New developments on exact solutions for PDEs with Boundary Conditions

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New developments on exact solutions for PDEs with Boundary Conditions

Posted: ecterrab 14200 Product: Maple

June 27 2016

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).

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

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

(1.1)

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

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

(1.2)

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

(1.3)

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:

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

(1.1.1)

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

(1.1.2)

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

(1.1.3)

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

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

(1.1.4)

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

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

(1.1.5)

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

(1.1.6)

Or, in the new variable U,

(1.1.7)

And by applying it to bounded_solution_U, we find the relationship

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

(1.1.8)

(1.1.9)

so that our solution now becomes

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

(1.1.10)

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

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

(1.1.11)

Four other related examples

A few other examples:

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

(1.2.1)

(1.2.2)

(1.2.3)

(1.2.4)

(1.2.5)

(1.2.6)

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)

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)

(1.2.7)

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

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

(2.1)

(2.2)

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))

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)

(2.3)

(2.4)

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)

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)

(2.5)

(2.6)

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

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)

(2.7)

(2.8)

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

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)

(2.9)

(2.10)

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):

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)

(2.11)

Current pdetest is unable to verify that this solution cancells the 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

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

(2.12)

Consider a heat absorption-radiation problem in the bounded domain :

pde__12 := diff(u(x, t), t) = diff(u(x, t), x, x)

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)

(2.13)

(2.14)

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)

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)

(2.15)

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

(2.16)

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)

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)

(2.17)

(2.18)

Another method has been implemented for linear PDE&BC

This method is for problems of the form

", 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 .

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

(3.1)

(3.2)

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))

(3.3)

(3.4)

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))

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)

(3.5)

(3.6)

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

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

(4.1)

and we can check this answer by using pdetest:

(4.2)

How it works, step by step:

The general solution for just the PDE is:

(4.1.1)

Substituting in the condition , we get:

(4.1.2)

(4.1.3)

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

(4.1.4)

(4.1.5)

Three other related examples

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

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)

(4.2.1)

(4.2.2)

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

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)

(4.2.3)

(4.2.4)

pde__21 := diff(u(r, t), r, r)+(diff(u(r, t), r))/r+(diff(u(r, t), t, t))/r^2 = 0

(4.2.5)

(4.2.6)

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

Consider the following problem with an initial condition:

pdsolve can solve this problem directly:

(5.1)

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

(5.2)

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:

(5.1.1)

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

(5.1.2)

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

(5.1.3)

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

(5.1.4)

(5.1.5)

Or, in the new variable U,

(5.1.6)

Now, we evaluate solution_U at t = 0:

(5.1.7)

and substitute the transformed initial condition into it:

>	$eval(solution_U_at_IC, {trasnformed_IC_U})$

(5.1.8)

Putting this into our solution_U, we get

>	$eval(solution_U, {(rhs = lhs)(IPi(Dirac(s+1)-Dirac(s-1)) = _F1(s))})$

(5.1.9)

Finally, we apply the inverse Fourier transformation to this,

(5.1.10)

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

(6.1)

(6.2)

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

(6.3)

(6.4)

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

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