I want to solve the "modified" heat equation $$ \frac{\partial y}{\partial t}=a\frac{\partial^2 y}{\partial x^2} +b\frac{\partial y}{\partial x} +cy+d $$

I assumed that a, b, c and d are all constant

I tried using Laplace transform. $$ L(y(x,t))=\int_{0}^{\infty}e^{-st}u(x,t)dt=U(x,s) $$ $$ L(\frac{\partial y(x,t)}{\partial t})=sU(x,s)-u(x,0) $$ $$ L(\frac{\partial^2 y(x,t)}{\partial x^2})=\frac{\partial^2 U(x,s)}{\partial x^2} =U_{xx}(x,s) $$ $$ L(\frac{\partial y(x,t)}{\partial x})=\frac{\partial U(x,s)}{\partial x} =U_{x}(x,s) $$ Now equation looks like $$ sU(x,s)-u(x,0) =aU_{xx}(x,s)+bU_{x}(x,s)+cU(x,s)+d $$ or defining $f(x)=-d-u(x,0)$ $$ aU_{xx}(x,s)+bU_{x}(x,s)+(c-s)U(x,s)=f(x) $$ Now, homogeneous solution is $$ U_h(x,s)=U_1+U_2=c_1e^{\lambda_1x}+c_2e^{\lambda_2x} $$ where $\lambda_1=\frac{-b+\sqrt{b^2-4a(c-s)}}{2a}$, $\lambda_2=\frac{-b-\sqrt{b^2-4a(c-s)}}{2a}$,

and for particular solution, I used variation of parameters $$ U_p=-U_1\int\frac{U_2}{U_1U'_2-U_2U'_1}f(x)dx+U_2\int\frac{U_1}{U_1U'_2-U_2U'_1}f(x)dx $$

Then something hit me. It is wise to see if other people have already done this.

So, is there any literature that deals with this partial differential equation with the general solution?

(Anyway,does this have solution? just like integral of heat kernel in usual "heat equation"?)

  • $\begingroup$ What about $x$? $\endgroup$ Feb 23, 2016 at 5:31
  • $\begingroup$ What's the range of $x$? $\endgroup$ Feb 23, 2016 at 5:37
  • $\begingroup$ You can simplify your problem slightly by setting $y(x,t) = u(x + bt, t) = u(z, t)$ and applying the chain rule to get $$u_{t} = au_{zz} + cu + d$$ $\endgroup$ Feb 23, 2016 at 5:59
  • $\begingroup$ One possible approach: You can easily get rid of $d$ by setting $y(t)=exp[dt]v(t)$. Then take the Fourier transform wrt. $x$, find the solution and transform back. $\endgroup$
    – Urgje
    Feb 23, 2016 at 10:40
  • $\begingroup$ yeah, Fourier Transform works! $\endgroup$
    – user65452
    Feb 25, 2016 at 22:46


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