### 4.4 Helmholtz in 2D

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##### 4.4.0.1 [330] In rectangle

problem number 330

Taken from Mathematica DSolve help pages.

Solve for $$u\left ( x,y\right )$$ \begin {align*} u_{xx}+ u_{yy} + 5 u(x,y) & = 0 \end {align*}

Boundary conditions \begin {align*} u(x,0) &= \text {UnitTriangle[x-2]} \\ u(x,2) &= 0 \\ u(0,y) &= 0 \\ u(4,y) &=0 \end {align*}

Mathematica

ClearAll["Global*"];
pde =  {Laplacian[u[x, y], {x, y}] + 5*u[x, y] == 0};
bc  = {u[x, 0] == Piecewise[{{-1 + x, x > 1 && x < 2}, {3 - x, x > 2 && x < 3}}], u[x, 2] == 0, u[0, y] == 0, u[4, y] == 0};
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}], 60*10]];
sol =  sol /. K[1] -> n



$\left \{\left \{u(x,y)\to \frac {1}{2} \sum _{n=1}^{\infty }\frac {128 \left (\cos \left (\frac {n \pi }{8}\right )+\cos \left (\frac {3 n \pi }{8}\right )\right ) \text {csch}\left (\frac {1}{2} \sqrt {n^2 \pi ^2-80}\right ) \sin ^3\left (\frac {n \pi }{8}\right ) \sin \left (\frac {n \pi x}{4}\right ) \sinh \left (\frac {1}{4} \sqrt {n^2 \pi ^2-80} (2-y)\right )}{n^2 \pi ^2}\right \}\right \}$

Maple

restart;
pde := diff(u(x,y),x$2)+diff(u(x,y),y$2)+5*u(x,y)=0;
bc  := u(x,0)=piecewise( x>1 and x<2, -1+x,x>2 and x<3 ,3-x),
u(x,2)=0,
u(0,y)=0,
u(4,y)=0;
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve([pde,bc],u(x,y))),output='realtime'));



$u \left ( x,y \right ) =\sum _{n=1}^{\infty }32\,{\frac { \left ( 1/2\, \left ( \sin \left ( 1/2\,\pi \,n \right ) -1/2\,\sin \left ( 1/4\,\pi \,n \right ) -1/2\,\sin \left ( 3/4\,\pi \,n \right ) \right ) \sin \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) \cos \left ( 1/4\,\sqrt {-{\pi }^{2}{n}^{2}+80}y \right ) +\cos \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) \sin \left ( 1/4\,\pi \,n \right ) \cos \left ( 1/4\,\pi \,n \right ) \sin \left ( 1/4\,\sqrt {-{\pi }^{2}{n}^{2}+80}y \right ) \left ( \cos \left ( 1/4\,\pi \,n \right ) -1 \right ) \right ) \sin \left ( 1/4\,n\pi \,x \right ) }{\sin \left ( 1/2\,\sqrt {-{\pi }^{2}{n}^{2}+80} \right ) {n}^{2}{\pi }^{2}}}$

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##### 4.4.0.2 [331] On whole plane

problem number 331

Solve for $$u\left ( x,y\right )$$ \begin {align*} u_{xx}+u_{yy} + 5 u(x,y) & = 0 \end {align*}

Mathematica

ClearAll["Global*"];
pde =  {Laplacian[u[x, y], {x, y}] + 5*u[x, y] == 0};
sol =  AbsoluteTiming[TimeConstrained[DSolve[pde, u[x, y], {x, y}], 60*10]];



Failed why? It solved earlier with BC?

Maple

restart;
pde := diff(u(x,y),x$2)+diff(u(x,y),y$2)+5*u(x,y)=0;
cpu_time := timelimit(60*10,CodeTools[Usage](assign('sol',pdsolve(pde,u(x,y),'build')),output='realtime'));



$u \left ( x,y \right ) ={\frac { \left ( \left ( {{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}} \right ) ^{2}{\it \_C1}+{\it \_C2} \right ) \left ( {\it \_C3}\,\sin \left ( \sqrt {{\it \_c}_{{1}}+5}y \right ) +{\it \_C4}\,\cos \left ( \sqrt {{\it \_c}_{{1}}+5}y \right ) \right ) }{{{\rm e}^{\sqrt {{\it \_c}_{{1}}}x}}}}$

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##### 4.4.0.3 [332] Reduced Helmholtz Inside square

problem number 332

Solve for $$u\left ( x,y\right )$$ \begin {align*} \frac {\partial ^{2}u}{\partial x^{2}}+\frac {\partial ^{2}u}{\partial y^2} - k u(x,y) & = 0 \end {align*}

With $$k>0$$. It is called reduced Helmholtz, because of the minus sign above. Otherwise, standard Helmholtz has a positive sign.

Boundary conditions \begin {align*} u(x,0) &= 0 \\ u(x,\pi ) &= 0 \\ u(0,y) &= 1 \\ u(\pi ,y) &=0 \end {align*}

Mathematica

ClearAll["Global*"];
pde =  Laplacian[u[x, y], {x, y}] - k*u[x, y] == 0;
bc  = {u[x, 0] == 0, u[x, Pi] == 0, u[0, y] == 1, u[Pi, y] == 0};
sol =  AbsoluteTiming[TimeConstrained[DSolve[{pde, bc}, u[x, y], {x, y}, Assumptions -> k > 0], 60*10]];



Failed

Maple

restart;
pde := diff(u(x, y), x$2)+diff(u(x, y), y$2)-k*u(x, y) = 0;
bc_left_edge:=u(0, y) = 1;
bc_lower_edge:=u(x, 0) = 0;
bc_top_edge:=u(x,Pi)=0;
bc_right_edge:=u(Pi,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 k>0),output='realtime'));

`

$u \left ( x,y \right ) =\sum _{n=1}^{\infty }2\,{\frac {\sin \left ( ny \right ) \left ( -1+ \left ( -1 \right ) ^{n} \right ) \left ( -{{\rm e}^{- \left ( -2\,\pi +x \right ) \sqrt {{n}^{2}+k}}}+{{\rm e}^{\sqrt {{n}^{2}+k}x}} \right ) }{ \left ( {{\rm e}^{2\,\sqrt {{n}^{2}+k}\pi }}-1 \right ) \pi \,n}}$