Paper 2, Section II, A

Differential Equations | Part IA, 2021

For an n×nn \times n matrix AA, define the matrix exponential by

exp(A)=m=0Amm!\exp (A)=\sum_{m=0}^{\infty} \frac{A^{m}}{m !}

where A0IA^{0} \equiv I, with II being the n×nn \times n identity matrix. [You may assume that exp((s+t)A)=exp(sA)exp(tA)\exp ((s+t) A)=\exp (s A) \exp (t A) for real numbers s,ts, t and you do not need to consider issues of convergence.] Show that

ddtexp(tA)=Aexp(tA)\frac{\mathrm{d}}{\mathrm{d} t} \exp (t A)=A \exp (t A)

Deduce that the unique solution to the initial value problem

dydt=Ay,y(0)=y0, where y(t)=(y1(t)yn(t))\frac{\mathrm{d} \mathbf{y}}{\mathrm{d} t}=A \mathbf{y}, \quad \mathbf{y}(0)=\mathbf{y}_{0}, \quad \text { where } \mathbf{y}(t)=\left(\begin{array}{c} y_{1}(t) \\ \vdots \\ y_{n}(t) \end{array}\right)

is y(t)=exp(tA)y0\mathbf{y}(t)=\exp (t A) \mathbf{y}_{0}.

Let x=x(t)\mathbf{x}=\mathbf{x}(t) and f=f(t)\mathbf{f}=\mathbf{f}(t) be vectors of length nn and AA a real n×nn \times n matrix. By considering a suitable integrating factor, show that the unique solution to

dxdtAx=f,x(0)=x0\frac{\mathrm{d} \mathbf{x}}{\mathrm{d} t}-A \mathbf{x}=\mathbf{f}, \quad \mathbf{x}(0)=\mathbf{x}_{0}

is given by

x(t)=exp(tA)x0+0texp[(ts)A]f(s)ds\mathbf{x}(t)=\exp (t A) \mathbf{x}_{0}+\int_{0}^{t} \exp [(t-s) A] \mathbf{f}(s) \mathrm{d} s

Hence, or otherwise, solve the system of differential equations ()(*) when

A=(222513153),f(t)=(sint3sint0),x0=(112)A=\left(\begin{array}{ccc} 2 & 2 & -2 \\ 5 & 1 & -3 \\ 1 & 5 & -3 \end{array}\right), \quad \mathbf{f}(t)=\left(\begin{array}{c} \sin t \\ 3 \sin t \\ 0 \end{array}\right), \quad \mathbf{x}_{0}=\left(\begin{array}{l} 1 \\ 1 \\ 2 \end{array}\right)

[Hint: Compute A2A^{2} and show that A3=0.]\left.A^{3}=0 .\right]

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