I tried to solve the heat equation here, but I am not sure this works out.

We have

\begin{equation} \frac{\partial^2u}{\partial x^2}=4\frac{\partial u}{\partial y}~, \qquad 0<x<\pi\,,~t>0 \end{equation}

I.C. are

\begin{cases} \dfrac{\partial u}{\partial x}(0,y) = 0 \\ \dfrac{\partial u}{\partial x}(\pi,y) = 0 \\ u(x,0)= \cos5x \end{cases}

So we know that

\begin{align} &fu_x=f_xg \\ &u_{xx}=f_{xx}g \\ &u_y=g_yf \end{align}

So the equation by separation of variables is:

\begin{equation} f_{xx}g-4g_yf=0 \end{equation}

But since we can write

\begin{equation} f_{xx}g=\frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}f\bigg)g \end{equation}

and I.C.s can be written as:

\begin{cases} \dfrac{\partial }{\partial x}f(0)g(y) = 0 \\ \dfrac{\partial }{\partial x}f(0)g(y) = 0 \\ f(x)g(0)= \cos5x \end{cases}

Then we can substitute $\frac{\partial}{\partial x}f(0)g(y)=0$ right into the equation:

\begin{equation} \frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}f(0)g\bigg)-4g_yf=0 \end{equation}

and that will zero out the first term, and we are left with

\begin{equation} g_yf=0 \end{equation}

Knowing that $f(x)= \cos5x$, we have at $u(x,0)$:

\begin{equation} g_y \cos 5x=0 \end{equation}

which unfortunately, has only the trivial solution. Is there any way to use this simplification to solve this system? If not, what is the better way?



1 Answer 1



We can guess that $u(x,y) = X(x)Y(y)$ Then we get

$$u_{xx} = 4u_y = X^{''}(x)*Y(y) = 4Y^{'}(y)X(x)$$

From there, we isolate the functions to get


However, a function of x can only equal a function of y if both functions are constant. Let's call this function $-\lambda$




See if you can solve from here!

  • $\begingroup$ Got it, please check that typo of $u_{xx}$ instead of $u_xx$ $\endgroup$ Dec 18, 2021 at 15:31
  • 1
    $\begingroup$ Done, thanks for the feedback! $\endgroup$
    – mode_er
    Dec 18, 2021 at 15:32

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