Solution of nonhomogeneous heat equation problem

In summary, the conversation discusses a change in notation from $\frac{\partial}{\partial t}$ to $-\frac{\partial}{\partial s}$ and from $\Delta_x$ to $\Delta_y$. This change is related to 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 conversation also includes a discussion on how to divide the terms $I_\epsilon, J_\epsilon,$ and $K$, and the calculation of $I_\epsilon$ using integration by parts.
  • #1
Shackleford
1,656
2
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*}
 
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  • #2
I am not sure how four variables x,y,s,t relate but say (x,y) and (t,s) are independent
[tex]\frac{\partial}{\partial t}:=\frac{\partial}{\partial t}|_s[/tex]
[tex]\triangle_x:=\frac{\partial^2}{\partial^2 x}|_{y}[/tex]
,
[tex]\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}[/tex]
[tex]\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}[/tex]
They are different only in sign. Similar for ##\triangle_x## and ##\triangle_x## but same sign because of squared differential operator.
 
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  • #3
mitochan said:
I am not sure how four variables x,y,s,t relate but say (x,y) and (t,s) are independent
[tex]\frac{\partial}{\partial t}:=\frac{\partial}{\partial t}|_s[/tex]
[tex]\triangle_x:=\frac{\partial^2}{\partial^2 x}|_{y}[/tex]
,
[tex]\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}[/tex]
[tex]\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}[/tex]
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.
 
  • #4
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?
 
  • #5
You give sum of ##I_\epsilon,J_\epsilon,K##. How should we divide it to get ##I_\epsilon## ?
 
  • #6
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*}
 
  • #7
mitochan said:
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.
 

1. What is the nonhomogeneous heat equation problem?

The nonhomogeneous heat equation problem is a mathematical model used to describe the behavior of heat in a non-uniform medium. It takes into account both the diffusion of heat and any external heat sources or sinks that may be present.

2. How is the nonhomogeneous heat equation problem solved?

The nonhomogeneous heat equation problem is typically solved using separation of variables, which involves breaking down the problem into simpler equations that can be solved individually. The final solution is then obtained by combining these individual solutions.

3. What are the boundary conditions for the nonhomogeneous heat equation problem?

The boundary conditions for the nonhomogeneous heat equation problem specify the temperature at the boundaries of the system. These conditions are necessary to fully define the problem and obtain a unique solution.

4. Can the nonhomogeneous heat equation problem be solved analytically?

In most cases, the nonhomogeneous heat equation problem cannot be solved analytically and requires numerical methods for approximation. However, there are some special cases where an analytical solution can be obtained.

5. What are some real-world applications of the nonhomogeneous heat equation problem?

The nonhomogeneous heat equation problem has many applications in fields such as engineering, physics, and chemistry. It is used to model heat transfer in various systems, including buildings, electronic devices, and chemical reactions. It is also used in weather forecasting and climate modeling.

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