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Run the following command in Maple:

Explore(plot(x^k), k = 1 .. 3);


Once you’ve run the command, move the slider from side to side. Neat, isn’t it?

With this single line of code, you have built an interactive application that shows the graph of x to the power of various exponent powers.


The Explore command is an application builder. More specifically, the Explore command can programmatically generate interactive content in Maple worksheets.

Programmatically generated content is inserted into a Maple worksheet by executing Maple commands. For example, when you run the Explore command on an expression, it inserts a collection of input and output controllers, called Embedded Components, into your Maple worksheet. In the preceding example, the Explore command inserts a table containing:

  • a Slider component, which corresponds to the value for the exponent k
  • a Plot component, which shows the graph of x raised to the power for k

Together these components form an interactive application that can be used to visualize the effect of changing parameter values.

Explore can be viewed as an easy application creator that generates simple applications with input and output components. Recently added packages for programmatic content generation broaden Maple’s application authoring abilities to form a full development framework for creating customized interactive content in a Maple worksheet. The DocumentTools package contains many of these new tools. Components and Layout are two sub-packages that generate XML using function calls that represents GUI elements, such as embedded components, tables, input, or output. For example, the DocumentTools:-Components:-Plot command creates a new Plot component. These key pieces of functionality provide all of the building blocks needed to create customizable interfaces inside of the Maple worksheet. For me, this new functionality has completely altered my approach to building Maple worksheets and made it much easier to create new applications that can explore hundreds of data sets, visualize mathematical functions, and more.

I would go so far as to say that the ability to programmatically generate content is one of the most important new sources of functionality over the past few years, and is something that has the potential to significantly alter the way in which we all use Maple. Programmatic content generation allows you to create applications with hundreds of interactive components in a very short period of time when compared to building them manually using embedded components. As an illustration of this, I will show you how I easily created a table with over 180 embedded components—and the logic to control them.


Building an interface for exploring data sets:

In my previous blog post on working with data sets in Maple, I demonstrated a simple customized interface for exploring country data sets. That post only hinted at the much bigger story of how the Maple programming language was used to author the application. What follows is the method that I used, and a couple of lessons that I learned along the way.

When I started building an application to explore the country data sets, I began with an approach that I had used to build several MathApps in the past. I started with a blank Maple worksheet and manually added embedded components for controlling input and output. This included checkbox components for each of the world’s countries, drop down boxes for available data sets, and a couple of control buttons for retrieving data to complete my application.

This manual, piece-by-piece method seemed like the most direct approach, but building my application by hand proved time-consuming, given that I needed to create 180 checkboxes to house all available countries with data. What I really needed was a quicker, more scriptable way to build my interface.


So jumping right into it, you can view the code that I wrote to create the country data application here:PECCode.txt

Note that you can download a copy of the associated Maple worksheet at the bottom of this page.


I won’t go into too much detail on how to write this code, but the first thing to note is the length of the code; in fewer than 70 lines, this code generates an interface with all of the required underlying code to drive interaction for 180+ checkboxes, 2 buttons and a plot. In fact, if you open up the application, you’ll see that every check box has several lines of code behind it. If you tried to do this by hand, the amount of effort would be multiplied several times over.

This is really the key benefit to the world of programmatic content generation. You can easily build and rebuild any kind of interactive application that you need using the Maple programming language. The possibilities are endless.


Some tips and tricks:

There are a few pitfalls to be aware of when you learn to create content with Maple code. One of the first lessons I learned was that it is always important to consider embedded component name collision and name resolution.

For those that have experimented with embedded components, you may have noticed that Maple’s GUI gives unique names to components that are copied (or added) in a Maple worksheet. For example, the first TextArea component that you add to a worksheet usually has the default name TextArea0. If you next add another TextArea, this new TextArea gets the name TextArea1, so as to not collide with the first component. Similar behaviour can be observed with any other component and even within some component properties such as ‘group’ name.

Many of the options for commands in the DocumentTools sub-packages can have “action code”, or code that is run when the component is interacted with. When building action code for a generated component, the action code is specified using a long string that encapsulates all of the code. Due to this code being provided as a long string, one trick that I quickly picked up is that it is important to separate out the names for any components into sub-strings inside of a longer cat statement.

For example, here is a line that is contained within a longer cat statement in the preceding code:

cat( "DocumentTools:-SetProperty( \"", "ComboBox_0", "\", 
'value', \"Internet Users\" );\n" )

It is necessary to enclose “ComboBox_0” in quotes, as well as to add in escaped quotes in order to have the resulting action code look like (also note the added new line at the end):

“DocumentTools:-SetProperty( “ComboBox_0”, ‘value’, “Internet Users” );”

Doing so ensures that when the components are created, the names are not hard-coded to always just look for a given name. This means that the GUI can scrape through the code and update any newly generated components with a new name when needed. This is important if “ComboBox_0” already exists so that the GUI can instead create “ComboBox_1”.


Another challenge for coding applications is adding a component state. One of the most common problems encountered with running any interactive content in Maple is that if state is not persistent, errors can occur when, for example, a play button is clicked but the required procedures have not been run. This is a very challenging problem, which often require solutions like the use of auto-executing start-up code or more involved component programming. Some features in Maple 2016 have started working to address this, but state is still something that usually needs to be considered on an application by application basis.

In my example, I needed to save the state of a table containing country names so that the interface retains the information for check box state (checked or unchecked) after restart. That is, if I saved the application with two countries selected, I wanted to ensure that when I opened the file again those same two countries would still be selected, both in the interface as well as in the table that is used to generate the plot. Now accomplishing this was a more interesting task: my hack was to insert a DataTable component, which stored my table as an entry of a 1x1 Matrix rtable. Since the rtable that underlies a DataTable is loaded into memory on Maple load, this gave me a way to ensure that the checked country table was loaded on open.

Here, for example, is the initial creation of this table:

"if not eval( :-_SelectedCountries )::Matrix then\n",
"    :-_SelectedCountries := Matrix(1,1,[table([])]):\n",
"end if;\n",

For more details, just look for the term: “:-_SelectedCountries” in the preceding code.

I could easily devote separate posts to discussing in detail each of these two quick tips. Similarly, there’s much more that can be discussed with respect to authoring an interface using programmatic tools from the DocumentTools packages, but I found the best way to learn more about a command is to try it out yourself. Once you do, you’ll find that there are an endless number of combinations for the kinds of interfaces that can be quickly built using programmatic content generation. Several commands in Maple have already started down the path of inserting customized content for their output (see DataSets:-InsertSearchBox and AudioTools:-Preview as a couple of examples) and I can only see this trend growing.

Finally, I would like to say that getting started with programmatic content generation was intimidating at first, but with a little bit of experimentation, it was a rewarding experience that has changed the way in which I work in Maple. In many cases, I now view output as something that can be customized for any command. More often than not, I turn to commands like ‘Explore’ to create an interface to see how sweeping through parameters effects my results, and any time I want to perform a more complex analysis or visualization for my data, I write code to create interfaces that can more easily be customized and re-used for other applications.

If you are interested in learning more about this topic, some good examples to get started with are the examples page for programmatic content generation as well as the help pages for the DocumentTools:-Components and DocumentTools:-Layout sub-packages.

To download a copy of the worksheet used in this post, click here (note that the code can be found in the start-up code of this worksheet): CountryDataPEC.mw To create the datasets interface, simply run the CountrySelection(); command.


New developments (after the release of Maple 2016) happened in the project on exact solutions for "Partial Differential Equations & Boundary Conditions". This is work in collaboration with Katherina von Bulow and the improvements are of wide range, representing a noticeable step forward in the capabilities of the Maple system for this kind of problem. As usual, these improvements can be installed in current Maple 2016 by downloading the updated library from the Maplesoft R&D webpage for Differential Equations and Mathematical functions (the update is distributed merged with the updates of the Physics package)


The improvements cover:



PDE&BC in semi-infinite domains for which a bounded solution is sought


PDE & BC problems in bounded spatial domains via eigenfunction (Fourier) expansions


Implementation of another algebraic method for tackling linear PDE & BC


Improvements in solving PDE & BC solutions by first finding the PDE's general solution.


Improvements in solving PDE & BC problems by using a Fourier transform.


PDE & BC problems that used to require the option HINT = `+` are now solved automatically


What follows is a set of examples solved now with these new developments, organized in sections according to the kind of problem. Where relevant, the sections include a subsection on "How it works step by step".

PDE&BC in semi-infinite domains for which a bounded solution is sought can now also be solved via Laplace transforms


Maple is now able to solve more PDE&BC problems via Laplace transforms.


How it works: Laplace transforms act to change derivatives with respect to one of the independent variables of the domain into multiplication operations in the transformed domain. After applying a Laplace transform to the original problem, we can simplify the problem using the transformed BC, then solve the problem in the transformed domain, and finally apply the inverse Laplace transform to arrive at the final solution. It is important to remember to give pdsolve any necessary restrictions on the variables and constants of the problem, by means of the "assuming" command.


A new feature is that we can now tell pdsolve that the dependent variable is bounded, by means of the optional argument HINT = boundedseries.




Consider the problem of a falling cable lying on a table that is suddenly removed (cf. David J. Logan's Applied Partial Differential Equations p.115).

pde[1] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))-g
iv[1] := u(x, 0) = 0, u(0, t) = 0, (D[2](u))(x, 0) = 0


If we ask pdsolve to solve this problem without the condition of boundedness of the solution, we obtain:

`assuming`([pdsolve([pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

u(x, t) = -invlaplace(exp(s*x/c)*_F1(s), s, t)+invlaplace(exp(-s*x/c)*_F1(s), s, t)-(1/2)*g*t^2+invlaplace(exp(s*x/c)/s^3, s, t)*g


New: If we now ask for a bounded solution, by means of the option HINT = boundedseries, pdsolve simplifies the problem accordingly.

ans[1] := `assuming`([pdsolve([pde[1], iv[1]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

u(x, t) = (1/2)*g*(Heaviside(t-x/c)*(c*t-x)^2-c^2*t^2)/c^2



And we can check this answer against the original problem, if desired:

`assuming`([pdetest(ans[1], [pde[1], iv[1]])], [0 < t, 0 < x, 0 < c])

[0, 0, 0, 0]


How it works, step by step


 Let us see the process this problem undergoes to be solved by pdsolve, step by step.


First, the Laplace transform is applied to the PDE:


transformed_PDE := laplace((lhs-rhs)(pde[1]), t, s)

s^2*laplace(u(x, t), t, s)-(D[2](u))(x, 0)-s*u(x, 0)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s


and the result is simplified using the initial conditions:

simplified_transformed_PDE := eval(transformed_PDE, {iv[1]})

s^2*laplace(u(x, t), t, s)-c^2*(diff(diff(laplace(u(x, t), t, s), x), x))+g/s


Next, we call the function "laplace(u(x,t),t,s)" by the new name U:

eq_U := subs(laplace(u(x, t), t, s) = U(x, s), simplified_transformed_PDE)

s^2*U(x, s)-c^2*(diff(diff(U(x, s), x), x))+g/s


And this equation, which is really an ODE, is solved:

solution_U := dsolve(eq_U, U(x, s))

U(x, s) = exp(-s*x/c)*_F2(s)+exp(s*x/c)*_F1(s)-g/s^3


Now, since we want a BOUNDED solution, the term with the positive exponential must be zero, and we are left with:

bounded_solution_U := subs(coeff(rhs(solution_U), exp(s*x/c)) = 0, solution_U)

U(x, s) = exp(-s*x/c)*_F2(s)-g/s^3


Now, the initial solution must also be satisfied. Here it is, in the transformed domain:

Laplace_BC := laplace(u(0, t), t, s) = 0

laplace(u(0, t), t, s) = 0


Or, in the new variable U,

Laplace_BC_U := U(0, s) = 0

U(0, s) = 0


And by applying it to bounded_solution_U, we find the relationship

simplify(subs(x = 0, rhs(bounded_solution_U))) = 0

(_F2(s)*s^3-g)/s^3 = 0


isolate((_F2(s)*s^3-g)/s^3 = 0, indets((_F2(s)*s^3-g)/s^3 = 0, unknown)[1])

_F2(s) = g/s^3


so that our solution now becomes

bounded_solution_U := subs(_F2(s) = g/s^3, bounded_solution_U)

U(x, s) = exp(-s*x/c)*g/s^3-g/s^3


to which we now apply the inverse Laplace transform to obtain the solution to the problem:

`assuming`([u(x, t) = invlaplace(rhs(bounded_solution_U), s, t)], [0 < x, 0 < t, 0 < c])

u(x, t) = (1/2)*g*(-t^2+Heaviside(t-x/c)*(c*t-x)^2/c^2)


Four other related examples


A few other examples:

pde[2] := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
iv[2] := u(x, 0) = 0, u(0, t) = g(t), (D[2](u))(x, 0) = 0

ans[2] := `assuming`([pdsolve([pde[2], iv[2]], HINT = boundedseries)], [0 < t, 0 < x, 0 < c])

u(x, t) = Heaviside(t-x/c)*g((c*t-x)/c)


`assuming`([pdetest(ans[2], [pde[2], iv[2]])], [0 < t, 0 < x, 0 < c])

[0, 0, 0, 0]


pde[3] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[3] := u(x, 0) = 0, u(0, t) = 1

ans[3] := `assuming`([pdsolve([pde[3], iv[3]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

u(x, t) = 1-erf((1/2)*x/(t^(1/2)*k^(1/2)))


pdetest(ans[3], [pde[3], iv[3][2]])

[0, 0]


pde[4] := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv[4] := u(x, 0) = mu, u(0, t) = lambda

ans[4] := `assuming`([pdsolve([pde[4], iv[4]], HINT = boundedseries)], [0 < t, 0 < x, 0 < k])

u(x, t) = (-lambda+mu)*erf((1/2)*x/(t^(1/2)*k^(1/2)))+lambda


pdetest(ans[4], [pde[4], iv[4][2]])

[0, 0]



The following is an example from page 76 in Logan's book:

pde[5] := diff(u(x, t), t) = diff(u(x, t), x, x)
iv[5] := u(x, 0) = 0, u(0, t) = f(t)

ans[5] := `assuming`([pdsolve([pde[5], iv[5]], HINT = boundedseries)], [0 < t, 0 < x])

u(x, t) = (1/2)*x*(int(f(_U1)*exp(-x^2/(4*t-4*_U1))/(t-_U1)^(3/2), _U1 = 0 .. t))/Pi^(1/2)


More PDE&BC problems in bounded spatial domains can now be solved via eigenfunction (Fourier) expansions


The code for solving PDE&BC problems in bounded spatial domains has been expanded. The method works by separating the variables by product, so that the problem is transformed into an ODE system (with initial and/or boundary conditions) problem, one of which is a Sturm-Liouville problem (a type of eigenvalue problem) which has infinitely many solutions - hence the infinite series representation of the solutions.



Here is a simple example for the heat equation:

pde__6 := diff(u(x, t), t) = k*(diff(u(x, t), x, x)); iv__6 := u(0, t) = 0, u(l, t) = 0

ans__6 := `assuming`([pdsolve([pde__6, iv__6])], [0 < l])

u(x, t) = Sum(_C1*sin(_Z1*Pi*x/l)*exp(-k*Pi^2*_Z1^2*t/l^2), _Z1 = 1 .. infinity)


pdetest(ans__6, [pde__6, iv__6])

[0, 0, 0]



Now, consider the displacements of a string governed by the wave equation, where c is a constant (cf. Logan p.28).

pde__7 := diff(u(x, t), t, t) = c^2*(diff(u(x, t), x, x))
iv__7 := u(0, t) = 0, u(l, t) = 0

ans__7 := `assuming`([pdsolve([pde__7, iv__7])], [0 < l])

u(x, t) = Sum(sin(_Z2*Pi*x/l)*(sin(c*_Z2*Pi*t/l)*_C1+cos(c*_Z2*Pi*t/l)*_C5), _Z2 = 1 .. infinity)


pdetest(ans__7, [pde__7, iv__7])

[0, 0, 0]


Another wave equation problem (cf. Logan p.130):

pde__8 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = 0; iv__8 := u(0, t) = 0, (D[2](u))(x, 0) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

ans__8 := `assuming`([pdsolve([pde__8, iv__8], u(x, t))], [0 <= x, x <= l])

u(x, t) = Sum(2*(Int(f(x)*sin((1/2)*Pi*(2*_Z3+1)*x/l), x = 0 .. l))*sin((1/2)*Pi*(2*_Z3+1)*x/l)*cos((1/2)*c*Pi*(2*_Z3+1)*t/l)/l, _Z3 = 1 .. infinity)


pdetest(ans__8, [pde__8, iv__8[1 .. 3]])

[0, 0, 0, 0]



Here is a problem with periodic boundary conditions (cf. Logan p.131). The function u(x, t) stands for the concentration of a chemical dissolved in water within a tubular ring of circumference 2*l. The initial concentration is given by f(x), and the variable x is the arc-length parameter that varies from 0 to 2*l.

pde__9 := diff(u(x, t), t) = M*(diff(u(x, t), x, x))
iv__9 := u(0, t) = u(2*l, t), (D[1](u))(0, t) = (D[1](u))(2*l, t), u(x, 0) = f(x)

ans__9 := `assuming`([pdsolve([pde__9, iv__9], u(x, t))], [0 <= x, x <= 2*l])

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*sin(_Z4*Pi*x/l), x = 0 .. 2*l))*sin(_Z4*Pi*x/l)+(Int(f(x)*cos(_Z4*Pi*x/l), x = 0 .. 2*l))*cos(_Z4*Pi*x/l))*exp(-M*Pi^2*_Z4^2*t/l^2)/l, _Z4 = 1 .. infinity)


pdetest(ans__9, [pde__9, iv__9[1 .. 2]])

[0, 0, 0]



The following problem is for heat flow with both boundaries insulated (cf. Logan p.166, 3rd edition)

pde__10 := diff(u(x, t), t) = k*(diff(u(x, t), x, x))
iv__10 := (D[1](u))(0, t) = 0, (D[1](u))(l, t) = 0, u(x, 0) = f(x)

ans__10 := `assuming`([pdsolve([pde__10, iv__10], u(x, t))], [0 <= x, x <= l])

u(x, t) = Sum(2*(Int(f(x)*cos(_Z6*Pi*x/l), x = 0 .. l))*cos(_Z6*Pi*x/l)*exp(-k*Pi^2*_Z6^2*t/l^2)/l, _Z6 = 1 .. infinity)


pdetest(ans__10, [pde__10, iv__10[1 .. 2]])

[0, 0, 0]



This is a problem in a bounded domain with the presence of a source. A source term represents an outside influence in the system and leads to an inhomogeneous PDE (cf. Logan p.149):

pde__11 := diff(u(x, t), t, t)-c^2*(diff(u(x, t), x, x)) = p(x, t)
iv__11 := u(0, t) = 0, u(Pi, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

ans__11 := pdsolve([pde__11, iv__11], u(x, t))

u(x, t) = Int(Sum(2*(Int(p(x, tau1)*sin(_Z7*x), x = 0 .. Pi))*sin(_Z7*x)*sin(c*_Z7*(t-tau1))/(Pi*_Z7*c), _Z7 = 1 .. infinity), tau1 = 0 .. t)


Current pdetest is unable to verify that this solution cancells the pde__11 mainly because it currently fails in identifying that there is a fourier expansion in it, but its subroutines for testing the boundary conditions work well with this problem

pdetest_BC := `pdetest/BC`

pdetest_BC({ans__11}, [iv__11], [u(x, t)])

[0, 0, 0, 0]




Consider a heat absorption-radiation problem in the bounded domain 0 <= x and x <= 2, t >= 0:

pde__12 := diff(u(x, t), t) = diff(u(x, t), x, x)
iv__12 := u(x, 0) = f(x), (D[1](u))(0, t)+u(0, t) = 0, (D[1](u))(2, t)+u(2, t) = 0

ans__12 := `assuming`([pdsolve([pde__12, iv__12], u(x, t))], [0 <= x and x <= 2, 0 <= t])

u(x, t) = (1/2)*_C8+Sum(((Int(f(x)*cos((1/2)*_Z8*Pi*x), x = 0 .. 2))*cos((1/2)*_Z8*Pi*x)+(Int(f(x)*sin((1/2)*_Z8*Pi*x), x = 0 .. 2))*sin((1/2)*_Z8*Pi*x))*exp(-(1/4)*Pi^2*_Z8^2*t), _Z8 = 1 .. infinity)


pdetest(ans__12, pde__12)



Consider the nonhomogeneous wave equation problem (cf. Logan p.213, 3rd edition):

pde__13 := diff(u(x, t), t, t) = A*x+diff(u(x, t), x, x)
iv__13 := u(0, t) = 0, u(1, t) = 0, u(x, 0) = 0, (D[2](u))(x, 0) = 0

ans__13 := pdsolve([pde__13, iv__13])

u(x, t) = Int(Sum(2*A*(Int(x*sin(Pi*_Z9*x), x = 0 .. 1))*sin(Pi*_Z9*x)*sin(Pi*_Z9*(t-tau1))/(Pi*_Z9), _Z9 = 1 .. infinity), tau1 = 0 .. t)


pdetest_BC({ans__13}, [iv__13], [u(x, t)])

[0, 0, 0, 0]



Consider the following Schrödinger equation with zero potential energy (cf. Logan p.30):

pde__14 := I*h*(diff(f(x, t), t)) = -h^2*(diff(f(x, t), x, x))/(2*m)
iv__14 := f(0, t) = 0, f(d, t) = 0

ans__14 := `assuming`([pdsolve([pde__14, iv__14])], [0 < d])

f(x, t) = Sum(_C1*sin(_Z10*Pi*x/d)*exp(-((1/2)*I)*h*Pi^2*_Z10^2*t/(d^2*m)), _Z10 = 1 .. infinity)


pdetest(ans__14, [pde__14, iv__14])

[0, 0, 0]


Another method has been implemented for linear PDE&BC


This method is for problems of the form


 "(&PartialD;w)/(&PartialD;t)=M[w]"", w(`x__i`,0) = f(`x__i`)" or


"((&PartialD;)^2w)/((&PartialD;)^( )t^2)="M[w]", w(`x__i`,0) = f(`x__i`), (&PartialD;w)/(&PartialD;t)() ? ()|() ? (t=0) =g(`x__i`)"


where M is an arbitrary linear differential operator of any order which only depends on the spatial variables x__i.


Here are some examples:

pde__15 := diff(w(x1, x2, x3, t), t)-(diff(w(x1, x2, x3, t), x2, x1))-(diff(w(x1, x2, x3, t), x3, x1))-(diff(w(x1, x2, x3, t), x3, x3))+diff(w(x1, x2, x3, t), x3, x2) = 0
iv__15 := w(x1, x2, x3, 0) = x1^5*x2*x3NULL

pdsolve([pde__15, iv__15])

w(x1, x2, x3, t) = 20*(((1/20)*x2*x3-(1/20)*t)*x1^2+(1/4)*t*(x2+x3)*x1+t^2)*x1^3


pdetest(%, [pde__15, iv__15])

[0, 0]



Here are two examples for which the derivative with respect to t is of the second order, and two initial conditions are given:

pde__16 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
iv__16 := w(x1, x2, x3, 0) = x1^3*x2^2+x3, (D[4](w))(x1, x2, x3, 0) = -x2*x3+x1

pdsolve([pde__16, iv__16])

w(x1, x2, x3, t) = x1^3*x2^2+x3-t*x2*x3+t*x1+3*t^2*x2*x1^2+(1/6)*t^3+(1/2)*t^4*x1


pdetest(%, [pde__16, iv__16])

[0, 0, 0]


pde__17 := diff(w(x1, x2, x3, t), t, t) = diff(w(x1, x2, x3, t), x2, x1)+diff(w(x1, x2, x3, t), x3, x1)+diff(w(x1, x2, x3, t), x3, x3)-(diff(w(x1, x2, x3, t), x3, x2))
iv__17 := w(x1, x2, x3, 0) = x1^3*x3^2+sin(x1), (D[4](w))(x1, x2, x3, 0) = cos(x1)-x2*x3

pdsolve([pde__17, iv__17])

w(x1, x2, x3, t) = (1/2)*t^4*x1+t^2*x1^3+3*t^2*x1^2*x3+x1^3*x3^2+(1/6)*t^3-t*x2*x3+cos(x1)*t+sin(x1)


pdetest(%, [pde__17, iv__17])

[0, 0, 0]


More PDE&BC problems are now solved via first finding the PDE's general solution.


The following are examples of PDE&BC problems for which pdsolve is successful in first calculating the PDE's general solution, and then fitting the initial or boundary condition to it.

pde__18 := diff(u(x, y), x, x)+diff(u(x, y), y, y) = 0
iv__18 := u(0, y) = sin(y)/y

If we ask pdsolve to solve the problem, we get:

ans__18 := pdsolve([pde__18, iv__18])

u(x, y) = (sin(-y+I*x)+_F2(y-I*x)*(y-I*x)+(-y+I*x)*_F2(y+I*x))/(-y+I*x)


and we can check this answer by using pdetest:

pdetest(ans__18, [pde__18, iv__18])

[0, 0]


How it works, step by step:


The general solution for just the PDE is:

gensol := pdsolve(pde__18)

u(x, y) = _F1(y-I*x)+_F2(y+I*x)


Substituting in the condition iv__18, we get:

u(0, y) = sin(y)/y


gensol_with_condition := eval(rhs(gensol), x = 0) = rhs(iv__18)

_F1(y)+_F2(y) = sin(y)/y


We then isolate one of the functions above (we can choose either one, in this case), convert it into a function operator, and then apply it to gensol

_F1 = unapply(solve(_F1(y)+_F2(y) = sin(y)/y, _F1(y)), y)

_F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc)


eval(gensol, _F1 = (proc (y) options operator, arrow; (-_F2(y)*y+sin(y))/y end proc))

u(x, y) = (-_F2(y-I*x)*(y-I*x)-sin(-y+I*x))/(y-I*x)+_F2(y+I*x)




Three other related examples


pde__19 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
iv__19 := u(0, y) = sin(y)/y

pdsolve([pde__19, iv__19])

u(x, y) = (2*sin(-y+((1/2)*I)*2^(1/2)*x)+(-I*2^(1/2)*x+2*y)*_F2(y-((1/2)*I)*2^(1/2)*x)+(I*2^(1/2)*x-2*y)*_F2(y+((1/2)*I)*2^(1/2)*x))/(I*2^(1/2)*x-2*y)


pdetest(%, [pde__19, iv__19])

[0, 0]


pde__20 := diff(u(x, y), x, x)+(1/2)*(diff(u(x, y), y, y)) = 0
iv__20 := u(x, 0) = sin(x)/x

pdsolve([pde__20, iv__20])

u(x, y) = (sinh((1/2)*(I*2^(1/2)*x-2*y)*2^(1/2))*2^(1/2)-(I*2^(1/2)*x-2*y)*(_F2(-y+((1/2)*I)*2^(1/2)*x)-_F2(y+((1/2)*I)*2^(1/2)*x)))/(I*2^(1/2)*x-2*y)


pdetest(%, [pde__20, iv__20])

[0, 0]


pde__21 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
iv__21 := u(3, t) = sin(6*t)

ans__21 := pdsolve([pde__21, iv__21])

u(r, t) = -_F2(-(2*I)*ln(3)+I*ln(r)+t)+sin(-(6*I)*ln(3)+(6*I)*ln(r)+6*t)+_F2(-I*ln(r)+t)


pdetest(ans__21, [pde__21, iv__21])

[0, 0]


More PDE&BC problems are now solved by using a Fourier transform.



Consider the following problem with an initial condition:

pde__22 := diff(u(x, t), t) = diff(u(x, t), x, x)+m
iv__22 := u(x, 0) = sin(x)


pdsolve can solve this problem directly:

ans__22 := pdsolve([pde__22, iv__22])

u(x, t) = sin(x)*exp(-t)+m*t


And we can check this answer against the original problem, if desired:

pdetest(ans__22, [pde__22, iv__22])

[0, 0]


How it works, step by step


Similarly to the Laplace transform method, we start the solution process by first applying the Fourier transform to the PDE:


transformed_PDE := fourier((lhs-rhs)(pde__22) = 0, x, s)

-2*m*Pi*Dirac(s)+s^2*fourier(u(x, t), x, s)+diff(fourier(u(x, t), x, s), t) = 0


Next, we call the function "fourier(u(x,t),x,s1)" by the new name U:

transformed_PDE_U := subs(fourier(u(x, t), x, s) = U(t, s), transformed_PDE)

-2*m*Pi*Dirac(s)+s^2*U(t, s)+diff(U(t, s), t) = 0


And this equation, which is really an ODE, is solved:

solution_U := dsolve(transformed_PDE_U, U(t, s))

U(t, s) = (2*m*Pi*Dirac(s)*t+_F1(s))*exp(-s^2*t)


Now, we apply the Fourier transform to the initial condition iv__22:

u(x, 0) = sin(x)


transformed_IC := fourier(iv__22, x, s)

fourier(u(x, 0), x, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))


Or, in the new variable U,

trasnformed_IC_U := U(0, s) = rhs(transformed_IC)

U(0, s) = I*Pi*(Dirac(s+1)-Dirac(s-1))


Now, we evaluate solution_U at t = 0:

solution_U_at_IC := eval(solution_U, t = 0)

U(0, s) = _F1(s)


and substitute the transformed initial condition into it:

eval(solution_U_at_IC, {trasnformed_IC_U})

I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s)


Putting this into our solution_U, we get

eval(solution_U, {(rhs = lhs)(I*Pi*(Dirac(s+1)-Dirac(s-1)) = _F1(s))})

U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)


Finally, we apply the inverse Fourier transformation to this,

solution := u(x, t) = invfourier(rhs(U(t, s) = (2*m*Pi*Dirac(s)*t+I*Pi*(Dirac(s+1)-Dirac(s-1)))*exp(-s^2*t)), s, x)

u(x, t) = sin(x)*exp(-t)+m*t


PDE&BC problems that used to require the option HINT = `+` to be solved are now solved automatically


The following two PDE&BC problems used to require the option HINT = `+` in order to be solved. This is now done automatically within pdsolve.

pde__23 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0
iv__23 := u(1, t) = 0, u(2, t) = 5

ans__23 := pdsolve([pde__23, iv__23])

u(r, t) = 5*ln(r)/ln(2)


pdetest(ans__23, [pde__23, iv__23])

[0, 0, 0]


pde__24 := diff(u(x, y), y, y)+diff(u(x, y), x, x) = 6*x-6*y

iv__24 := u(x, 0) = x^3+11*x+1, u(x, 2) = x^3+11*x-7, u(0, y) = -y^3+1, u(4, y) = -y^3+109

ans__24 := pdsolve([pde__24, iv__24])

u(x, y) = x^3-y^3+11*x+1


pdetest(ans__24, [pde__24, iv__24])

[0, 0, 0, 0, 0]



Download PDE_and_BC_update.mw

Edgardo S. Cheb-Terrab
Physics, Differential Equations and Mathematical Functions, Maplesoft

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