____________________________________________________________________________________
This is problem 2.5.1 part (a) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq H\), with following boundary conditions
\begin {align*} \frac {\partial u}{\partial x}(0,y) &= 0 \\ \frac {\partial u}{\partial x}(L,y) &= 0 \\ u(x,0)&=0 \\ u(x,H)&=f(x) \\ \end {align*}
Mathematica ✓
ClearAll[u, t, k, x, L, H, f]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {Derivative[1, 0][u][0, y] == 0, Derivative[1, 0][u][L, y] == 0, u[x, 0] == 0, u[x, H] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x <= L && 0 <= y <= H}], 60*10]]; sol = sol /. {K[1] -> n};
\[ \left \{\left \{u(x,y)\to \sum _{n=1}^{\infty }\frac {2 \cos \left (\frac {n \pi x}{L}\right ) \text {csch}\left (\frac {H n \pi }{L}\right ) \left (\int _0^L \cos \left (\frac {n \pi x}{L}\right ) f(x) \, dx\right ) \sinh \left (\frac {n \pi y}{L}\right )}{L}+\frac {y \int _0^L f(x) \, dx}{H L}\right \}\right \} \]
Maple ✓
H:='H';L:='L'; u:='u'; y:='y'; x:='x';f:='f'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; assume(L>0 and H>0); bc:=D[1](u)(0,y)=0,D[1](u)(L,y)=0,u(x,0)=0,u(x,H)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,bc],u(x,y)) assuming(0<=x and x<=L and 0<=y and y<=H)),output='realtime')); #these simplifications below to convert answer to one that match standard; sol:=convert(sol,trigh); sol:=simplify(expand(sol));
\[ u \left ( x,y \right ) ={\frac {1}{H\,L} \left ( 4\,\sum _{n=1}^{\infty } \left ( 1/2\,{1\cos \left ( {\frac {n\pi \,x}{L}} \right ) \int _{0}^{L}\!f \left ( x \right ) \cos \left ( {\frac {n\pi \,x}{L}} \right ) \,{\rm d}x\sinh \left ( {\frac {n\pi \,y}{L}} \right ) \left ( \sinh \left ( {\frac {n\pi \,H}{L}} \right ) \right ) ^{-1}} \right ) H+\int _{0}^{L}\!f \left ( x \right ) \,{\rm d}xy \right ) } \]
____________________________________________________________________________________
This is problem 2.5.1 part (b) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq H\), with following boundary conditions
\begin {align*} \frac {\partial u}{\partial x}(0,y) &= g(y) \\ \frac {\partial u}{\partial x}(L,y) &= 0 \\ u(x,0)&=0 \\ u(x,H)&=0 \\ \end {align*}
Mathematica ✓
ClearAll[u, t, k, x, L, H, g, f]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {Derivative[1, 0][u][0, y] == g[y], Derivative[1, 0][u][L, y] == 0, u[x, 0] == 0, u[x, H] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x <= L && 0 <= y <= H}], 60*10]]; sol = sol /. {K[1] -> n};
\[ \left \{\left \{u(x,y)\to \sum _{n=1}^{\infty }-\frac {2 \cosh \left (\frac {n \pi (L-x)}{H}\right ) \text {csch}\left (\frac {L n \pi }{H}\right ) \left (\int _0^H g(y) \sin \left (\frac {n \pi y}{H}\right ) \, dy\right ) \sin \left (\frac {n \pi y}{H}\right )}{n \pi }\right \}\right \} \]
Maple ✓
H:='H';L:='L'; u:='u'; y:='y'; x:='x';f:='f';g:='g'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; assume(L>0 and H>0): bc:=D[1](u)(0,y)=g(y),D[1](u)(L,y)=0,u(x,0)=0,u(x,H)=0: cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,bc],u(x,y)) assuming(0<=x and x<=L and 0<=y and y<=H)),output='realtime')); sol:=convert(sol,trigh);
\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( -2\,{\frac {1}{n\pi }\sin \left ( {\frac {n\pi \,y}{H}} \right ) \int _{0}^{H}\!\sin \left ( {\frac {n\pi \,y}{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( \cosh \left ( {\frac {n\pi \, \left ( 2\,L-x \right ) }{H}} \right ) +\sinh \left ( {\frac {n\pi \, \left ( 2\,L-x \right ) }{H}} \right ) +\cosh \left ( {\frac {n\pi \,x}{H}} \right ) +\sinh \left ( {\frac {n\pi \,x}{H}} \right ) \right ) \left ( \cosh \left ( 2\,{\frac {n\pi \,L}{H}} \right ) +\sinh \left ( 2\,{\frac {n\pi \,L}{H}} \right ) -1 \right ) ^{-1}} \right ) \]
____________________________________________________________________________________
This is problem 2.5.1 part (c) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq H\), with following boundary conditions
\begin {align*} \frac {\partial u}{\partial x}(0,y) &= 0 \\ u(L,y) &= g(y) \\ u(x,0)&=0 \\ u(x,H)&=0 \\ \end {align*}
Mathematica ✗
ClearAll[u, t, k, x, L, H, g, f]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {Derivative[1, 0][u][0, y] == 0, u[L, y] == g[y], u[x, 0] == 0, u[x, H] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x <= L && 0 <= y <= H}], 60*10]];
\[ \text {Failed} \]
Maple ✓
H:='H';L:='L'; u:='u'; y:='y'; x:='x';f:='f';g:='g'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; assume(L>0 and H>0); bc:=D[1](u)(0,y)=0,u(L,y)=g(y),u(x,0)=0,u(x,H)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol', pdsolve([pde,bc],u(x,y)) assuming(0<=x and x<=L and 0<=y and y<=H)),output='realtime')); sol:=convert(sol,trigh);
\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( 4\,{\frac {1}{H}\sin \left ( {\frac {n\pi \,y}{H}} \right ) \int _{0}^{H}\!\sin \left ( {\frac {n\pi \,y}{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( \cosh \left ( {\frac {n\pi \,L}{H}} \right ) +\sinh \left ( {\frac {n\pi \,L}{H}} \right ) \right ) \cosh \left ( {\frac {n\pi \,x}{H}} \right ) \left ( \cosh \left ( 2\,{\frac {n\pi \,L}{H}} \right ) +\sinh \left ( 2\,{\frac {n\pi \,L}{H}} \right ) +1 \right ) ^{-1}} \right ) \]
____________________________________________________________________________________
This is problem 2.5.1 part (d) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq H\), with following boundary conditions
\begin {align*} u(0,y) &= g(y) \\ u(L,y) &= 0 \\ \frac {\partial u}{\partial y}u(x,0)&=0 \\ u(x,H)&=0 \\ \end {align*}
Mathematica ✗
ClearAll[u, x, L, H, g, f]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {u[0, y] == 0, u[L, y] == 0, Derivative[0, 1][u][x, 0] == 0, u[x, H] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x <= L && 0 <= y <= H}], 60*10]];
\[ \text {Failed} \]
Maple ✓
H:='H';L:='L'; u:='u'; y:='y'; x:='x';f:='f';g:='g'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; assume(L>0 and H>0); bc:=u(0,y)=g(y),u(L,y)=0,D[2](u)(x,0)=0,u(x,H)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y)) assuming(0<=x and x<=L and 0<=y and y<=H)),output='realtime')); sol:=convert(sol,trigh);
\[ u \left ( x,y \right ) =\sum _{n=0}^{\infty } \left ( 2\,{\frac {1}{H}\sin \left ( 1/2\,{\frac {\pi \, \left ( 2\,ny+H+y \right ) }{H}} \right ) \int _{0}^{H}\!\sin \left ( 1/2\,{\frac {\pi \, \left ( 2\,ny+H+y \right ) }{H}} \right ) g \left ( y \right ) \,{\rm d}y \left ( \cosh \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \, \left ( 2\,L-x \right ) }{H}} \right ) +\sinh \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \, \left ( 2\,L-x \right ) }{H}} \right ) -\cosh \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{H}} \right ) -\sinh \left ( 1/2\,{\frac { \left ( 1+2\,n \right ) \pi \,x}{H}} \right ) \right ) \left ( \cosh \left ( {\frac { \left ( 1+2\,n \right ) \pi \,L}{H}} \right ) +\sinh \left ( {\frac { \left ( 1+2\,n \right ) \pi \,L}{H}} \right ) -1 \right ) ^{-1}} \right ) \]
____________________________________________________________________________________
This is problem 2.5.1 part (e) from Richard Haberman applied partial differential equations, 5th edition
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq H\), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(L,y) &= 0 \\ u(x,0) - \frac {\partial u}{\partial y}u(x,0)&=0 \\ u(x,H)&= f(x) \\ \end {align*}
Mathematica ✗
ClearAll[u, x, L, H, g, f]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {u[0, y] == 0, u[L, y] == 0, u[x, 0] - Derivative[0, 1][u][x, 0] == 0, u[x, H] == f[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x <= L && 0 <= y <= H}], 60*10]];
\[ \text {Failed} \]
Maple ✓
H:='H';L:='L'; u:='u'; y:='y'; x:='x';f:='f';g:='g'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; assume(L>0 and H>0); bc:=u(0,y)=0,u(L,y)=0,u(x,0)-D[2](u)(x,0)=0,u(x,H)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y)) assuming(0<=x and x<=L and 0<=y and y<=H)),output='realtime')); sol:=convert(sol,trigh);
\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty } \left ( 4\,{\frac {1}{L}\int _{0}^{L}\!\sin \left ( {\frac {n\pi \,x}{L}} \right ) f \left ( x \right ) \,{\rm d}x \left ( \cosh \left ( {\frac {n\pi \,H}{L}} \right ) +\sinh \left ( {\frac {n\pi \,H}{L}} \right ) \right ) \sin \left ( {\frac {n\pi \,x}{L}} \right ) \left ( \pi \,\cosh \left ( {\frac {n\pi \,y}{L}} \right ) n+L\,\sinh \left ( {\frac {n\pi \,y}{L}} \right ) \right ) \left ( \pi \,\sinh \left ( 2\,{\frac {n\pi \,H}{L}} \right ) n+\pi \,\cosh \left ( 2\,{\frac {n\pi \,H}{L}} \right ) n+L\,\sinh \left ( 2\,{\frac {n\pi \,H}{L}} \right ) +L\,\cosh \left ( 2\,{\frac {n\pi \,H}{L}} \right ) +\pi \,n-L \right ) ^{-1}} \right ) \]
Hand solution
Let \(u\left ( x,y\right ) =X\left ( x\right ) Y\left ( x\right ) \). Substituting this into the PDE \(\frac {\partial ^{2}u}{\partial x^{2}}+\frac {\partial ^{2}u}{\partial y^{2}}=0\) and simplifying gives\[ \frac {X^{\prime \prime }}{X}=-\frac {Y^{\prime \prime }}{Y}\] Each side depends on different independent variable and they are equal, therefore they must be equal to same constant.\[ \frac {X^{\prime \prime }}{X}=-\frac {Y^{\prime \prime }}{Y}=\pm \lambda \] Since the boundary conditions along the \(x\) direction are the homogeneous ones, \(-\lambda \) is selected in the above. Two ODE’s (1,2) are obtained as follows\begin {equation} X^{\prime \prime }+\lambda X=0 \tag {1} \end {equation} With the boundary conditions\begin {align*} X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
And\begin {equation} Y^{\prime \prime }-\lambda Y=0 \tag {2} \end {equation} With the boundary conditions\begin {align*} Y\left ( 0\right ) & =Y^{\prime }\left ( 0\right ) \\ Y\left ( H\right ) & =f\left ( x\right ) \end {align*}
In all these cases \(\lambda \) will turn out to be positive. This is shown for this problem only and not be repeated again.
Case \(\lambda <0\)
The solution to (1) us
\[ X=A\cosh \left ( \sqrt {\left \vert \lambda \right \vert }x\right ) +B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }x\right ) \] At \(x=0\), the above gives \(0=A\). Hence \(X=B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }x\right ) \). At \(x=L\) this gives \(X=B\sinh \left ( \sqrt {\left \vert \lambda \right \vert }L\right ) \). But \(\sinh \left ( \sqrt {\left \vert \lambda \right \vert }L\right ) =0\) only at \(0\) and \(\sqrt {\left \vert \lambda \right \vert }L\neq 0\), therefore \(B=0\) and this leads to trivial solution. Hence \(\lambda <0\) is not an eigenvalue.
Case \(\lambda =0\)
\[ X=Ax+B \] Hence at \(x=0\) this gives \(0=B\) and the solution becomes \(X=B\). At \(x=L\), \(B=0\). Hence the trivial solution. \(\lambda =0\) is not an eigenvalue.
Case \(\lambda >0\)
Solution is \[ X=A\cos \left ( \sqrt {\lambda }x\right ) +B\sin \left ( \sqrt {\lambda }x\right ) \] At \(x=0\) this gives \(0=A\) and the solution becomes \(X=B\sin \left ( \sqrt {\lambda }x\right ) \). At \(x=L\) \[ 0=B\sin \left ( \sqrt {\lambda }L\right ) \] For non-trivial solution \(\sin \left ( \sqrt {\lambda }L\right ) =0\) or \(\sqrt {\lambda }L=n\pi \) where \(n=1,2,3,\cdots \), therefore\[ \lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2}\qquad n=1,2,3,\cdots \] Eigenfunctions are\begin {equation} X_{n}\left ( x\right ) =B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \qquad n=1,2,3,\cdots \tag {3} \end {equation} For the \(Y\) ODE, the solution is\begin {align*} Y_{n} & =C_{n}\cosh \left ( \frac {n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac {n\pi }{L}y\right ) \\ Y_{n}^{\prime } & =C_{n}\frac {n\pi }{L}\sinh \left ( \frac {n\pi }{L}y\right ) +D_{n}\frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) \end {align*}
Applying B.C. at \(y=0\) gives\begin {align*} Y\left ( 0\right ) & =Y^{\prime }\left ( 0\right ) \\ C_{n}\cosh \left ( 0\right ) & =D_{n}\frac {n\pi }{L}\cosh \left ( 0\right ) \\ C_{n} & =D_{n}\frac {n\pi }{L} \end {align*}
The eigenfunctions \(Y_{n}\) are\begin {align*} Y_{n} & =D_{n}\frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +D_{n}\sinh \left ( \frac {n\pi }{L}y\right ) \\ & =D_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +\sinh \left ( \frac {n\pi }{L}y\right ) \right ) \end {align*}
Now the complete solution is produced\begin {align*} u_{n}\left ( x,y\right ) & =Y_{n}X_{n}\\ & =D_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +\sinh \left ( \frac {n\pi }{L}y\right ) \right ) B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Let \(D_{n}B_{n}=B_{n}\) since a constant. (no need to make up a new symbol).\[ u_{n}\left ( x,y\right ) =B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +\sinh \left ( \frac {n\pi }{L}y\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \] Sum of eigenfunctions is the solution, hence\[ u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +\sinh \left ( \frac {n\pi }{L}y\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \] The nonhomogeneous boundary condition is now resolved. At \(y=H\)\[ u\left ( x,H\right ) =f\left ( x\right ) \] Therefore\[ f\left ( x\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}H\right ) +\sinh \left ( \frac {n\pi }{L}H\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \] Multiplying both sides by \(\sin \left ( \frac {m\pi }{L}x\right ) \) and integrating gives\begin {align*} \int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx & =\int _{0}^{L}\sin \left ( \frac {m\pi }{L}x\right ) \sum _{n=1}^{\infty }B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}H\right ) +\sinh \left ( \frac {n\pi }{L}H\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\\ & =\sum _{n=1}^{\infty }B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}H\right ) +\sinh \left ( \frac {n\pi }{L}H\right ) \right ) \int _{0}^{L}\sin \left ( \frac {n\pi }{L}x\right ) \sin \left ( \frac {m\pi }{L}x\right ) dx\\ & =B_{m}\left ( \frac {m\pi }{L}\cosh \left ( \frac {m\pi }{L}H\right ) +\sinh \left ( \frac {m\pi }{L}H\right ) \right ) \frac {L}{2} \end {align*}
Hence\begin {equation} B_{n}=\frac {2}{L}\frac {\int _{0}^{L}f\left ( x\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx}{\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}H\right ) +\sinh \left ( \frac {n\pi }{L}H\right ) \right ) } \tag {4} \end {equation} This completes the solution. In summary\[ u\left ( x,y\right ) =\sum _{n=1}^{\infty }B_{n}\left ( \frac {n\pi }{L}\cosh \left ( \frac {n\pi }{L}y\right ) +\sinh \left ( \frac {n\pi }{L}y\right ) \right ) \sin \left ( \frac {n\pi }{L}x\right ) \] With \(B_{n}\) given by (4).
____________________________________________________________________________________
Taken from Mathematica DSolve help pages.
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
inside a rectangle \(0 \leq x \leq 1, 0 \leq y \leq 2\), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(1,y) &= 0 \\ u(x,0) &= \text {UnitTriagle(2 x-1)} \\ u(x,2) &= \text {UnitTriagle(2 x-1)} \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; L0 = 1; H0 = 2; bc = DirichletCondition[u[x, y] == Piecewise[{{UnitTriangle[2*x - L0], y == 0 || y == H0}}, 0], True]; domain = Rectangle[{0, 0}, {L0, H0}]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], Element[{x, y}, domain]]], 60*10]]; sol = sol /. K[1] -> n;
\[ \left \{\left \{u(x,y)\to \sum _{n=1}^{\infty }\frac {8 \text {csch}(2 n \pi ) \sin \left (\frac {n \pi }{2}\right ) \sin (n \pi x) (\sinh (n \pi (2-y))+\sinh (n \pi y))}{n^2 \pi ^2}\right \}\right \} \]
Maple ✓
u:='u'; y:='y'; x:='x'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; f:=x-> piecewise(x>0 and x<1/2, 2*x, x>1/2 and x<1, 2-2*x); bc:=u(0,y)=0,u(1,y)=0,u(x,0)=f(x),u(x,2)=f(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y)) assuming x>0,y>0),output='realtime'));
\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty }8\,{\frac {\sin \left ( 1/2\,\pi \,n \right ) {{\rm e}^{2\,\pi \,n}}\sin \left ( n\pi \,x \right ) \left ( -{{\rm e}^{\pi \,n \left ( -2+y \right ) }}+{{\rm e}^{-\pi \,n \left ( -2+y \right ) }}+{{\rm e}^{n\pi \,y}}-{{\rm e}^{-n\pi \,y}} \right ) }{{n}^{2}{\pi }^{2} \left ( {{\rm e}^{4\,\pi \,n}}-1 \right ) }} \]
____________________________________________________________________________________
Added December 20, 2018.
Example 21, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve Laplace equation \[ u_{xx}+u_{yy} = 0 \]
Inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq \infty \), with following boundary conditions
\begin {align*} u(0,y) &= A \\ u(L,y) &= 0 \\ u(x,0) &= 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, L, A]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == A, u[L, y] == 0, u[x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {x > 0, y > 0, L > 0}]], 60*10]];
\[ \text {Failed} \]
Maple ✓
u:='u'; y:='y'; x:='x';L:='L';A:='A'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge := u(0, y) = A; bc_right_edge:= u(L, y) = 0; bc_bottom_edge:= u(x, 0) = 0; bc:=bc_left_edge ,bc_right_edge,bc_bottom_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], HINT = boundedseries(y = infinity)) assuming x>0,y>0,L>0),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {1}{L} \left ( \sum _{n=1}^{\infty }-2\,{\frac {A}{\pi \,n}\sin \left ( {\frac {n\pi \,x}{L}} \right ) {{\rm e}^{-{\frac {n\pi \,y}{L}}}}}L+A \left ( L-x \right ) \right ) } \]
Hand solution
Let \begin {equation} u=U+v\tag {1} \end {equation} Where \(U\) satisfies \(\nabla ^{2}U=0\) but with right edge boundary conditions zero, and \(v\left ( x\right ) \) satisfies the nonhomogeneous boundary conditions \(v\left ( 0\right ) =A,v\left ( L\right ) =0\). This implies \[ v\left ( x\right ) =A\left ( 1-\frac {x}{L}\right ) \] Hence \(u=U+A\left ( 1-\frac {x}{L}\right ) \). Substituting this back in \(\nabla ^{2}u=0\) gives\[ \nabla ^{2}U=0 \] But with boundary condition on right edge being zero now. Let \(U=X\left ( x\right ) Y\left ( x\right ) \). Substituting this in the above gives\[ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y}=0 \] We want the eigenvalue problem to be in the \(X\) direction. Hence \begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
This has eigenvalues \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,\cdots \) with eigenfunctions \(X_{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \). The \(Y\) ode is\begin {align*} Y_{n}^{\prime \prime }-\lambda _{n}Y_{n} & =0\\ Y_{n}\left ( 0\right ) & =0 \end {align*}
Since \(\lambda _{n}>0\) then the solution is \(Y_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt {\lambda _{n}}y}+c_{2_{n}}e^{-\sqrt {\lambda _{n}}y}\). Since \(Y_{n}\left ( y\right ) \) is bounded, then \(c_{1_{n}}=0\) and the \(Y_{n}\left ( y\right ) =c_{2_{n}}e^{-\sqrt {\lambda _{n}}y}\). Hence\begin {align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-\sqrt {\lambda _{n}}y}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y} \end {align*}
Using the above in (1) gives the solution\begin {equation} u\left ( x,y\right ) =A\left ( 1-\frac {x}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y}\tag {2} \end {equation} At \(y=0\) the above gives\begin {align*} 0 & =A\left ( 1-\frac {x}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \\ A\left ( \frac {x}{L}-1\right ) & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Therefore \(B_{n}\) are the Fourier sine coefficients of \(\frac {A}{L}x\)\begin {align*} B_{n} & =\frac {2}{L}\int _{0}^{L}A\left ( \frac {x}{L}-1\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\\ & =\frac {2A}{L}\int _{0}^{L}\left ( \frac {x}{L}-1\right ) \sin \left ( \frac {n\pi }{L}x\right ) dx\\ & =-\frac {2A}{L}\frac {L}{n\pi }\\ & =-\frac {2A}{n\pi } \end {align*}
Hence the solution (2) becomes\[ u\left ( x,y\right ) =A\left ( 1-\frac {x}{L}\right ) -2\frac {A}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y}\]
____________________________________________________________________________________
Added March 19, 2019
Solve Laplace equation \[ u_{xx} + u_{yy} = 0 \]
Inside a rectangle \(0 \leq x \leq L, 0 \leq y \leq \infty \), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(L,y) &= A \\ u(x,0) &= 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, L, A]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == 0, u[L, y] == A, u[x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {x > 0, y > 0, L > 0}]], 60*10]];
\[ \text {Failed} \]
Maple ✓
u:='u'; y:='y'; x:='x';L:='L';A:='A'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge := u(0, y) = 0; bc_right_edge:= u(L, y) = A; bc_bottom_edge:= u(x, 0) = 0; bc:=bc_left_edge ,bc_right_edge,bc_bottom_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], HINT = boundedseries(y = infinity)) assuming x>0,y>0,L>0),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {1}{L} \left ( Ax+\sum _{n=1}^{\infty }2\,{\frac { \left ( -1 \right ) ^{n}A}{\pi \,n}\sin \left ( {\frac {n\pi \,x}{L}} \right ) {{\rm e}^{-{\frac {n\pi \,y}{L}}}}}L \right ) } \]
Hand solution
Let \begin {equation} u=U+v\tag {1} \end {equation} Where \(U\) satisfies \(\nabla ^{2}U=0\) but with right edge boundary conditions zero, and \(v\left ( x\right ) \) satisfies the nonhomogeneous boundary conditions \(v\left ( 0\right ) =0A,v\left ( L\right ) =A\). This implies \[ v\left ( x\right ) =A\frac {x}{L}\] Hence \(u=U+\frac {A}{L}x\). Substituting this back in \(\nabla ^{2}u=0\) gives\[ \nabla ^{2}U=0 \] But with boundary condition on right edge being zero now. Let \(U=X\left ( x\right ) Y\left ( x\right ) \). Substituting this in the above gives\[ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y}=0 \] We want the eigenvalue problem to be in the \(X\) direction. Hence \begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ X\left ( L\right ) & =0 \end {align*}
This has eigenvalues \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,\cdots \) with eigenfunctions \(X_{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}x\right ) \). The \(Y\) ode is\begin {align*} Y_{n}^{\prime \prime }-\lambda _{n}Y_{n} & =0\\ Y_{n}\left ( 0\right ) & =0 \end {align*}
Since \(\lambda _{n}>0\) then the solution is \(Y_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt {\lambda _{n}}y}+c_{2_{n}}e^{-\sqrt {\lambda _{n}}y}\). Since \(Y_{n}\left ( y\right ) \) is bounded, then \(c_{1_{n}}=0\) and the \(Y_{n}\left ( y\right ) =c_{2_{n}}e^{-\sqrt {\lambda _{n}}y}\). Hence\begin {align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}x\right ) e^{-\sqrt {\lambda _{n}}y}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y} \end {align*}
Using the above in (1) gives the solution\begin {equation} u\left ( x,y\right ) =\frac {A}{L}x+\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y}\tag {2} \end {equation} At \(y=0\) the above gives\begin {align*} 0 & =\frac {A}{L}x+\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \\ -\frac {A}{L}x & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}x\right ) \end {align*}
Therefore \(B_{n}\) are the Fourier sine coefficients of \(-\frac {A}{L}x\)\begin {align*} B_{n} & =-\frac {2}{L}\int _{0}^{L}\frac {A}{L}x\sin \left ( \frac {n\pi }{L}x\right ) dx\\ & =-\frac {2A}{L^{2}}\int _{0}^{L}x\sin \left ( \frac {n\pi }{L}x\right ) dx\\ & =-\frac {2A}{L^{2}}\frac {\left ( -1\right ) ^{n+1}L^{2}}{n\pi }\\ & =\frac {2A}{n\pi }\left ( -1\right ) ^{n} \end {align*}
Hence the solution (2) becomes\[ u\left ( x,y\right ) =\frac {A}{L}x+\frac {2A}{\pi }\sum _{n=1}^{\infty }\frac {\left ( -1\right ) ^{n}}{n}\sin \left ( \frac {n\pi }{L}x\right ) e^{-\frac {n\pi }{L}y}\]
____________________________________________________________________________________
Added March 19, 2019.
Solve Laplace equation \[ u_{xx}+u_{yy}= 0 \]
Inside a rectangle \(0 \leq y \leq L, 0 \leq x \leq \infty \), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(x,0) &= A \\ u(x,L) &= 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, L, A]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == 0, u[x, 0] == A, u[x, L] == 0}; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {x > 0, y > 0, L > 0}]], 60*10]];
\[ \text {Failed} \]
Maple ✗
u:='u'; y:='y'; x:='x';L:='L';A:='A'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge := u(0, y) = 0; bc_top_edge:= u(x, L) = 0; bc_bottom_edge:= u(x, 0) = A; bc:=bc_left_edge ,bc_top_edge,bc_bottom_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], HINT = boundedseries(x = infinity)) assuming x>0,y>0,L>0),output='realtime'));
\[ \text { sol=() } \]
Hand solution
Let \begin {equation} u\left ( x,y\right ) =U\left ( x,y\right ) +v\left ( y\right ) \tag {1} \end {equation} Where \(U\) satisfies \(\nabla ^{2}U=0\) but with bottom edge boundary conditions zero, and \(v\left ( y\right ) \) satisfies the nonhomogeneous boundary conditions \(v\left ( 0\right ) =A,v\left ( L\right ) =0\). This implies \[ v\left ( y\right ) =A\left ( 1-\frac {y}{L}\right ) \] Substituting (1) back in \(\nabla ^{2}u=0\) results in\[ \nabla ^{2}U=0 \] But with boundary condition on bottom edge as \(U=0\). Now we can use separation of variables. Let \(U=X\left ( x\right ) Y\left ( x\right ) \). Substituting this in the above gives\[ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y}=0 \] We want the eigenvalue problem to be in the \(Y\) direction. Hence\[ \frac {Y^{\prime \prime }}{Y}=-\frac {X^{\prime \prime }}{X}=-\lambda \] Therefore the eigenvalue problem is\begin {align*} Y^{\prime \prime }+\lambda Y & =0\\ Y\left ( 0\right ) & =0\\ Y\left ( L\right ) & =0 \end {align*}
This has eigenvalues \(\lambda _{n}=\left ( \frac {n\pi }{L}\right ) ^{2},n=1,2,\cdots \) with eigenfunctions \(Y_{n}\left ( x\right ) =\sin \left ( \sqrt {\lambda _{n}}y\right ) \). The \(X\) ode is\begin {align*} X_{n}^{\prime \prime }-\lambda _{n}X_{n} & =0\\ X_{n}\left ( 0\right ) & =0 \end {align*}
Since \(\lambda _{n}>0\) then the solution is \(X_{n}\left ( y\right ) =c_{1_{n}}e^{\sqrt {\lambda _{n}}x}+c_{2_{n}}e^{-\sqrt {\lambda _{n}}x}\). Since \(X_{n}\left ( x\right ) \) is bounded, then \(c_{1_{n}}=0\) and the \(X_{n}\left ( x\right ) =c_{2_{n}}e^{-\sqrt {\lambda _{n}}x}\). Hence by superposition the solution is\begin {align*} U\left ( x,y\right ) & =\sum _{n=1}^{\infty }X_{n}\left ( x\right ) Y_{n}\left ( y\right ) \\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \sqrt {\lambda _{n}}y\right ) e^{-\sqrt {\lambda _{n}}x}\\ & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}y\right ) e^{-\frac {n\pi }{L}x} \end {align*}
Substituting the above in (1) gives\begin {equation} u\left ( x,y\right ) =A\left ( 1-\frac {y}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}y\right ) e^{-\frac {n\pi }{L}x}\tag {2} \end {equation} At \(x=0\) the above gives\begin {align*} 0 & =A\left ( 1-\frac {y}{L}\right ) +\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}y\right ) \\ A\left ( \frac {y}{L}-1\right ) & =\sum _{n=1}^{\infty }B_{n}\sin \left ( \frac {n\pi }{L}y\right ) \end {align*}
Therefore \(B_{n}\) are the Fourier sine coefficients of \(A\left ( \frac {y}{L}-1\right ) \)\begin {align*} B_{n} & =\frac {2}{L}\int _{0}^{L}A\left ( \frac {y}{L}-1\right ) \sin \left ( \frac {n\pi }{L}y\right ) dy\\ & =\frac {2A}{L}\int _{0}^{L}\left ( \frac {y}{L}-1\right ) \sin \left ( \frac {n\pi }{L}y\right ) dy\\ & =-\frac {2A}{L}\frac {L}{n\pi }\\ & =-\frac {2A}{n\pi } \end {align*}
Hence the solution (2) becomes\[ u\left ( x,y\right ) =A\left ( 1-\frac {y}{L}\right ) -\frac {2A}{\pi }\sum _{n=1}^{\infty }\frac {1}{n}\sin \left ( \frac {n\pi }{L}y\right ) e^{-\frac {n\pi }{L}x}\]
____________________________________________________________________________________
Added March 20, 2019.
Solve Laplace equation \[ u_{xx}+u_{yy} = 0 \]
Inside a rectangle \(0 \leq y \leq L, 0 \leq x \leq \infty \), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(x,L) &= e^{-x} \\ u(x,0) &= 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, L, A]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == 0, u[x, L] == Exp[-x], u[x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {x > 0, y > 0, L > 0}]], 60*10]];
\[ \text {Failed} \]
Maple ✓
u:='u'; y:='y'; x:='x';L:='L';A:='A'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge := u(0, y) = 0; bc_top_edge:= u(x, L) = exp(-x); bc_bottom_edge:= u(x, 0) = 0; bc:=bc_left_edge ,bc_top_edge,bc_bottom_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc],u(x,y)) assuming x>0,y>0,L>0),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {1}{L} \left ( \sum _{n=1}^{\infty }-{\frac {1}{\pi \,n \left ( \pi \,n+L \right ) }\sin \left ( {\frac {n\pi \,y}{L}} \right ) \left ( {\it \_C5} \left ( n \right ) \left ( -{\pi }^{2}{n}^{2}+{L}^{2} \right ) {{\rm e}^{-{\frac {n\pi \,x}{L}}}}+ \left ( \left ( 2\,\pi \,n+4\,L \right ) \left ( -1 \right ) ^{n+1}+{\it \_C5} \left ( n \right ) \left ( \pi \,n+L \right ) ^{2} \right ) {{\rm e}^{{\frac {n\pi \,x}{L}}}}-2\,L \left ( - \left ( -1 \right ) ^{n}{{\rm e}^{-x}}+{\it \_C5} \left ( n \right ) \left ( \pi \,n+L \right ) \right ) \right ) }L+y{{\rm e}^{-x}} \right ) } \] I need to find out how Maple obtained the above solution. It seems to have unknown constant in it
Hand solution
Let \(u=X\left ( x\right ) Y\left ( x\right ) \). Substituting this in \(\nabla ^{2}u=0\)\[ \frac {X^{\prime \prime }}{X}+\frac {Y^{\prime \prime }}{Y}=0 \] We want the eigenvalue problem to be in the \(X\) direction. Hence\[ \frac {X^{\prime \prime }}{X}=-\frac {Y^{\prime \prime }}{Y}=-\lambda \] Therefore the eigenvalue problem is\begin {align*} X^{\prime \prime }+\lambda X & =0\\ X\left ( 0\right ) & =0\\ \left \vert X\left ( x\right ) \right \vert & <\infty \end {align*}
case \(\lambda <0\)
Solution is \(X\left ( x\right ) =c_{1}\cosh \left ( \sqrt {-\lambda }x\right ) +c_{2}\sinh \left ( \sqrt {-\lambda }x\right ) \). Since \(X\left ( 0\right ) =0\) then \(c_{1}=0\). Solution becomes \(X\left ( x\right ) =c_{2}\sinh \left ( \sqrt {-\lambda }x\right ) \). Since \(\sinh \) is not bounded on \(x>0\) as \(x\rightarrow \infty \) then \(c_{2}=0\). Therefore \(\lambda <0\) is not eigenvalue.
case \(\lambda =0\)
Solution is \(X\left ( x\right ) =c_{1}x+c_{2}\). At \(x=0\) this gives \(c_{2}=0\). Hence solution is \(X\left ( x\right ) =c_{1}x\). This is bounded as \(x\rightarrow \infty \) only when \(c_{1}=0\). Therefore \(\lambda =0\) is not eigenvalue.
case \(\lambda >0\)
Let \(\lambda =\alpha ^{2},\alpha >0\). Then solution is \(X\left ( x\right ) =c_{1}\cos \left ( \alpha x\right ) +c_{2}\sin \left ( \alpha x\right ) \). At \(x=0\) this results in \(0=c_{1}\). Hence the eigenvalues are \(\lambda =\alpha ^{2}\) for all real positive real numbers and eigenfunctions are \[ X_{\alpha }\left ( x\right ) =\sin \left ( \alpha x\right ) \] For the \(Y\) ode, \begin {align*} Y^{\prime \prime }-\alpha ^{2}Y & =0\\ Y\left ( 0\right ) & =0 \end {align*}
The solution is \(Y_{\alpha }\left ( y\right ) =c_{1}e^{\alpha y}+c_{2}e^{-\alpha y}\). Since \(Y\left ( 0\right ) =0\) then \(c_{2}=-c_{1}\) and the solution becomes \(Y_{\alpha }\left ( y\right ) =c_{1}\left ( e^{\alpha y}-e^{-\alpha y}\right ) =c_{1}\sinh \left ( \alpha y\right ) \). Hence the solution is generalized linear combination of \(Y\left ( y\right ) X\left ( x\right ) \) given by Fourier integral (since eigenvalues are continuous now and not discrete)\begin {align} u\left ( x,y\right ) & =\int _{0}^{\infty }A\left ( \alpha \right ) Y_{\alpha }\left ( y\right ) X_{\alpha }\left ( x\right ) d\alpha \nonumber \\ & =\int _{0}^{\infty }A\left ( \alpha \right ) \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) d\alpha \tag {1} \end {align}
When \(y=L\), then above becomes\[ e^{-x}=\int _{0}^{\infty }\left ( A\left ( \alpha \right ) \sinh \left ( \alpha L\right ) \right ) \sin \left ( \alpha x\right ) d\alpha \] Hence the coefficient\(\ A\left ( \alpha \right ) \sinh \left ( \alpha L\right ) \) is given by\begin {align*} A\left ( \alpha \right ) \sinh \left ( \alpha L\right ) & =\frac {2}{\pi }\int _{0}^{\infty }e^{-x}\sin \left ( \alpha x\right ) dx\\ & =\frac {2}{\pi }\frac {\alpha }{1+\alpha ^{2}} \end {align*}
Therefore \(A\left ( \alpha \right ) =\frac {2}{\pi \sinh \left ( \alpha L\right ) }\frac {\alpha }{1+\alpha }\). The solution (1) becomes\[ u\left ( x,y\right ) =\frac {2}{\pi }\int _{0}^{\infty }\frac {\alpha \sinh \left ( \alpha y\right ) \sin \left ( \alpha x\right ) }{\left ( 1+\alpha ^{2}\right ) \sinh \left ( \alpha L\right ) }d\alpha \]
____________________________________________________________________________________
Added April 4, 2019.
Exam problem, Math 4567, UMN. Spring 2019.
Solve Laplace equation \[ u_{xx}+u_{yy} = 0 \]
Inside a rectangle \(0 \leq y \leq 1, 0 \leq x \leq \infty \), with following boundary conditions
\begin {align*} u(0,y) &= 0 \\ u(x,1) &= f(x) \\ u(x,0) &= 0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, L, A]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == 0, u[x, 1] == f[x], u[x, 0] == 0}; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {x > 0, y > 0}]], 60*10]];
\[ \text {Failed} \]
Maple ✗
unassign('u,y,x,f'); pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge := u(0, y) = 0; bc_top_edge:= u(x, 1) = f(x); bc_bottom_edge:= u(x, 0) = 0; bc:=bc_left_edge ,bc_top_edge,bc_bottom_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc],u(x,t)) assuming x>0,y>0),output='realtime'));
\[ \text { Exception } \] Maple can not solve it when using boundedseries(x = infinity)
____________________________________________________________________________________
Added December 20, 2018
Solve Laplace equation for \(u(x,y)\)
\[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2} = 0 \]
With boundary condition
\begin {align*} u(x,0) &= \delta (x) \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = u[x, 0] == DiracDelta[x]; sol = AbsoluteTiming[TimeConstrained[Simplify[DSolve[{pde, bc}, u[x, y], x, y]], 60*10]];
\[ \left \{\left \{u(x,y)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {y}{\pi \left (x^2+y^2\right )} & y\geq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✓
u:='u'; y:='y'; x:='x'; pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; bc := u(x, 0) = Dirac(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc],u(x,y),method=Fourier)),output='realtime')); sol:=convert(sol,Int);
\[ u \left ( x,y \right ) =1/2\,{\frac {\int _{-\infty }^{\infty }\!{{\rm e}^{-sy+isx}}\,{\rm d}s}{\pi }} \]
____________________________________________________________________________________
Added December 20, 2018
Solve Laplace equation for \(u(x,y)\)
\[ u_{xx}+u_{yy} = 0 \]
With boundary condition
\begin {align*} u(0,y)&=0 \\ u(\pi ,y) &= \sinh (\pi ) \cos (y) \\ u(x,0) &= \sin (x) \\ u(x,\pi ) &= -\sinh (x) \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[0, y] == 0, u[Pi, y] == Sinh[Pi]*Cos[y], u[x, 0] == Sin[x], u[x, Pi] == -Sinh[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], x, y], 60*10]];
\[ \left \{\left \{u(x,y)\to \sum _{K[1]=1}^{\infty }\left (\frac {2 \left (1+(-1)^{K[1]}\right ) \text {csch}(\pi K[1]) K[1] \sin (y K[1]) \sinh (\pi ) \sinh (x K[1])}{\pi \left (K[1]^2-1\right )}+\text {csch}(\pi K[1]) \delta (K[1]-1) \sin (x K[1]) \sinh ((\pi -y) K[1])+\frac {2 (-1)^{K[1]} \text {csch}(\pi K[1]) K[1] \sin (x K[1]) \sinh (\pi ) \sinh (y K[1])}{\pi K[1]^2+\pi }\right )\right \}\right \} \]
Maple ✓
u:='u'; y:='y'; x:='x'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_side:= u(0,y)=0; bc_right_side:= u(Pi,y)=sinh(Pi)*cos(y); bc_bottom_side:= u(x,0)=sin(x); bc_top_side:= u(x,Pi)=-sinh(x); bc:= bc_left_side,bc_right_side,bc_bottom_side,bc_top_side; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {1}{{{\rm e}^{2\,\pi }}-1} \left ( \left ( {{\rm e}^{2\,\pi }}-1 \right ) \sum _{n=1}^{\infty }{\frac { \left ( -1 \right ) ^{n}n \left ( {{\rm e}^{2\,\pi }}-1 \right ) {{\rm e}^{n \left ( \pi -y \right ) -\pi }}\sin \left ( nx \right ) \left ( {{\rm e}^{2\,ny}}-1 \right ) }{\pi \, \left ( {n}^{2}+1 \right ) \left ( {{\rm e}^{2\,\pi \,n}}-1 \right ) }}+ \left ( {{\rm e}^{2\,\pi }}-1 \right ) \sum _{n=2}^{\infty }2\,{\frac {\sin \left ( ny \right ) {{\rm e}^{n \left ( \pi -x \right ) }}n\sinh \left ( \pi \right ) \left ( \left ( -1 \right ) ^{n}+1 \right ) \left ( {{\rm e}^{2\,nx}}-1 \right ) }{\pi \, \left ( {{\rm e}^{2\,\pi \,n}}-1 \right ) \left ( {n}^{2}-1 \right ) }}+\sin \left ( x \right ) \left ( {{\rm e}^{-y+2\,\pi }}-{{\rm e}^{y}} \right ) \right ) } \]
____________________________________________________________________________________
PDE example 18 from Maple help page
see march_20_2019_11_pm.tex for start of solution. Not completed yet
Solve Laplace equation \[ u_{xx} + u_{yy} = 0 \]
With boundary conditions
\begin {align*} u(0,y) &= \frac {\sin y}{y} \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = u[0, y] == Sin[y]/y; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> {0 <= x && 0 <= y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \frac {(\sinh (x)-\cosh (x)) (x \cos (y)-y \sin (y))+x}{x^2+y^2}\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; interface(showassumed=0); pde:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; bc:=u(0,y)=sin(y)/y; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {\sin \left ( -y+ix \right ) +{\it \_F2} \left ( y-ix \right ) \left ( y-ix \right ) + \left ( -y+ix \right ) {\it \_F2} \left ( y+ix \right ) }{-y+ix}} \]
____________________________________________________________________________________
Solve Laplace equation \[ \frac {\partial ^2 u}{\partial x^2} + \frac {\partial ^2 u}{\partial y^2}=0 \]
With boundary conditions
\begin {align*} u(0,y) &= \sin y \\ u(x,0) &= 0 \\ u(x,a) &= 0 \\ u(\infty ,y) &= 0 \end {align*}
Mathematica ✗
ClearAll[x, y, a]; ode = D[u[x, y], {x, 2}] + D[u[x, y], {y, 2}] == 0; bc = {u[x, 0] == 0, u[x, a] == 0, u[0, y] == Sin[y], u[Infinity, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{ode, bc}, u[x, y], {x, y}, Assumptions -> a > 0], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; y:='y'; a:='a'; interface(showassumed=0); ode:=diff(u(x,y),x$2)+diff(u(x,y),y$2)=0; bc:=u(x,0)=0, u(x,a)=0, u(0,y)=sin(y), u(infinity,y)=0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve({ode, bc}, u(x,y)) assuming a>0),output='realtime'));
\[ \text {Bad latex generated} \]
____________________________________________________________________________________
Solve Laplace equation in polar coordinates inside a disk
Solve for \(u\left ( r,\theta \right ) \) \begin {align*} u_{rr} + \frac {1}{r} u_r + \frac {1}{r^2} u_{\theta \theta }=0 \end {align*}
With \(0 \leq r\leq a, 0 < \theta \leq 2\pi \)
Boundary conditions \begin {align*} u(a,\theta ) & = f(\theta ) \\ \left \vert u\left ( 0,\theta \right ) \right \vert & <\infty \\ u\left ( r,0\right ) & =u\left ( r,2\pi \right ) \\ \frac {\partial u}{\partial \theta }\left ( r,0\right ) & =\frac {\partial u}{\partial \theta }\left ( r,2\pi \right ) \end {align*}
Mathematica ✓
ClearAll[u, theta, r, a, f]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r])/r + (1*D[u[r, theta], {theta, 2}])/r^2 == 0; bc = u[a, theta] == f[theta]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[r, theta], {r, theta}, Assumptions -> a < r && a > 0 && Inequality[0, Less, theta, LessEqual, 2*Pi]], 60*10]]; sol = sol /. K[1] -> n;
\[ \left \{\left \{u(r,\theta )\to \sum _{n=1}^{\infty }r^n \left (\frac {\cos (n \theta ) \left (\int _{-\pi }^{\pi } \cos (n \theta ) f(\theta ) \, d\theta \right ) a^{-n}}{\pi }+\frac {\left (\int _{-\pi }^{\pi } f(\theta ) \sin (n \theta ) \, d\theta \right ) \sin (n \theta ) a^{-n}}{\pi }\right )+\frac {\int _{-\pi }^{\pi } f(\theta ) \, d\theta }{2 \pi }\right \}\right \} \]
Maple ✓
r:='r'; theta:='theta'; a:='a'; r:='r';f:='f'; interface(showassumed=0); pde := (diff(r*(diff(u(r, theta), r)), r))/r +(diff(u(r, theta), theta, theta))/r^2 = 0; bc := u(a, theta) = f(theta), u(r, -Pi) = u(r, Pi), (D[2](u))(r, -Pi) = (D[2](u))(r, Pi); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc], u(r, theta), HINT = boundedseries(r=0))),output='realtime'));
\[ u \left ( r,\theta \right ) =1/2\,{\frac {1}{\pi } \left ( 2\,\sum _{n=1}^{\infty } \left ( {\frac {\cos \left ( n\theta \right ) \int _{-\pi }^{\pi }\!f \left ( \theta \right ) \cos \left ( n\theta \right ) \,{\rm d}\theta +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \sin \left ( n\theta \right ) \,{\rm d}\theta \sin \left ( n\theta \right ) }{\pi } \left ( {\frac {a}{r}} \right ) ^{-n}} \right ) \pi +\int _{-\pi }^{\pi }\!f \left ( \theta \right ) \,{\rm d}\theta \right ) } \]
____________________________________________________________________________________
Taken from Mathematica DSolve help pages
Solve for \(u( x,y) \)
\begin {align*} u_{xx}+ y_{yy} =0 \end {align*}
Boundary conditions \(u(x,0)=1\) for \(-\frac {1}{2}\leq x \leq \frac {1}{2}\) and \(x=0\) otherwise. This is called UnitBox in Mathematica.
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = u[x, 0] == UnitBox[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {\begin {array}{cc} \{ & \begin {array}{cc} \tan ^{-1}\left (\frac {\frac {1}{2}-x}{y}\right )+\tan ^{-1}\left (\frac {x+\frac {1}{2}}{y}\right ) & y>0\lor x>\frac {1}{2}\lor x<-\frac {1}{2} \\ 0 & \text {True} \\\end {array} \\\end {array}}{\pi } & y\geq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde:=diff(u(x,y),x$2)+ diff(u(x,y),y$2)=0; bc := u(x,0) =piecewise( x< -1/2 or x>1/2,0, 1); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime')); sol:=convert(sol,Int);
\[ u \left ( x,y \right ) =i \left ( 1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{-1/2\,s \left ( 2\,y+i \right ) +isx}}}{s}}\,{\rm d}s}-1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{1/2\,s \left ( -2\,y+i \right ) +isx}}}{s}}\,{\rm d}s} \right ) \]
____________________________________________________________________________________
Taken from Mathematica DSolve help pages
Solve for \(u( x,y) \)
\begin {align*} u_{xx}+ y_{yy} =0 \end {align*}
Boundary conditions \(u(0,y)=\sinc (y)\).
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = u[0, y] == Sinc[y]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {x+(x \cos (y)-y \sin (y)) (\sinh (x)-\cosh (x))}{x^2+y^2} & x\geq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde:=diff(u(x,y),x$2)+ diff(u(x,y),y$2)=0; bc := u(0,y) =sin(y)/y; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) ={\frac {\sin \left ( -y+ix \right ) +{\it \_F2} \left ( y-ix \right ) \left ( y-ix \right ) + \left ( -y+ix \right ) {\it \_F2} \left ( y+ix \right ) }{-y+ix}} \]
____________________________________________________________________________________
Taken from Mathematica DSolve help pages
Solve for \(u( x,y) \)
\begin {align*} u_{xx}+ y_{yy} =0 \end {align*}
Boundary conditions
\begin {align*} u(x,0) &= - \frac {1}{(x-2)^2+3}\\ u(0,y) &= \frac {1}{(y-3)^2+1}\\ \end {align*}
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[x, 0] == -((x - 2)^2 + 3)^(-1), u[0, y] == 1/((y - 3)^2 + 1)}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {2 \left (\frac {3 \left (y \left (3 \pi (x+1) \left (x^4-4 x^3+2 \left (y^2+12\right ) x^2-4 \left (y^2+10\right ) x+y^4-16 y^2+100\right )+x \left (\left (6 \tan ^{-1}(3)-\log (10)\right ) x^4+2 \left (\left (6 \tan ^{-1}(3)-\log (10)\right ) y^2+10 \left (6 \tan ^{-1}(3)+\log (10)\right )\right ) x^2-20 y^2 \left (6 \tan ^{-1}(3)+\log (10)\right )+260 \log (10)+y^4 \left (6 \tan ^{-1}(3)-\log (10)\right )+360 \tan ^{-1}(3)\right )+x \left (x^4+2 \left (y^2-10\right ) x^2+y^4+20 y^2-260\right ) \log \left (x^2+y^2\right )\right )-\left (x^6+\left (y^2+6\right ) x^4-\left (y^4-92 y^2+60\right ) x^2-y^6+6 y^4+60 y^2-1000\right ) \tan ^{-1}\left (\frac {y}{x}\right )\right )}{\left (x^2-2 x+y^2-6 y+10\right ) \left (x^2+2 x+y^2-6 y+10\right ) \left (x^2-2 x+y^2+6 y+10\right ) \left (x^2+2 x+y^2+6 y+10\right )}-\frac {3 \left (x^6+\left (y^2+5\right ) x^4-\left (y^4+46 y^2-35\right ) x^2-y^6+5 y^4-35 y^2+343\right ) \tan ^{-1}\left (\frac {x}{y}\right )+x \left (2 \pi \left (\sqrt {3} y^5-9 y^4+14 \sqrt {3} y^3-6 y^2-35 \sqrt {3} y+x^4 \left (\sqrt {3} y+3\right )+2 x^2 \left (\sqrt {3} y^3-3 y^2-7 \sqrt {3} y-3\right )+147\right )+y \left (\left (4 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )-3 \log (7)\right ) x^4+2 \left (y^2 \left (4 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )-3 \log (7)\right )-7 \left (4 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )+\log (343)\right )\right ) x^2+14 y^2 \left (4 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )+\log (343)\right )+189 \log (7)+y^4 \left (4 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )-3 \log (7)\right )-140 \sqrt {3} \tan ^{-1}\left (\frac {2}{\sqrt {3}}\right )\right )+3 y \left (x^4+2 \left (y^2+7\right ) x^2+y^4-14 y^2-63\right ) \log \left (x^2+y^2\right )\right )}{\left (x^4-8 x^3+2 \left (y^2+15\right ) x^2-8 \left (y^2+7\right ) x+y^4+2 y^2+49\right ) \left (x^4+8 x^3+2 \left (y^2+15\right ) x^2+8 \left (y^2+7\right ) x+y^4+2 y^2+49\right )}\right )}{3 \pi } & x\geq 0\land y\geq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✗
x:='x'; y:='y'; u:='u'; pde:=diff(u(x,y),x$2)+ diff(u(x,y),y$2)=0; bc:=u(x, 0) = (-1/((x - 2)^2 + 3)), u(0, y) = 1/((y - 3)^2 + 1); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[ \text { sol=() } \]
____________________________________________________________________________________
Taken from Mathematica DSolve help pages
Solve for \(u( x,y) \)
\begin {align*} \frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {align*}
Boundary conditions \(\frac {u}{y}(x,0)=\text {UnitBox[x]}\) where \(\text {UnitBox[x]}\) is \(1\) for \(-\frac {1}{2}\leq x \leq \frac {1}{2}\) and \(0\) otherwise. This is called UnitBox in Mathematica.
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = Derivative[0, 1][u][x, 0] == UnitBox[x]; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \left \{\left \{u(x,y)\to \begin {array}{cc} \{ & \begin {array}{cc} \frac {-4 y \tan ^{-1}\left (\frac {x-\frac {1}{2}}{y}\right )+4 y \tan ^{-1}\left (\frac {x+\frac {1}{2}}{y}\right )-2 x \log \left (4 x^2-4 x+4 y^2+1\right )+\log \left (4 x^2-4 x+4 y^2+1\right )+2 x \log \left (4 x^2+4 x+4 y^2+1\right )+\log \left (4 x^2+4 x+4 y^2+1\right )-2 \log (4)-4}{4 \pi } & y\geq 0 \\ \text {Indeterminate} & \text {True} \\\end {array} \\\end {array}\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde:=diff(u(x,y),x$2)+ diff(u(x,y),y$2)=0; bc:=(D[2](u))(x, 0) = piecewise( x< -1/2 or x>1/2,0, 1); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime')); sol:=convert(sol,Int);
\[ u \left ( x,y \right ) =i \left ( 1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{1/2\,s \left ( -2\,y+i \right ) +isx}}}{{s}^{2}}}\,{\rm d}s}-1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {{{\rm e}^{-1/2\,s \left ( 2\,y+i \right ) +isx}}}{{s}^{2}}}\,{\rm d}s} \right ) \] used convert(sol,Int).
____________________________________________________________________________________
Taken from Mathematica DSolve help pages
Solve for \(u( x,y) \)
\begin {align*} \frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {align*}
Boundary conditions \(u(x, 0) = x^2 (1 - x), u(x, 2) = 0, u(0, y) = 0, u(1, y) = 0\).
Mathematica ✓
ClearAll[u, x, y]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[x, 0] == x^2*(1 - x), u[x, 2] == 0, u[0, y] == 0, u[1, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]]; sol = sol /. K[1] -> n
\[ \left \{\left \{u(x,y)\to \sum _{n=1}^{\infty }-\frac {4 \left (1+2 (-1)^n\right ) \text {csch}(2 n \pi ) \sin (n \pi x) \sinh (n \pi (2-y))}{n^3 \pi ^3}\right \}\right \} \]
Maple ✓
x:='x'; y:='y'; u:='u'; pde:=diff(u(x,y),x$2)+ diff(u(x,y),y$2)=0; bc:=u(x, 0) = x^2*(1 - x),u(x, 2) = 0, u(0, y) = 0, u(1, y) = 0; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));
\[ u \left ( x,y \right ) =\sum _{n=1}^{\infty }4\,{\frac {\sin \left ( n\pi \,x \right ) \left ( 2\, \left ( -1 \right ) ^{1+n}{{\rm e}^{-\pi \,n \left ( y-4 \right ) }}-{{\rm e}^{-\pi \,n \left ( y-4 \right ) }}+2\,{{\rm e}^{n\pi \,y}} \left ( -1 \right ) ^{n}+{{\rm e}^{n\pi \,y}} \right ) }{{n}^{3}{\pi }^{3} \left ( {{\rm e}^{4\,\pi \,n}}-1 \right ) }} \]
____________________________________________________________________________________
Added December 20, 2018.
Solve for \(u( x,y) \)
\begin {align*} \frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {align*}
Boundary conditions \begin {align*} u(x, 0) &= 0 \\ u(x, b) &= h(x) \end {align*}
Mathematica ✗
ClearAll[u, x, y, h, b]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[x, 0] == 0, u[x, b] == h[x]}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; y:='y'; u:='u';h:='h'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2)=0; bc:=u(x,0)=0,u(x,b)=h(x); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc],u(x,y))),output='realtime')); sol:=convert(sol,Int);
\[ u \left ( x,y \right ) =-1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {\int _{-\infty }^{\infty }\!h \left ( x \right ) {{\rm e}^{-isx}}\,{\rm d}x{{\rm e}^{s \left ( b-y \right ) +isx}}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}s}+1/2\,{\frac {1}{\pi }\int _{-\infty }^{\infty }\!{\frac {\int _{-\infty }^{\infty }\!h \left ( x \right ) {{\rm e}^{-isx}}\,{\rm d}x{{\rm e}^{s \left ( b+y \right ) +isx}}}{{{\rm e}^{2\,sb}}-1}}\,{\rm d}s} \]
____________________________________________________________________________________
Added December 20, 2018.
Example 23, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve for \(u( x,y) \)
\begin {align*} \frac {\partial ^2 u}{\partial x^2}+\frac {\partial ^2 u}{\partial y^2}=0 \end {align*}
Boundary conditions \begin {align*} u(x, 0) &= 0 \\ u(x, a) &= 0\\ u(0,y) &= \sin (y) \\ u(\infty ,y) &=0 \end {align*}
Mathematica ✗
ClearAll[u, x, y, a]; pde = Laplacian[u[x, y], {x, y}] == 0; bc = {u[x, 0] == 0, u[x, a] == 0, u[0, y] == Sin[y], u[Infinity, y] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> a > 0], 60*10]];
\[ \text {Failed} \]
Maple ✓
x:='x'; y:='y'; u:='u';a:='a'; pde := diff(u(x, y), x$2)+diff(u(x, y), y$2) = 0; bc_left_edge:=u(0, y) = sin(y); bc_lower_edge:=u(x, 0) = 0; bc_top_edge:=u(x,a)=0; bc_right_edge:=u(infinity,y)=0; bc:=bc_left_edge,bc_lower_edge,bc_top_edge,bc_right_edge; cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc ], u(x, y)) assuming a>0),output='realtime'));
\[ \text {Bad latex generated} \]
____________________________________________________________________________________
Added December 20, 2018.
Example 20, Taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018
Solve Laplace equation in polar coordinates inside quarter disk with \(0<r<1\) and \(0<\theta <\frac {\pi }{2}\)
Solve for \(u\left ( r,\theta \right ) \)
\begin {align*} \frac {\partial ^{2}u}{\partial r^{2}}+\frac {1}{r}\frac {\partial u}{\partial r} +\frac {1}{r^{2}}\frac {\partial ^{2}u}{\partial \theta ^{2}} & =0\\ \end {align*}
Boundary conditions
\begin {align*} u(r,0) &= 0 \\ u(r,\frac {\pi }{2}) &=0 \ \frac {\partial u}{\partial r}(1,\theta ) &= f(\theta ) \end {align*}
Mathematica ✗
ClearAll[u, theta, r, f]; pde = D[u[r, theta], {r, 2}] + (1*D[u[r, theta], r])/r + (1*D[u[r, theta], {theta, 2}])/r^2 == 0; bcOnR = {Derivative[1, 0][u][1, theta] == f[theta]}; bcOnTheta = {u[r, 0] == 0, u[r, Pi/2] == 0}; sol = AbsoluteTiming[TimeConstrained[DSolve[{pde, bcOnR, bcOnTheta}, u[r, theta], {r, theta}, Assumptions -> {r > 0, r < 1, theta > 0, theta < Pi/2}], 60*10]];
\[ \text {Failed} \]
Maple ✓
r:='r'; theta:='theta'; r:='r';f:='f'; pde := diff(u(r, theta), r$2)+1/r* diff(u(r, theta), r)+1/r^2* diff(u(r, theta), theta$2)= 0; bc_on_theta:=u(r, 0) = 0, u(r,Pi/2) = 0; bc_on_r:= eval( diff(u(r,theta),r),r=1)=f(theta); cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde, bc_on_theta,bc_on_r], u(r, theta), HINT = boundedseries(r = [0])) assuming theta>0, theta < (1/2)*Pi, r>0, r < 1),output='realtime'));
\[ u \left ( r,\theta \right ) =\sum _{n=1}^{\infty } \left ( 2\,{\frac {\int _{0}^{\pi /2}\!f \left ( \theta \right ) \sin \left ( 2\,n\theta \right ) \,{\rm d}\theta {r}^{2\,n}\sin \left ( 2\,n\theta \right ) }{\pi \,n}} \right ) \]