Question

solve ∂u/∂t=K∂^2u/∂x^2 under the condition (1) u≠∞ if f->∞

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Answer 1

Step-by-step explanation:

So, we’ve seen that our solution from the first example will satisfy at least a small number of highly specific initial conditions.

Now, let’s extend the idea out that we used in the second part of the previous example a little to see how we can get a solution that will satisfy any sufficiently nice initial condition. The Principle of Superposition is, of course, not restricted to only two solutions. For instance, the following is also a solution to the partial differential equation.

u(x,t)=M∑n=1Bnsin(nπxL)e−k(nπL)2tu(x,t)=∑n=1MBnsin⁡(nπxL)e−k(nπL)2t

and notice that this solution will not only satisfy the boundary conditions but it will also satisfy the initial condition,

u(x,0)=M∑n=1Bnsin(nπxL)u(x,0)=∑n=1MBnsin⁡(nπxL)

Let’s extend this out even further and take the limit as M→∞M→∞. Doing this our solution now becomes,

u(x,t)=∞∑n=1Bnsin(nπxL)e−k(nπL)2tu(x,t)=∑n=1∞Bnsin⁡(nπxL)e−k(nπL)2t

This solution will satisfy any initial condition that can be written in the form,

u(x,0)=f(x)=∞∑n=1Bnsin(nπxL)u(x,0)=f(x)=∑n=1∞Bnsin⁡(nπxL)