5.4 Helmholtz in 2D

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5.4.0.1 [331] In rectangle

problem number 331

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*}

pict
Figure 5.51:PDE specification

Mathematica


\[\left \{\left \{u(x,y)\to \underset {n=1}{\overset {\infty }{\sum }}\frac {64 \left (\cos \left (\frac {n \pi }{8}\right )+\cos \left (\frac {3 n \pi }{8}\right )\right ) \text {csch}\left (\frac {1}{2} \sqrt {n^2 \pi ^2-80}\right ) \sin ^3\left (\frac {n \pi }{8}\right ) \sin \left (\frac {n \pi x}{4}\right ) \sinh \left (\frac {1}{4} \sqrt {n^2 \pi ^2-80} (2-y)\right )}{n^2 \pi ^2}\right \}\right \}\]

Maple


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

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5.4.0.2 [332] On whole plane

problem number 332

Added December 27, 2018.

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

pict
Figure 5.52:PDE specification

Mathematica


\[\left \{\left \{u(x,y)\to e^{\sqrt {c_5} (-x)} \left (c_1 e^{2 \sqrt {c_5} x}+c_2\right ) \left (c_4 \cos \left (\sqrt {5+c_5} y\right )+c_3 \sin \left (\sqrt {5+c_5} y\right )\right )\right \}\right \}\] why? It solved earlier with BC?

Maple


\[u \left (x , y\right ) = \left (c_{1} {\mathrm e}^{2 x \sqrt {\textit {\_c}_{1}}}+c_{2}\right ) \left (c_{3} \sin \left (\sqrt {\textit {\_c}_{1}+5}\, y \right )+c_{4} \cos \left (\sqrt {\textit {\_c}_{1}+5}\, y \right )\right ) {\mathrm e}^{-x \sqrt {\textit {\_c}_{1}}}\]

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5.4.0.3 [333] Reduced Helmholtz Inside square

problem number 333

Added December 20, 2018.

Example 24, taken from https://www.mapleprimes.com/posts/209970-Exact-Solutions-For-PDE-And-Boundary--Initial-Conditions-2018

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*}

pict
Figure 5.53:PDE specification

Mathematica


\[\left \{\left \{u(x,y)\to \underset {K[1]=1}{\overset {\infty }{\sum }}-\frac {2 \left (-1+(-1)^{K[1]}\right ) \text {csch}\left (\pi \sqrt {K[1]^2+k}\right ) \sin (y K[1]) \sinh \left ((\pi -x) \sqrt {K[1]^2+k}\right )}{\pi K[1]}\right \}\right \}\]

Maple


\[u \left (x , y\right ) = \moverset {\infty }{\munderset {n =1}{\sum }}\frac {2 \left (\left (-1\right )^{n}-1\right ) \left ({\mathrm e}^{\sqrt {n^{2}+k}\, x}-{\mathrm e}^{-\left (x -2 \pi \right ) \sqrt {n^{2}+k}}\right ) \sin \left (n y \right )}{\left ({\mathrm e}^{2 \sqrt {n^{2}+k}\, \pi }-1\right ) \pi n}\]