Solution of $\frac{\partial f}{\partial t}(t,x) = 2 \frac{\partial^2f}{\partial x^2}(t,x)$

Question

Consider the PDE $$\frac{\partial f}{\partial t}(t,x) = 2 \frac{\partial^2f}{\partial x^2}(t,x)\tag{1} $$ with $t\ge0,\ x\in\mathbb R,\ f(0,x)=e^x$. I want to find $f(t,x)$.

I know that the heat equation $$\frac{\partial p}{\partial t}(t,x) = \frac{1}{2}\frac{\partial^2p}{\partial x^2}(t,x)\tag{2}$$ with $t\ge0,\ x\in\mathbb R,\ p(0,x)=h(x)$ has the solution $p(t,x) =\mathbb E[h(x+W_t)]$ where $W_t$ is a Brownian motion.

I have tried things like setting $p(t,x)=f(2t,x)$, but I do not seem to be able to put $(1)$ into the form of $(2)$. How can I find $f(t,x)$ from using the general solution of the heat equation?

Answer 1

Following the suggtestion on Cameron Williams' comment:

We set $u(t,y)=f(t,2y)$ with $y=x/2$ and $\overline h(y)=e^{2y}$. Then

$$\frac{\partial u}{\partial t}(t,y)=\frac{\partial f}{\partial t}(t,x), \quad \quad \quad \frac{\partial^2 u}{\partial y^2}(t,y)=4\frac{\partial^2 f}{\partial x^2}(t,x)$$

So $$\frac{\partial u}{\partial t}(t,y)=\frac12\frac{\partial^2 u}{\partial y^2}(t,y) \quad \text{with}\quad u(0,y)=\overline h(y)$$ $$\iff \frac{\partial f}{\partial t}(t,x)=\frac42\frac{\partial^2 f}{\partial x^2}(t,x)=2\frac{\partial^2 f}{\partial y^2}(t,x) \quad \text{with}\quad f(0,x)=u(0,y)=\overline h(y)=h(x)$$

Now the solution of $u$ is given by $u(t,y)=\mathbb E[\overline h(y+W_t)]=e^{2y}\mathbb E[e^{2W_t}]=e^{2y+2t}$ and therefore $f(t,x)=u(t,x/2)=e^{2t+x}$.

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Follow-up question: I am a relativ beginner using change of variables. So how does one actually spot this particular change of variables in the first place (i.e. what tips you off), and also, is there a way to perform this change of variables maybe quicker (less awkward than I did)?

Answer 2

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Let's define $\ds{\tilde{\fermi}\pars{s,x} = \int_{0}^{\infty}\fermi\pars{t,x}\expo{-st}\,\dd s}$. Then, $$ -\fermi\pars{0,x} + s\tilde{\fermi}\pars{s,x} = 2\,\partiald[2]{\tilde{\fermi}\pars{s,x}}{x}\quad\imp\quad \pars{\partiald[2]{}{x} - \half\,s}\tilde{\fermi}\pars{s,x} = -\,\half\,\expo{x} $$ Then, $\tilde{\fermi}\pars{s,x} = A\expo{x}$ such that $A - sA/2 = -1/2\quad\imp\quad A = -1/\bracks{2\pars{1 - s/2}} = 1/\pars{s - 2}$ which leads to: $$ \tilde{\fermi}\pars{s,x} = {\expo{x} \over s - 2} \quad\mbox{and}\quad \fermi\pars{t,x} = \expo{x}\int_{\gamma - \ic\infty}^{\gamma + \ic\infty} {\expo{st} \over s - 2}\,{\dd s \over 2\pi\ic}\quad\mbox{with}\quad\gamma > 2 $$ $$ \color{#0000ff}{\large\fermi\pars{t,x} = \expo{x\ +\ 2t}} $$

Even more simple: Write $\fermi\pars{t,x} \equiv \expo{x}\varphi\pars{t}$ and you get $\dot{\varphi}\pars{t} = 2\varphi\pars{t}$ with $\varphi\pars{0} = 1$. Then, $\varphi\pars{t} = \expo{2t}$