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To solve the problem: $$(1):u_t=u_{xx}, 0<x<L, t>0$$ $$u(0,t)=T_1, u(L,t)=T_2, t>0$$ $$u(x,0)=f(x), 0<x<L$$ we do the following: $$(2):u(x,t)=v(x,t)+s(x)$$ where $s(x)$ is a stationary of the initial boundary value problem $(1)$ while $t$ goes to infinity, that means that $s(x)=\lim_{t \rightarrow \infty} u(x,t), 0<x<L$

$$\lim_{t \rightarrow \infty} u(x,t)= \lim_{t \rightarrow \infty} v(x,t)+ \lim_{t \rightarrow \infty} s(x) \Rightarrow \lim_{t \rightarrow \infty} v(x,t)=0, 0<x<L$$

Replacing the relation $(2)$ into the relation $(1)$, we get: $$(5):v_t=v_{xx}+s''(x), 0<x<L, t>0$$ $$v(x,0)+(x)=f(x), 0<x<L$$ $$v(0,t)+s(0)=T_1$$ $$v(L,t)+S(L)=T_2$$

Taking the limit $t \rightarrow \infty$ at the relations $(5)$, we have: $$s''=0, 0<x<L$$ $$s(0)=T_1$$ $$s(L)=T_2$$

$$\lim_{t \rightarrow \infty} v_t= \lim_{t \rightarrow \infty} v_{xx}+s''(x)$$ $$v(x,0)+s(x)=f(x) \Rightarrow v(x,t)=f(x)-s(x)$$ $$\lim_{t \rightarrow \infty}v(0,t)+s(0)=T_1 \Rightarrow 0+s(0)=T_1$$ $$\lim_{t \rightarrow \infty} v(L,t)+s(L)=T_2 \Rightarrow 0+s(L)=T_2$$

But how do we get get from $$\lim_{t \rightarrow \infty} v_t= \lim_{t \rightarrow \infty} v_{xx}+s''(x), 0<x<L$$ to $$s''(x)=0, 0<x<L$$???

Mary Star
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1 Answers1

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Since $\lim\limits_{t\to \infty} v = 0$, it follows that $\lim\limits_{t\to \infty} v_t = 0$ and also that $\lim\limits_{t\to \infty} v_{xx} = 0$.

Edit: This follows because for instance: $$\lim_{t\to \infty} v_t = \lim_{t\to \infty} \lim_{h\to 0} \frac{v(x,t+h) -v(x,t)}{h} = \lim_{h\to 0} \lim_{t\to \infty} \frac{v(x,t+h) -v(x,t)}{h} = \lim_{h\to 0} \frac{0-0}{h} = 0$$

Klaas van Aarsen
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