- #1

Phys pilot

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Hello, I have to solve this problem:

$$u_t=ku_{xx}+h \; \;\; \; \; 0<x<1 \; \; \,\; t>0$$

$$u(x,0)=u_0(1-\cos{\pi x}) \; \;\; \; \; 0\leq x \leq 1$$

$$u(0,t)=0 \; \;\; \; \; u(1,t)=2u_0 \; \;\; \; \; t\geq0$$

So I know that I can split the solution in two (I don't know the reason. I would appreciate a short explanation because we introduce a new function depending just on x and because the sum of the both solutions are also a solution):

$$u(x,t)=v(x,t)+U(x)$$

This give us two different problems.

$$kU''(x)+h=0$$

and

$$v_t=kv_{xx}$$

$$\text{I.C}\begin{cases} v(x,0)=u_0(1-\cos{\pi x})-U(x)\end{cases}$$

$$\text{B.C}\begin{cases} v(0,t)=0 \\ v(1,t)=0 \end{cases}$$

First I can sol the U(x) which is giving me:

$$kU''(x)+h=0$$

$$U(x)=-\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k} \right)x$$

Now I solve the PDE with the new boundary conditions using the separation of variables and I get:

$$X_n(x)=B_n\sin{n\pi x}$$

$$T_n(t)=C_ne^{-kn^2\pi^2 t}$$

$$v(x,t)=\sum_{n=1}^\infty{=a_ne^{-kn^2\pi^2 t}\sin{n\pi x}}$$

If I apply the initial condition:

$$v(x,0)=\sum_{n=1}^\infty{=a_n\sin{n\pi x}}=u_0(1-cos(\pi x))-U(x)$$

So finally the coefficients should be given by the expression:

$$a_n=2\int_0^1 (u_0(1-cos(\pi x))-U(x))\sin(n \pi x)dx$$

But I'm not gettin the supposed solution which is:

$$a_n=\frac{2u_0}{n\pi}[1+(-1)^n]+\frac{2u_0n}{(n^2-1)\pi}[1-(-1)^n]+\frac{2h}{k\pi^3 n^3}[(-1)^n-1]$$

Actually the supposed solution to the problem is:

$$u(x,t)=-\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k}\right)x-\frac{4h}{k\pi}e^{-k\pi^2 t}\sin{\pi x}+ \sum_{n=2}^\infty{a_n e^{-kn^2\pi^2 t}\sin{n\pi x}} $$

As you can see the expansion series starts at 2 instead of one so how am I supposed to get the term for n=1 because you can not get it from a_n

$$u_t=ku_{xx}+h \; \;\; \; \; 0<x<1 \; \; \,\; t>0$$

$$u(x,0)=u_0(1-\cos{\pi x}) \; \;\; \; \; 0\leq x \leq 1$$

$$u(0,t)=0 \; \;\; \; \; u(1,t)=2u_0 \; \;\; \; \; t\geq0$$

So I know that I can split the solution in two (I don't know the reason. I would appreciate a short explanation because we introduce a new function depending just on x and because the sum of the both solutions are also a solution):

$$u(x,t)=v(x,t)+U(x)$$

This give us two different problems.

$$kU''(x)+h=0$$

and

$$v_t=kv_{xx}$$

$$\text{I.C}\begin{cases} v(x,0)=u_0(1-\cos{\pi x})-U(x)\end{cases}$$

$$\text{B.C}\begin{cases} v(0,t)=0 \\ v(1,t)=0 \end{cases}$$

First I can sol the U(x) which is giving me:

$$kU''(x)+h=0$$

$$U(x)=-\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k} \right)x$$

Now I solve the PDE with the new boundary conditions using the separation of variables and I get:

$$X_n(x)=B_n\sin{n\pi x}$$

$$T_n(t)=C_ne^{-kn^2\pi^2 t}$$

$$v(x,t)=\sum_{n=1}^\infty{=a_ne^{-kn^2\pi^2 t}\sin{n\pi x}}$$

If I apply the initial condition:

$$v(x,0)=\sum_{n=1}^\infty{=a_n\sin{n\pi x}}=u_0(1-cos(\pi x))-U(x)$$

So finally the coefficients should be given by the expression:

$$a_n=2\int_0^1 (u_0(1-cos(\pi x))-U(x))\sin(n \pi x)dx$$

But I'm not gettin the supposed solution which is:

$$a_n=\frac{2u_0}{n\pi}[1+(-1)^n]+\frac{2u_0n}{(n^2-1)\pi}[1-(-1)^n]+\frac{2h}{k\pi^3 n^3}[(-1)^n-1]$$

Actually the supposed solution to the problem is:

$$u(x,t)=-\frac{hx^2}{2k}+\left(2u_0+\frac{h}{2k}\right)x-\frac{4h}{k\pi}e^{-k\pi^2 t}\sin{\pi x}+ \sum_{n=2}^\infty{a_n e^{-kn^2\pi^2 t}\sin{n\pi x}} $$

As you can see the expansion series starts at 2 instead of one so how am I supposed to get the term for n=1 because you can not get it from a_n

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