,24 +57,27 @@ length of the microwave pulse that heats the tissue is much shorter than the time it takes a wave to cross the domain). In that case, the heating rate $H(t,\mathbf r)$ can be written as $H(t,\mathbf r) = a(\mathbf r)\delta(t)$ (where $a(\mathbf r)$ is a map of absorption strengths for -microwave energy), which together with the first equation above will yield +microwave energy and $\delta(t)$ is the Dirac delta function), which together +with the first equation above will yield an instantaneous jump in the temperature $T(\mathbf r)$ at time $t=0$. Using this assumption, and taking all equations together, we can rewrite and combine the above as follows: @f[ -\Delta p-\frac{1}{c_0^2} \frac{\partial^2 p}{\partial^2 t} = \lambda +\Delta p-\frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} = \lambda a(\mathbf r)\frac{d\delta(t)}{dt} @f] -where $\lambda = - \frac{\beta}{C_p}$. This corresponds to a wave equation -with initial conditions as follows: +where $\lambda = - \frac{\beta}{C_p}$. + +This somewhat strange equation with the derivative of a Dirac delta function +on the right hand side can be rewritten as an initial value problem as follows: @f{eqnarray*} -\Delta \bar{p}- \frac{1}{c_0^2} \frac{\partial^2 \bar{p}}{\partial^2 t} & = & -f(t,\mathbf r) \\ +\Delta \bar{p}- \frac{1}{c_0^2} \frac{\partial^2 \bar{p}}{\partial t^2} & = & +0 \\ \bar{p}(0,\mathbf r) &=&\lambda a(\mathbf r) = b(\mathbf r) \\ \frac{\partial\bar{p}(0,\mathbf r)}{\partial t} &=& 0. @f} -(With $f=0$, though we usually keep it around to derive formulas that -are valid even for the case that $f$ was non-zero.) +(A derivation of this transformation into an initial value problem is given at +the end of this introduction as an appendix.) In the inverse problem, it is the initial condition $b(\mathbf r) = \lambda a(\mathbf r)$ that one would like to recover, since it is a map of absorption strengths for @@ -123,8 +126,11 @@ two seperate equations: with initial conditions: @f{eqnarray*} \bar{p}(0,\mathbf r) & = & b(r) \\ -v(0,\mathbf r)=\bar{p}_t(0,\mathbf r) & = & 0 +v(0,\mathbf r)=\bar{p}_t(0,\mathbf r) & = & 0. @f} +Note that we have introduced a right hand side $f(t,\mathbf r)$ here to show +how to derive these formulas in the general case, although in the application +to the thermoacoustic problem $f=0$. The semi-discretized, weak version of this model, using the general $\theta$ scheme introduced in @ref step_23 "step-23" is then: @@ -207,3 +213,111 @@ from actual experimental measurements. To this end, we need to evaluate the solution at points at which the experiment also evaluates a real pressure field. We will see how to do that using the VectorTools::point_value function further down below. + + + +

Appendix: PDEs with Dirac delta functions as right hand side and their +transformation to an initial value problem

+ +In the derivation of the initial value problem for the wave equation, we +initially found that the equation had the derivative of a Dirac delta function +as a right hand side: +@f[ +\Delta p-\frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} = \lambda +a(\mathbf r)\frac{d\delta(t)}{dt}. +@f] +In order to see how to transform this single equation into the usual statement +of a PDE with initial conditions, let us make the assumption that the +physically quite reasonable medium is at rest initially, i.e. $p(t,\mathbf +r)=\frac{\partial p(t,\mathbf r)}{\partial t}=0$ for $t<0$. Next, let us form +the indefinite integral with respect to time of both sides: +@f[ +\int^t \Delta p\; dt -\int^t \frac{1}{c_0^2} \frac{\partial^2 p}{\partial t^2} +\; dt += +\int^t \lambda a(\mathbf r)\frac{d\delta(t)}{dt} \;dt. +@f] +This immediately leads to the statement +@f[ +P(t,\mathbf r) - \frac{1}{c_0^2} \frac{\partial p}{\partial t} +\; dt += +\lambda a(\mathbf r) \delta(t), +@f] +where $P(t,\mathbf r)$ is such that $\frac{dP(t,\mathbf r)}{dt}=\Delta +p$. Next, we form the (definite) integral over time from $t=-\epsilon$ to +$t=+\epsilon$ to find +@f[ +\int_{-\epsilon}^{\epsilon} P(t,\mathbf r)\; dt +- \frac{1}{c_0^2} \left[ p(\epsilon,\mathbf r) - p(-\epsilon,\mathbf r) \right] += +\int_{-\epsilon}^{\epsilon} \lambda a(\mathbf r) \delta(t) \; dt. +@f] +If we use the property of the delta function that $\int_{-\epsilon}^{\epsilon} +\delta(t)\; dt = 1$, and assume that $P$ is a smooth function in time, we find +as we let $\epsilon$ go to zero that +@f[ +- \frac{1}{c_0^2} \left[ p(\epsilon,\mathbf r) - p(-\epsilon,\mathbf r) \right] += +\lambda a(\mathbf r). +@f] +In other words, using that $p(-\epsilon,\mathbf r)=0$, we retrieve the initial +condition +@f[ + \frac{1}{c_0^2} p(0,\mathbf r) + = + \lambda a(\mathbf r). +@f] +At the same time, we know that for every $t>0$ the delta function is zero, so +for $0