I would call it a nonhomogeneous initial condition, but initial conditions have nothing to do with your assumption. Only with nonhomogeneous boundary conditions, this assumption can be applied (or if the PDE is nonhomogeneous). In PDEs, the boundary conditions are conditions in which the substitution is done in the spatial variable such as [imath]x, y, \text{or} \ z[/imath]. On the other hand, a substitution done in the time variable, [imath]t[/imath], is called an initial condition.
[math]v_t(x, t)-v_{xx} (x, t)=0 , x\in \mathbb{R}, t>0, v (x, 0) = sin^2(3x), x\in \mathbb{R}[/math] can this be solved using separation of variables?
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I don't think separation of variables works here. This is a 1D heat equation with [imath]t > 0[/imath], [imath]x \in \mathbb R[/imath], so maybe try using the heat kernel [imath]S(x, t) = \dfrac1{\sqrt{4\pi t}} \exp\left(-\dfrac{x^2}{4t}\right)[/imath]. For such heat equation, the solution is given by [math]v(x, t) = \displaystyle \int_{\mathbb R} S(x - y, t) v(y, 0)\,dy[/math]. Plugging in the heat kernel and the initial conditions gives
[math]v(x, t) = \dfrac1{\sqrt{4\pi t}}\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(x - y)^2}{4t}\right) \sin^2(3y)\,dy[/math]The computation is not simple at all, but using the fact that [imath]\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{x^2}{a^2}\right)\,dx =\sqrt \pi |a|[/imath] and [imath]\displaystyle \int_{\mathbb R} \exp(-x^2) \cos(kx)\,dx = \sqrt \pi \exp\left(-\dfrac{k^2}4\right)[/imath], we have:
[math] \begin{array}{rcl} v(x, t) &=& \dfrac1{\sqrt{4\pi t}}\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(x - y)^2}{4t}\right) \sin^2(3y)\,dy\\ &=& \dfrac12 - \dfrac1{2\sqrt{4\pi t}} \displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(y - x)^2}{4t}\right) \cos (6y)\,dy\\ &=& \dfrac12 - \dfrac1{2\sqrt \pi} \displaystyle \int_{\mathbb R} \exp(-y^2) \cos(12 \sqrt t y + 6x)\,dy\\ &=& \dfrac12 - \dfrac{\cos 6x}{2\sqrt \pi} \displaystyle \int_{\mathbb R} \exp(-y^2) \cos(12 \sqrt ty)\,dy\\ &=& \dfrac12 - \dfrac12 \exp(-36t) \cos(6x) \end{array} [/math]
You can check that this indeed satisfies the PDE.
[math]v(x, t) = \dfrac1{\sqrt{4\pi t}}\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(x - y)^2}{4t}\right) \sin^2(3y)\,dy[/math]The computation is not simple at all, but using the fact that [imath]\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{x^2}{a^2}\right)\,dx =\sqrt \pi |a|[/imath] and [imath]\displaystyle \int_{\mathbb R} \exp(-x^2) \cos(kx)\,dx = \sqrt \pi \exp\left(-\dfrac{k^2}4\right)[/imath], we have:
[math] \begin{array}{rcl} v(x, t) &=& \dfrac1{\sqrt{4\pi t}}\displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(x - y)^2}{4t}\right) \sin^2(3y)\,dy\\ &=& \dfrac12 - \dfrac1{2\sqrt{4\pi t}} \displaystyle \int_{\mathbb R} \exp\left(-\dfrac{(y - x)^2}{4t}\right) \cos (6y)\,dy\\ &=& \dfrac12 - \dfrac1{2\sqrt \pi} \displaystyle \int_{\mathbb R} \exp(-y^2) \cos(12 \sqrt t y + 6x)\,dy\\ &=& \dfrac12 - \dfrac{\cos 6x}{2\sqrt \pi} \displaystyle \int_{\mathbb R} \exp(-y^2) \cos(12 \sqrt ty)\,dy\\ &=& \dfrac12 - \dfrac12 \exp(-36t) \cos(6x) \end{array} [/math]
You can check that this indeed satisfies the PDE.
You are a Genius. In my solution, I had to use the Fourier transform to convert the PDE to an ordinary differential equation and solve it. It is later when I was trying to get the inverse Fourier transform, I had to use the Convolution theorem to arrive to your heat kernel integral. It was just amazing how the PDE could be solved directly by the kernel.
Everyday, we learn something new!
I'm not a genius. I just learnt this fact from our university's PDE course and then memorized it.
You probably know more about PDE than I do.