# A inhomogeneous heat equation with mixed conditions

I want to solve the given problem:

$$$$u_t-u_{xx}=2 \ \ \ \ \ 00 \\ u(0,t)=0, \ \ u_x(1,t)=1, \ \ \ t>0\\ u(x,0)=-x^2, \ \ \ \ 0

but I am not sure I have done it correctly. This is what I did:

Step 1. Homogenize, by solving the stationary problem $$u_t=0$$:

$$$$-u_{xx}=2 \rightarrow u(x)-x^2+Cx+D \\ I.C. give \rightarrow u(x)=-x^2+3x$$$$

The homogenized PDE is now:

$$$$u_t-u_{xx}=0 \ \ \ \ \ 00 \\ u(0,t)=0, \ \ u_x(1,t)=1, \ \ \ t>0\\ u(x,0)=-x^2+x^2-3x \rightarrow u(x,0)=-3x, \ \ \ \ 0

Step 2. Solve the homogenous PDE.

Since we have mixed conditions, we need to look for a linear combination of $$u(x)=A\sin\lambda x+B\cos\lambda x$$ which satisfies the I.C.

We get:

$$$$u(x)=A\sin\lambda x+ B\cos\lambda x \rightarrow \ \ IC: u(0)=0 \rightarrow u(x)=A\sin\lambda x\\ u(x)=A\sin\lambda x, \rightarrow IC2: u'(1)=1,\ \ \ u'(x)= \lambda A\cos\lambda x \rightarrow 1=\lambda A\cos\lambda \rightarrow A=\frac{1}{\lambda\cos\lambda}\\ u(x)=\frac{1}{\lambda\cos\lambda}\sin\lambda x$$$$

So now we have a first candidate of the function

$$$$u(x,t)=\frac{\sin\lambda x}{\lambda\cos\lambda}u(t)$$$$

We find out u(t) by plugging this in the PDE, where each of the following is:

$$$$u(x,t)=\frac{\sin\lambda x}{\lambda\cos\lambda}u(t)\\ u_{xx}(x,t)=-\frac{\lambda^2\sin\lambda x}{\lambda\cos\lambda}u(t)\\ u_t(x,t)=\frac{\sin\lambda x}{\lambda\cos\lambda}u_t$$$$

By inserting eacah part in the homogenized PDE we get:

$$$$\frac{\sin\lambda x}{\lambda\cos\lambda}u_t+\frac{\lambda^2\sin\lambda x}{\lambda\cos\lambda}u(t)=0\\ u(t)=C_ne^{-\lambda^2t}$$$$

Step 3. Find the coefficients

So since we now have the full form of $$u(x,t)$$, we can use the third IC for the homogenized problem, $$u(x,0)=-3x$$ and use the Fourier series method to find the coefficient:

$$$$u(x,0)=-3x=\sum_{n=1}^\infty\frac{C_n}{\lambda\cos\lambda}\sin\lambda xe^0$$$$

This coefficient we find by using the Fourier series form for $$\beta_n=\frac{2}{L}\int_0^Lu(x,0)\sin\lambda xdx$$. Here we have both $$u(x,0)=-3x$$ and $$L=1$$ so we obtain that $$\frac{C_n}{\lambda\cos\lambda}=\beta_n$$: This is the famous "Fourier trick" used to find the coefficients of the heat, Laplace and wave equations. So this "trick" gives:

$$$$\frac{C_n}{\lambda\cos\lambda}=2\int_0^1(-3x)\sin\lambda xdx$$$$

Solving the L.H.S we get $$-\frac{\cos\lambda}{\lambda}$$, so the equation becomes

$$$$\frac{C_n}{\lambda\cos\lambda}=-\frac{\cos\lambda}{\lambda}$$$$

Hence,

$$C_n=-\cos^2\lambda$$

Since the system was inhomogenous, we need to add the function $$u(x,0)=-3x$$

This gives the final form of $$u(x,t)$$

$$$$u(x,t)=-3x+\sum_{n=1}^\infty\frac{\cos\lambda}{\lambda}\sin\lambda x e^{-\lambda^2 t}$$$$

But I am not sure about the last step, to add the function $$u(x,0)=-3x$$. Is it right to do, or not?

Thanks

The problem in question is $$u_t-u_{xx}=2 \\ u(0,t)=0,\; u_x(1,t)=1 \\ u(x,0)=-x^2$$ This needs to be transformed to a homogeneous problem in order for separation of variables to work. One way to do this is to add a function $$f(x)$$ to $$u$$ so that the differential equation and endpoint conditions are homogeneous. That requires finding a function $$f$$ to satisfy the following: $$(u+f)_t-(u+f)_{xx}=0 \implies f''(x)=2 \\ u(0,t)+f(0)=0 \implies f(0)=0 \\ u_x(1,t)+f'(1)=0 \implies f'(1)=-1$$

$$f(x)=x^2-3x$$ is such a solution. (Thank you @Luthier415Hz for pointing out my errors and confusion about this.)

The original problem for $$u$$ has now been transformed to a problem in $$v=u+f$$ that satisfies the following: $$v_t=v_{xx} \\ v(0,t)=0,\;\; v_x(1,t)=0, \\ v(x,0) = -3x.$$ Separation of variables can be used to directly solve this problem because of the homogenous endpoint conditions.

The desired solution is $$u=v-f$$. To solve for $$v$$ using separation of variables, assume $$v(t,x)=T(t)X(x)$$. This will work because of the homogeneous endpoint conditions in $$x$$: $$\frac{T'(t)}{T(t)}=\lambda = \frac{X''(x)}{X(x)} \\ X(0)=0,\;\; X'(1)=0.$$ The $$X$$ equation has solutions that are determined only up to a constant $$C$$: $$X_n(x) = C_n\sin((n+1/2)\pi x),\;\; n=0,1,2,3,\cdots.$$ The corresponding eigenvalue parameter $$\lambda$$ is $$\lambda_n = -(n+1/2)^2\pi^2,\;\; n=0,1,2,3,\cdots.$$ And the corresponding solution $$T_n$$ is any constant multiple of $$T_n(t) = \exp(-(n+1/2)^2\pi^2 t)$$ This leads to the general solution for $$v$$: $$v(x,t) = \sum_{n=0}^{\infty}C_n \exp(-(n+1/2)^2\pi^2 t)\sin((n+1/2)\pi x).$$ The constants $$C_n$$ are determined by the condition $$v(x,0)=-3x$$, through the orthogonality of the eigenfunctions $$\{\sin((n+1/2)\pi x)\}_{n=0}^{\infty}$$: $$-3x = v(x,0)= \sum_{n=0}^{\infty}C_n\sin((n+1/2)\pi x) \\ \frac{\int_0^1 (-3x)\sin((n+1/2)\pi x)dx}{\int_0^1\sin^2((n+1/2)\pi x)dx}= C_n,\;\;\; n=0,1,2,3,\cdots.$$ What remains is to determine the constants $$C_n$$ by evaluating the integrals in the corresponding fractions above. The numerator for $$C_n$$ is \begin{align} &\int_0^1 (-3x)\sin((n+1/2)\pi x)dx \\ &= \left.\frac{3x\cos((n+1/2)\pi x)}{(n+1/2)\pi}\right|_{0}^{1} \\ & - \int_0^1 3\frac{\cos((n+1/2)\pi x)}{(n+1/2)\pi}dx \\ &= \left.-3\frac{\sin((n+1/2)\pi x)}{(n+1/2)^2\pi^2}\right|_{x=0}^{1} \\ &= \frac{3(-1)^{n+1}}{(n+1/2)^2\pi^2} \end{align} The denominator for $$C_n$$ is \begin{align} &\int_0^1\sin^2((n+1/2)\pi x)dx \\ & = \left.\frac{-\cos((n+1/2)\pi x)}{(n+1/2)\pi}\sin((n+1/2)\pi x)\right|_{x=0}^{1} \\ & + \int_0^1\cos^2((n+1/2)^2\pi x)dx \\ & = \int_0^1\cos^2((n+1/2)^2\pi x)dx \end{align} Therefore, \begin{align} &\int_0^1\sin^2((n+1/2)\pi x)dx \\ &=\frac{1}{2} \int_0^1\sin^2((n+1/2)\pi x)+\cos^2((n+1/2)^2\pi x) dx \\ &=\frac{1}{2}. \end{align} Finally, $$C_n = \frac{6(-1)^{n+1}}{(n+1/2)^2\pi^2}$$ The full expression for $$v$$ is \begin{align} &v(x,t)=\sum_{n=0}^{\infty}C_n \exp(-(n+1/2)^2\pi^2 t)\sin((n+1/2)\pi x) \\ &= \sum_{n=0}^{\infty}\frac{6(-1)^{n+1}}{(n+1/2)^2\pi^2}\exp(-(n+1/2)^2\pi^2 t)\sin((n+1/2)\pi x). \end{align} The desired solution is $$u=v-f=v-(x^2-3x)$$: \begin{align} %% u(x,t)&=v(x,t)-(x^2-3x) \\ &u(x,t)= -x^2+3x \\ &+\sum_{n=0}^{\infty}\frac{6(-1)^{n+1}}{(n+1/2)^2\pi^2}\exp(-(n+1/2)^2\pi^2 t)\sin((n+1/2)\pi x) \end{align} NOTE: If I have all the details for this right, I'll be shocked!

• @Luthier415Hz : What does your final sum add up to when you evaluate at $t=0$? Does your final answer satisfy $u(x,0)=-3x$? It looks like it does to me. And what about $u_x(1,t)$? Does that sum equal $1$, as required? You just have to check the conditions of the original problem, one by one. Let me know what you find. Sep 1, 2022 at 11:03
• I found the error, I added an answer. Thanks Disintegrating Sep 1, 2022 at 12:13
• @Luthier415Hz : I've expanded the solution to a full solution above. Everything should check out. Sep 6, 2022 at 16:55
• Great , it looks good, but you didn't calculate the integrals. Sep 7, 2022 at 7:11
• I chcked and it doesnt seem right. wolframalpha.com/…*%5Cexp%28-%281%2B1%2F2%29%5E2%5Cpi%5E2+t%29*%5Csin%28%281%2B1%2F2%29%5Cpi+x%29 and Sep 12, 2022 at 19:14

$$u(x)=-x^2+3x$$ to the solution of the homogeneous problem.

That is

$$$$u(x,t)=\bigg[\sum_{n=1}^\infty\frac{\cos n\pi} {n\pi}\sin n\pi x e^{-(n\pi)^2 t}\bigg]-x^2+3x)$$$$

which gives

$$$$u(x,t)=-x^2+3x+\bigg[\sum_{n=1}^\infty\frac{\cos n\pi}{n\pi}\sin n\pi x e^{-(n\pi)^2 t}\bigg]$$$$

This satisfies the non-homogeneous solution, giving, when inserted in the PDE with $$n=1$$

$$$$\frac{d}{dt}\bigg[-x^2+3x+e^{-(n\pi)^2 t} \cos n\pi \frac{\sin n\pi x}{n\pi}\bigg]-\frac{d^2}{dx^2}\bigg[-x^2+3x+e^{-(n\pi)^2t} \cos n\pi\frac{\sin n\pi x}{n\pi}\bigg]=2\\ -e^{-(n\pi)^2 t} n\pi \cos n\pi \sin n\pi x - \bigg[-e^{-(n\pi)^2 t } n\pi \cos n\pi \sin n\pi x-2\bigg]=2\\ 2=2$$$$

• You're looking like a pro at these problems. Did you check $u_x(1,t)=1$? That seems to be a tough one to check. Sep 1, 2022 at 12:19
• No, I am not sure on that one. Sep 1, 2022 at 12:22
• Maybe you could check my solution to see if it satisfies all of the required conditions. Sep 6, 2022 at 11:02
• Thanks Disintegration. To do that, I would need the whole solution, not only $−2x^2+x$. Or is the rest the same as mine? Sep 6, 2022 at 11:13
• @Luther415Hz : How do your values of $\lambda$ depend on the sum index $n$? Sep 12, 2022 at 19:45