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Mathematics LibreTexts

13.1E: Boundary Value Problems (Exercises)

  • Page ID
    18268
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    [exer:13.1.1] Verify that \(B_{1}\) and \(B_{2}\) are linear operators; that is, if \(c_{1}\) and \(c_{2}\) are constants then \[B_{i}(c_{1}y_{1}+c_{2}y_{2})=c_{1}B_{i}(y_{1})+c_{2}B_{i}(y_{2}),\quad i=1,2.\]

    [exer:13.1.2] \(y''-y=x\), \(y(0)=-2\),\(y(1)=1\)

    [exer:13.1.3] \(y''=2-3x\), \(y(0)=0\),\(y(1)-y'(1)=0\)

    [exer:13.1.4] \(y''-y=x\), \(y(0)+y'(0)=3\),\(y(1)-y'(1)=2\)

    [exer:13.1.5] \(y''+4y=1\), \(y(0)=3\), \(y(\pi/2)+y'(\pi/2)=-7\)

    [exer:13.1.6] \(y''-2y'+y=2e^{x}\), \(y(0)-2y'(0)=3\), \(y(1)+y'(1)=6e\)

    [exer:13.1.7] \(y''-7y'+12y=4e^{2x}\), \(y(0)+y'(0)=8\), \(y(1)=-7e^{2}\) (see Example [example:13.1.5})

    [exer:13.1.8] State a condition on \(F\) such that the boundary value problem \[y''=F(x), \quad y(0)=0, \quad y(1)-y'(1)=0\] has a solution, and find all solutions.

    [exer:13.1.9]

    State a condition on \(a\) and \(b\) such that the boundary value problem

    \[y''+y=F(x),\quad y(a)=0,\quad y(b)=0 \tag{A}\]

    has a unique solution for every continuous \(F\), and find the solution by the method used to prove Theorem [thmtype:13.1.3}

    In the case where \(a\) and \(b\) don’t satisfy the condition you gave for (a), state necessary and sufficient on \(F\) such that (A) has a solution, and find all solutions by the method used to prove Theorem [thmtype:13.1.4}.

    [exer:13.1.10] Follow the instructions in Exercise [exer:13.1.9} for the boundary value problem \[y''+y=F(x),\quad y(a)=0,\quad y'(b)=0.\]

    [exer:13.1.11] Follow the instructions in Exercise [exer:13.1.9} for the boundary value problem \[y''+y=F(x),\quad y'(a)=0,\quad y'(b)=0.\]

    [exer:13.1.12] \(y''-y=F(x)\),\(y(a)=0\),\(y(b)=0\)

    [exer:13.1.13] \(y''-y=F(x)\),\(y(a)=0\),\(y'(b)=0\)

    [exer:13.1.14] \(y''-y=F(x)\),\(y'(a)=0\),\(y'(b)=0\)

    [exer:13.1.15] \(y''-y=F(x)\),\(y(a)-y'(a)=0\),\(y(b)+y'(b)=0\)

    [exer:13.1.16] \(y''+ \omega^{2}y=F(x)\),\(y(0)=0\),\(y(\pi)=0\)

    [exer:13.1.17] \(y''+ \omega^{2}y=F(x)\),\(y(0)=0\),\(y'(\pi)=0\)

    [exer:13.1.18] \(y''+ \omega^{2}y=F(x)\),\(y'(0)=0\),\(y(\pi)=0\)

    [exer:13.1.19] \(y''+ \omega^{2}y=F(x)\),\(y'(0)=0\),\(y'(\pi)=0\)

    [exer:13.1.20] Let \(\{z_{1},z_{2}\}\) be a fundamental set of solutions of \(Ly=0\). Given that the homogeneous boundary value problem \[Ly=0, \quad B_{1}(y)=0,\quad B_{2}(y)=0\] has a nontrivial solution, express it explicity in terms of \(z_{1}\) and \(z_{2}\).

    [exer:13.1.21] If the boundary value problem has a solution for every continuous \(F\), then find the Green’s function for the problem and use it to write an explicit formula for the solution. Otherwise, if the boundary value problem does not have a solution for every continuous \(F\), find a necessary and sufficient condition on \(F\) for the problem to have a solution, and find all solutions. Assume that \(a<b\).

    \(y''=F(x)\), \(y(a)=0\),\(y(b)=0\)

    \(y''=F(x)\), \(y(a)=0\),\(y'(b)=0\)

    \(y''=F(x)\), \(y'(a)=0\),\(y(b)=0\)

    \(y''=F(x)\), \(y'(a)=0\),\(y'(b)=0\)

    [exer:13.1.22] Find the Green’s function for the boundary value problem \[y''=F(x), \quad y(0)-2y'(0)=0, \quad y(1)+2y'(1)=0. \tag{A}\] Then use the Green’s function to solve (A) with

    1. \(F(x)=1\),
    2. \(F(x)=x\), and
    3. \(F(x)=x^{2}\).

    [exer:13.1.23] Find the Green’s function for the boundary value problem \[x^{2}y''+xy'+(x^{2}-1/4)y=F(x), \quad y(\pi/2)=0,\quad y(\pi)=0, \tag{A}\] given that \[y_{1}(x)=\frac{\cos x}{\sqrt{x}} \text{\; and\; \; } y_{2}(x)=\frac{\sin x}{\sqrt{x}}\] are solutions of the complementary equation. Then use the Green’s function to solve (A) with

    1. \(F(x)=x^{3/2}\) and
    2. \(F(x)=x^{5/2}\).

    [exer:13.1.24] Find the Green’s function for the boundary value problem \[x^{2}y''-2xy'+2y=F(x), \quad y(1)=0,\quad y(2)=0, \tag{A}\] given that \(\{x,x^{2}\}\) is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with

    1. \(F(x)=2x^{3}\) and
    2. \(F(x)~=~6x^{4}\).

    [exer:13.1.25] Find the Green’s function for the boundary value problem \[x^{2}y''+xy'-y=F(x), \quad y(1)-2y'(1)=0,\quad y'(2)=0, \tag{A}\] given that \(\{x,1/x\}\) is a fundamental set of solutions of the complementary equation. Then use the Green’s function to solve (A) with

    1. \(F(x)=1\),
    2. \(F(x)=x^{2}\), and
    3. \(F(x)=x^{3}\).

    [exer:13.1.26] \(y''=F(x)\), \(\alpha y(0)+\beta y'(0)=0\), \(\rho y(1)+\delta y'(1)=0\)

    [exer:13.1.27] \(y''+y=F(x)\), \(\alpha y(0)+\beta y'(0)=0\), \(\rho y(\pi)+\delta y'(\pi)=0\)

    [exer:13.1.28] \(y''+y=F(x)\), \(\alpha y(0)+\beta y'(0)=0\), \(\rho y(\pi/2)+\delta y'(\pi/2)=0\)

    [exer:13.1.29] \(y''-2y'+2y=F(x)\), \(\alpha y(0)+\beta y'(0)=0\), \(\rho y(\pi)+\delta y'(\pi)=0\)

    [exer:13.1.30] \(y''-2y'+2y=F(x)\), \(\alpha y(0)+\beta y'(0)=0\), \(\rho y(\pi/2)+\delta y'(\pi/2)=0\)

    [exer:13.1.31] Find necessary and sufficient conditions on \(\alpha\), \(\beta\), \(\rho\), and \(\delta\) for the boundary value problem \[y''-y=F(x), \quad \alpha y(a)+\beta y'(a)=0, \quad \rho y(b)+\delta y'(b)=0 \tag{A}\] to have a unique solution for every continuous \(F\), and find the Green’s function for (A). Assume that \(a<b\).

    [exer:13.1.32] Show that the assumptions of Theorem [thmtype:13.1.3} imply that the unique solution of \[Ly=F, \quad B_{1}(y)=k_{1},\quad B_{2}(y)=f_{2}\] is \[y=\int_{a}^{b} G(x,t)F(t)\,dt +\frac{k_{2}}{B_{2}}(y_{1})y_{1} +\frac{k_{1}}{B_{1}(y_{2})}y_{2}.\]