Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

12.3E: Laplace's Equation in Rectangular Coordinates (Exercises)

  • Page ID
    18306
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    In Exercises [exer:12.3.1}- [exer:12.3.16} apply the definition developed in Example 1 to solve the boundary value problem. (Use Theorem [thmtype:11.3.5} where it applies.) Where indicated by , graph the surface \(u=u(x,y)\), \(0\le x\le a\), \(0\le y\le b\).

    [exer:12.3.1] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<1,\)
    \(u(x,0)=x(1-x),\quad u(x,1)=0,\quad 0\le x\le 1,\)
    \(u(0,y)=0,\quad u(1,y)=0,\quad 0\le y\le 1\)

    [exer:12.3.2] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<3,\)
    \(u(x,0)=x^2(2-x),\quad u(x,3)=0,\quad 0\le x\le 2,\)
    \(u(0,y)=0,\quad u(2,y)=0,\quad 0\le y\le 3\)

    [exer:12.3.3] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<2,\)
    \(u(x,0)= \left\{\begin{array}{cl} x,& 0\le x\le 1,\\ 2-x,&1\le x\le2, \end{array}\right. \quad u(x,2)=0,\quad 0\le x\le 2,\)
    \(u(0,y)=0,\quad u(2,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.4] \(u_{xx}+u_{yy}=0,\quad 0<x<\pi,\quad 0<y<1,\)
    \(u(x,0)=x\sin x,\quad u(x,\pi)=0,\quad 0\le x\le \pi,\)
    \(u(0,y)=0,\quad u(\pi,y)=0,\quad 0\le y\le 1\)

    [exer:12.3.5] \(u_{xx}+u_{yy}=0,\quad 0<x<3,\quad 0<y<2,\)
    \(u(x,0)=0,\quad u_y(x,2)=x^2,\quad 0\le x\le 3,\)
    \(u_x(0,y)=0,\quad u_x(3,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.6] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<2,\)
    \(u(x,0)=0,\quad u_y(x,2)=1-x,\quad 0\le x\le 1,\)
    \(u_x(0,y)=0,\quad u_x(1,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.7] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<2,\)
    \(u(x,0)=0,\quad u_y(x,2)=x^2-4,\quad 0\le x\le 2,\)
    \(u_x(0,y)=0,\quad u_x(2,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.8] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<1,\)
    \(u(x,0)=0,\quad u_y(x,1)=(x-1)^2,\quad 0\le x\le 1,\)
    \(u_x(0,y)=0,\quad u_x(1,y)=0,\quad 0\le y\le 1\)

    [exer:12.3.9] \(u_{xx}+u_{yy}=0,\quad 0<x<3,\quad 0<y<2,\)
    \(u(x,0)=0,\quad u_y(x,2)=0,\quad 0\le x\le 3,\)
    \(u(0,y)=y(4-y),\quad u_x(3,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.10] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<1,\)
    \(u(x,0)=0,\quad u_y(x,1)=0,\quad 0\le x\le 2,\)
    \(u(0,y)=y^2(3-2y),\quad u_x(2,y)=0,\quad 0\le y\le 1\)

    [exer:12.3.11] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<2,\)
    \(u(x,0)=0,\quad u_y(x,2)=0,\quad 0\le x\le 2,\)
    \(u(0,y)=(y-2)^3+8,\quad u_x(2,y)=0,\quad 0\le y\le 2\)

    [exer:12.3.12] \(u_{xx}+u_{yy}=0,\quad 0<x<3,\quad 0<y<1,\)
    \(u(x,0)=0,\quad u_y(x,1)=0,\quad 0\le x\le 3,\)
    \(u(0,y)=y(2y^2-9y+12),\quad u_x(3,y)=0,\quad 0\le y\le 1\)

    [exer:12.3.13] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<\pi,\)
    \(u_y(x,0)=0,\quad u(x,\pi)=0,\quad 0\le x\le 1,\)
    \(u_x(0,y)=0,\quad u_x(1,y)=\sin y,\quad 0\le y\le \pi\)

    [exer:12.3.14] \(u_{xx}+u_{yy}=0,\quad 0<x<2,\quad 0<y<3,\)
    \(u_y(x,0)=0,\quad u(x,3)=0,\quad 0\le x\le 2,\)
    \(u_x(0,y)=0,\quad u_x(2,y)=y(3-y),\quad 0\le y\le 3\)

    [exer:12.3.15] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<\pi,\)
    \(u_y(x,0)=0,\quad u(x,\pi)=0,\quad 0\le x\le 1,\)
    \(u_x(0,y)=0,\quad u_x(1,y)=\pi^2-y^2,\quad 0\le y\le \pi\)

    [exer:12.3.16] \(u_{xx}+u_{yy}=0,\quad 0<x<1,\quad 0<y<1,\)
    \(u_y(x,0)=0,\quad u(x,1)=0,\quad 0\le x\le 1,\)
    \(u_x(0,y)=0,\quad u_x(1,y)=1-y^3,\quad 0\le y\le 1\)

    [exer:12.3.17] \(u(x,0)=0,\quad u(x,b)=f(x),\quad 0<x<a,\\ u(0,y)=0,\quad u(a,y)=0,\quad 0<y<b\)
    \(a=3\),\(b=2\),\(f(x)=x(3-x)\)

    [exer:12.3.18] \(u(x,0)=f(x),\quad u(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=0,\quad u_x(a,y)=0,\quad 0<y<b\)
    \(a=2\),\(b=1\),\(f(x)=x^2(x-2)^2\)

    [exer:12.3.19] \(u(x,0)=f(x),\quad u(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=0,\quad u(a,y)=0,\quad 0<y<b\)
    \(a=1\),\(b=2\),\(f(x)=3x^3-4x^2+1\)

    [exer:12.3.20] \(u(x,0)=f(x),\quad u(x,b)=0,\quad 0<x<a,\\ u(0,y)=0,\quad u_x(a,y)=0,\quad 0<y<b\)
    \(a=3\),\(b=2\),\(f(x)=x(6-x)\)

    [exer:12.3.21] \(u(x,0)=f(x),\quad u_y(x,b)=0,\quad 0<x<a,\\ u(0,y)=0,\quad u(a,y)=0,\quad 0<y<b\)
    \(a=\pi\),\(b=2\),\(f(x)=x(\pi^2-x^2)\)

    [exer:12.3.22] \(u_y(x,0)=0,\quad u(x,b)=f(x),\quad 0<x<a,\\ u_x(0,y)=0,\quad u_x(a,y)=0,\quad 0<y<b\)
    \(a=\pi\),\(b=1\),\(f(x)=x^2(x-\pi)^2\)

    [exer:12.3.23] \(u_y(x,0)=f(x),\quad u(x,b)=0,\quad 0<x<a,\\ u(0,y)=0,\quad u(a,y)=0,\quad 0<y<b\)
    \(a=\pi\),\(b=1\), \(f(x)= \left\{\begin{array}{cl} x,&0\le x\le{\pi\over2},\\\pi-x,&{\pi\over2}\le x\le \pi \end{array}\right.\)

    [exer:12.3.24] \(u(x,0)=0,\quad u(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=0,\quad u(a,y)=g(y),\quad 0<y<b\)
    \(a=1\),\(b=1\),\(g(y)=y(y^3-2y^2+1)\)

    [exer:12.3.25] \(u_y(x,0)=0,\quad u(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=0,\quad u(a,y)=g(y),\quad 0<y<b\)
    \(a=2\),\(b=2\),\(g(y)=4-y^2\)

    [exer:12.3.26] \(u(x,0)=0,\quad u(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=0,\quad u_x(a,y)=g(y),\quad 0<y<b\)
    \(a=1\), \(b=4\), \(g(y)= \left\{\begin{array}{cl} y,&0\le y\le2,\\4-y,&2\le y\le 4 \end{array}\right.\)

    [exer:12.3.27] \(u(x,0)=0,\quad u_y(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=g(y),\quad u_x(a,y)=0,\quad 0<y<b\)
    \(a=1\),\(b=\pi\),\(g(y)=y^2(3\pi-2y)\)

    [exer:12.3.28] \(u_y(x,0)=0,\quad u_y(x,b)=0,\quad 0<x<a,\\ u_x(0,y)=g(y),\quad u(a,y)=0,\quad 0<y<b\)
    \(a=2\),\(b=\pi\),\(g(y)=y\)

    [exer:12.3.29] \(u(x,0)=f(x),\quad 0<x<a\),
    \(u_x(0,y)=0,\quad u_x(a,y)=0,\quad y>0\)
    \(a=\pi\) \(f(x)=x^2(3\pi-2x)\)

    [exer:12.3.30] \(u(x,0)=f(x),\quad 0<x<a\),
    \(u_x(0,y)=0,\quad u(a,y)=0,\quad y>0\)
    \(a=3\),\(f(x)=9-x^2\)

    [exer:12.3.31] \(u(x,0)=f(x),\quad 0<x<a\),
    \(u(0,y)=0,\quad u_x(a,y)=0,\quad y>0\)
    \(a=\pi\),\(f(x)=x(2\pi-x)\)

    [exer:12.3.32] \(u_y(x,0)=f(x),\quad 0<x<a\),
    \(u(0,y)=0,\quad u(a,y)=0,\quad y>0\)
    \(a=\pi\),\(f(x)=x^2(\pi-x)\)

    [exer:12.3.33] \(u_y(x,0)=f(x),\quad 0<x<a\),
    \(u_x(0,y)=0,\quad u(a,y)=0,\quad y>0\)
    \(a=7\),\(f(x)=x(7-x)\)

    [exer:12.3.34] \(u_y(x,0)=f(x),\quad 0<x<a\),
    \(u(0,y)=0,\quad u_x(a,y)=0,\quad y>0\)
    \(a=5\),\(f(x)=x(5-x)\)

    [exer:12.3.35] Define the formal solution of the Dirichlet problem

    \[\begin{array}{c} u_{xx}+u_{yy}=0,\quad 0<x<a,\quad 0<y<b,\\ u(x,0)=f_0(x),\quad u(x,b)=f_1(x),\quad 0\le x\le a,\\ u(0,y)=g_0(y),\quad u(a,y)=g_1(y),\quad 0\le y\le b \end{array}\]

    [exer:12.3.36] Show that the Neumann Problem

    \[\begin{array}{c} u_{xx}+u_{yy}=0,\quad 0<x<a,\quad 0<y<b,\\ u_y(x,0)=f_0(x),\quad u_y(x,b)=f_1(x),\quad 0\le x\le a,\\ u_x(0,y)=g_0(y),\quad u_x(a,y)=g_1(y),\quad 0\le y\le b \end{array}\]

    has no solution unless

    \[\int_0^af_0(x)\,dx= \int_0^af_1(x)\,dx= \int_0^bg_0(y)\,dy= \int_0^bg_1(y)\,dy=0.\]

    In this case it has infinitely many formal solutions. Find them.

    [exer:12.3.37] In this exercise take it as given that the infinite series \(\sum_{n=1}^\infty n^pe^{-qn}\) converges for all \(p\) if \(q>0\), and, where appropriate, use the comparison test for absolute convergence of an infinite series.

    Let

    \[u(x,y)=\sum_{n=1}^\infty \alpha_n {\sinh n\pi(b-y)/a\over\sinh n\pi b/a} \sin{n\pi x\over a},\]

    where

    \[\alpha_n={2\over a}\int_0^a f(x)\sin{n\pi x\over a}\,dx\]

    and \(f\) is piecewise smooth on \([0,a]\).

    Verify the approximations

    \[{\sinh n\pi(b-y)/a\over\sinh n\pi b/a}\approx e^{-n\pi y/a},\quad y<b, \eqno{\rm(A)}\]

    and

    \[{\cosh n\pi(b-y)/a\over\sinh n\pi b/a}\approx e^{-n\pi y/a},\quad y<b \eqno{\rm(B)}\]

    for large \(n\).

    Use (A) to show that \(u\) is defined for \((x,y)\) such that \(0<y<b\).

    For fixed \(y\) in \((0,b)\), use (A) and Theorem [thmtype:12.1.2} with \(z=x\) to show that

    \[u_x(x,y)={\pi\over a}\sum_{n=1}^\infty n\alpha_n {\sinh n\pi(b-y)/a\over\sinh n\pi b/a} \cos{n\pi x\over a},\quad -\infty<x< \infty.\]

    Starting from the result of

    b

    , use (A) and Theorem [thmtype:12.1.2} with \(z=x\) to show that, for a fixed \(y\) in \((0,b)\),

    \[u_{xx}(x,y)=-{\pi^2\over a^2}\sum_{n=1}^\infty n^2\alpha_n {\sinh n\pi(b-y)/a\over\sinh n\pi b/a} \sin{n\pi x\over a},\quad -\infty<x< \infty.\]

    For fixed but arbitrary \(x\), use (B) and Theorem [thmtype:12.1.2} with \(z=y\) to show that

    \[u_y(x,y)=-{\pi\over a}\sum_{n=1}^\infty n\alpha_n {\cosh n\pi(b-y)/a\over\sinh n\pi b/a} \sin{n\pi x\over a}\]

    if \(0<y_0<y<b\), where \(y_0\) is an arbitrary number in \((0,b)\). Then argue that since \(y_0\) can be chosen arbitrarily small, the conclusion holds for all \(y\) in \((0,b)\).

    Starting from the result of

    e

    , use (A) and Theorem [thmtype:12.1.2} to show that

    \[u_{yy}(x,y)={\pi^2\over a^2}\sum_{n=1}^\infty n^2\alpha_n {\sinh n\pi(b-y)/a\over\sinh n\pi b/a} \sin{n\pi x\over a},\quad 0<y<b.\]

    Conclude that \(u\) satisfies Laplace’s equation for all \((x,y)\) such that \(0<y<b\).

    By repeatedly applying the arguments in

    c

    f

    , it can be shown that \(u\) can be differentiated term by term any number of times with respect to \(x\) and/or \(y\) if \(0<y<b\).