We now turn to the solving of differential equations in which the solution is a function that depends on several independent variables. One such equation is called a partial differential equation (PDE, plural: PDEs).The subject of PDEs is enormous. At the same time, it is very important, since so many phenomena in nature and technology find their mathematical formulation through such equations. Knowing how to solve at least some PDEs is therefore of great importance to engineers.

A Partial Differential Equation Solver for the Classroom. CHUNG-YAU LAM and F. ALAN KOH School of Mechanical and Aerospace Engineering, Nanyang Technological University, Nanyang Avenue, Singapore 639798. E-mail: mcylam@ntu.edu.sg This paper presents a solver for partial differential equations that was developed in Microsoft Excel. PDE using $theta$ method in Matlab. Ask Question Asked 5 years, 6 months ago. Active 5 years, 6 months ago. First order partial differential equation - method of characteristics. Does the command `reboot` hurt a Mac computer?

In an introductory book like this, nowhere near full justice to the subject can be made. However, we still find it valuable to give the reader a glimpse of the topic by presenting a few basic and general methods that we will apply to a very common type of PDE.

Let us look at a specific application and how the diffusion equation with initial and boundary conditions then appears. We consider the evolution of temperature in a one-dimensional medium, more precisely a long rod, where the surface of the rod is covered by an insulating material. The heat can then not escape from the surface, which means that the temperature distribution will only depend on a coordinate along the rod, x, and time t.

At one end of the rod, x = L, we also assume that the surface is insulated, but at the other end, x = 0, we assume that we have some device for controlling the temperature of the medium. Here, a function s( t) tells what the temperature is in time. We therefore have a boundary condition u(0, t) = s( t).

At the other insulated end, x = L, heat cannot escape, which is expressed by the boundary condition ∂u( L, t)∕ ∂x = 0. The surface along the rod is also insulated and hence subject to the same boundary condition (here generalized to ∂u∕ ∂n = 0 at the curved surface).

However, since we have reduced the problem to one dimension, we do not need this physical boundary condition in our mathematical model. In one dimension, we can set Ω = 0, L. Boundary and initial conditions are needed!The initial and boundary conditions are extremely important. Without them, the solution is not unique, and no numerical method will work. Unfortunately, many physical applications have one or more initial or boundary conditions as unknowns. Such situations can be dealt with if we have measurements of u, but the mathematical framework is much more complicated.What about the source term g in our example with temperature distribution in a rod? G( x, t) models heat generation inside the rod.

One could think of chemical reactions at a microscopic level in some materials as a reason to include g. However, in most applications with temperature evolution, g is zero and heat generation usually takes place at the boundary (as in our example with u(0, t) = s( t)). 9.1.1 A Particular CaseBefore continuing, we may consider an example of how the temperature distribution evolves in the rod. At time t = 0, we assume that the temperature is 10 ∘C. Then we suddenly apply a device at x = 0 that keeps the temperature at 50 ∘C at this end. What happens inside the rod?

Intuitively, you think that the heat generation at the end will warm up the material in the vicinity of x = 0, and as time goes by, more and more of the rod will be heated, before the entire rod has a temperature of 50 ∘C (recall that no heat escapes from the surface of the rod).Mathematically, (with the temperature in Kelvin) this example has I( x) = 283 K, except at the end point: I(0) = 323 K, s( t) = 323 K, and g = 0. The figure below shows snapshots from two different times in the evolution of the temperature. 9.2 Finite Difference Methods. We shall now construct a numerical method for the diffusion equation.

We know how to solve ODEs, so in a way we are able to deal with the time derivative. Very often in mathematics, a new problem can be solved by reducing it to a series of problems we know how to solve. In the present case, it means that we must do something with the spatial derivative ∂ 2∕ ∂x 2 in order to reduce the PDE to ODEs. One important technique for achieving this, is based on finite difference discretization of spatial derivatives. 9.2.1 Reduction of a PDE to a System of ODEs.

At this point, it is tempting to implement a real physical case and run it. However, PDEs constitute a non-trivial topic where mathematical and programming mistakes come easy.

A better start is therefore to address a carefully designed test example where we can check that the method works. The most attractive examples for testing implementations are those without approximation errors, because we know exactly what numbers the program should produce. It turns out that solutions u( x, t) that are linear in time and in space can be exactly reproduced by most numerical methods for PDEs. A candidate solution might be. This loop can be replaced by a vectorized expression with the following reasoning. We want to set all the inner points at once: rhs1:N-1 (this goes from index 1 up to, but not including, N).

As the loop index i runs from 1 to N-1, the ui+1 term will cover all the inner u values displaced one index to the right (compared to 1:N-1), i.e., u2:N. Similarly, ui-1 corresponds to all inner u values displaced one index to the left: u0:N-2.

Finally, ui has the same indices as rhs: u1:N-1. The vectorized loop can therefore be written in terms of slices. The very nice thing is that we can now easily experiment with many different integration methods. Trying out some simple ones first, like RK2 and RK4, quickly reveals that the time step limitation of the Forward Euler scheme also applies to these more sophisticated Runge-Kutta methods, but their accuracy is better.

However, the Odespy package offers also adaptive methods. We can then specify a much larger time step in timepoints, and the solver will figure out the appropriate step.

Above we indicated how to use the adaptive Runge-Kutta-Fehlberg 4-5 solver. While the Δt corresponding to the Forward Euler method requires over 8000 steps for a simulation, we started the RKFehlberg method with 100 times this time step and in the end it required just slightly more than 2500 steps, using the default tolerance parameters. Lowering the tolerance did not save any significant amount of computational work.

Figure shows a comparison of the length of all the time steps for two values of the tolerance. We see that the influence of the tolerance is minor in this computational example, so it seems that the blow-up due to instability is what governs the time step size. The nice feature of this adaptive method is that we can just specify when we want the solution to be computed, and the method figures out on its own what time step that has to be used because of stability restrictions. 9.2Time steps used by the Runge-Kutta-Fehlberg method: error tolerance 10 −3 (left) and 10 −6 (right)We have seen how easy it is to apply sophisticated methods for ODEs to this PDE example. We shall take the use of Odespy one step further in the next section.

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9.2.7 Implicit MethodsA major problem with the stability criterion ( ) is that the time step becomes very small if Δx is small. For example, halving Δx requires four times as many time steps and eight times the work. Now, with N = 40, which is a reasonable resolution for the test problem above, the computations are very fast. What takes time, is the visualization on the screen, but for that purpose one can visualize only a subset of the time steps. However, there are occasions when you need to take larger time steps with the diffusion equation, especially if interest is in the long-term behavior as t → ∞. You must then turn to implicit methods for ODEs.

These methods require the solutions of linear systems, if the underlying PDE is linear, and systems of nonlinear algebraic equations if the underlying PDE is non-linear. (9.33) We see that A = I − Δt K.To implement the Backward Euler scheme, we can either fill a matrix and call a linear solver, or we can apply Odespy. We follow the latter strategy. Sonar platinum | download. Implicit methods in Odespy need the K matrix above, given as an argument jac (Jacobian of f) in the call to odespy.BackwardEuler. Here is the Python code for the right-hand side of the ODE system ( rhs) and the K matrix ( K) as well as statements for initializing and running the Odespy solver BackwardEuler (in the file ). Odespy solvers apply dense matrices!Looking at the entries of the K matrix, we realize that there are at maximum three entries different from zero in each row.

Therefore, most of the entries are zeroes. The Odespy solvers expect dense square matrices as input, here with ( N + 1) × ( N + 1) elements. When solving the linear systems, a lot of storage and work are spent on the zero entries in the matrix. It would be much more efficient to store the matrix as a tridiagonal matrix and apply a specialized Gaussian elimination solver for tridiagonal systems. Actually, this reduces the work from the order N 3 to the order N.In one-dimensional diffusion problems, the savings of using a tridiagonal matrix are modest in practice, since the matrices are very small anyway. In two- and three-dimensional PDE problems, however, one cannot afford dense square matrices.

Rather, one must resort to more efficient storage formats and algorithms tailored to such formats, but this is beyond the scope of the present text. 9.3 Exercises. Exercise 9.2: Compute Temperature Variations in the GroundThe surface temperature at the ground shows daily and seasonal oscillations. When the temperature rises at the surface, heat is propagated into the ground, and the coefficient β in the diffusion equation determines how fast this propagation is. It takes some time before the temperature rises down in the ground.

At the surface, the temperature has then fallen. We are interested in how the temperature varies down in the ground because of temperature oscillations on the surface.Assuming homogeneous horizontal properties of the ground, at least locally, and no variations of the temperature at the surface at a fixed point of time, we can neglect the horizontal variations of the temperature. Then a one-dimensional diffusion equation governs the heat propagation along a vertical axis called x. The surface corresponds to x = 0 and the x axis point downwards into the ground. There is no source term in the equation (actually, if rocks in the ground are radioactive, they emit heat and that can be modeled by a source term, but this effect is neglected here).

Solve this heat propagation problem numerically for some days and animate the temperature. You may use the Forward Euler method in time. Plot both the numerical and analytical solution. As initial condition for the numerical solution, use the exact solution during program development, and when the curves coincide in the animation for all times, your implementation works, and you can then switch to a constant initial condition: u( x, 0) = T 0. For this latter initial condition, how many periods of oscillations are necessary before there is a good (visual) match between the numerical and exact solution (despite differences at t = 0)?

Experiment to see if you can use larger time steps than what is required by the Forward Euler method and get solutions with the same order of accuracy.HintTo avoid oscillations in the solutions when using the RKFehlberg method, the rtol and atol parameters to RKFFehlberg must be set no larger than 0.001 and 0.0001, respectively. You can print out solverRKF.tall to see all the time steps used by the RKFehlberg solver (if solver is the RKFehlberg object). You can then compare the number of time steps with what is required by the other methods.Filename: groundtempadaptive.py.

Filename: rodThetaRule.py.RemarksDespite the fact that the Crank-Nicolson method, or the θ rule with θ = 1∕2, is theoretically more accurate than the Backward Euler and Forward Euler schemes, it may exhibit non-physical oscillations as in the present example if the solution is very steep. The oscillations are damped in time, and decreases with decreasing Δt. To avoid oscillations one must have Δt at maximum twice the stability limit of the Forward Euler method. This is one reason why the Backward Euler method (or a 2-step backward scheme, see Exercise ) are popular for diffusion equations with abrupt initial conditions. Exercise 9.8: Explore SymmetryOne can observe (and also mathematically prove) that the solution u( x, t) of the problem in Exercise is symmetric around x = 0: u(− x, t) = u( x, t). In such a case, we can split the domain in two and compute u in only one half, −1, 0 or 0, 1.

At the symmetry line x = 0 we have the symmetry boundary condition ∂u∕ ∂x = 0. Reformulate the problem in Exercise such that we compute only for x ∈ 0, 1. (9.38) which is known as a two-point boundary value problem.

This is nothing but the stationary limit of the diffusion problem in Sect. How can we solve such a stationary problem ( )? The simplest strategy, when we already have a solver for the corresponding time-dependent problem, is to use that solver and simulate until t → ∞, which in practice means that u( x, t) no longer changes in time (within some tolerance).A nice feature of implicit methods like the Backward Euler scheme is that one can take one very long time step to “infinity” and produce the solution of ( ). Filename: rodstationary.py.RemarksIf the interest is in the stationary limit of a diffusion equation, one can either solve the associated Laplace or Poisson equation directly, or use a Backward Euler scheme for the time-dependent diffusion equation with a very long time step.

Using a Forward Euler scheme with small time steps is typically inappropriate in such situations because the solution changes more and more slowly, but the time step must still be kept small, and it takes “forever” to approach the stationary state. This is yet another example why one needs implicit methods like the Backward Euler scheme. Open Access This chapter is licensed under the terms of the Creative Commons Attribution 4.0 International License (which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license and indicate if changes were made.The images or other third party material in this chapter are included in the chapter's Creative Commons license, unless indicated otherwise in a credit line to the material. If material is not included in the chapter's Creative Commons license and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder. Authors and Affiliations.