# Solve the heat equation - by splitting up the second order derivative

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

We have

$$$$\frac{\partial^2u}{\partial x^2}=4\frac{\partial u}{\partial y}~, \qquad 00$$$$

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:

$$$$f_{xx}g-4g_yf=0$$$$

But since we can write

$$$$f_{xx}g=\frac{\partial}{\partial x}\bigg(\frac{\partial}{\partial x}f\bigg)g$$$$

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:

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

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

$$$$g_yf=0$$$$

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

$$$$g_y \cos 5x=0$$$$

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?

Thanks

Hint:

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

$$\frac{X^{''}(x)}{X(x)}=4\frac{Y^{'}(y)}{Y(y)}$$

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

Then

$$\frac{X^{''}(x)}{X(x)}=-\lambda$$

$$4\frac{Y^{'}(y)}{Y(y)}=-\lambda$$

See if you can solve from here!

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