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 x,y,orz. On the other hand, a substitution done in the time variable, t, is called an initial condition.
I don't think separation of variables works here. This is a 1D heat equation with t>0, x∈R, so maybe try using the heat kernel S(x,t)=4πt1exp(−4tx2). For such heat equation, the solution is given by v(x,t)=∫RS(x−y,t)v(y,0)dy. Plugging in the heat kernel and the initial conditions gives v(x,t)=4πt1∫Rexp(−4t(x−y)2)sin2(3y)dyThe computation is not simple at all, but using the fact that ∫Rexp(−a2x2)dx=π∣a∣ and ∫Rexp(−x2)cos(kx)dx=πexp(−4k2), we have:
v(x,t)=====4πt1∫Rexp(−4t(x−y)2)sin2(3y)dy21−24πt1∫Rexp(−4t(y−x)2)cos(6y)dy21−2π1∫Rexp(−y2)cos(12ty+6x)dy21−2πcos6x∫Rexp(−y2)cos(12ty)dy21−21exp(−36t)cos(6x)
You can check that this indeed satisfies the PDE.
I don't think separation of variables works here. This is a 1D heat equation with t>0, x∈R, so maybe try using the heat kernel S(x,t)=4πt1exp(−4tx2). For such heat equation, the solution is given by v(x,t)=∫RS(x−y,t)v(y,0)dy. Plugging in the heat kernel and the initial conditions gives v(x,t)=4πt1∫Rexp(−4t(x−y)2)sin2(3y)dyThe computation is not simple at all, but using the fact that ∫Rexp(−a2x2)dx=π∣a∣ and ∫Rexp(−x2)cos(kx)dx=πexp(−4k2), we have:
v(x,t)=====4πt1∫Rexp(−4t(x−y)2)sin2(3y)dy21−24πt1∫Rexp(−4t(y−x)2)cos(6y)dy21−2π1∫Rexp(−y2)cos(12ty+6x)dy21−2πcos6x∫Rexp(−y2)cos(12ty)dy21−21exp(−36t)cos(6x)
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.
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.
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