Initial Value Problem for an Inhomogeneous Equation

Question

Find the solution of this initial value problem.

$u_t = 2u_{xx} + 6u$, $0<x<\pi$, $t>0$

$u(0,t) = 0 = u(\pi,t)$, $t>0$

$u(x,0)=5\sin3x-2\sin4x+3\sin10x$.

Can someone help get me started?

Answer 1

Let us use separation of variables. This means we assume that $u(x,t)$ takes the form $$ u(x,t) = X(x)T(t) $$ Plugging in to the original equation gives: $$ X(x)T'(t) = 2X''(x)T(t) + 6X(x)T(t) $$ Dividing through by $X(x)T(t)$: $$ \frac{T'(t)}{T(t)} = 2\frac{X''(x)}{X(x)} + 6 $$ The left hand side is dependent only on $t$ and the right hand side is dependent only on $x$, yet they equal each other for all $(x,t)$. The only way this can be true is if each side is constant: $$ 2\frac{X''(x)}{X(x)} + 6 = -\lambda $$ $$ \frac{T'(t)}{T(t)} = -\lambda $$ Reordering the first equation gives $$ X''(x) + \frac{\lambda+6}{2}X(x) = 0 $$ If we let $k^2=\frac{\lambda+6}{2}$, the general solution to this equation is $$ X(x) = c_1\sin(kx)+c_2\cos(kx) $$ Now because $u(0,t)=u(\pi,t)=0$, this translates to $X(0)=X(\pi)=0$. This is only possible if the $\cos$ term vanishes and $k$ takes on certain discrete values to match those boundary conditions. We get a family of solutions $$ X_n(x) = c_n\sin(k_n x),\;\;k_n=n $$ Now we return to the time equation, this time knowing that the values of the unknown constant are restricted to discrete values $\lambda_n=2k_n^2-6$; reordering that gives $$ T_n'(t) + \lambda_n T_n(t) = 0 $$ With solution family $$ T_n(t) = e^{-\lambda_n t} $$ Now $u$ has a solution family $$ u_n(x,t) = X_n(x)T_n(t) = c_n\sin(k_n x)e^{-\lambda_n t} $$ That is a solution for any integer value of $n$, so the most general solution is a superposition of those solutions $$ u(x,t) = \sum_{n=-\infty}^\infty u_n(x,t) = \sum_{n=-\infty}^\infty c_n\sin(k_n x)e^{-\lambda_n t} $$ To determine the unknown $c_n$ coefficients, we have to apply the initial condition $$ 5\sin(3x)-2\sin(4x)+3\sin(10x) = u(x,0) = \sum_{n=-\infty}^\infty c_n\sin(k_n x) $$ Recalling that $k_n=n$, it is clear that the sequence of $c_n$ values is mostly zero, except that $c_3=5,\;c_4=-2,\;c_{10}=3$. So the final solution is $$ u(x,t) = 5\sin(3 x)e^{-12 t}-2\sin(4 x)e^{-26 t}+3\sin(10 x)e^{-194 t} $$

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ {\rm u}\pars{x,t}= {\rm A}_{3}\pars{t}\sin\pars{3x} - {\rm A}_{4}\pars{t}\sin\pars{4x} + {\rm A}_{10}\pars{t}\sin\pars{10x} $$

\begin{align} &\dot{{\rm A}}_{3}\pars{t}\sin\pars{3x} - \dot{{\rm A}}_{4}\pars{t}\sin\pars{4x} + \dot{{\rm A}}_{10}\pars{t}\sin\pars{10x} \\[3mm]&= \bracks{-18{\rm A}_{3}\pars{t}\sin\pars{3x} + 32{\rm A}_{4}\pars{t}\sin\pars{4x} -200 {\rm A}_{10}\pars{t}\sin\pars{10x}} \\[3mm]&\phantom{=}+ \bracks{6{\rm A}_{3}\pars{t}\sin\pars{3x} - 6{\rm A}_{4}\pars{t}\sin\pars{4x} + 6{\rm A}_{10}\pars{t}\sin\pars{10x}} \end{align}

$$\left\lbrace% \begin{array}{rcrclrcr} \dot{\rm A}_{3}\pars{t} & + & 12\,{\rm A}_{3}\pars{t} & = & 0\,,\qquad &{\rm A}_{3}\pars{0} & = & 5 \\[1mm] \dot{\rm A}_{4}\pars{t} & + & 26\,{\rm A}_{4}\pars{t} & = & 0\,,\qquad &{\rm A}_{4}\pars{0} & = & -2 \\[1mm] \dot{\rm A}_{3}\pars{t} & + & 194\,{\rm A}_{3}\pars{t} & = & 0\,,\qquad &{\rm A}_{10}\pars{0} & = & 3 \end{array}\right. $$

$$ {\rm A}_{3}\pars{t} = 4\expo{-12t}\,,\qquad {\rm A}_{4}\pars{t} = -2\expo{-26t}\,,\qquad {\rm A}_{3}\pars{t} = 3\expo{-194t} $$

$$ \color{#00f}{\large{\rm u}\pars{x,t}= 4\sin\pars{3x}\expo{-12t} + 2\sin\pars{4x}\expo{-26t} + 3\sin\pars{10x}\expo{-194t}} $$