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Expanded Ability to solve PDEs with Boundary Conditions

For Partial differential equations with boundary condition (PDE and BC), problems in three independent variables can now be solved, and more problems in two independent variables are now solved.

 

PDE&BC problems in three independent variables for bounded spatial domains can now be solved

More PDE&BC problems in two independent variables for bounded spatial domains can be solved

PDE&BC problems in three independent variables for bounded spatial domains can now be solved

We are now able to solve 3-variable PDE&BC problems in bounded spatial domains through separation of variables by product and eigenfunction expansion. The example problems and corresponding descriptions below are taken from: Articolo, George A. Partial Differential Equations and Boundary Value Problems with Maple. 2nd ed., Elsevier Academic Press, 2009.

 

A homogeneous diffusion PDE in two bounded spatial dimensions

 

The temperature distribution in thin plates whose lateral surfaces are insulated, in the bounded domain x=0..1,y=0..1, is sought. The problem is represented as follows (Articolo, p.396):

pde1tux,y,t=1102x2ux,y,t+2y2ux,y,t:

 

with the boundary and initial conditions:

 

iv1u0,y,t=0,u1,y,t=0,ux,0,t=0,ux,1,t=0,ux,y,0=x1x1yy:

pdsolvepde1,iv1,ux,y,t

ux,y,t=n1=1n=116−1n1+n+−1n1+−1n1sinnπxsinn1πyⅇtπ2n2+n1210n3π6n13

(1)

 

One way to verify that this solution satisfies the initial condition is to plot ux,y,0=x1x1yy together with its series representation (below we use the first 10000 terms in the solution) in the corresponding range, in this case x=0..1,y=0..1.

 

mapplot3d,subst=0,infinity=100,rhs,x1x1yy,x=0..1,y=0..1

PLOT3D...,PLOT3D...

(2)

 

A homogeneous diffusion PDE with heat loss in two bounded spatial dimensions

 

The temperature distribution in thin plates whose lateral surfaces are experiencing heat loss proportional to the plate and surrounding temperatures in a bounded domain is sought. The problem is represented as follows (Articolo, p. 401):

 

pde2tux,y,t=1102x2ux,y,t+1102y2ux,y,t15ux,y,t:

 

with the boundary and initial conditions:

 

iv2D1u0,y,t=0,u1,y,t=0,ux,0,t=0,D2ux,1,t=0,ux,y,0=x2+1112yy:

pdsolvepde2,iv2,ux,y,t

ux,y,t=n1=0n=0512−1ncos1+2nπx2sin1+2n1πy2ⅇ2+n2+n12+n+n1+12π2t101+2n3π61+2n13

(3)

 

To verify that this solution satisfies the initial condition, we plot ux,y,0=x2+1112yy and its series representation in the domain, which in this case is again x=0..1,y=0..1:

 

mapplot3d,subst=0,infinity=100,rhs,x2+1112yy,x=0..1,y=0..1

PLOT3D...,PLOT3D...

(4)

 

A homogeneous wave PDE in two bounded spatial dimensions

 

The wave amplitude for the transverse wave propagation on a thin rectangular membrane in a medium with no damping over a finite domain is sought. The problem is represented as follows (Articolo, p. 425):

 

pde32t2ux,y,t=142x2ux,y,t+142y2ux,y,t:

 

with the boundaries x=0,x=π unsecured, boundaries y=0,y=π secured, and initial conditions as follows:

 

iv3D1u0,y,t=0,D1uπ,y,t=0,ux,0,t=0,ux,π,t=0,D3ux,y,0=0,ux,y,0=xyπy:

pdsolvepde3,iv3,ux,y,t

ux,y,t=n=12−1n1sinnycosnt2n3+n=1n1=18−1n1+n+−1n1+−1n1cosn1xsinnycosn2+n12t2π2n12n3

(5)

 

We verify the initial condition by comparing plots of both the initial condition function ux,y,0=xyπy and its series representation:

 

mapplot3d,subst=0,infinity=100,rhs,xyπy,x=0..π,y=0..π

PLOT3D...,PLOT3D...

(6)

 

A homogeneous wave PDE (with damping) in two bounded spatial dimensions

 

The wave amplitude for the transverse wave propagation on a thin rectangular membrane in a medium with damping over a finite domain is sought. The problem is represented as follows (Articolo, p. 468):

 

pde42t2ux,y,t=142x2ux,y,t+2y2ux,y,t110tux,y,t:

 

with the boundaries x=0,y=0 held secure, boundaries x=1,y=1 unsecure, and initial conditions as follows:

 

iv4u0,y,t=0,D1u1,y,t=0,ux,0,t=0,D2ux,1,t=0,ux,y,0=0,D3ux,y,0=x112x112yy:

pdsolvepde4,iv4,ux,y,t

ux,y,t=n1=0n=05120sin1+2nπx2sin1+2n1πy2ⅇt20sint1+100n2+100n12+100n+100n1+50π2201+100n2+100n12+100n+100n1+50π2π61+2n131+2n3

(7)

 

To verify the initial condition, we compare the plots of both the initial condition function D3ux,y,0=x112x112yy and its series representation:

 

mapplot3d,subst=0,infinity=100,trhs,x112x112yy,x=0..1,y=0..1

PLOT3D...,PLOT3D...

(8)

 

 

More PDE&BC problems in two independent variables for bounded spatial domains can be solved

The code for solving PDE&BC problems in bounded spatial domains through separation by product and eigenfunction expansion continues to grow.

 

Solving more problems with periodic boundary conditions

 

A problem involving the diffusion PDE:

 

pde5tux,t=k2x2ux,t:

iv5ux,0=fx,ul,t=ul,t,D1ul,t=D1ul,t:

pdsolvepde5&comma;iv5&comma;ux&comma;tassuming0<l

ux&comma;t=2n=1llfxsinnπxl&DifferentialD;xsinnπxl+llfxcosnπxl&DifferentialD;xcosnπxl&ExponentialE;kπ2n2tl2ll+llfx&DifferentialD;x2l

(9)

 

A problem involving the Laplace PDE in cylindrical coordinates:

 

pde6rrrur&comma;θr+2θ2ur&comma;θr2=0&colon;

iv6ua&comma;θ=fθ,ur&comma;π=ur&comma;π,D2ur&comma;π=D2ur&comma;π&colon;

pdsolvepde6&comma;iv6&comma;ur&comma;θ&comma;HINT=boundedseriesassuming0<a

ur&comma;θ=2n=1rnππfθsinnθ&DifferentialD;θsinnθ+ππfθcosnθ&DifferentialD;θcosnθanππ+ππfθ&DifferentialD;θ2π

(10)

 

 

Another problem involving the heat PDE:

 

pde7tux&comma;t=2x2ux&comma;t&colon;

iv7ux&comma;0=fx,u1&comma;t=0,u1&comma;t=0&colon;

pdsolvepde7&comma;iv7&comma;ux&comma;t

ux&comma;t=n=1−11fxsinnπx&DifferentialD;xsinnπx&ExponentialE;tπ22n124+−11fxcosπx2n12&DifferentialD;xcosπx2n12&ExponentialE;tπ2n2&ExponentialE;tπ28n24n+14

(11)

 

Solving more problems with non-homogeneous boundary conditions

 

A problem involving the heat PDE:

 

pde8tux&comma;t=2x2ux&comma;t&colon;

iv8u0&comma;t=20,u1&comma;t=50,ux&comma;0=0&colon;

pdsolvepde8&comma;iv8&comma;ux&comma;t

ux&comma;t=20+30x+n=140+100−1nsinnπx&ExponentialE;tπ2n2nπ

(12)

 

A problem involving the heat PDE with a source:

 

pde9tux&comma;t+k2x2ux&comma;t+k&colon;

iv9ux&comma;0=fx,u0&comma;t=A,uL&comma;t=B&colon;

pdsolvepde9&comma;iv9&comma;ux&comma;t

ux&comma;t=2n=10L2fxL+L2x2+x22+ALxABsinπnxL&DifferentialD;xsinπnxL&ExponentialE;kπ2n2tL2L2L+L2x+x2+2AL2xAB2L

(13)

Solving more problems with a source term in the PDE

 

A diffusion problem with a source function fx&comma;t:

 

pde10tux&comma;t=k2x2ux&comma;t+fx&comma;t&colon;

iv10u0&comma;t=0,ul&comma;t=0,ux&comma;0=gx&colon;

pdsolvepde10&comma;iv10&comma;ux&comma;tassuming 0x l&semi;

ux&comma;t=n1=120lgxsinn1πxl&DifferentialD;xsinn1πxl&ExponentialE;kπ2n12tl2l+0tn=120lfx&comma;&tau;1sinnπxl&DifferentialD;xsinnπxl&ExponentialE;kπ2n2t&tau;1l2l&DifferentialD;&tau;1

(14)

Improvements to the ability to solve problems with the Laplace PDE

 

In cylindrical coordinates:

pde11rur&comma;θ+r2r2ur&comma;θr+2θ2ur&comma;θr2=0&colon;

iv11u0&comma;θ=0,u1&comma;θ=θ13πθ,ur&comma;0=0,ur&comma;13π=0&colon;

pdsolvepde11&comma;iv11&comma;ur&comma;θassuming 0<r,0<  &theta;<13π

ur&comma;θ=n=1_C5nπn34−1n9+49r3n+_C5nr3nπn3sin3nθπn3

(15)

Improvements to the solving of problems for which a bounded solution is sought

 

Solutions for this Laplace PDE problem are shown for the general problem, and then for the bounded problem (with u0&comma;θ<&rpar;:

pde12rur&comma;θr+2r2ur&comma;θ+2θ2ur&comma;θr2=0&colon;

iv12u1&comma;θ=fθ,ur&comma;0=0,ur&comma;13π=0&colon;

pdsolvepde12&comma;iv12assuming0<r1&comma;0< &theta;13π

ur&comma;θ=n=1sin3nθ6r3n0π3fθsin3nθ&DifferentialD;θ+π_C5nr3nr3nπ

(16)

pdsolvepde12&comma;iv12&comma;HINT=boundedseriesassuming 0<r1&comma;0<&theta;13π

ur&comma;θ=n=160π3fθsin3nθ&DifferentialD;θr3nsin3nθπ

(17)

 

Another problem with a Laplace PDE for which a bounded solution is sought:

 

pde13rur&comma;θ+r2r2ur&comma;θr+2θ2ur&comma;θr2=0&colon;

iv13u1&comma;θ=θ132θπ,ur&comma;0=0,D2ur&comma;13π=0&colon;

pdsolvepde13&comma;iv13&comma;ur&comma;θ&comma;HINT=boundedseries

ur&comma;θ=n=016r32+3nsin3nθ+32θ3π21+2n3

(18)

 

Solving problems for which there are both homogeneous and non-homogeneous boundary conditions for each independent variable

 

In this kind of problem, we don't directly arrive at a Sturm-Liouville problem after separation of variables, but instead must solve two simplified versions of the problem and then combine those answers to solve the original problem:

 

pde142x2ux&comma;y+2y2ux&comma;y=0&colon;

iv14u0&comma;y=y2+y,D1u1&comma;y=0,ux&comma;0=0,ux&comma;1=x112x&colon;

pdsolvepde14&comma;iv14&comma;ux&comma;yassuming 0<  x1

ux&comma;y=n=14−1n1sinπyn&ExponentialE;πnx2+&ExponentialE;nπxn3π3&ExponentialE;2nπ+1+n1=016sinπ1+2n1x2&ExponentialE;1+2n1πy1&ExponentialE;1+2n1πy12&ExponentialE;π1+2n111+2n13π3

(19)

 

General improvements

 

A non-linear PDE:

 

pde15xux&comma;y+ux&comma;yyux&comma;y=0&colon;

 

iv15ux&comma;0=1x+1&colon;

pdsolvepde15&comma;iv15&comma;ux&comma;y

ux&comma;y=y+1x+1

(20)

 

 

A wave PDE problem:

 

pde161002x2ux&comma;t=2t2ux&comma;t&colon;

iv16u0&comma;t=0,u2&comma;t=0,ux&comma;0=32sinπx+e2sin3πx+25sin6πx,D2ux&comma;0=6sin2πx16sin5πx12&colon;

pdsolvepde16&comma;iv16&comma;ux&comma;tassuming 0< x2

ux&comma;t=50sin3πxe2cos30πtπ+1600sinπxcos10πtπ+1250sin6πxcos60πtπ32sin5πx2sin25πt+15sin2πxsin20πt50π

(21)

 

Another wave PDE problem:

 

pde1742x2ux&comma;t=2t2ux&comma;t&colon;

iv17u0&comma;t=0,uπ&comma;t=0,ux&comma;0=0,D2ux&comma;0=6&colon;

pdsolvepde17&comma;iv17&comma;ux&comma;tassuming 0< x&pi;

ux&comma;t=n=16−1n1sinnxsin2ntπn2

(22)