Solution of nonhomogeneous heat equation problem

[Shackleford]

This is from Evans page 50. I'm sure it's something simple, but I don't follow the change from $$ \frac{\partial}{\partial t} \quad \text{to} \quad -\frac{\partial}{\partial s}$$
and from $$ \Delta_x \quad \text{to} \quad \Delta_y$$.
\begin{gather*}
\begin{split}
u_t(x,t) - \Delta u(x,t) & = \int_{0}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(\frac{\partial}{\partial t}-\Delta_x)f(x-y,t-s)] \; dyds \\
& + \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& + \int_{0}^{\varepsilon} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& + \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy
\end{split}
\end{gather*}

 

[mitochan]

I am not sure how four variables x,y,s,t relate but say (x,y) and (t,s) are independent
\frac{\partial}{\partial t}:=\frac{\partial}{\partial t}|_s
\triangle_x:=\frac{\partial^2}{\partial^2 x}|_{y}
,
\frac{\partial}{\partial t}f(x-y,t-s)=\frac{\partial}{\partial (t-s)}f(x-y,t-s)|_{x-y}\frac{\partial (t-s)}{\partial t}
\frac{\partial}{\partial s}f(x-y,t-s)=\frac{\partial}{\partial (t-s)}f(x-y,t-s)|_{x-y}\frac{\partial (t-s)}{\partial s}
They are different only in sign. Similar for $\triangle_x$ and $\triangle_x$ but same sign because of squared differential operator.

 

[Shackleford]

I am not sure how four variables x,y,s,t relate but say (x,y) and (t,s) are independent
\frac{\partial}{\partial t}:=\frac{\partial}{\partial t}|_s
\triangle_x:=\frac{\partial^2}{\partial^2 x}|_{y}
,
\frac{\partial}{\partial t}f(x-y,t-s)=\frac{\partial}{\partial (t-s)}f(x-y,t-s)|_{x-y}\frac{\partial (t-s)}{\partial t}
\frac{\partial}{\partial s}f(x-y,t-s)=\frac{\partial}{\partial (t-s)}f(x-y,t-s)|_{x-y}\frac{\partial (t-s)}{\partial s}
They are different only in sign. Similar for $\triangle_x$ and $\triangle_x$ but same sign because of squared differential operator.

I thought that it might have something to do with the fact that the equation on the left is a function of x and t while the integral on the right is with respect to y and s. The substitution in the integral is obvious from the chain rule.

For the spatial variable, the function is just being shifted around it seems when you vary y or x.

 

[Shackleford]

There's another part I'm not clear on.

\begin{gather*}
\begin{split}
u_t(x,t) - \Delta u(x,t) & = \int_{0}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(\frac{\partial}{\partial t}-\Delta_x)f(x-y,t-s)] \; dyds \\
& + \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& + \int_{0}^{\varepsilon} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& + \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = I_\varepsilon + J_\varepsilon + K.
\end{split}
\end{gather*}
Integrating by parts, we also find
\begin{gather*}
\begin{split}
I_\varepsilon & = \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} [(\frac{\partial}{\partial s}-\Delta_y) \Phi(y,s) ]f(x-y,t-s) \; dyds \\
& + \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy \\
& - \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy - K, \\
\end{split}
\end{gather*}

On what piece exactly did he apply integration by parts?

 

[mitochan]

You give sum of $I_\epsilon,J_\epsilon,K$. How should we divide it to get $I_\epsilon$ ?

 

[mitochan]

I take

\begin{gather*}
\begin{split}
&I_\varepsilon= \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& J_\varepsilon= \int_{0}^{\varepsilon} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
&K=\int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
\end{split}
\end{gather*}

Calculation of $I_\varepsilon$
Integrating by parts once for $\frac{\partial}{\partial s}$ and twice for $\triangle_y$ we get

\begin{gather*}
\begin{split}
I_\varepsilon
& = \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} [(\frac{\partial}{\partial s}-\Delta_y) \Phi(y,s) ]f(x-y,t-s) \; dyds +A\\
\end{split}
\end{gather*}

A comes from values at infinities. Say at $R^n$ infinity the values which comes from integration by y are zero, the remaining comes from integration by s

\begin{gather*}
\begin{split}

& A= \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy - \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy - K \\
\end{split}
\end{gather*}

 

[Shackleford]

I take

\begin{gather*}
\begin{split}
&I_\varepsilon= \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
& J_\varepsilon= \int_{0}^{\varepsilon} \int_{\mathbb{R}^n}^{} \Phi(y,s) [(-\frac{\partial}{\partial s}-\Delta_y)f(x-y,t-s)] \; dyds \\
&K=\int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
\end{split}
\end{gather*}

Calculation of $I_\varepsilon$
Integrating by parts once for $\frac{\partial}{\partial s}$ and twice for $\triangle_y$ we get

\begin{gather*}
\begin{split}
I_\varepsilon
& = \int_{\varepsilon}^{t} \int_{\mathbb{R}^n}^{} [(\frac{\partial}{\partial s}-\Delta_y) \Phi(y,s) ]f(x-y,t-s) \; dyds +A\\
\end{split}
\end{gather*}

A comes from values at infinities. Say at $R^n$ infinity the values which comes from integration by y are zero, the remaining comes from integration by s

\begin{gather*}
\begin{split}

& A= \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy - \int_{\mathbb{R}^n}^{} \Phi(y,t) f(x-y,0) \; dy \\
& = \int_{\mathbb{R}^n}^{} \Phi(y,\varepsilon) f(x-y,t-\varepsilon) \; dy - K \\
\end{split}
\end{gather*}

Thanks for the reply. I'll get around to doing the integration by parts myself and will post again if something isn't clear to me.