# Combining solutions to PDE's

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1. Feb 1, 2016

### Panphobia

1. The problem statement, all variables and given/known data

If you have the heat equation

$$u_{t}-u_{xx}=a \\ u(0,t)=b\\u(1,t)=c\\u(x,0)=d$$

Show that the solution to the above equation can be made up of a linear combination of solutions to
$$u_{t}-u_{xx}=a_i \\ u(0,t)=b_i\\u(1,t)=c_i\\u(x,0)=d_i$$
$$i=1,2,3,4$$
if the following matrix has no zero eigenvalues.
$$A =\begin{bmatrix} a_1 & a_2 & a_3 & a_4 \\ b_1 & b_2 & b_3 & b_4 \\ c_1 & c_2 & c_3 & c_4 \\ d_1 & d_2 & d_3 & d_4 \end{bmatrix}$$

So I know that if a matrix has a zero eigenvalue, then it is singular. A matrix is only singular if it has at least two rows that are not linearly independent. Can someone give me a hint on how to continue this proof? I know that because it mentions zero eigenvalues that it has to do with linear independence, but I have no clue.

2. Feb 2, 2016

### Samy_A

What can you tell about the columns of a non-singular matrix? Are they linearly independent?

Last edited: Feb 2, 2016
3. Feb 2, 2016

### Panphobia

Well yeah the columns are also linearly independent. That means that all of the coefficients for the problem descriptions are linearly independent. But how does that correlate to combining their solutions?

4. Feb 2, 2016

### Samy_A

Consider the following four vectors: $\vec {v_i}=(a_i,b_i,c_i,d_i)$, for $i=1,2,3,4$ (the columns of matrix A).
Can you write the vector $(a, b, c, d)$ as a linear combination of $(\vec {v_i})_i$? If so, can you then concoct a linear combination of the solutions $(u_i)_i$ to the 4 equations with the $(a_i,b_i,c_i,d_i)_i$ that solves the first heat equation?

Last edited: Feb 2, 2016
5. Feb 2, 2016

### Panphobia

Is it necessary for those four vectors to span $R^4$? Also is that a corollary from some theorem? That if you can write the problem description as a linear combination of other problems, then the solution can also be written as a linear combination of those other problems?

6. Feb 2, 2016

### Samy_A

Yes, in a n-dimensional vector space, n linearly independent vectors will span the whole space.
It's the same linear operator.
Say $T: X \to Y$ is a linear operator between two vector spaces. Then if $T\vec x_1=\vec y_1,\ T\vec x_2=\vec y_2$, the linearity of $T$ implies that $T(c_1\vec x_1+c_2\vec x_2)=c_1\vec y_1+c_2\vec y_2$.

7. Feb 2, 2016

### Panphobia

Ohhhh ok that makes a lot of sense, thank you very much!

8. Feb 2, 2016

### Panphobia

So the linear operator $T$ in this case just takes in the vector of coefficients and it outputs the corresponding solution? Just wondering how the solution is a vector, it's probably a stupid question.

9. Feb 2, 2016

### Samy_A

No, it's not a stupid question. The key point is that the four equations (or boundary conditions) are linear in u.
If you want, you could represent the whole setup as a linear operator from some vector space of suitable functions X to X4 (or to Y4, where Y is some other vector space of functions).
$T:X \to X^4\ : u \mapsto (u_{t}-u_{xx}, u(0,t),u(1,t),u(x,0))$.
Using the same notation as above, we then have $Tu_i=\vec v_i$ (for i=1,2,3,4), where we interpret the constants as constant functions.
Then if $(a,b,c,d)=\sum_{i=1}^4 \lambda_i\vec v_i$,
$\displaystyle T(\sum_{i=1}^4 \lambda_iu_i)=\sum_{i=1}^4\lambda_iTu_i=\sum_{i=1}^4\lambda_i\vec v_i=(a,b,c,d)$.
But again, this representation is not so important. What is important is that the equation and boundary conditions are linear.